Problems And Solutions In Quantum Computing And Quantum Information [4 ed.] 2018001917, 9789813238404, 9813238402, 9789813239289, 981323928X

Table of contents :
Contents
Preface
Notation
I Finite-Dimensional Hilbert Spaces
1 Qubits
1.1 Introduction
1.2 Solved Problems
1.3 Supplementary Problems
2 Kronecker Product and Tensor Product
2.1 Introduction
2.2 Solved Problems
2.3 Supplementary Problems
3 Matrix Properties
3.1 Introduction
3.2 Solved Problems
3.3 Supplementary Problems
4 Density Operators
4.1 Introduction
4.2 Solved Problems
4.3 Supplementary Problems
5 Trace and Partial Trace
5.1 Introduction
5.2 Solved Problems
5.3 Supplementary Problems
6 Boolean Functions and Quantum Gates
6.1 Introduction
6.2 Solved Problems
6.3 Supplementary Problems
7 Unitary Transforms and Quantum Gates
7.1 Introduction
7.2 Solved Problems
7.3 Supplementary Problems
8 Entropy
8.1 Introduction
8.2 Solved Problems
8.3 Supplementary Problems
9 Measurement
9.1 Introduction
9.2 Solved Problems
9.3 Supplementary Problems
10 Entanglement
10.1 Introduction
10.2 Solved Problems
10.3 Supplementary Problems
11 Bell Inequality
11.1 Introduction
11.2 Solved Problems
11.3 Supplementary Problems
12 Teleportation
12.1 Introduction
12.2 Solved Problems
12.3 Supplementary Problems
13 Cloning
13.1 Introduction
13.2 Solved Problems
13.3 Supplementary Problems
14 Quantum Algorithms
14.1 Introduction
14.2 Solved Problems
14.3 Supplementary Problems
15 Quantum Error Correction
15.1 Introduction
15.2 Solved Problems
15.3 Supplementary Problem
16 Quantum Cryptography
16.1 Introduction
16.2 Solved Problems
16.3 Supplementary Problem
Chapter 17 Quantum Channels
17.1 Introduction
17.2 Solved Problems
17.3 Supplementary Problems
II Infinite-Dimensional Hilbert Spaces
18 Bose Operators and Number States
18.1 Introduction
18.2 Solved Problems
18.3 Supplementary Problems
19 Coherent States
19.1 Introduction
19.2 Solved Problems
19.3 Supplementary Problems
20 Squeezed States
20.1 Introduction
20.2 Solved Problems
20.3 Supplementary Problems
21 Trace and Partial Trace
21.1 Introduction
21.2 Solved Problems
21.3 Supplementary Problems
22 Entanglement
22.1 Introduction
22.2 Solved Problems
22.3 Supplementary Problems
23 Continuous Variable Teleportation
23.1 Introduction
23.2 Solved Problems
23.3 Supplementary Problem
24 Swapping and Cloning
24.1 Introduction
24.2 Solved Problems
24.3 Supplementary Problems
25 Homodyne Detection
25.1 Introduction
25.2 Solved Problems
25.3 Supplementary Problems
26 Hamilton Operators
26.1 Introduction
26.2 Solved Problems
26.3 Supplementary Problems
Bibliography
Index

Citation preview

PROBLEMS 4th

Edition

AND

SOLUTIONS IN

QUANTUM COMPUTING AND

QUANTUM INFORMATION

10943hc_9789813238404_tp.indd 1

30/1/18 2:00 PM

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Willi-Hans Steeb

University of Johannesburg, South Africa

Yorick Hardy University of the Witwatersrand, South Africa

PROBLEMS 4th

Edition

AND

SOLUTIONS IN

QUANTUM COMPUTING AND

QUANTUM INFORMATION

World Scientific NEW JERSEY

•

LONDON

10943hc_9789813238404_tp.indd 2

•

SINGAPORE

•

BEIJING

•

SHANGHAI

•

HONG KONG

•

TA I P E I

•

CHENNAI

30/1/18 2:00 PM

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Names: Steeb, W.-H., author. | Hardy, Yorick, 1976– author. Title: Problems and solutions in quantum computing and quantum information / Willi-Hans Steeb and Yorick Hardy, University of Johannesburg, South Africa. Description: 4th edition. | Singapore ; Hackensack, NJ : World Scientific, [2018] | Includes bibliographical references and index. Identifiers: LCCN 2018001917| ISBN 9789813238404 (hardcover ; alk. paper) | ISBN 9813238402 (hardcover ; alk. paper) | ISBN 9789813239289 (pbk ; alk. paper) | ISBN 981323928X (pbk ; alk. paper) Subjects: LCSH: Quantum computers. | Quantum theory. Classification: LCC QA76.889 .S74 2018 | DDC 006.3/843--dc23 LC record available at https://lccn.loc.gov/2018001917

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 2018 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

For any available supplementary material, please visit http://www.worldscientific.com/worldscibooks/10.1142/10943#t=suppl

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10943 - Problems and Solutions in Quantum Computing and Quantum Information

Preface The purpose of this book is to supply a collection of problems in quantum computing and quantum information together with their detailed solutions which will prove to be valuable to graduate students as well as to research workers in these fields. All the important concepts and topics such as quantum gates and quantum circuits, quantum channels, entanglement, teleportation, Bell states, Bell inequality, Schmidt decomposition, quantum Fourier transform, magic gate, von Neumann entropy, quantum channels, quantum cryptography, quantum error correction, coherent states, coherent Bell states, squeezed states, POVM measurement, beam splitter, homodyne detection and Kerr Hamilton operator are included. The topics range in difficulty from elementary to advanced. Almost all problems are solved in detail and most of the problems are self-contained. All relevant definitions are given. Students can learn important principles and strategies required for problem solving. Teachers will also find this text useful as a supplement, since important concepts and techniques are developed in the problems. The book can also be used as a text or a supplement for linear and multilinear algebra or matrix theory. Each chapter also includes supplementary problems. Most chapters also include programming problems in Maxima and SymbolicC++. The material was tested in our lectures given around the world. Any useful suggestions and comments are welcome. The International School for Scientific Computing (ISSC) provides certificate courses for this subject. Please contact the first author if you want to do this course. More quantum computing exercises can be found on the web page given below. e-mail addresses of the authors: [email protected] [email protected] [email protected] [email protected] Home page of the authors: http://issc.uj.ac.za v

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Contents Preface

v

Notation

xi

I

1

Finite-Dimensional Hilbert Spaces

1 Qubits 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . .

3 3 5 27

2 Kronecker Product and Tensor Product 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . .

31 31 33 59

3 Matrix Properties 65 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 67 3.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 108 4 Density Operators 115 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 4.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 117 4.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 139 5 Trace and Partial Trace 141 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 5.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 143 5.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 153 6 Boolean Functions and Quantum Gates 155 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 6.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 157 6.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 173 vii

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Contents

7 Unitary Transforms and Quantum Gates 175 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 7.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 177 7.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 208 8 Entropy 209 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 8.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 211 8.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 225 9 Measurement 227 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 9.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 229 9.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 249 10 Entanglement 251 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 10.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 253 10.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 293 11 Bell 11.1 11.2 11.3

Inequality 295 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 297 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 306

12 Teleportation 307 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 12.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 308 12.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 318 13 Cloning 319 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 13.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 320 13.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 324 14 Quantum Algorithms 325 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 14.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 327 14.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 349 15 Quantum Error Correction 351 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 15.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 353 15.3 Supplementary Problem . . . . . . . . . . . . . . . . . . . . 358

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Contents

ix

16 Quantum Cryptography 359 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 16.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 360 16.3 Supplementary Problem . . . . . . . . . . . . . . . . . . . . 368 17 Quantum Channels 369 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 17.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 371 17.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 384

II

Infinite-Dimensional Hilbert Spaces

18 Bose Operators and Number 18.1 Introduction . . . . . . . . . 18.2 Solved Problems . . . . . . 18.3 Supplementary Problems . .

385

States 387 . . . . . . . . . . . . . . . . . . 387 . . . . . . . . . . . . . . . . . . 389 . . . . . . . . . . . . . . . . . . 417

19 Coherent States 421 19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 19.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 423 19.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 437 20 Squeezed States 439 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 439 20.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 441 20.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 457 21 Trace and Partial Trace 461 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 21.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 462 21.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 468 22 Entanglement 469 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 22.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 471 22.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 487 23 Continuous Variable Teleportation 489 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 23.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 489 23.3 Supplementary Problem . . . . . . . . . . . . . . . . . . . . 493

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24 Swapping and Cloning 495 24.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 24.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 495 24.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 500 25 Homodyne Detection 501 25.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 25.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 502 25.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 504 26 Hamilton Operators 505 26.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 505 26.2 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . 506 26.3 Supplementary Problems . . . . . . . . . . . . . . . . . . . . 519 Bibliography

521

Index

535

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Notation ∅ N N0 Z Q R R+ C Rn Cn H i ωn := exp(2πi/n) 0. Are the normalized eigenvectors orthonormal to each other? Solution 27.

In matrix form we have the (hermitian) 2 × 2 matrix   ε0 + ~ω ∆1 − i∆2 ˆ = H . ∆1 + i∆2 ε0 − ~ω

From ˆ − EI2 ) = (ε + ~ω − E)(ε − ~ω − E) − (∆1 + i∆2 )(∆1 − i∆2 ) = 0 det(H we obtain the characteristic equation E 2 − 2ε0 E + ε20 = ~2 ω 2 + ∆21 + ∆22 . Thus the two eigenvalues E+ , E− are q E± = ε0 ± ~2 ω 2 + ∆21 + ∆22 . Let S :=

p ~2 ω 2 + ∆21 + ∆22 . For the eigenvector of E+ we have to solve      ε0 + ~ω ∆1 − i∆2 v1 v1 = E+ ∆1 + i∆2 ε0 − ~ω v2 v2

or (ε0 + ~ω)v1 + (∆1 − i∆2 )v2 = E+ v1 = (ε0 + S)v1 (∆1 + i∆2 )v1 + (ε0 − ~ω)v2 = (ε0 + S)v2 .

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10943 - Problems and Solutions in Quantum Computing and Quantum Information

22 Problems and Solutions We can set v1 = S + ~ω. Thus v2 = ∆1 + i∆2 . After normalizing we have the eigenvector   1 S + ~ω p . (S + ~ω)2 + ∆2 + ∆2 ∆1 + i∆2 1

2

Analogously for E− we obtain the normalized eigenvector   1 ∆1 − i∆2 p . (S + ~ω)2 + ∆21 + ∆22 −S − ~ω Obviously the two eigenvectors are orthonormal to each other, i.e. the scalar product vanishes. The eigenvectors do not depend on ε0 . ˆ be n × n hermitian matrices. Let |ψi be a Problem 28. Let Aˆ and B normalized state in the Hilbert space Cn . Then we have the inequality ˆ ˆ ≥ (∆A)(∆ B)

1 ˆ ˆ |h[A, B]i| 2

where ∆Aˆ :=

q

ˆ 2, hAˆ2 i − hAi

ˆ := ∆B

q

ˆ 2 i − hBi ˆ 2 hB

and ˆ := hψ|A|ψi, ˆ hAi

ˆ := hψ|B|ψi. ˆ hBi

Consider the hermitian spin- 21 matrices S1 =

1 2



0 1

1 0

 ,

S2 =

1 2



0 i

−i 0

 ,

S3 =

1 2



1 0

0 −1

ˆ = S2 . Find states |ψi such that Let Aˆ = S1 and B ˆ ˆ = 1 |h[A, ˆ B]i| ˆ (∆A)(∆ B) 2 i.e. the inequality given above should be an equality. Solution 28.

For the commutator we find [S1 , S2 ] = iS3 . Now S12 = S22 = S32 =

1 I2 . 4

We set  |ψi =

c1 c2

 ,

hψ| = ( c∗1

c∗2 )

 .

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10943 - Problems and Solutions in Quantum Computing and Quantum Information

Qubits

23

with c1 c∗1 + c2 c∗2 = 1 (normalization). Thus we have for the right-hand side of the equality     1 1 1 ˆ ˆ 1 0 c1 1 2 |h[A, B]i| = |hψ|iS3 |ψi = ( c∗1 c∗2 ) = |r − r22 | 0 −1 c2 4 1 2 2 4 where we set c1 = r1 eiφ1 , c2 = r2 eiφ2 . Now we have r q 1 1 ∆S1 = h I2 i − hS1 i2 = 1 − (c∗1 c2 + c1 c∗2 )2 4 2 r q 1 1 ∆S2 = h I2 i − hS2 i2 = 1 − (c1 c∗2 − c∗1 c2 )2 . 4 2 Thus for the left-hand side we find q q 1 1 (∆S1 )(∆S2 ) = √ 1 − (c1 c∗2 )2 − (c∗1 c2 )2 = √ 1 − 2r12 r22 cos(2(φ1 − φ2 )). 2 2 2 2 Thus the condition from the equality is q 1 1 √ 1 − 2r12 r22 cos(2(φ1 − φ2 )) = |r12 − r22 |. 4 2 2

Problem 29. Consider a d-dimensional Hilbert space with two orthonormal bases |b11 i, |b12 i, . . . |b1d i ∈ B1 |b21 i,

|b22 i,

...

|b2d i ∈ B2 .

The two bases are said to be mutually unbiased bases if 1 |hb2j |b1k i| = √ d for all j, k = 1, . . . , d and h | i denotes the scalar product in the Hilbert space. Consider the Hilbert space M2 (C) of 2 × 2 matrices over C, where the scalar product is defined as hA|Bi = tr(AB ∗ ),

A, B ∈ M2 (C).

Thus d = dim(M2 (C)) = 4. The standard basis in this Hilbert space is given by         0 0 1 0 0 1 0 0 . , E12 = , E21 = , E22 = E11 = 0 0 1 0 0 1 0 0

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24 Problems and Solutions Let UH be the Hadamard matrix   1 1 1 UH = √ , 2 1 −1

∗ UH = UH .

ejk (j, k = 1, 2) Show that the matrices E ejk = UH Ejk U ∗ , E H

j, k = 1, 2

and the standard basis form mutually unbiased bases. Solution 29.

Straightforward calculations yield   e11 = UH E11 U ∗ = 1 1 1 E H 2 1 1   1 1 −1 ∗ e12 = UH E12 UH E = 2 1 −1   1 1 e21 = UH E21 U ∗ = 1 E H 2 −1 −1   1 1 −1 ∗ e22 = UH E22 UH E = . 2 −1 1

It follows that

e2k )| = 1 for all j, k = 1, 2. |tr(E1j E 2 Thus we have mutually unbiased bases. Apply the vec-operator to the ejk (j, k = 1, 2) to find mutually unbiased bases in the matrices Ejk and E 4 Hilbert space C . Problem 30. Let x0 = ct. We define a linear bijection, h, between R4 and H(2), the set of complex 2 × 2 hermitian matrices, by   x0 + x1 x2 − ix3 (x0 , x1 , x2 , x3 ) → . x2 + ix3 x0 − x1 We denote the matrix on the right hand side by H. (i) Show that the matrix can be written as a linear combination of the Pauli spin matrices and the identity matrix I2 . (ii) Find the inverse map. (iii) Calculate the determinant of the 2 × 2 hermitian matrix H. Discuss. Solution 30. (i) We have H = x0 I2 + x1 σ3 + x2 σ1 + x3 σ2 . (ii) Consider (a, b ∈ R)     a c x0 + x1 x2 − ix3 = . c∗ b x2 + ix3 x0 − x1

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Qubits

25

Comparing the entries we obtain a+b , 2

x0 =

x1 =

a−b , 2

x2 =

c + c∗ , 2

x3 =

c∗ − c . 2i

(iii) We obtain det(H) = x20 − x21 − x22 − x23 . This is the Lorentz metric. det(U HU ∗ ) = det(H).

Let U be a unitary 2 × 2 matrix.

Then

Programming Problems Problem 1.

Consider the unary gates (2 × 2 unitary matrices)     1 1 1 0 1 , N= , H=√ 1 0 2 1 −1     1 0 1 0 V = , W = 0 eiπ/2 0 eiπ/4

and the normalized state 1 |ψi = √ 2

  1 . 1

Calculate the state N HV W |ψi and the expectation value hψ|N HV W |ψi. Solution 1.

Applying the Maxima program

/* unary.mac */ N: matrix([0,1],[1,0]); H: matrix([1/sqrt(2),1/sqrt(2)],[1/sqrt(2),-1/\sqrt(2)]); V: matrix([1,0],[0,exp(%i*%pi/2)]); W: matrix([1,0],[0,exp(%i*%pi/4)]); psi: matrix([1/sqrt(2)],[1/sqrt(2)]); psiT: matrix([1/sqrt(2),1/sqrt(2)]); R1: N . H . V . W; R1: ratsimp(R1); R2: R1 . psi; R2: ratsimp(R2); R3: psiT . R2; R3: ratsimp(R3);

we find the unitary matrix 1 N HV W = √ 2



1 1

−ei3π/4 ei3π/4





1 − ei3π/4 1 + ei3π/4



the normalized state N HV W |ψi =

1 2

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26 Problems and Solutions and

1 hψ|N HV W |ψi = √ . 2

Problem 2. in C2

Consider the Pauli spin matrix σ1 and the normalized state   cos(θ) |ψi = . sin(θ)

Calculate the variance Vσ1 (ψ) := hψ|σ12 |ψi − (hψ|σ1 |ψi)2 and discuss the dependence on θ Solution 2.

The following Maxima program

/* Variancesig1 */ sig1: matrix([0,1],[1,0]); sig2: matrix([0,-%i],[%i,0]); sig3: matrix([1,0],[0,-1]); psi: matrix([cos(theta)],[sin(theta)]); psiT: transpose(psi); Vs1: psiT . (sig1 . sig1) . psi - (psiT . sig1 . psi)^2; Vs1: trigsimp(Vs1); D1: diff(Vs1,theta); list: solve(D1=0,theta); theta1: rhs(part(list,1)); theta2: rhs(part(list,2)); theta3: rhs(part(list,3)); theta4: rhs(part(list,4)); r11: subst(theta1,theta,Vs1); r12: subst(theta2,theta,Vs1); r13: subst(theta3,theta,Vs1); r14: subst(theta4,theta,Vs1);

provides Vσ1 (ψ) := hψ|σ12 |ψi − (hψ|σ1 |ψi)2 = 1 + 4(cos4 (θ) − cos2 (θ)). Differentiation with respect to θ and solving the resulting equation provides that the variance is 1 for θ = 0 and θ = π/2. The variance is 0 for θ = π/4 and θ = 3π/4. Problem 3. Find the eigenvalues and normalized eigenvectors of the Hamilton operator   ˆ 1 H 1 1 ˆ =√ . K= ~ω 2 1 −1 Solution 3.

The Maxima program

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Qubits

27

/* Hamilton1.mac */ H: matrix([1,1],[1,-1])/sqrt(2); list: eigenvectors(H); p1: part(list,1); p11: part(p1,1); lam1: part(p11,1); lam2: part(p11,2); p2: part(list,2); v1: part(p2,1); v1: part(v1,1); v2: part(p2,2); v2: part(v2,1); v2T: transpose(v2); scalar: v1 . v2T; scalar: ratsimp(scalar);

provides the eigenvalues λ1 = −1, λ2 = 1 with the corresponding (nonnormalized) eigenvectors     1 1 √ √ v1 = , v2 = . − 2−1 2−1 The two eigenvectors are orthogonal to each other, i.e. scalar=0.

1.3

Supplementary Problems

Problem 1.

Consider the map f : C2 → R3 defined by     sin(2θ) cos(φ) cos(θ) f : 7→  sin(2θ) sin(φ)  . eiφ sin(θ) cos(2θ)

Are the vectors in C2 and R3 normalized? Consider the four normalized vectors in C2         1 1 1 1 1 1 1 1 √ , √ , √ , √ . 1 −1 i −i 2 2 2 2 Find the vectors in R3 . Problem 2. (i) Let x1 , x2 , x3 ∈ R and σ1 , σ2 , σ3 be the Pauli spin matrices. Show that sin(r) ei(x1 σ1 +x2 σ2 +x3 σ3 ) = cos(r)I2 + i (x1 σ1 + x2 σ2 + x3 σ3 ) r   cos(r) + ix3 sin(r)/r i(x1 − ix2 ) sin(r)/r = i(x1 + ix2 ) sin(r)/r cos(r) − ix3 sin(r)/r

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28 Problems and Solutions p where r := x21 + x22 + x23 . (ii) Let y1 , y2 , y3 ∈ R and X := x1 σ1 + x2 σ2 + x3 σ3 ,

Y := y1 σ1 + y2 σ2 + y3 σ3 .

Consider the maps 

 x1 X ↔ x =  x2  , x3



 y1 Y ↔ y =  y2  . y3

Let x · y := x1 y1 + x2 y2 + x3 y3 (scalar product). Show that x·y =

1 tr(XY ). 2

(iii) Show that   x2 y3 − x3 y2 i − [X, Y ] ↔ x × y =  x3 y1 − x1 y3  . 2 x1 y2 − x2 y1

ˆ be a 2 × 2 hermitian matrix. Consider the normalized Problem 3. Let H state  iφ  e cos(θ) |ψi = sin(θ) in the Hilbert space C2 . Assume that ˆ hψ|H|ψi = ~ω cos(φ) sin(2θ),

ˆ 2 |ψi = ~2 ω 2 . hψ|H

ˆ from these three assumptions. Note Reconstruct the hermitian matrix H that cos(θ) sin(θ) ≡

1 sin(2θ), 2

eiφ = cos(φ)+i sin(φ),

e−iφ = cos(φ)−i sin(φ).

Show that H = ~ωσ1 . Problem 4. Let A, B be n × n matrices over C. Let v be a normalized (column) vector in Cn . Let hAi := v∗ Av and hBi := v∗ Bv. We have the identity AB ≡ (A − hAiIn )(B − hBiIn ) + AhBi + BhAi − hAihBiIn . We approximate AB as AB ≈ AhBi + BhAi − hAihBiIn .

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Qubits

29

(i) Let n = 2 and A = σ1 ,

B = σ2 ,

1 v= √ 2

  1 . 1

Find AB and AhBi + BhAi − hAihBiI2 and the distance (Frobenius norm) between the two matrices. (ii) Let n = 2 and   1 1 A = σ1 , B = σ2 , v = √ . 2 −1 Find AB and AhBi + BhAi − hAihBiI2 and the distance (Frobenius norm) between the two matrices. (iii) Consider the case   cos(θ) A = σ1 , B = σ2 , v = . sin(θ) Find AB and AhBi + BhAi − hAihBiI2 and the distance (Frobenius norm) between the two matrices. Let α ∈ R. Show that the vectors     1 1 cosh(α) sinh(α) , v2 = q v1 = q sinh(α) − cosh(α) 1 + 2 sinh2 (α) 1 + 2 sinh2 (α)

Problem 5.

form an orthonormal basis in C2 . Find v1 v1T , v2 v2T , v1 v2T . Problem 6.

Let σ1 , σ2 , σ3 be the Pauli spin matrices. Show that Π+ =

1 (I2 + σj ), 2

Π− =

1 (I2 − σj ) 2

(j = 1, 2, 3) are projection matrices. Find the vectors     1 1 1 1 Π+ √ , Π− √ . 1 2 2 1 Are the vectors normalized? Problem 7. Given two arbitrary normalized states |ψi and |φi in C2 . Find a 2 × 2 unitary matrix U such that |ψi = U |φi, i.e. U must be expressed in terms of the components of the states |ψi and φi. Since U is unitary we have U −1 = U ∗ .

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30 Problems and Solutions Problem 8.

Let A, B be 2 × 2 hermitian matrices and   cos(θ) |ψi = . sin(θ)

Find the minima of the function f (θ) = kAB − Ahψ|B|ψi − hψ|A|ψiB + hψ|A|ψihψ|B|ψiI2 k where k.k denotes the norm. Problem 9. Let |0i, |1i be an orthonormal basis in C2 and z00 , z01 , z10 , z11 be complex numbers. Calculate exp(z00 |0ih0| + z01 |0ih1| + z10 |1ih0| + z11 |1ih1|). Then set z00 = −i~ω1 ,

z01 = −z10 = −i~ω2 ,

z11 = −i~ω3 .

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Chapter 2

Kronecker Product and Tensor Product

2.1

Introduction

Let H1 and H2 be two Hilbert spaces and H be a third Hilbert space defined in terms of H1 and H2 with the following specifications. For each pair of vectors f1 , f2 in H1 , H2 , respectively, there are vectors in H denoted by f1 ⊗ f2 and g1 ⊗ g2 , respectively such that hf1 ⊗ f2 |g1 ⊗ g2 i = hf1 |g1 iH1 hf2 |g2 iH2 where hf1 |g1 i is the scalar product in the Hilbert space H1 . The vector f1 ⊗ f2 is called the tensor product of the vectors f1 and f2 . The Hilbert space H consists of the linear combinations of the vectors f1 ⊗ f2 together with the strong limits of their Cauchy sequences. We term H the tensor product of H1 and H2 and denote it by H1 ⊗H2 . Given a basis { |φi i : i ∈ I } in the Hilbert H1 and a basis { |ψj i : j ∈ J } in the Hilbert space H2 we can construct a basis { |φi i ⊗ |ψj i : i ∈ I, j ∈ J } in the product Hilbert space. The tensor product is associative and distributive. If Aˆ1 and Aˆ2 are linear operators in H1 and H2 , respectively, we define the operator Aˆ1 ⊗ Aˆ2 in H1 ⊗ H2 by the formula (Aˆ1 ⊗ Aˆ2 )(f1 ⊗ f2 ) = (Aˆ1 f1 ) ⊗ (Aˆ2 f2 ) . 31

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32 Problems and Solutions Aˆ1 ⊗ Aˆ2 is called the tensor product of Aˆ1 and Aˆ2 . Similarly we can define the tensor product of n Hilbert spaces. For the finite-dimensional Hilbert spaces Cn and Rn the tensor product is realized by the Kronecker product. Let A be an m×n matrix and B be an r ×s matrix. The Kronecker product of A and B is defined as the (m · r) × (n · s) matrix a11 B  a21 B A ⊗ B :=   ...

a12 B a22 B .. .

... ... .. .

 a1n B a2n B  . ..  . 

am1 B

am2 B

...

amn B



We have the following properties. 1) If vj (j = 1, . . . , n) is an orthonormal basis in Cn , then vj ⊗ vk ,

(j, k = 1, . . . , n)

2

is an orthonormal basis in Cn . 2) If A, B are normal matrices, then A ⊗ B is a normal matrix. 3) If A, B are hermitian matrices, then A ⊗ B is a hermitian matrix. 4) If A, B are unitary matrices, then A ⊗ B is a unitary matrix. 5) If A, B are projection matrices, then A ⊗ B is a projection matrix. 6) If A, B are nilpotent matrices, then A ⊗ B is a nilpotent matrix. 7) If P1 , P2 are permutation matrices, then P1 ⊗P2 is a permutation matrix. 8) If A and B are invertible, then A ⊗ B is invertible with (A ⊗ B)−1 ≡ A−1 ⊗ B −1 . Let A, B, C, D be matrices and assume that the matrix products AC and BD exist. Then (A ⊗ B)(C ⊗ B) = (AC) ⊗ (BD). Let A be an m × m matrix and B be an n × n matrix. The underlying field is C. Let Im , In be the m × m and n × n unit matrix, respectively. Then tr(A ⊗ B) = tr(A)tr(B) and tr(A ⊗ In + Im ⊗ B) = n tr(A) + m tr(B).

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Kronecker Product and Tensor Product

2.2

33

Solved Problems

Problem 1.

(i) Let |φ1 i :=

  1 , 0

|φ2 i :=

  0 . 1

Thus the set { |φ1 i, |φ2 i } forms a basis in C2 (the standard basis). Calculate the vectors in C4 |φ1 i ⊗ |φ1 i,

|φ1 i ⊗ |φ2 i,

|φ2 i ⊗ |φ1 i,

|φ2 i ⊗ |φ2 i

and interpret the result. (ii) Consider the Pauli matrices σ1 and σ3 . Find σ1 ⊗ σ3 and σ3 ⊗ σ1 and discuss. Both σ1 and σ3 are hermitian. Are σ1 ⊗ σ3 and σ3 ⊗ σ1 hermitian? Both σ1 and σ3 are unitary. Is σ1 ⊗ σ3 and σ3 ⊗ σ1 unitary? Solution 1.

(i) We obtain   1 0 |φ1 i ⊗ |φ1 i =   , 0 0   0 0 |φ2 i ⊗ |φ1 i =   , 1 0

  0 1 |φ1 i ⊗ |φ2 i =   , 0 0   0 0 |φ2 i ⊗ |φ2 i =   . 0 1

Thus we find the standard basis in C4 from the standard basis in C2 . (ii) We obtain   0 0 1 0  0 0 0 −1  σ1 ⊗ σ3 =   1 0 0 0 0 −1 0 0 and

0 1 σ3 ⊗ σ1 =  0 0 

1 0 0 0

0 0 0 −1

 0 0  . −1 0

We note that σ1 ⊗ σ3 6= σ3 ⊗ σ1 . σ1 ⊗ σ3 and σ3 ⊗ σ1 are hermitian. σ1 ⊗ σ3 and σ3 ⊗ σ1 are unitary. Problem 2.

Given the orthonormal basis  iφ    e cos(θ) − sin(θ) |ψ1 i = , |ψ2 i = sin(θ) e−iφ cos(θ)

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34 Problems and Solutions in the Hilbert space C2 . Use this basis to find a basis in C4 . Solution 2.

A basis in C4 is given by

{ |ψ1 i ⊗ |ψ1 i,

|ψ1 i ⊗ |ψ2 i,

|ψ2 i ⊗ |ψ1 i,

|ψ2 i ⊗ |ψ2 i }

since (hψj | ⊗ hψk |)(|ψm i ⊗ |ψn i) = δjm δkn where j, k, m, n = 1, 2. Problem 3. A system of n-qubits can be represented by a finite-dimensional Hilbert space over the complex numbers of dimension 2n . A state |ψi of the system is a superposition of the basic states |ψi =

1 X

cj1 j2 ...jn |j1 i ⊗ |j2 i ⊗ · · · ⊗ |jn i.

j1 ,j2 ,...,jn =0

In a short cut notation this state is written as |ψi =

1 X

cj1 j2 ...jn |j1 j2 . . . jn i.

j1 ,j2 ,...,jn =0

Consider as a special case the state |ψi =

1 1 (|0i⊗|0i+|0i⊗|1i+|1i⊗|0i+|1i⊗|1i) ≡ (|00i+|01i+|10i+|11i) 2 2

in the Hilbert space H = C4 (n = 2). Can this state be written as a product state? Solution 3.

Yes, the state can be written as product state. We have 1 1 √ (|0i + |1i) ⊗ √ (|0i + |1i). 2 2

Problem 4. The single-bit Walsh-Hadamard transform is the unitary map W given by 1 W |0i = √ (|0i + |1i), 2

1 W |1i = √ (|0i − |1i). 2

The n-bit Walsh-Hadamard transform Wn is defined as Wn := W ⊗ W ⊗ · · · ⊗ W

(n − times).

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Kronecker Product and Tensor Product

35

Consider n = 2. Find the normalized state W2 (|0i ⊗ |0i). Solution 4.

We have

W2 (|0i ⊗ |0i) = (W ⊗ W )(|0i ⊗ |0i) = W |0i ⊗ W |0i. Thus W2 (|0i ⊗ |0i) =

1 ((|0i + |1i) ⊗ (|0i + |1i)). 2

Finally W2 (|0i ⊗ |0i) =

1 (|0i ⊗ |0i + |0i ⊗ |1i + |1i ⊗ |0i + |1i ⊗ |1i). 2

W2 generates a linear combination of all states. This also applies to Wn . Problem 5.

Consider the spin matrix S1 for spin- 21   1 1 0 1 S1 = σ1 = 2 2 1 0

with the eigenvalues 1/2 and −1/2 and the corresponding normalized eigenvectors     1 1 1 1 √ √ e1/2 = , e−1/2 = . 2 1 2 −1 Do the four vectors 1 √ (e1/2 ⊗ e1/2 + e−1/2 ⊗ e−1/2 ), 2

1 √ (e1/2 ⊗ e1/2 − e−1/2 ⊗ e−1/2 ), 2

1 √ (e1/2 ⊗ e−1/2 + e−1/2 ⊗ e1/2 ), 2

1 √ (e1/2 ⊗ e−1/2 − e−1/2 ⊗ e1/2 ) 2

form a basis in C4 ? Prove or disprove. Solution 5. We obtain the Bell basis         1 0 1 0 1 0 1 1 1  0  1  −1  v1 = √   , v2 = √   , v3 = √   , v4 = √   0 1 2 0 2 1 2 2 1 0 −1 0 which forms an orthonormal basis in C4 Problem 6.

Let A be an arbitrary n × n matrix over C. Show that exp(A ⊗ In ) ≡ exp(A) ⊗ In .

(1)

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36 Problems and Solutions Solution 6.

Using the expansion

exp(A ⊗ In ) =

∞ X (A ⊗ In )k k=0

k!

= In ⊗ In +

1 1 1 (A ⊗ In ) + (A ⊗ In )2 + (A ⊗ In )3 + · · · 1! 2! 3!

and (A ⊗ In )k = Ak ⊗ In , k ∈ N we find identity (1). Problem 7. Let A, B be arbitrary n × n matrices over C. Let In be the n × n unit matrix. Show that exp(A ⊗ In + In ⊗ B) ≡ exp(A) ⊗ exp(B).

Solution 7. and

The proof of this identity relies on [A ⊗ In , In ⊗ B] = 0n2

(A ⊗ In )r (In ⊗ B)s ≡ (Ar ⊗ In )(In ⊗ B s ) ≡ Ar ⊗ B s ,

r, s ∈ N.

Thus exp(A ⊗ In + In ⊗ B) =

∞ X (A ⊗ In + In ⊗ B)j

j!

j=0

=

  j ∞ X X 1 j (A ⊗ In )k (In ⊗ B)j−k j! k j=0 k=0

j ∞ X X

  1 j (Ak ⊗ B j−k ) j! k j=0 k=0   ! ∞ ∞ j X X Bk A   = ⊗ j! k! j=0 =

k=0

= exp(A) ⊗ exp(B).

Problem 8. Let A and B be arbitrary n × n matrices over C. Prove or disprove the equation eA⊗B = eA ⊗ eB . Solution 8. Obviously this is not true in general. For example, let A = B = In . Then eA⊗B = eIn2

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Kronecker Product and Tensor Product

37

and eA ⊗ eB = eIn ⊗ eIn 6= eIn2 . Problem 9. Let A be an m × m normal matrix and B be an n × n normal matrix. The underlying field is C. The eigenvalues and eigenvectors of A are given by λ1 , λ2 , . . . , λm and u1 , u2 , . . . , um . The eigenvalues and eigenvectors of B are given by µ1 , µ2 , . . . , µn and v1 , v2 , . . . , vn . Let 1 , 2 and 3 be real parameters. Find the eigenvalues and eigenvectors of the matrix 1 A ⊗ B + 2 A ⊗ In + 3 Im ⊗ B. Solution 9.

Let x ∈ Cm and y ∈ Cn . Then we have (A ⊗ B)(x ⊗ y) = (Ax) ⊗ (By),

(A ⊗ In )(x ⊗ y) = (Ax) ⊗ y,

(Im ⊗ B)(x ⊗ y) = x ⊗ (By).

Thus the eigenvectors of the matrix are uj ⊗ vk ,

j = 1, 2, . . . , m

k = 1, 2, . . . , n.

The corresponding eigenvalues are given by 1 λj µk + 2 λj + 3 µk . Problem 10. Let A, B be n × n matrices over C. The n × n matrices form a vector space. A scalar product can be defined as hA, Bi := tr(AB ∗ ). This provides a Hilbert space. The scalar product implies a norm kAk2 = hA, Ai = tr(AA∗ ). This norm is called the Hilbert-Schmidt norm. (i) Consider the Dirac matrices    0 1 0 0 0 0   0 0 1 0 γ1 :=  γ0 :=  , 0 0 0 −1 0 −1 0 0 0 −1

0 0 −1 0

 0 1 1 0 . 0 0 0 0

Calculate the scalar product hγ0 , γ1 i. Discuss. (ii) Let U be a unitary n × n matrix. Find the scalar product hU A, U Bi. (iii) Let C, D be m × m matrices over C. Find hA ⊗ C, B ⊗ Di. Solution 10. (i) We find hγ0 , γ1 i = tr(γ0 γ1∗ ) = 0. Thus γ0 and γ1 are orthogonal to each other.

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38 Problems and Solutions (ii) Since tr(U A(U B)∗ ) = tr(U AB ∗ U ∗ ) = tr(U ∗ U AB ∗ ) = tr(AB ∗ ) where we used the cyclic invariance for matrices, we find that hU A, U Bi = hA, Bi. Thus the scalar product is invariant under unitary transformations. (iii) Since tr((A ⊗ C)(B ⊗ D)∗ ) = tr((A ⊗ C)(B ∗ ⊗ D∗ )) = tr((AB ∗ ) ⊗ (CD∗ )) = tr(AB ∗ )tr(CD∗ ) we find hA ⊗ C, B ⊗ Di = hA, BihC, Di. Problem 11.

Let T be the 4 × 4 matrix   3 X T := I2 ⊗ I2 + t j σj ⊗ σj  j=1

where σj , j = 1, 2, 3 are the Pauli spin matrices and −1 ≤ tj ≤ +1, j = 1, 2, 3. Find the matrix T 2 . Solution 11.

We have

T 2 = I2 ⊗ I2 + 2

3 X

t j σj ⊗ σj +

j=1

3 X 3 X

tj tk σj σk ⊗ σj σk .

j=1 k=1

Since σ1 σ2 = iσ3 ,

σ2 σ1 = −iσ3 ,

σ2 σ3 = iσ1 ,

σ3 σ2 = −iσ1 ,

σ3 σ1 = iσ2 ,

σ1 σ3 = −iσ2

and σ12 = I2 , σ22 = I2 , σ32 = I2 , we find 3 X

tj tk σj σk ⊗σj σk ≡ I2 ⊗I2

3 X

t2j −2(t1 t2 σ3 ⊗σ3 +t2 t3 σ1 ⊗σ1 +t3 t1 σ2 ⊗σ2 ).

j=1

j,k=1

Therefore  T 2 = (I2 ⊗ I2 ) 1 +

3 X

 t2j 

j=1

+2(t1 − t2 t3 )σ1 ⊗ σ1 + 2(t2 − t3 t1 )σ2 ⊗ σ2 + 2(t3 − t1 t2 )σ3 ⊗ σ3 .

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Kronecker Product and Tensor Product

39

Problem 12. Let { |0i, |1i, . . . , |n − 1i } be an orthonormal basis in the Hilbert space Cn . Is   n−2 1 X |ψi = √ |ji ⊗ |j + 1i + |n − 1i ⊗ |0i n j=0 independent of the chosen orthonormal basis? Prove or disprove. Solution 12.

Consider the special case R2 . Let     1 0 |0i = , |1i = . 0 1

Thus   0         1 1 1 1 1 0 0 1 ⊗ +√ ⊗ = √  . |ψi = √ 1 0 2 0 2 1 2 1 0 Now let

1 |0i = √ 2

  1 , 1

1 |1i = √ 2



1 −1

 .

Then   1         1 1 1 1 1 1  0  1 1 1 |ψi = √ ⊗ +√ ⊗ =√  . 0 1 −1 −1 1 2 2 2 2 2 −1 Thus, |ψi depends on the chosen basis. Problem 13. are given by

In the product Hilbert space C4 ∼ = C2 ⊗ C2 the Bell states

1 |Φ+ i = √ (|0i ⊗ |0i + |1i ⊗ |1i), 2

1 |Φ− i = √ (|0i ⊗ |0i − |1i ⊗ |1i), 2

1 |Ψ+ i = √ (|0i ⊗ |1i + |1i ⊗ |0i), 2

1 |Ψ− i = √ (|0i ⊗ |1i − |1i ⊗ |0i) 2

and form an orthonormal basis in C4 . Here, { |0i, |1i } is an arbitrary orthonormal basis in the Hilbert space C2 . Let   iφ   iφ e cos(θ) −e sin(θ) , |1i = . |0i = cos(θ) sin(θ)

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40 Problems and Solutions (i) Find |Φ+ i, |Φ− i, |Ψ+ i, |Ψ− i for this basis. (ii) Consider the special case when φ = 0 and θ = 0. Solution 13.

(i) We obtain  2iφ   2iφ  e e cos(2θ) iφ 1  e sin(2θ)  1  0  |Φ− i = √  iφ |Φ+ i = √  , , 0 e sin(2θ) 2 2 1 − cos(2θ)  2iφ    −e sin(2θ) 0 1  eiφ cos(2θ)  1  eiφ  |Ψ+ i = √  iφ |Ψ− i = √  iφ  . , e cos(2θ) 2 2 −e sin(2θ) 0

(ii) If we choose φ = 0 and θ = 0 which simply means we choose the standard basis for |0i and |1i (i.e. |0i = (1 0)T and |1i = (0 1)T ), we find that the Bell states take the form     1 1 1 0 1  0  |Φ+ i = √   , |Φ− i = √  , 0 2 0 2 1 −1     0 0 1 1 1  1  |Ψ− i = √  |Ψ+ i = √   , . 1 2 2 −1 0 0 Problem 14. Let HA and HB be two p-dimensional Hilbert spaces over C, where p is a prime number. Let { |0iA , |1iA , . . . , |(p − 1)iA },

{ |0iB , |1iB , . . . , |(p − 1)iB }

be orthonormal bases in these Hilbert spaces. We define the states p−1

1 X |ψ(a, b)i := (Ip ⊗ X Z ) √ |jiA ⊗ |jiB p j=0 a

b

in the Hilbert space HA ⊗ HB , where a, b ∈ { 0, 1, . . . , p − 1 }. The p × p matrices X and Z are defined as X|ji = |j + 1 mod pi,

Z|ji = ω j |ji,

j = 0, 1, . . . , p

with a complex primitive pth root ω of 1 and { |0i, |1i, . . . , |p − 1i } is the orthonormal basis given above for the Hilbert space HB . Calculate the states |ψ(0, 0)i and |ψ(1, 1)i.

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Kronecker Product and Tensor Product Solution 14.

41

Since

X 0 = Z 0 = Ip ,

Ip |jiA = |jiA ,

we obtain

Ip |jiB = |jiB

p−1

1 X |ψ(0, 0)i = √ |jiA ⊗ |jiB . p j=0 Using (Ip ⊗ XZ)(|jiA ⊗ |jiB = |jiA ⊗ (XZ|jiB ) = |jiA ⊗ ω j X|jiB = |jiA ⊗ ω j |jB + 1 mod pi = ω j |jiA ⊗ |jB + 1 mod pi we find

p−1

1 X j |ψ(1, 1)i = √ ω |jiA ⊗ |jB + 1 mod pi. p j=0 The states |ψ(a, b)i are maximally entangled states in the Hilbert space HA ⊗ HB . Problem 15.

Consider the Pauli matrices σ1 and σ2 and the GHZ state             1 1 1 1 0 0 0 √ |ψi = ⊗ ⊗ + ⊗ ⊗ . 0 0 0 1 1 1 2

(i) Show that |ψi is an eigenvector of the operator σ2 ⊗ σ2 ⊗ σ1 . What is the eigenvalue? (ii) Is             1 1 1 1 0 0 0 |φi = √ ⊗ ⊗ + eiφ ⊗ ⊗ 0 0 0 1 1 1 2 an eigenvector of σ2 ⊗ σ2 ⊗ σ1 ? Solution 15. (i) We have                 1 0 0 1 1 0 0 −i σ1 = , σ1 = , σ2 = , σ2 = . 0 1 1 0 0 i 1 0 Thus        1 1 1 σ2 ⊗ σ2 ⊗ σ1 0 0 0       0 0 0 ⊗ σ1 + σ2 ⊗ σ2 1 1 1

1 (σ2 ⊗ σ2 ⊗ σ1 )|ψi = √ 2

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42 Problems and Solutions             1 0 0 0 −i −i 1 =√ ⊗ ⊗ + ⊗ ⊗ i i 1 0 0 0 2             1 0 0 0 1 1 1 ⊗ ⊗ + ⊗ ⊗ = −√ 1 1 1 0 0 0 2 = −|ψi. Consequently, the eigenvalue is −1. (ii) We find using the calculation from (i)              1 1 1 1 0 0 0 eiφ ⊗ ⊗ + ⊗ ⊗ . (σ2 ⊗σ2 ⊗σ1 )|φi = − √ 0 0 0 1 1 1 2 Thus, in general |φi is not an eigenvector of σ2 ⊗ σ2 ⊗ σ1 , for example if φ = π/4. Problem 16.

Consider the three-qubit GHZ state 1 |ψi = √ (|0i ⊗ |0i ⊗ |0i + |1i ⊗ |1i ⊗ |1i) 2

with the standard basis   1 |0i = , 0

  0 |1i = . 1

Let σ1 , σ2 be the Pauli spin matrices. (i) Calculate the expectation values hψ|(σ2 ⊗ σ2 ⊗ σ2 )|ψi,

hψ|(σ1 ⊗ σ1 ⊗ σ2 )|ψi,

hψ|(σ1 ⊗ σ2 ⊗ σ1 )|ψi,

hψ|(σ2 ⊗ σ1 ⊗ σ1 )|ψi.

(ii) Calculate the expectation values

Solution 16.

hψ|(σ1 ⊗ σ2 ⊗ σ2 )|ψi,

hψ|(σ2 ⊗ σ1 ⊗ σ2 )|ψi,

hψ|(σ2 ⊗ σ2 ⊗ σ1 )|ψi,

hψ|(σ1 ⊗ σ1 ⊗ σ1 )|ψi.

(i) We find

hψ|(σ2 ⊗ σ2 ⊗ σ2 )|ψi = 0,

hψ|(σ1 ⊗ σ1 ⊗ σ2 )|ψi = 0,

hψ|(σ1 ⊗ σ2 ⊗ σ1 )|ψi = 0,

hψ|(σ2 ⊗ σ1 ⊗ σ1 )|ψi = 0.

(ii) We obtain hψ|(σ1 ⊗ σ2 ⊗ σ2 )|ψi = −1,

hψ|(σ2 ⊗ σ1 ⊗ σ2 )|ψi = −1,

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Kronecker Product and Tensor Product hψ|(σ2 ⊗ σ2 ⊗ σ1 )|ψi = −1, Problem 17.

43

hψ|(σ1 ⊗ σ1 ⊗ σ1 )|ψi = 1.

Consider the Bell state         1 1 0 0 1 |Ψ− i = √ ⊗ − ⊗ . 0 1 1 0 2

Let n, m be unit vectors in R3 . Calculate the expectation value E(n, m) = hΨ− |(n · σ) ⊗ (m · σ)|Ψ− i.

Solution 17.

Straightforward calculation yields

E(n, m) = −n1 m1 −n2 m2 −n3 m3 = −n·m = −|n|·|m| cos(θ) = − cos(θm,n ) since |m| = |n| = 1. We write cos(θm,n ) instead of cos(θ) in order to indicate that θn,m is the angle between the quantization directions m and n. Problem 18. Let A1 , A2 , . . . , Am and B1 , B2 , . . . , Bn be two sets of 4 × 4 matrices over C. Assume that Aj Bk = Bk Aj for all j, k with j = 1, 2, . . . , m and k = 1, 2, . . . , n. Find two such sets of matrices using the Kronecker product of 2 × 2 matrices. Solution 18. If Aj = Cj ⊗ I2 , Bk = I2 ⊗ Dk , where Cj , Dk are arbitrary 2 × 2 matrices we have [Aj , Bk ] = [Cj ⊗ I2 , I2 ⊗ Dk ] = Cj ⊗ Dk − Cj ⊗ Dk = 04 for all j, k. Problem 19.

Let A, B be n × n hermitian matrices over C and H = ~ω(A ⊗ B)

be a Hamilton operator, where ~ is the Planck constant and ω the frequency. The Heisenberg equation of motion for the operator B ⊗ A is given by i~

d(B ⊗ A) = [B ⊗ A, H](t). dt

(i) Assume that [A, B] = 0n . Simplify the Heisenberg equation of motion using this condition. (ii) Assume that [A, B]+ = 0n Simplify the Heisenberg equation of motion using this condition. Give an example for such matrices.

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44 Problems and Solutions (iii) Assume that [A, B]+ = In . Simplify the Heisenberg equation of motion using this condition. Give an example for such matrices. Solution 19.

(i) We have

ˆ [B ⊗ A, H](t) = ~ω[B ⊗ A, A ⊗ B](t) = ~ω((B ⊗ A)(A ⊗ B) − (A ⊗ B)(B ⊗ A))(t) = ~ω((BA) ⊗ (AB) − (AB) ⊗ (BA))(t). In general, (AB) ⊗ (BA) 6= (BA) ⊗ (AB). If [A, B] = 0n we find i~

d(B ⊗ A) = 0n2 . dt

Thus B ⊗ A is a constant of motion. (ii) From [A, B]+ = 0n we have AB = −BA. Thus we also have i~

d(B ⊗ A) = 0n2 dt

in this case. An example are the Pauli spin matrices. For example, let A = σ1 and B = σ2 . Then [A, B]+ = 02 . Another example is given by Fermi operators. They have the matrix representation (j = 1, 2, . . . , N )   1 σ+ ⊗ I2 ⊗ · · · ⊗ I2 c†j = σ3 ⊗ · · · ⊗ σ3 ⊗ 2   1 σ− ⊗ I2 · · · ⊗ I2 cj = σ3 ⊗ · · · ⊗ σ3 ⊗ 2 where σ+ and σ− appears in the j-th place and where σ+ := σ1 + iσ2 , σ− := σ1 − iσ2 . We have [c†k , cj ]+ = δjk I and [c†k , c†j ]+ = [ck , cj ]+ = 0. (iii) Since BA = −AB + In we find i~

d(B ⊗ A) = ~ω(In ⊗ (AB) − (AB) ⊗ In )(t). dt

Problem 20. The four Bell states with spin J (J = 1/2, 1, 3/2, 2, . . .) are given by |B1 i = √

2J X 1 |ki ⊗ |ki 2J + 1 k=0

|B2 i = √

2J X 1 (−1)k |ki ⊗ |ki 2J + 1 k=0

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Kronecker Product and Tensor Product

|B3 i = √

2J X 1 |ki ⊗ |2J − ki 2J + 1 k=0

|B4 i = √

2J X 1 (−1)k |ki ⊗ |2J − ki. 2J + 1 k=0

45

For J = 1/2 we obtain the standard Bell basis in C4 . Note that for J = 1/2 the four Bell states form an orthonormal basis in C4 . Show that for spin J = 1 the Bell states are linearly dependent. Solution 20.

For J = 1 the Bell states are 1 |B1 i = √ (|0i ⊗ |0i + |1i ⊗ |1i + |2i ⊗ |2i) 3 1 |B2 i = √ (|0i ⊗ |0i − |1i ⊗ 1i + |2i ⊗ |2i) 3 1 |B3 i = √ (|0i ⊗ |2i + |1i ⊗ |1i + |2i ⊗ |0i) 3 1 |B4 i = √ (|0i ⊗ |2i − |1i ⊗ |1i + |2i ⊗ |0i). 3

Let c1 , c2 , c3 , c4 ∈ C. Then the equation c1 |B1 i + c2 |B2 i + c3 |B3 i + c4 |B4 i = 0 provides the solution c2 = −c1 , c3 = −c1 , c4 = c1 with c1 arbitrary. Thus the four states are not linearly independent. We also have hB1 |B2 i =

1 , 3

1 hB2 |B3 i = − , 3 Problem 21.

1 , 3

hB1 |B3 i =

hB2 |B4 i =

1 hB1 |B4 i = − , 3

1 , 3

hB3 |B4 i =

Consider the state in C4 |ψi =

1 X

cj1 j2 |j1 i ⊗ |j2 i

j1 ,j2 =0

where |0i =

  1 , 0

Let S, T be 2 × 2 matrices over C   s11 s12 S= , s21 s22

|1i =

  0 . 1

 T =

t11 t21

t12 t22



1 . 3

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10943 - Problems and Solutions in Quantum Computing and Quantum Information

46 Problems and Solutions where det(S) = 1, det(T ) = 1. This means that S and T are elements of the Lie group SL(2, C). Let 1 X

dj1 j2 |j1 i ⊗ |j2 i = (S ⊗ T )

j1 ,j2 =0

1 X

cj1 j2 |j1 i ⊗ |j2 i.

j1 ,j2 =0

Show that d00 d11 − d01 d10 = c00 c11 − c01 c10 .

(1)

Owing to (1) the quantity c00 c11 − c01 c10 is called an invariant. Solution 21.

Since s11 t11  s11 t21 S⊗T = s21 t11 s21 t21 

s11 t12 s11 t22 s21 t12 s21 t22

s12 t11 s12 t21 s22 t11 s22 t21

 s12 t12 s12 t22   s22 t12 s22 t22

we have        s11 t11 s11 t12 s12 t11 s12 t12 s t  s t  s t  s t  (S⊗T )|ψi = c00  11 21 +c01  11 22 +c10  12 21 +c11  12 22  . s21 t11 s21 t12 s22 t11 s22 t12 s21 t21 s21 t22 s22 t21 s22 t22 

Thus d00 = c00 s11 t11 + c01 s11 t12 + c10 s12 t11 + c11 s12 t12 d01 = c00 s11 t21 + c01 s11 t22 + c10 s12 t21 + c11 s12 t22 d10 = c00 s21 t11 + c01 s21 t12 + c10 s22 t11 + c11 s22 t12 d11 = c00 s21 t21 + c01 s21 t22 + c10 s22 t21 + c11 s22 t22 . From det(S) = 1 and det(T ) = 1 it follows that s11 s22 − s12 s21 = 1 and t11 t22 − t12 t21 = 1. Thus s11 s22 t11 t22 + s12 s21 t12 t21 − s11 s22 t12 t21 − s12 s21 t11 t22 = 1. Using this result we obtain (1). Problem 22. Let I2 be the 2 × 2 identity matrix and let σ1 , σ2 and σ3 be the Pauli spin matrices. (i) Let A be an m × m matrix over C and B be an n × n matrix over C. Find all solutions for m, n, A and B satisfying 1 1 1 1 I2 ⊗ I2 + σ1 ⊗ σ1 + σ2 ⊗ σ2 + σ3 ⊗ σ3 = A ⊗ B. 2 2 2 2

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47

(ii) Let C, D, E, and F be 2 × 2 matrices over C. Find all solutions for C, D, E and F satisfying 1 1 1 1 I2 ⊗ I2 + σ1 ⊗ σ1 + σ2 ⊗ σ2 + σ3 ⊗ σ3 = C ⊗ D + E ⊗ F. 2 2 2 2 (iii) Find the set stabilized by S = { σ1 ⊗ I2 , I2 ⊗ σ1 } i.e. find  u ∈ C4 | ∀ A ∈ S : Au = u .

Solution 22.

(i) Straightforward calculation yields  1 0 1 1 1 1 0 0 I2 ⊗ I2 + σ1 ⊗ σ1 + σ2 ⊗ σ2 + σ3 ⊗ σ3 =  0 1 2 2 2 2 0 0

0 1 0 0

 0 0  = A ⊗ B. 0 1

Since A ⊗ B is an mn × mn matrix we have mn = 4. For m = n = 2 we consider     a1 a2 b1 b2 A= , B= . a3 a4 b3 b4 Thus we have a1 b1 = 1, a1 b4 = 0 and a4 b4 = 1. Consequently a1 6= 0 and b4 6= 0, so that a1 b4 6= 0. Thus m = n = 2 does not yield a solution. For m = 1 and n = 4 we find the solution  1 0 0 1 0 0 1 A = (a), B=  a 0 1 0 0 0 0 For m = 4 and n = 1 we find  1 0 0 1 0 0 1 A=  b 0 1 0 0 0 0

 0 0 , 0 1

the solution  0 0 B = (b), , 0 1

a ∈ C/{0}.

b ∈ C/{0}.

(ii) Clearly C and E, and D and F must be linearly independent, otherwise we would have the case m = n = 2 discussed in (a) which has no solution. Let     c1 c2 e1 e2 C= , E= . c3 c4 e3 e4 We have to satisfy the equations   1 0 c1 D + e1 F = , 0 0

 c2 D + e2 F =

0 1

0 0

 ,

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48 Problems and Solutions  c3 D + e3 F =

0 0

1 0



 ,

c4 D + e4 F =

0 0

0 1

 .

It follows that {D, F } should be a basis for the 2 × 2 matrices over C, however the span of {D, F } is a two dimensional space whereas the 2 × 2 matrices over C forms a four dimensional space. Thus we have a contradiction. There are no solutions. (iii) Thus the vector u must simultaneously be an eigenvector of σ1 ⊗I2 and I2 ⊗ σ1 with eigenvalue 1. The eigenspace corresponding to the eigenvalue 1 from σ1 ⊗ I2 yields         1 1 1 0 u=α ⊗ +β ⊗ 1 0 1 1 where α, β ∈ C. We must now satisfy         1 1 1 0 (I2 ⊗ σ1 )u = α(I2 ⊗ σ1 ) ⊗ + β(I2 ⊗ σ1 ) ⊗ 1 0 1 1         1 0 1 1 =α ⊗ +β ⊗ 1 1 1 0 =u so that α = β. Thus         1 1 1 0 u=α ⊗ +α ⊗ , 1 0 1 1

α ∈ C.

N

Problem 23. Let N ≥ 1. Consider the Hilbert space C2 . The (N + 1) Dicke states are defined by |N/2, ` − N/2i := p

1 NC

`

(|0i ⊗ · · · ⊗ |0i ⊗ |1i ⊗ · · · ⊗ |1i +permutations) {z } | {z } | `

N −`

where ` = 0, 1, . . . , N and N

C` = N !/(`!(N − `)!).

Write down the Dicke states for N = 2 and N = 3. Which of the states are entangled? Solution 23.

For N = 2 we have the three states in the Hilbert space C4

|1, −1i = |1i ⊗ |1i,

1 |1, 0i = √ (|0i ⊗ |1i + |1i ⊗ |0i), 2

|1, 1i = |0i ⊗ |0i.

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49

The first and the last states are product states. The second state is a Bell state and fully entangled. For N = 3 we find the four states in the Hilbert space C8 3 3 , − i = |1i ⊗ |1i ⊗ |1i 2 2 3 1 , − i = √1 (|0i ⊗ |1i ⊗ |1i + |1i ⊗ |0i ⊗ |1i + |1i ⊗ |1i ⊗ |0i) 2 2 3 3 1 1 , i = √ (|0i ⊗ |0i ⊗ |1i + |0i ⊗ |1i ⊗ |0i + |1i ⊗ |0i ⊗ |0i) 2 2 3 3 3 , i = |0i ⊗ |0i ⊗ |0i. 2 2 Obviously the first and last states are product states. The other two states are entangled. Problem 24. Can we find 2 × 2 matrices A, B, C with det(A) = 1, det(B) = 1 and det(C) = 1 such that     0 1 1 0     1 0    1 0 1  0 √   = (A ⊗ B ⊗ C) √   ? 1 0 3 2     0 0     0 0 0 1 On the left-hand side we have the W state and on the right-hand side we have the GHZ state. Solution 24.

Let  a11 A= a21

a12 a22

 det(A) = a11 a22 − a12 a21 = 1

,

etc. Thus the condition yields 8 equations and we have the three constraints a11 a22 − a12 a21 = 1,

b11 b22 − b12 b21 = 1,

c11 c22 − c12 c21 = 1.

There is no solution for this system. Problem 25.

Consider the vector v=

1 (1 2

0

1

1

0

T

1 ) ∈ C6 .

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50 Problems and Solutions Find a Schmidt decomposition of v over C6 = C2 ⊗ C3 and over C6 = C3 ⊗ C2 . Over C2 ⊗ C3 we have             1 1 1 1 1 1 1 1 0 1 v= ⊗ 0 + ⊗ 0 = √ ⊗ √ 0 2 0 2 1 2 1 2 1 1 1

Solution 25.

which is trivially a Schmidt decomposition of v. Thus, over C2 ⊗ C3 , v has Schmidt rank 1. Over C3 ⊗ C2 we have             1 0 0 1  1  1  1 1 0 0 ⊗ 1 ⊗ 0 ⊗ + + . v= 0 1 1 2 2 2 0 0 1 Identifying C3 with the columns of a matrix and C2 with the rows of a matrix, we rearrange v to yield the matrix   1 0 1 1 1 A= 2 0 1 which has the singular value decomposition  1  √  √ √1 √1 3 6 2 3 0 2  √2  1  0 − √13   0 A= 6 2 1 1 1 √ √ 0 0 −√ 6

2

√1 2 √1 2

√1 2 − √12

!∗ .

3

From the singular value decomposition we obtain the Schmidt decomposition     √     1 1 3 1   1 1 1 1 1 1 v= ·√ 2 ⊗√ + · √  0 ⊗ √ . 2 2 6 1 2 1 2 −1 2 −1 Thus, over C3 ⊗ C2 , v has Schmidt rank 2. Problem 26. Let A be an m × m hermitian matrix and let B be an n × n hermitian matrix. Then A ⊗ B, A ⊗ In , Im ⊗ B are also hermitian matrices, where Im is the m × m identity matrix. Let 1 , 2 and 3 be real parameters. Consider the Hamilton operator H = ~ω(1 A ⊗ B + 2 A ⊗ In + 3 Im ⊗ B). The partition function Z(β) is given by Z(β) = tr(exp(−βH)), where H is the (hermitian) Hamilton operator and tr denotes the trace. From the

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51

partition function we obtain the Helmholtz free energy, entropy and specific heat. (i) Calculate Z(β) for the Hamilton operator given above. (ii) Consider the special case that n = m = 2 and A, B are any of the Pauli spin matrices σ1 , σ2 , σ3 . Solution 26. Since A is an m × m matrix we can find a m × m unitary e = U ∗ AUA is a diagonal matrix. We set diag(A) e = matrix UA such that A A (λ1 , λ2 , . . . , λm ). Analogously for the n × n hermitian matrix B we find a e = U ∗ BUB is a diagonal matrix. We n×n hermitian matrix UB such that B B e = (µ1 , µ2 , . . . , µn ). Since A and B are hermitian the diagonal set diag(B) e and B e are real. Since UA and UB are unitary matrices we elements of A find that UA ⊗ UB is also a unitary matrix and (UA ⊗ UB )∗ = UA∗ ⊗ UB∗ . Now we find tr(e−βH ) = tr((UA∗ ⊗ UB∗ )e−βH (UA ⊗ UB )) ∗

∗

= tre−β(UA ⊗UB )H(UA ⊗UB ) ∗

∗

∗

∗

= tre−β~ω(1 (UA AUA )⊗(UB BUB )+2 (UA AUA )⊗In +3 (Im ⊗(UB BUB )) = tre−β~ω(1 A⊗B+2 A⊗In +3 Im ⊗B) m X n X = e−β~ω(1 λj µk +2 λj +3 µk ) . e

e

e

e

j=1 k=1

This calculation can be extended straightforward to the matrix A1 ⊗ A2 ⊗ A3 + A1 ⊗ In2 ⊗ In3 + In1 ⊗ A2 ⊗ In3 + In1 ⊗ In2 ⊗ A3 and so on, where A1 , A2 , A3 are n1 × n1 , n2 × n2 , n3 × n3 matrices, respectively. (ii) The eigenvalues of any Pauli spin matrix are +1 and −1. Thus for any combination σj ⊗ σk for A ⊗ B we find Z(β) = e−β~ω1 2 cosh(β~ω(2 + 3 )) + eβ~ω1 2 cosh(β~ω(2 − 3 )). Problem 27. Cn . Is

(i) Let { |0i, |1i, . . . , |n − 1i } be an orthonormal basis in n−1 1 X |ji ⊗ |ji |ψi := √ n j=0

independent of the chosen orthonormal basis? (ii) Find the density matrix |ψihψ|. (iii) Consider the linear operator Π :=

n−1 n−1 1 XX (|jki − |kji)(hjk| − hkj|) 4 j=0 k=0 k6=j

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52 Problems and Solutions where we used the short-cut notation |jki ≡ |ji ⊗ |ki. Calculate Π∗ and Π2 . What is the use of this operator? Solution 27. (i) Let { |φ0 i, |φ1 i, . . . , |φn−1 i } be an orthonormal basis in Cn . Then we have the expansion for the state |ji |ji =

n−1 X

hj|φk i|φk i.

k=0

Thus |ψi can be written as n−1 1 X |ψi = √ n j=0

!

n−1 X

hj|φk i|φk i

⊗

k=0

!!

n−1 X

hj|φl i|φl i

l=0

n−1 n−1 n−1 1 XXX hφk |jihj|φl i|φk i ⊗ |φl i =√ n j=0 k=0 l=0   n−1 n−1 n−1 1 X X X =√ hφk |jihj|φl i |φk i ⊗ |φl i n j=0 k=0 l=0

where we used hj|φk i = hφk |ji. Note that for the sum n−1 X

hφk |jihj|φl i

j=0

we cannot apply Parseval’s relation. Parseval’s relation would apply to n−1 X

hφk |jihφl |ji = hφk |φl i = δkl .

j=0

Thus the Bell state |ψi is dependent on the chosen basis. However, if all scalar products hj|φk i are real numbers then |ψi is independent of the chosen basis. (ii) We have |ψihψ| =

n−1 n−1 n−1 n−1 1 XX 1 XX (|ji ⊗ |ji) (hk| ⊗ hk|) = |jihk| ⊗ |jihk|. n j=0 n j=0 k=0

k=0

(iii) Clearly, Π∗ = Π. Furthermore Π2 =

n−1 n−1 1 XXX X (|jki − |kji)(hjk| − hkj|)(|lmi − |mli)(hlm| − hml|) 16 j=0 j6=k l=0 l6=m

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=

53

n−1 n−1 1 XXX X (|jki − |kji)(2δjl δkm − 2δjm δlk )(hlm| − hml|) 16 j=0 j6=k l=0 l6=m

= Π. Thus Π is a projection matrix. It projects onto the space spanned by   1 √ (|jki − |kji) : j, k ∈ {0, 1, . . . , n − 1}, k > j . 2

Problem 28. Let HA and HB be two finite-dimensional Hilbert spaces. The Schmidt rank of a linear operator L : HA ⊗ HB → HA ⊗ HB over HA ⊗ HB is the smallest non-negative integer Sch(L, HA , HB ) such that L can be written as Sch(L,HA ,HB ) X L= Lj,A ⊗ Lj,B j=1

where Lj,A : HA → HA and Lj,B : HB → HB are linear operators. Let {|0i, |1i} denote an orthonormal basis in C2 . Find the Schmidt rank Sch(UCN OT , C2 , C2 ) and Sch(USW AP , C2 , C2 ) where UCN OT = |00ih00| + |01ih01| + |11ih10| + |10ih11| USW AP = |00ih00| + |10ih01| + |01ih10| + |11ih11|.

Solution 28.

We note that UCN OT = |0ih0| ⊗ I2 + |1ih1| ⊗ UN OT

where UN OT := |0ih1| + |1ih0|. In other words 0 < Sch(UCN OT , C2 , C2 ) ≤ 2. Now suppose UCN OT can be written as the product A ⊗ B where A := a0 |0ih0| + a1 |0ih1| + a2 |1ih0| + a3 |1ih1| B := b0 |0ih0| + b1 |0ih1| + b2 |1ih0| + b3 |1ih1|. This yields the conditions a0 b0 = 1, a0 b1 = 0 and a3 b1 = 1. These equations are inconsistent, i.e. Sch(UCN OT , C2 , C2 ) 6= 1. Thus Sch(UCN OT , C2 , C2 ) = 2.

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54 Problems and Solutions The operator USW AP has the eigenvalue 1 (three times) with corresponding orthonormal eigenvectors   1 |00i, |11i, √ (|01i + |10i) 2 and the eigenvalue −1 with corresponding eigenvector Defining 1 |φ1 i := √ (|01i + |10i), 2

√1 (|01i 2

− |10i).

1 |φ2 i := √ (|01i − |10i) 2

we find that USW AP := |00ih00| + |φ1 ihφ1 | − |φ2 ihφ2 | + |11ih11| where {|00i, |φ1 i, |φ2 i, |11i} forms an orthonormal basis in C4 . In this basis USW AP is the diagonal matrix   1 0 0 0 0 1 0 0 USW AP =  . 0 0 −1 0 0 0 0 1 Clearly, the matrices |00ih00|, |11ih11|, |φ1 ihφ1 | and |φ2 ihφ2 | are linearly independent. Thus Sch(USW AP , C2 , C2 ) = 4. Problem 29. The operator-Schmidt decomposition of a linear operator Q acting in the product Hilbert space H = H1 ⊗ H2 of two finite-dimensional Hilbert spaces (dim(H1 ) = m, dim(H2 ) = n) with H1 = Cm and H2 = Cn can be constructed as follows. Let X, Y be d × d matrices over C. Then we can define a scalar product or inner product hX, Y i := tr(XY ∗ ). Using this inner product we can define an orthonormal set of d × d matrices { Xj : j = 1, 2, . . . , d2 } which satisfies the condition hXj , Xk i = tr(Xj Xk∗ ) = δjk . Thus we can write the matrix Q as 2

Q=

2

n m X X

cjk Aj ⊗ Bk

j=1 k=1

where { Aj : j = 1, 2, . . . , m2 } and { Bk : k = 1, 2, . . . , n2 } are fixed orthonormal bases of m × m and n × n matrices in the Hilbert spaces Cm

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55

and Cn respectively, and cjk are complex coefficients. Thus C = (cjk ), with j = 1, 2, . . . , m2 and k = 1, 2, . . . , n2 is an m2 × n2 matrix. The singular value decomposition theorem states that the matrix C can be written as C = U ΣV ∗ where U is an m2 × m2 unitary matrix, V is an n2 × n2 unitary matrix and Σ is an m2 × n2 diagonal matrix. The matrix Σ is of the form   s1 . . . 0 ..   .. . . .  . .     0 . . . sn2  Σ= .  0 ... 0   . . ..  ..  .. .  0 ... 0 It is assumed that C, U and V are calculated in orthonormal bases, for example the standard basis. Thus we obtain 2

Q=

2

2

m X n X n X

Uj` s` V`k Aj ⊗ Bk

j=1 k=1 `=1

where s` is the `-th diagonal entry of the m2 × n2 diagonal matrix Σ. Defining m2 n2 X X H` := Uj` Aj , K` := V`k Bk j=1

k=1

2

where ` = 1, 2, . . . , n we find the operator-Schmidt decomposition 2

Q=

n X

s` H` ⊗ K` .

`=1

(i) Consider the CNOT gate 1 0 = 0 0 

UCN OT

0 1 0 0

Find the operator-Schmidt decomposition (ii) Consider the SWAP operator  1 0 0 0 USW AP =  0 1 0 0

0 0 0 1

 0 0 . 1 0

of UCN OT . 0 1 0 0

 0 0 . 0 1

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56 Problems and Solutions Find the operator-Schmidt decomposition of USW AP . (iii) Let p  p  √ √ Z := 1 − pI2 ⊗ I2 + i pσ1 ⊗ σ1 1 − pI2 ⊗ I2 + i pσ3 ⊗ σ3 where σ1 , σ2 and σ3 are the Pauli spin matrices. Find the operator-Schmidt decomposition of Z. Solution 29.

(i) We have        1 0 1 0 0 0 0 UCN OT = ⊗ + ⊗ 0 0 0 1 0 1 1     1 0 0 0 = ⊗ I2 + ⊗ σ1 . 0 0 0 1

1 0



(ii) We have USW AP =

1 (I2 ⊗ I2 + σ1 ⊗ σ1 + σ2 ⊗ σ2 + σ3 ⊗ σ3 ). 2

(iii) We have Z = (1−p)I2 ⊗I2 +pσ2 ⊗σ2 +

h    i p p(1 − p) eiπ/4 σ1 ⊗ σ1 + eiπ/4 σ3 ⊗ σ3 .

Programming Problems Problem 1.

Consider the Hadamard basis in C2     1 1 1 1 , v2 = √ . v1 = √ 2 1 2 −1

Apply the Kronecker product to find a basis in C4 . Solution 1.

The following Maxima program

/* hadamardbasis.mac */ v1: matrix([1/sqrt(2)],[1/sqrt(2)]); v1T: transpose(v1); v2: matrix([1/sqrt(2)],[-1/sqrt(2)]); v2T: transpose(v2); r1: v1T . v1; r2: v1T . v2; r3: v2T . v2;

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Kronecker Product and Tensor Product v1v1: kronecker_product(v1,v1); v1v2: kronecker_product(v1,v2); v2v1: kronecker_product(v2,v1); v2v2: kronecker_product(v2,v2); v1v1T: transpose(v1v1); v1v2T: transpose(v1v2); v2v1T: transpose(v2v1); v2v2T: transpose(v2v2); r4: v1v1T . v2v2T;

provides the orthonormal basis     1 1 1 1 1  −1   ,  , 1 2 1 2 1 −1

 1 1 1   , 2 −1 −1 

 1 1  −1   . 2 −1 1 

Let S1 , S2 , S3 be the spin matrices for spin- 12       1 0 1 1 0 −i 1 1 0 S1 = , S2 = , S3 = . 2 1 0 2 i 0 2 0 −1

Problem 2.

Consider the Hamilton operators ˆ e = H = S1 ⊗ S1 + S2 ⊗ S2 + S3 ⊗ S3 H ~ω ˆ e = H = S1 ⊗ S2 + S2 ⊗ S3 + S3 ⊗ S1 . K ~ω e and K. e Find the eigenvalues and eigenvectors of H Solution 2.

Applying the Maxima program

/* eigenS1S2S3.mac */ I2: matrix([1,0],[0,1]); S1: matrix([0,1/2],[1/2,0]); S2: matrix([0,-%i/2],[%i/2,0]); S3: matrix([1/2,0],[0,-1/2]); T1: kronecker_product(S1,S1); T2: kronecker_product(S2,S2); T3: kronecker_product(S3,S3); H: T1 + T2 + T3; EH: eigenvectors(H); X1: kronecker_product(S1,S2); X2: kronecker_product(S2,S3); X3: kronecker_product(S3,S1); K: X1 + X2 + X3; EK: eigenvectors(K);

57

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58 Problems and Solutions e the eigenvalues we find for H 3 − (1×), 4 with the corresponding eigenvectors     0 1 1  1  0 √  ,  , 0 2 −1 0 0

1 (3×) 4   0 1 1 √  , 2 1 0

  0 0   0 1

e the eigenvalues and for K 3 − (1×), 4

1 (3×) 4

with the corresponding eigenvectors       1 1 0 1 0 1  1  1  −1  , √  , √  ,  0 2 −i 2 0 2 i i −i

 0 1  √ 0 2 1

 . −1

Problem 3. Let M be a 2 × 2 matrix and M T the transpose. Let |0i, |1i be the standard basis in C2 . Show that (M ⊗ I2 )(|0i ⊗ |0i + |1i ⊗ |1i) = (I2 ⊗ M T )(|0i ⊗ |0i + |1i ⊗ |1i). Solution 3.

The following Maxima program will do the job

/* M22.mac */ b1: matrix([1],[0]); b2: matrix([0],[1]); b1b1: kronecker_product(b1,b1); b2b2: kronecker_product(b2,b2); I2: matrix([1,0],[0,1]); M: matrix([m11,m12],[m21,m22]); MT: matrix([m11,m21],[m12,m22]); T1: kronecker_product(M,I2); T2: kronecker_product(I2,MT); R1: T1 . b1b1 + T1 . b2b2; R2: T2 . b1b1 + T2 . b2b2; F: R1 - R2;

Do we find the same result if we select the orthonormal basis     cos(θ) sin(θ) , ? sin(θ) − cos(θ)

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2.3

59

Supplementary Problems

Problem 1. Consider the standard basis in the vector space of 2 × 2 matrices         1 0 0 1 0 0 0 0 E00 = , E01 = , E10 = , E11 = 0 0 0 0 1 0 0 1 and the mutually unbiased basis   1 1 0 µ0 = √ , 2 0 1   1 0 −i , µ2 = √ 2 i 0

  1 0 1 µ1 = √ , 2 1 0   1 1 0 µ3 = √ . 2 0 −1

Express the Bell matrix 1 1 0 B=√  2 0 1 

0 1 1 0

0 1 −1 0

 1 0   0 −1

with the basis given by µj ⊗ µk (j, k = 0, 1, 2, 3). Problem 2. The following states form an orthonormal basis in the Hilbert space C3       0 1 1 1 1 |π + i = √  0  , |π 0 i =  1  , |π − i = √  0  . 2 1 2 −1 0 These states play a role for the π-mesons. Show that the states |π + i ⊗ |π + i, 1 √ (|π + i ⊗ |π 0 i + |π 0 i ⊗ |π + i), 2 1 √ (|π + i ⊗ |π 0 i − |π 0 i ⊗ |π + i), 2

|π − i ⊗ |π − i 1 √ (|π 0 i ⊗ |π − i + |π − i ⊗ |π 0 i) 2 1 √ (|π + i ⊗ |π − i − |π − i ⊗ |π + i), 2

1 √ (|π 0 i ⊗ |π − i − |π − i ⊗ |π 0 i) 2 1 √ (2|π 0 i ⊗ |π 0 i + |π + i ⊗ |π − i + |π − i ⊗ |π + i), 6

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60 Problems and Solutions 1 √ (|π + i ⊗ |π − i + |π − i ⊗ |π + i − |π 0 i ⊗ |π 0 i) 3 form an orthonormal basis in the Hilbert space C9 . Which of these states are entangled? Problem 3. Let v1 , v2 , v3 be elements of C2 . Find the conditions on v1 , v2 , v3 such that v1 ⊗ v2 ⊗ v3 = v3 ⊗ v2 ⊗ v1 . Problem 4. Let σ1 , σ2 , σ3 be the Pauli spin matrices. Consider the 8 × 8 matrices K = σ1 ⊗ σ2 ⊗ σ3 , S = σ1 ⊗ σ1 ⊗ σ1 . Note that the matrices K and S are unitary and hermitian. Show that [K, S] = 08 . Show that Π1 =

1 (I8 + S), 2

Π2 =

1 (I8 − S) 2

are projection matrices and Π1 Π2 = 08 . Show that Π3 =

1 (I8 − K), 2

Π4 =

1 (I8 + K) 2

are projection matrices. Problem 5.

Let σ1 , σ2 , σ3 be the Pauli spin matrices. We have σ1 σ2 = iσ3 ,

σ2 σ3 = iσ1 ,

σ3 σ1 = iσ2

and σ2 σ1 = −iσ3 ,

σ3 σ2 = −iσ1 ,

σ1 σ3 = −iσ2 .

Show that [σ1 ⊗ σ1 , σ2 ⊗ σ2 ] = 04 and [σ1 ⊗ σ1 , σ3 ⊗ σ3 ] = 04 . Problem 6.

Can the Z4 Fourier matrix  1 1 1  1 i −1  1 −1 1 1 −i −1

 1 −i   −1 i

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61

be written as the Kronecker product of two 2 × 2 unitary matrices? Problem 7. Let |ai, |bi be normalized states in Cn and X, Y be n × n matrices over C. Show that (ha| ⊗ hb|)((X ⊗ In )(In ⊗ Y ))(|ai ⊗ |bi) = ha|X|aihb|Y |bi. Note that (|ai ⊗ |bi)∗ = ha| ⊗ hb|. Problem 8.

(i) Consider the standard basis     1 0 |0i = , |1i = 0 1

in C2 . Calculate

 |0ih0| ⊗ I2 + |1ih1| ⊗

(ii) Consider the orthonormal basis   cos(θ) |0i = , sin(θ)

0 1

 |1i =

in C2 . Calculate

 |0ih0| ⊗ I2 + |1ih1| ⊗

1 0

 .

sin(θ) − cos(θ)

0 1

1 0



 .

Discuss. Problem 9.

Consider the states in C2     cos(θ) sin(θ) |ψ1 i = , |ψ2 i = sin(θ) − cos(θ)

which form an orthonormal basis in C2 . Find (hψ1 | ⊗ hψ2 |)(σj ⊗ I2 + I2 ⊗ σj )(|ψ1 i ⊗ |ψ2 i), and discuss the dependence on θ. Problem 10.

Consider the four Bell states 1 |ψ1 i = √ (|0i ⊗ |0i + |1i ⊗ |1i) 2 1 |ψ2 i = √ (|0i ⊗ |1i + |1i ⊗ |0i) 2 1 |ψ3 i = √ (|0i ⊗ |1i − |1i ⊗ |0i) 2 1 |ψ4 i = √ (|0i ⊗ |0i − |1i ⊗ |1i) 2

j = 1, 2, 3

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62 Problems and Solutions and the three 4 × 4 matrices Tj := σj ⊗ I2 + I2 ⊗ σj ,

j = 1, 2, 3.

Calculate Tj |ψk i,

j = 1, 2, 3

k = 1, 2, 3, 4.

Which of these expressions is an eigenvalue equation? Calculate hψk |Tj |ψk i.

Problem 11. Let A be an m × m and B be an n × n matrix. Study the conditions on A and B such that (A ⊗ In + Im ⊗ B)v = 0 · 0 = 0 where v 6= 0, i.e. we have an eigenvalue problem with eigenvalue 0. As a consequence we have det(A ⊗ In + Im ⊗ B) = 0. Note that tr(A ⊗ In + Im ⊗ B) = ntr(A) + mtr(B). Study first the case with m = n = 2 and A and B the Pauli spin matrices. Problem 12. Given the hermitian matrices    0 1 0 0 −i 1  1 L1 = √ 1 0 1  , L2 = √  i 0 2 0 1 0 2 0 i and the hermitian matrices  0 0 W1 =  0 0 1 0 

0 1 V1 = √  1 2 0

of the three dipole   1 0 −i  , L3 =  0 0 0

of five quadrupole operators    1 0 0 −i 0, W2 =  0 0 0  , 0 i 0 0

 0 −1  , V2 = 0  1 0 1  Q0 = √ 0 −2 3 0 0 1 0 −1



0 1 √ i 2 0  0 0. 1

−i 0 −i

 0 i , 0

operators  0 0 0 0  0 −1

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63

Show that multiplying these eight hermitian matrices by i we obtain a basis for the semi-simple Lie algebra su(3). Consider the Hamilton operator ˆ = κ0 Q0 ⊗ Q0 + κ1 (V1 ⊗ V1 + V2 ⊗ V2 ) + κ2 (W1 ⊗ W1 + W2 ⊗ W2 ). H ˆ Find the eigenvalues and eigenvectors of H. Problem 13.

Let w1 =

  1 , 0

w2 =

  0 1

and 2 X

u=

tj1 j2 j3 wj1 ⊗ wj2 ⊗ wj3 = w1 ⊗ w2 ⊗ w2 + w2 ⊗ w1 ⊗ w1

j1 ,j2 ,j3 =1

=(0

0

0

1

1

0

0

0)

T

i.e. t122 = 1, t211 = 1 and all other coefficients are equal to 0. Can the vector u ∈ C8 be written as the Kronecker product of a vector in C2 and a vector in C4 ? Consider both cases C2 ⊗ C4 and C4 ⊗ C2 . Problem 14. Let σ1 , σ2 , σ3 be the Pauli spin matrices. Consider the Hamilton operator ˆ = ~ω1 σ1 ⊗ σ1 + ~ω2 σ2 ⊗ σ2 + ~ω3 σ3 ⊗ σ3 H acting in the Hilbert space C4 . Show that ˆ

ˆ

eiHt/~ (σ1 ⊗ I2 )e−iHt/~ = (σ1 ⊗ I2 ) cos(ω2 t) cos(ω3 t) +(I2 ⊗ σ1 ) sin(ω2 t) sin(ω3 t) −(σ2 ⊗ σ3 ) cos(ω2 t) sin(ω3 t) +(σ3 ⊗ σ2 ) sin(ω2 t) cos(ω3 t) ˆ

ˆ

eiHt/~ (σ2 ⊗ I2 )e−iHt/~ = (σ2 ⊗ I2 ) cos(ω3 t) cos(ω1 t) +(I2 ⊗ σ2 ) sin(ω3 t) sin(ω1 t) −(σ3 ⊗ σ1 ) cos(ω3 t) sin(ω1 t) +(σ1 ⊗ σ3 ) sin(ω3 t) cos(ω1 t) ˆ

ˆ

eiHt/~ (σ3 ⊗ I2 )e−iHt/~ = (σ3 ⊗ I2 ) cos(ω1 t) cos(ω2 t) +(I2 ⊗ σ3 ) sin(ω1 t) sin(ω2 t) −(σ1 ⊗ σ2 ) cos(ω1 t) sin(ω2 t) +(σ2 ⊗ σ1 ) sin(ω1 t) cos(ω2 t).

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64 Problems and Solutions Let S1 , S2 , S3 be the spin- 12 matrices      1 0 −i 1 1 1 0 1 , S2 = , S3 = S1 = 2 1 0 2 i 0 2 0

Problem 15.

0 −1

Solve the eigenvalue problem for the Hamilton operator ˆ = ~ω1 (S1 ⊗ S2 ⊗ S3 + S3 ⊗ S1 ⊗ S2 + S2 ⊗ S3 ⊗ S1 ) H +~ω2 (S3 ⊗ I2 ⊗ I2 + I2 ⊗ S3 ⊗ I2 + I2 ⊗ I2 ⊗ S3 ).

 .

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Chapter 3

Matrix Properties

3.1

Introduction

For finite-dimensional quantum systems finding the norm, eigenvalues, eigenvectors, Schmidt rank and inverse (if it exists) of square matrices is important. Let A be an n × n matrix over C. Then we can define the sup-norm kAk := sup kAxk kxk=1

where kAxk denotes the Euclidean norm in Cn . The Hilbert-Schmidt norm of a square matrix A is defined as kAk := (tr(AA∗ ))1/2 where tr denotes the trace. Let A be an n × n matrix over C. Then the eigenvalue equation is defined as Ax = λx where λ ∈ C is the eigenvalue and x ∈ Cn with x 6= 0 is a corresponding eigenvector. It follows that x∗ A∗ = λx∗ . The most important function in quantum computing is the exponential function of a square matrix A defined by  k ∞ X Aj A exp(A) := = lim I + . k→∞ j! k j=0 65

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66 Problems and Solutions An n × n matrix over C is called normal if A∗ A = AA∗ . Let A be an n × n normal matrix over C with eigenvalues λ1 , . . . , λn and corresponding pairwise orthonormal eigenvectors vj (j = 1, . . . , n). Then the matrix A can be written as (spectral decomposition) A=

n X

λj vj vj∗ .

j=1

Note that vj vj∗ ,

j = 1, . . . , n

(vj vj∗ )(vj vj∗ )

are projection matrices, i.e. = vj vj∗ , (vj vj∗ )∗ = vj vj∗ and ∗ ∗ vj vj = 1 for j = k and vj vk = 0 for j 6= k. Consider an n × n matrix A over C and the polynomial p(λ) = det(A − λIn ) with the characteristic equation p(λ) = 0. The Cayley-Hamilton theorem states that substituting the matrix A in the characteristic polynomial results in the n × n zero matrix, i.e. p(A) = 0n . Decompositions of square matrices such as the singular value decomposition, spectral decomposition, polar decomposition and Schur decomposition are necessary in quantum computing. Any unitary 2n × 2n matrix U can be decomposed as     U1 0 C S U3 0 U= 0 U2 −S C 0 U4 where U1 , U2 , U3 , U4 are 2n−1 × 2n−1 unitary matrices and C and S are the 2n−1 × 2n−1 diagonal matrices C = diag(cos(α1 ), cos(α2 ), . . . , cos(α2n /2 )), S = diag(sin(α1 ), sin(α2 ), . . . , sin(α2n /2 )) where αj ∈ R. This decomposition is called cosine-sine decomposition.

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3.2

67

Solved Problems

Problem 1.

Consider the hermitian 4 × 4 matrix (Hamilton operator)

ˆ = ~ω (σ1 ⊗ σ1 − σ2 ⊗ σ2 ) H 2 ˆ i.e. where ω is the frequency. Find the norm of H, ˆ := sup kHxk, ˆ kHk

x ∈ C4 .

kxk=1

ˆ In the Solution 1. There are two methods to find the norm of H. first method we use the Lagrange multiplier method where the constraint kxk = 1 can be written as x21 + x22 + x23 + x24 = 1. Since     0 0 0 1 0 0 0 −1 0 0 1 0  0 0 1 0  σ1 ⊗ σ1 =   , σ2 ⊗ σ2 =   0 1 0 0 0 1 0 0 1 0 0 0 −1 0 0 0 we have

0  ˆ = ~ω  0 H 0 1 

0 0 0 0

0 0 0 0

 1 0 . 0 0

Let x = (x1 , x2 , x3 , x4 )T ∈ C4 . We maximize ˆ 2 − λ(x21 + x22 + x23 + x24 − 1) f (x) := kHxk where λ is the Lagrange multiplier. To find the extrema we solve the four equations ∂f ∂x1 ∂f ∂x2 ∂f ∂x3 ∂f ∂x4

= 2~2 ω 2 x1 − 2λx1 = 0 = −2λx2 = 0 = −2λx3 = 0 = 2~2 ω 2 x4 − 2λx4 = 0

together with the constraint x21 + x22 + x23 + x24 = 1. The four equations can be written in the matrix form  2 2     0 0 ~ ω −λ 0 x1 0 0 −λ 0 0    x2   0     =  . 0 0 −λ 0 x3 0 0 0 0 ~2 ω 2 − λ x4 0

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68 Problems and Solutions ˆ If λ = 0 then x1 = x4 = 0 and kHxk = 0, which is a minimum. If λ 6= 0 ˆ then x2 = x3 = 0 and x21 + x24 = 1 so that kHxk = ~ω, which is the ˆ = ~ω. In the second method we calculate maximum. Thus we find kHk ˆ ∗H ˆ and find the square root of the largest the positive definite matrix H ∗ ∗ ˆ H. ˆ Since H ˆ =H ˆ we find the positive semi-definite eigenvalue of H 1  ∗ 2 2 ˆ H ˆ = ~ ω 0 H 0 0 

0 0 0 0

0 0 0 0

 0 0 . 0 1

ˆ = ~ω. Thus the maximum eigenvalue is ~2 ω 2 (twice degenerate) and kHk ˆ be a hermitian n×n matrix (Hamilton operator) with Problem 2. Let H eigenvalues E0 , E1 , . . . , En−1 with corresponding normalized eigenvectors |ψ0 i, ψ1 i, . . . , |ψn−1 i. The quantum correlation function of two n × n hermitian matrices A and B is given by Qk (t) :=

1 hψk |(A(t)B − AB(t) + BA(t) − B(t)A)|ψk i, 2

k = 0, 1, . . . , n − 1

where ˆ

ˆ

A(t) = eiHt/~ Ae−iHt/~ ,

ˆ

ˆ

B(t) = eiHt/~ Be−iHt/~ .

Note that Qk (t) is real valued. Find Qk (t) using the properties ˆ

eiHt/~ |ψk i = eiEk t/~ |ψk i,

n−1 X

|ψj ihψj | = In .

j=0

Solution 2.

Since

hψk |A(t)B|ψk i =

n−1 X

ei(Ek −Ej )t/~ hψk |A|ψj ihψj |B|ψk i

j=0

hψk |AB(t)|ψk i =

n−1 X

ei(Ej −Ek )t/~ hψk |A|ψj ihψj |B|ψk i

j=0

hψk |BA(t)|ψk i =

n−1 X

ei(Ej −Ek )t/~ hψk |B|ψj ihψj |A|ψk i

j=0

hψk |B(t)A|ψk i =

n−1 X j=0

ei(Ek −Ej )t/~ hψk |B|ψj ihψj |A|ψk i

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69

and utilizing the identity eiα − e−iα ≡ 2i sin(α) yields Qk (t) = i

n−1 X

sin((Ej −Ek )t/~) (hψj |A|ψk ihψk |B|ψj i − hψk |A|ψj ihψj |B|ψk i) .

j=0

Problem 3. (i) Let A and B be two n × n matrices over C. If there exists a non-singular n × n matrix X such that A = XBX −1 , then A and B are said to be similar matrices. Show that the spectra (eigenvalues) of two similar matrices are equal. (ii) Let A and B be n × n matrices over C. Show that the matrices AB and BA have the same set of eigenvalues. Solution 3.

(i) We have

det(A − λIn ) = det(XBX −1 − XλIn X −1 ) = det(X(B − λIn )X −1 ) = det(X)det(B − λIn )det(X −1 ) = det(B − λIn ). (ii) Consider first the case that A is invertible. Then we have AB = A(BA)A−1 . Thus AB and BA are similar and therefore have the same set of eigenvalues. If A is singular we apply the continuity argument: Consider the matrix A + In . We choose δ > 0 such that A + In is invertible for all , 0 <  < δ. Thus (A + In )B and B(A + In ) have the same set of eigenvalues for every  ∈ (0, δ). We equate their characteristic polynomials to obtain det(λIn − (A + In )B) = det(λIn − B(A + In )),

0 <  < δ.

Since both sides are analytic functions of  we find by letting  → 0+ that det(λIn − AB) = det(λIn − BA). Problem 4. Consider a square non-singular matrix A over C. The polar decomposition theorem states that A can be written as A = UP where U is a unitary matrix and P is a positive definite matrix. Thus P is hermitian. Show that A has a unique polar decomposition. Solution 4. Since A is invertible, so are A∗ and A∗ A. The positive square root P of A∗ A is also invertible. Set U := AP −1 . Then U is invertible and U ∗ U = P −1 A∗ AP −1 = P −1 P 2 P −1 = I

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70 Problems and Solutions so that U is unitary. Since P is invertible, it is obvious that AP −1 is the only possible choice for U . Problem 5.

Let A and B be n × n hermitian matrices. Suppose that A2 = In ,

B 2 = In

(1)

and [A, B]+ ≡ AB + BA = 0n

(2)

where 0n is the n × n zero matrix. Let x ∈ Cn be normalized, i.e. kxk = 1. Here x is considered as a column vector. (i) Show that (x∗ Ax)2 + (x∗ Bx)2 ≤ 1. (3) (ii) Give an example for the matrices A and B. Solution 5.

(i) Let a, b ∈ R and let r2 := a2 + b2 . The matrix C = aA + bB

is again hermitian. Then C 2 = a2 A2 + abAB + baBA + b2 B 2 . Using the properties (1) and (2) we find C 2 = a2 In + b2 In = r2 In . Therefore (x∗ C 2 x) = r2 and −r ≤ a(x∗ Ax) + b(x∗ Bx) ≤ r. Let a = x∗ Ax,

b = x∗ Bx

then a2 + b2 ≤ r or r2 ≤ r. This implies r ≤ 1 and r2 ≤ 1 from which (3) follows. (ii) An example is A = σ1 and B = σ2 since σ12 = I2 , σ22 = I2 and σ1 σ2 + σ2 σ1 = 02 . Problem 6. Let K be an n × n skew-hermitian matrix K = −K ∗ with eigenvalues µ1 , . . . , µn (counted according to multiplicity) and the corresponding normalized eigenvectors u1 , . . . , un , where u∗j uk = 0 for k 6= j. Then K can be written as K=

n X j=1

µj uj u∗j

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and uj u∗j uk u∗k = 0 for k 6= j and j, k = 1, 2, . . . , n. Note that the n × n matrices uj u∗j are projection matrices and n X

uj u∗j = In .

j=1

(i) Calculate exp(K). (ii) Every n × n unitary matrix can be written as U = exp(K), where K is a skew-hermitian matrix. Find U from a given K. (iii) Use the result from (ii) to find for a given U a possible K. (iv) Apply the result from (ii) and (iii) to the unitary 2 × 2 matrix   cos(θ) sin(θ) U (θ) = . − sin(θ) cos(θ) (v) Apply the result from (ii) and (iii) to the 2 × 2 unitary matrix   cos(θ) −eiφ sin(θ) V (θ, φ) = . e−iφ sin(θ) cos(θ) (vi) Every hermitian matrix H can be written as H = iK, where K is a skew-hermitian matrix. Find H for the examples given above. Solution 6.

(i) Using the properties of the n × n matrix uj u∗j we find   n n X X exp(K) = exp  µj uj u∗j  = eµj uj u∗j . j=1

j=1

(ii) From U = exp(K) we find U=

n X

eµj uj u∗j

j=1

where uj (j = 1, 2, . . . , n) are the normalized eigenvectors of U . (iii) The matrix K is given by K=

n X

ln(λj )uj u∗j

j=1

where λj (j = 1, 2, . . . , n) are the eigenvalues of U and uj are the normalized eigenvectors of U . Note that the eigenvalues of U are of the form exp(iα) with α ∈ R. Thus we have ln(eiα ) = iα.

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72 Problems and Solutions (iv) The eigenvalues of the matrix U (θ) are eiθ and e−iθ with the corresponding normalized eigenvectors     1 1 1 1 u1 = √ , u2 = √ . i −i 2 2 Thus K(θ) = ln(e

iθ

)u1 u∗1 +ln(eiθ )u2 u2

iθ = 2



1 i

  iθ −i 1 − 1 −i 2

i 1



 =

0 −θ

θ 0

(v) For the matrix V (θ, φ) the eigenvalues are e−iθ and eiθ with the corresponding normalized eigenvectors     1 1 1 1 √ √ , . −iφ −iφ 2 ie 2 −ie Thus K(θ, φ) = ln(e−iθ )u1 u∗1 + ln(eiθ )u2 u∗2 =



0 θe−iφ

−θeiφ 0

 .

(vi) For U (θ) we find the matrix  iθ

0 −1

For V (θ, φ) we find the matrix 

0 iθe−iφ

Problem 7.

1 0

 .

−iθeiφ 0

 .

Let A and B be n × n hermitian matrices. Suppose that A2 = A,

B2 = B

(1)

and [A, B]+ ≡ AB + BA = 0n

(2)

n

where 0n is the n × n zero matrix. Let x ∈ C be normalized, i.e. kxk = 1. Here x is considered as a column vector. Show that (x∗ Ax)2 + (x∗ Bx)2 ≤ 1.

Solution 7.

(3)

For an arbitrary n × n hermitian matrix M we have

0 ≤ (x∗ (M − (x∗ M x)In )2 x) = (x∗ (M 2 − 2(x∗ M x)M + (x∗ M x)2 In )x) = (x∗ M 2 x) − 2(x∗ M x)2 + (x∗ M x)2 = (x∗ M 2 x) − (x∗ M x)2 .

 .

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Thus 0 ≤ (x∗ M 2 x) − (x∗ M x)2

or

(x∗ M x)2 ≤ (x∗ M 2 x).

Thus for A = M we have using (1) (x∗ Ax)2 ≤ x∗ Ax and therefore 0 ≤ (x∗ Ax) ≤ 1. Similarly 0 ≤ (x∗ Bx) ≤ 1. Let a, b ∈ R, r2 := a2 + b2 and C := aA + bB. Then C 2 = a2 A2 + b2 B 2 + abAB + baBA. Using (1) and (2) we arrive at C 2 = a2 A + b2 B. Thus (x∗ Cx)2 ≤ (x∗ C 2 x) ≤ a2 + b2 . Let a := (x∗ Ax), b := (x∗ Bx) then (x∗ Cx) = a2 + b2 = r2 and therefore (r2 )2 ≤ r2 which implies that r2 ≤ 1 and thus (3) follows. Problem 8.

Let A, B be n × n matrices over C. Assume that [A, [A, B]] = [B, [A, B]] = 0n .

(1)

Show that 1

eA+B = eA eB e− 2 [A,B] 1

eA+B = eB eA e+ 2 [A,B] .

(2a) (2b)

Hint. Use the technique of parameter differentiation. Consider the matrixvalued function f () = eA eB where  is a real parameter and calculate the derivative df /d. Solution 8.

If we differentiate f () with respect to  we find df = AeA eB + eA eB B = (A + eA Be−A )f () d

since eA e−A = In . Owing to (1) we have eA B−A = B + [A, B]. Thus we obtain the linear matrix-valued differential equation df = ((A + B) + [A, B])f (). d

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74 Problems and Solutions Since the matrix A + B commutes with [A, B] we may treat A + B and [A, B] as ordinary commuting variables and integrate this linear differential equation with the initial conditions f (0) = In . We find 2

f () = e(A+B)+(

/2)[A,B]

= e(A+B) e(

2

/2)[A,B]

since A + B commutes with [A, B]. If we set  = 1 and multiply both sides by e−[A,B]/2 then (2a) follows. Likewise we can prove the second form of the identity (2b). Problem 9. Let A be an n × n matrix. Assume that the inverse matrix of A exists. The inverse matrix can be calculated as follows (Csanky’s algorithm). Let p(x) := det(xIn − A) (1) where In is the n × n unit matrix. The roots are, by definition, the eigenvalues λ1 , λ2 , . . . , λn of A. We write p(x) = xn + c1 xn−1 + · · · + cn−1 x + cn

(2)

where cn = (−1)n det(A). Since A is nonsingular we have cn 6= 0 and vice versa. The Cayley-Hamilton theorem states that p(A) = An + c1 An−1 + · · · + cn−1 A + cn In = 0n .

(3)

Multiplying this equation with A−1 we obtain A−1 =

1 (An−1 + c1 An−2 + · · · + cn−1 In ). −cn

(4)

If we have the coefficients cj we can calculate the inverse matrix A. Let sk :=

n X

λkj .

j=1

Then the sj and cj satisfy the following n × n lower triangular system of linear equations      1 0 0 ... 0 c1 −s1 2 0 . . . 0   c2   −s2   s1      s1 3 . . . 0   c3  =  −s3  .  s2  .     .. .. . . . . .   .. . ..   ..   ..  . . sn−1 sn−2 . . . s1 n cn −sn Since tr(Ak ) = λk1 + λk2 + · · · + λkn = sk

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75

we find sk for k = 1, 2, . . . , n. Thus we can solve the linear equation for cj . Finally, using (4) we obtain the inverse matrix of A. Apply Csanky’s algorithm to the 4 × 4 permutation matrix 0 0 U = 0 1

1 0 0 0



Solution 9.

Since 

0 0  U2 =  1 0

0 0 0 1

1 0 0 0

 0 1 , 0 0

0 1 0 0

 0 0 . 1 0

0 1  U3 =  0 0

0 0 1 0



0 0 0 1

 1 0  0 0

and U 4 = I4 we find tr(U 2 ) = 0 = s2 ,

tr(U ) = 0 = s1 ,

tr(U 3 ) = 0 = s3 ,

tr(U 4 ) = 4 = s4 .

We obtain the system of linear equations 1 0  0 0 

0 2 0 0

0 0 3 0

    0 c1 0 0   c2   0    =   0 c3 0 4 c4 −4

with the solution c1 = 0, c2 = 0, c3 = 0, of U is given by  0 1 U −1 = U 3 =  0 0

Problem 10. Let   0 1 + J := , 0 0

J

−

 :=

0 1 +

c4 = −1. Thus the inverse matrix

0 0

0 0 1 0

0 0 0 1

 1 0 . 0 0

 , −

(i) Let  ∈ R. Find the matrices eJ , eJ , e(J (ii) Let r ∈ R. Show that er(J

+

+J − )

≡ eJ

−

1 J3 := 2 +

+J − )



1 0

.

tanh(r) 2J3 ln(cosh(r)) J + tanh(r)

e

e

.

0 −1

 .

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76 Problems and Solutions (i) Using the expansion for an n × n matrix A

Solution 10.

exp(A) =

∞ j j X  A

j!

j=0

we find +

eJ =



1 0

0 1



 +

and e

(J + +J − )

0 0

 =

1 0

1 0



0 1

−

,

eJ =



 cosh() +



0 1

1 0 1 0

0 1



 +

0 1

0 0



 sinh().

(ii) Since e2J3 ln(cosh(r)) =



cosh(r) 0

0 1/ cosh(r)

 ,

1/ cosh(r) + tanh(r) sinh(r) ≡ cosh(r) and using the results from (i) we find the identity. Problem 11.

The Heisenberg commutation relation can be written as [ˆ p, qˆ] = −i~I

where pˆ := −i~∂/∂q and I is the identity operator. Let α, β ∈ R and U (α) = exp(iαˆ p),

V (β) = exp(iβ qˆ).

Then using the Baker-Campbell-Hausdorff formula we find U (α)V (β) = exp(iαβ)V (β)U (α). This is called the Weyl representation of Heisenberg’s commutation relation. Can we find finite-dimensional n × n unitary matrices U (U 6= In ) and V (V 6= In ) such that U V = ωV U with ω ∈ C, ω n = 1 ? Solution 11. Such matrices can be found, namely the permutation matrix   0 1 0 ... 0 0 0 1 ... 0 . . . .  . . . . . ...  U :=  . . .  0 0 0 ... 1 1

0

0

...

0

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77

and the diagonal matrix 1 0  0 V :=  .  ..

0 ω 0 .. .

0 0 ω2 .. .

... ... ... .. .

0 0 0 .. .

0

0

0

...

ω n−1



   .  

Let U be the n × n unitary matrix

Problem 12.

0 1 0 0 0 1 . . . . . . U :=  . . . 0 0 0 1 0 0 

... ... .. . ... ...

 0 0 ..  .  1 0

and V be the n × n unitary diagonal matrix (ω ∈ C) 1 0  0 V :=  .  ..

0 ω 0 .. .

0 0 ω2 .. .

... ... ... .. .

0 0 0 .. .

0

0

0

...

ω n−1



     

where ω n = 1 (ω 6= 1). Then the set of matrices { U j V k : j, k = 0, 1, 2, . . . , n − 1 } provide a basis in the Hilbert space for all n × n matrices with the scalar product 1 hA, Bi := tr(AB ∗ ) n for n × n matrices A and B. Write down the basis for n = 2. Solution 12.

For n = 2 we have the combinations (j, k) ∈ { (0, 0), (0, 1), (1, 0), (1, 1) }.

This yields the orthonormal basis (where ω = −1)  I2 =

1 0

0 1



 ,

σ1 =

0 1

1 0



 ,

σ3 =

1 0

0 −1



 ,

−iσ2 =

0 1

−1 0

 .

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78 Problems and Solutions Problem 13.

An n × n circulant matrix  c c1 c2 0 c0 c1  cn−1  cn−2 cn−1 c0 C :=   . .. ..  . . . . c1

c2

For example, the permutation matrix  0 1 0 0 . . . . P :=  . . 0 0 1 0

c3

C is given by  ... c n−1

. . . cn−2   . . . cn−3  . ..  ..  . . ... c0

0 1 .. .

... ... .. .

0 0

... ...

 0 0 ..  .  1 0

is a circulant matrix. It is also called the n×n primary permutation matrix. (i) Let C and P be the matrices given above. Let f (λ) = c0 + c1 λ + · · · + cn−1 λn−1 . Show that C = f (P ). (ii) Show that C is a normal matrix, that is C ∗ C = CC ∗ . (iii) Show that the eigenvalues of C are f (ω k ), k = 0, 1, . . . , n − 1, where ω is the nth primitive root of unity. (iv) Show that det(C) = f (ω 0 )f (ω 1 ) · · · f (ω n−1 ). (v) Show that F ∗ CF is a diagonal matrix, where F is the unitary matrix with (j, k)-entry equal to 1 √ ω (j−1)(k−1) , n

Solution 13.

j, k = 1, . . . , n.

(i) Direct calculation of f (P ) = c0 In + c1 P + c2 P 2 + · · · + cn−1 P n−1

yields the matrix C, where In is the n × n unit matrix. Notice that P 2 , P 3 , . . . , P n−1 are permutation matrices. (ii) We have P P ∗ = P ∗ P . If two n × n matrices A and B commute, then g(A) and h(B) commute, where g and h are polynomials. Thus C is a normal matrix. (iii) The characteristic polynomial of P is det(λIn − P ) = λn − 1 =

n−1 Y

(λ − ω k ).

k=0

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Thus the eigenvalues of P and P j are, respectively, ω k and ω jk , where k = 0, 1, . . . , n − 1. It follows that the eigenvalues of C = f (P ) are f (ω k ), k = 0, 1, . . . , n − 1. (iv) Using the result from (iii) we find det(C) =

n−1 Y

f (ω k ).

k=0

(v) For each k = 0, 1, . . . , n − 1, let xk = (1, ω k , ω 2k , . . . , ω (n−1)k )T where

T

denotes the transpose. If follows that P xk = (ω k , ω 2k , . . . , ω (n−1)k , 1)T = ω k xk

and Cxk = f (P )xk = f (ω k )xk . Thus the vectors xk are the eigenvectors of P and C corresponding to the respective eigenvalues ω k and f (ω k ), k = 0, 1, . . . , n − 1. Since hxj , xk i ≡ x∗j xk =

n−1 X

ω k` ω j` =

`=0

we find that



n−1 X

ω (j−k)` =

`=0

1 1 1 √ x0 , √ x1 , . . . , √ xn−1 n n n



0 j 6= k n j=k



is an orthonormal basis in the Hilbert space Cn . Thus we obtain the unitary matrix   1 1 1 ... 1 2 n−1 1  ω ω ... ω  1  2 4 2(n−1)   1 ω ω . . . ω F =√   .. .. .. n  .. ..  . . . . . n−1 2(n−1) (n−1)(n−1) 1 ω ω ... ω such that F ∗ CF = diag(f (ω 0 ), f (ω 1 ), . . . , f (ω n−1 )). The matrix F is unitary and is called the Fourier matrix. Problem 14. An n × n matrix A is called a Hadamard matrix if each entry of A is 1 or −1 and if the rows or columns of A are orthogonal, i.e. AAT = nIn

or AT A = nIn .

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80 Problems and Solutions Note that AAT = nIn and AT A = nIn are equivalent. Hadamard matrices Hn of order 2n can be generated recursively by defining     1 1 Hn−1 Hn−1 H1 = , Hn = 1 −1 Hn−1 −Hn−1 for n ≥ 2. Show that the eigenvalues of Hn are given by +2n/2 and −2n/2 each of multiplicity 2n−1 . Solution 14. We use induction on n. The case n = 1 is obvious. Now for n ≥ 2 we have λI − Hn−1 −Hn−1 det(λI − Hn ) = −Hn−1 λI + Hn−1 2 = det((λI − Hn−1 )(λI + Hn−1 ) − Hn−1 ).

Thus 2 det(λI − Hn ) = det(λ2 I − 2Hn−1 ) √ √ = det(λI − 2Hn−1 ) det(λI + 2Hn−1 ).

√ This shows that each eigenvalue µ of Hn−1 generates two eigenvalues ± 2µ of Hn . The assertion then follows by the induction hypothesis, for Hn−1 has eigenvalues +2(n−1)/2 and −2(n−1)/2 each of multiplicity 2n−2 . Problem 15. as

Let U be an n × n unitary matrix. Then U can be written U = V diag(λ1 , λ2 , . . . , λn )V ∗

where λ1 , λ2 , . . . , λn are the eigenvalues of U and V is an n × n unitary matrix. Let   0 1 U= . 1 0 Find the decomposition for U given above. Solution 15.

The eigenvalues of U are +1 and −1. Thus we have U = V diag(1, −1)V ∗

with 1 V =√ 2



1 1

1 −1

 .

Therefore V = V ∗ . The columns of V are the eigenvectors of U .

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Problem 16. An n × n hermitian matrix A over the complex numbers is called positive semidefinite (written as A ≥ 0), if x∗ Ax ≥ 0

for all x ∈ Cn .

Show that for every A ≥ 0, there exists a unique B ≥ 0 so that B 2 = A. Solution 16. Let A = U ∗ diag(λ1 , . . . , λn )U , where U is unitary. We take 1/2

B = U ∗ diag(λ1 , . . . , λ1/2 n )U. Then the matrix B is positive semidefinite and B 2 = A since U ∗ U = In . To show the uniqueness, suppose that C is an n × n positive semidefinite matrix satisfying C 2 = A. Since the eigenvalues of C are the nonnegative square roots of the eigenvalues of A, we can write 1/2

∗ C = V diag(λ1 , . . . , λ1/2 n )V

for some unitary matrix V . Then the identity C 2 = A = B 2 yields T diag(λ1 , . . . , λn ) = diag(λ1 , . . . , λn )T where T = U V . This yields tjk λk = λj tjk . Thus 1/2

tjk λk

1/2

= λj tjk .

Hence 1/2

1/2

1/2 T diag(λ1 , . . . , λ1/2 n ) = diag(λ1 , . . . , λn )T.

Since T = U V it follows that B = C. Problem 17. An n × n matrix A over the complex numbers is said to be normal if it commutes with its conjugate transpose A∗ A = AA∗ . The matrix A can be written n X A= λj Ej j=1

where λj ∈ C are the eigenvalues of A and Ej are n × n matrices satisfying Ej2 = Ej = Ej∗ ,

Ej Ek = 0n if j 6= k,

n X j=1

Let n = 2. Consider the NOT gate  0 A= 1

1 0

 .

Ej = In .

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82 Problems and Solutions Find the decomposition of A given above. Solution 17. The eigenvalues of A are given by λ1 = +1, λ2 = −1. The matrices Ej are constructed from the normalized eigenvectors of A. The normalized eigenvectors of A are given by     1 1 1 1 v1 = √ , v2 = √ . 2 1 2 −1 Thus E1 =

v1 v1∗

1 = 2



1 1

1 1

 ,

E2 =

v2 v2∗

1 = 2



−1 1

1 −1

 .

Problem 18. Let X be an n × n matrix over C with X 2 = In , where In is the n × n identity matrix. Let z ∈ C. (i) Show that ezX = In cosh(z) + X sinh(z). (ii) Let z = iα, where α ∈ R. Simplify the result from (i). Solution 18. ezX :=

(i) We have

∞ X (zX)k k=0

k!

=

∞ X zk X k k=0

= In + zX +

k!

1 2 2 1 z X + z3X 3 + · · · . 2! 3!

Since X 2 = In we have     1 1 1 1 ezX = In 1 + z 2 + z 4 + · · · + X z + z 3 + z 5 + · · · 2! 4! 3! 5! = In cosh(z) + X sinh(z). (ii) Since cosh(iα) = cos(α), sinh(iα) = i sin(α) we obtain eiαX = In cos(α) + iX sin(α).

Problem 19. (i) Consider

Let σ1 , σ2 be the Pauli spin matrices. V = exp(i(π/4)σ1 ),

Show that

1 V =√ 2



1 i

i 1

 ,

W = exp(i(π/4)σ2 ). 1 W =√ 2



1 −1

1 1

 .

(ii) Let V = exp(i(π/4)σ1 ),

W = exp(i(π/4)σ2 )

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83

be two unitary 2 × 2 matrices. Calculate the matrices V ∗ σ3 V , W ∗ σ3 W . (i) Using σ12 = I2 , σ22 = I2 we find

Solution 19.

1 1 V = In cos(π/4) + iσ1 sin(π/4) = √ (I2 + iσ1 ) = √ 2 2



1 i

i 1



and 1 1 W = In cos(π/4) + iσ2 sin(π/4) = √ (I2 + iσ2 ) = √ 2 2



1 −1

1 1

 .

(ii) We have    i 1 i 1 V σ3 V = √ I2 − √ σ1 σ3 √ I2 + √ σ1 2 2 2 2 1 i i 1 = σ3 + σ3 σ1 − σ1 σ3 + σ1 σ3 σ1 2 2 2 2 = −σ2 

∗

and    1 i i 1 √ I2 − √ σ2 σ3 √ I2 + √ σ2 2 2 2 2 1 i i 1 = σ3 + σ3 σ2 − σ2 σ3 + σ2 σ3 σ2 2 2 2 2 = σ1 .

W ∗ σ3 W =

Problem 20.



Find the matrices

e(iπ/4)σ2 σ1 e−(iπ/4)σ2 ,

e(iπ/4)σ2 σ3 e−(iπ/4)σ2 .

Use the technique of parameter differentiation f () = eσ2 σ1 e−σ2 with f ( = 0) = σ1 . Solution 20. −2iσ3 yields

Differentiation of f with respect to  and using [σ2 , σ1 ] =

df = −2ieσ2 σ3 e−σ2 d with df ( = 0)/d = −2iσ3 . The second derivative and using [σ2 , σ3 ] = 2iσ1 yields d2 f = 4f (). d2

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84 Problems and Solutions The solution of this second order linear differential equation with constant coefficients is f () = C1 cosh(2) + C2 sinh(2). Inserting the initial values provides f () = cosh(2)σ1 − i sinh(2)σ3 . Since  = iπ/4 we have cosh(iπ/2) = cos(π/2) = 0 and sinh(iπ/2) = i sin(π/2) = i. Thus f ( = iπ/4) = σ3 or e(iπ/4)σ2 σ1 e−(iπ/4)σ2 = σ3 . Analogously we find e(iπ/4)σ2 σ3 e−(iπ/4)σ2 = σ1 . Problem 21.

Let

V = exp(i(π/4)σ1 ) ⊗ exp(i(π/4)σ1 ),

W = exp(i(π/4)σ2 ) ⊗ exp(i(π/4)σ2 ).

Calculate V ∗ (σ3 ⊗ σ3 )V and W ∗ (σ3 ⊗ σ3 )W . Solution 21.

We have

V ∗ (σ3 ⊗ σ3 )V = (exp(−i(π/4)σ1 )σ3 exp(i(π/4)σ1 )) ⊗(exp(−i(π/4)σ1 )σ3 exp(i(π/4)σ1 )) = σ2 ⊗ σ2 . Analogously W ∗ (σ3 ⊗ σ3 )W = σ1 ⊗ σ1 . Problem 22. Let X be an n × n matrix over C. Assume that X 2 = In . Let Y be an arbitrary n × n matrix over C. Let z ∈ C. (i) Calculate exp(zX)Y exp(−zX) using the Baker-Campbell-Hausdorff formula ezX Y e−zX = Y + z[X, Y ] +

z3 z2 [X, [X, Y ]] + [X, [X, [X, Y ]]] + · · · . 2! 3!

(ii) Calculate exp(zX)Y exp(−zX) by first calculating exp(zX) and exp(−zX) and then doing the matrix multiplication. Compare the two methods. Solution 22.

(i) Using X 2 = In we find for the first three commutators

[X, [X, Y ]] = [X, XY − Y X] = 2(Y − XY X) [X, [X, [X, Y ]]] = 22 [X, Y ] [X, [X, [X, [X, Y ]]]] = 23 (Y − XY X).

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If the number of X’s in the commutator is even (say m, m ≥ 2) we have [X, [X, . . . [X, Y ] . . .] = 2m−1 (Y − XY X). If the number of X’s in the commutator is odd (say m, m ≥ 3) we have [X, [X, . . . [X, Y ] . . .] = 2m−1 [X, Y ]. Thus e

zX

Ye

−zX

    21 z 2 23 z 4 22 z 3 =Y 1+ + + · · · + [X, Y ] z + + ··· 2! 4! 3!  1 2  2 z 23 z 4 −XY X + + ··· . 2! 4!

Consequently ezX Y e−zX = Y cosh2 (z) + [X, Y ] sinh(z) cosh(z) − XY X sinh2 (z). (ii) Using the expansion ezX =

∞ X (zX)j j=0

j!

and X 2 = In we have ezX = In cosh(z) + X sinh(z),

e−zX = In cosh(z) − X sinh(z).

Matrix multiplication yields ezX Y e−zX = Y cosh2 (z) + [X, Y ] sinh(z) cosh(z) − XY X sinh2 (z).

Problem 23.

The definition of the Lie group SU (2) is

SU (2) := { A : A a 2 × 2 complex matrix, det(A) = 1, AA∗ = A∗ A = I2 } . In the name SU (2), the S stands for special and refers to the condition det(A) = 1 and the U stands for unitary and refers to the conditions AA∗ = A∗ A = I2 . Show that SU (2) can also be defined as SU (2) := { x0 I2 + ixT σ : (x0 , x)T ∈ R4 , x20 + kxk2 = 1 } where σ = (σ1 , σ2 , σ3 )T are the Pauli spin matrices. Here xT σ := x1 σ1 + x2 σ2 + x3 σ3 . Solution 23. Let A be any 2 × 2 complex matrix. Then A can be written as A = a0 I2 + iaT σ

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86 Problems and Solutions with a = (a1 , a2 , a3 )T ∈ C3 . Thus we have AA∗ = (a0 I2 + iaT σ)(a0 I2 − iaT σ) = |a0 |2 I2 + ia0 aT σ − ia0 aT σ + aT aI2 + i(a × a)T σ = (|a0 |2 + kak2 )I2 + i(a0 a − a0 a + a × a)T σ where we used that σj∗ = σj for j = 1, 2, 3. It follows that AA∗ = I2 ⇔ |a0 |2 + kak2 = 1,

a0 a − a0 a + a × a = 0.

First, suppose that a 6= 0. Since a × a is orthogonal to both a and a, the equation a0 a − a0 a + a × a = 0 can only be satisfied if a × a = 0. That is, only if a and a are parallel. Since a and a have the same length, this is the case only if a = e−2iθ a for some real number θ. This can be written as e−iθ a = e−iθ a which says that x = e−iθ a is real. Substituting a = eiθ x into a0 a − a0 a + a × a = 0 gives eiθ a0 x − e−iθ a0 x = 0. This forces a0 = eiθ x0 for some real x0 . If a = 0, we may still choose θ so that a0 = eiθ x0 . We have shown that AA∗ = I2 ⇔ A = eiθ (x0 I2 + ixT σ) for some (x0 , x)T ∈ R4 with |x0 |2 + kxk2 = 1 and some θ ∈ R. Since det(A) = det(eiθ (x0 I2 + ixT · σ)) = e2iθ (x20 + x21 + x22 + x23 ) = e2iθ we have that det(A) = 1 if and only if eiθ = ±1. If eiθ = −1, we can absorb the −1 into the vector (x0 , x)T . Problem 24. Let A, B be two n × n matrices over C. We introduce the scalar product 1 tr(AB ∗ ) = tr(AB ∗ ). hA, Bi := tr(In ) n The Lie group SU (N ) is defined by the complex n × n matrices U SU (N ) := { U : U ∗ U = U U ∗ = In , det(U ) = 1 }. The dimension is N 2 − 1. The semi-simple Lie algebra su(N ) is defined by the n × n matrices X su(N ) := { X : X ∗ = −X , tr(X) = 0 }.

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(i) Let U ∈ SU (N ). Calculate the scalar product hU, U i. (ii) Let A be an arbitrary complex n×n matrix. Let U ∈ SU (N ). Calculate the scalar product hU A, U Ai. (iii) Consider the Lie algebra su(2). Provide a basis. The elements of the basis should be orthogonal to each other with respect to the scalar product given above. Calculate the commutator of these matrices. Solution 24.

(i) We have hU, U i =

1 tr(U U ∗ ) = 1 n

where we used that U ∗ = U −1 . (ii) We obtain hU A, U Ai =

1 1 1 tr(U A(U A)∗ ) = tr(U AA∗ U ∗ ) = tr(AA∗ ) = hA, Ai n n n

where we used the cyclic invariance of the trace. (iii) We are looking for three linear independent 2 × 2 matrices which are traceless and skew-hermitian matrices. A choice is       i 0 0 −1 0 i τ1 = , τ2 = , τ3 = . 0 −i 1 0 i 0 The matrices are also orthogonal to each other using the scalar product given above. For the commutators we find [τ1 , τ2 ] = −2τ3 ,

[τ3 , τ1 ] = −2τ2 ,

[τ2 , τ3 ] = −2τ1 .

Problem 25. Let H be an n × n matrix which depends on n real parameters 1 , 2 ,. . . , n , where we assume that we can differentiate H with respect to all ’s. Let β > 0 and Z(β) := tr(exp(−βH)). (i) Show that Z β ∂ −βH ∂H −τ H e ≡− dτ e(τ −β)H e . (1) ∂j ∂j 0 (ii) Show that ∂ Z = −βtr ∂j Solution 25.



∂ He−βH ∂j

 .

(i) We set f (β, 1 , . . . , n ) := eβH

∂ −βH e . ∂j

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88 Problems and Solutions It follows that ∂f ∂ −βH ∂ = HeβH e + eβH ∂β ∂j ∂β



∂ −βH e ∂j

 .

Since ∂ ∂β

we obtain



∂ −βH e ∂j



  ∂ −βH ∂ ∂ e He−βH = =− ∂j ∂β ∂j ∂H −βH ∂ −βH =− e −H e ∂j ∂j

∂f ∂H −βH = −eβH e ∂β ∂j

with the initial value f (0, 1 , . . . , n ) = 0. Integrating provides Z β ∂H −τ H e dτ. f (β, 1 , . . . , n ) = − eτ H ∂j 0 Multiplying by exp(−βH) yields identity (1). (ii) We have   ∂Z ∂ ∂ = tr(exp(−βH)) = tr exp(−βH) ∂j ∂j ∂j ! Z β (τ −β)H ∂H −τ H = −tr dτ e e ∂j 0   −βH ∂H = −βtr e ∂j where we used the result from (i) and the cyclic invariance of the trace. Problem 26. To calculate exp(A) we can also use the Cayley-Hamilton theorem and the Putzer method. Using the Cayley-Hamilton theorem we can write f (A) = an−1 An−1 + an−2 An−2 + · · · + a2 A2 + a1 A + a0 In

(1)

where the complex numbers a0 , a1 , . . . , an−1 are determined as follows: Let r(λ) := an−1 λn−1 + an−2 λn−2 + · · · + a2 λ2 + a1 λ + a0 which is the right-hand side of (1) with Aj replaced by λj (j = 0, 1, . . . , n − 1). For each distinct eigenvalue λj of the matrix A, we consider the equation f (λj ) = r(λj ).

(2)

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If λj is an eigenvalue of multiplicity k, for k > 1, then we consider also the following equations . = r(k−1) (λ) f 0 (λ)|λ=λj = r0 (λ)|λ=λj , · · · , f (k−1) (λ) λ=λj

λ=λj

Any unitary matrix U can be written as U = exp(iK), where K is hermitian. Apply this method to find K for the Hadamard gate   1 1 1 UH = √ . 2 1 −1 Solution 26.

The hermitian 2 × 2 matrix K is given by   a b , a, c ∈ R, b ∈ C. K= b c

Then we find the condition on a, b and c such that eiK = UH . The eigenvalues of iK are given by i(a + c) 1 p λ1,2 = ± 2ac − a2 − c2 − 4bb. 2 2 We set in the following p ∆ := λ1 − λ2 = 2ac − a2 − c2 − 4bb. To apply the method given above we have r(λ) = α1 λ + α0 = f (λ) = eλ . Thus we obtain the two equations eλ1 = α1 λ1 + α0 , eλ2 = α1 λ2 + α0 . It follows that eλ1 − eλ2 eλ2 λ1 − eλ1 λ2 α1 = , α0 = . λ1 − λ2 λ1 − λ2 We have the condition     1 iα1 a + α0 iα1 b 1 1 iK e = α1 iK + α0 I2 = =√ . iα1 b iα1 c + α0 2 1 −1 We obtain the four equations 1 iα1 a + α0 = √ , 2

1 iα1 c + α0 = − √ , 2

1 iα1 b = √ , 2

1 iα1 b = √ . 2

From the last two equations we find that b = b, i.e. b is real. From the first two equations we find α0 = −iα1 (a + c)/2 and therefore, using the last two equations, c = a − 2b. Thus     iα1 a + α0 iα1 b iα1 b iα1 b = . iα1 b iα1 b −iα1 b iα1 c + α0

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90 Problems and Solutions From the eigenvalues of eiK we find eλ1 − eλ2 = 2 and p √ ∆ = 2ac − a2 − c2 − 4b2 = 2 2ib. √ √ Furthermore λ1 = i(a − b) + 2ib, λ2 = i(a − b) − 2ib. Thus we arrive at the equation √ √ ei(a−b)+ 2ib − ei(a−b)− 2ib = 2. √ It follows that iei(a−b) sin( 2b) = 1 and therefore √ √ i cos(a − b) sin( 2b) − sin(a − b) sin( 2b) = 1 with a solution π b= √ , 2 2

π a= 2



1 3+ √ 2

 ,

π c = a − 2b = 2

Then the matrix K is given by √ √    3π 1 π 3 + 1/ 2 1/ 2√ √ = K= 1/ 2 3 − 1/ 2 2 2 0

0 1





1 3− √ 2

π 1 + ·√ 2 2



1 1



1 −1

.

 .

We note that the second matrix on the right-hand side is the Hadamard gate again. Problem 27. Let A, B, C2 , ..., Cm , ... be n × n matrices over C. The Zassenhaus formula is given by exp(A + B) = exp(A) exp(B) exp(C2 ) · · · exp(Cm ) · · · The left-hand side is called the disentangled form and the right hand side is called the undisentangled form. Find C2 , C3 , . . . , using the comparison method. In the comparison method the disentangled and undisentangled form are expanded in terms of an ordering scalar α and matrix coefficients of equal powers of α are compared. From exp(α(A + B)) = exp(αA) exp(αB) exp(α2 C2 ) exp(α3 C3 ) · · · we obtain ∞ X αk k=0

k!

(A + B)k=

∞ X r0 ,r1 ,r2 ,r3

αr0 +r1 +2r2 +3r3 +... r0 r1 r2 r3 A B C2 C3 · · · r0 !r1 !r2 !r3 ! · · · ,...=0

(i) Find the matrices C2 and C3 . (ii) Assume that [A, [A, B]] = 0n and [B, [A, B]] = 0n . What conclusion can we draw for the Zassenhaus formula?

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Solution 27. (i) For α2 we have the decompositions (r0 , r1 , r2 ) = (2, 0, 0), (1, 1, 0), (0, 2, 0), (0, 0, 1). Thus we obtain (A + B)2 = A2 + 2AB + B 2 + 2C2 . Thus it follows that

1 C2 = − [A, B]. 2

For α3 we obtain (A + B)3 = A3 + 3A2 B + 3AB 2 + B 3 + 6AC2 + 6BC2 + 6C3 . Using C2 given above we obtain C3 =

1 1 [B, [A, B]] + [A, [A, B]]. 3 6

(ii) Since [B, [A, B]] = 0n and [A, [A, B]] = 0n we find that C3 = C4 = · · · = 0. Thus exp(α(A + B)) = exp(αA) exp(αB) exp(−α2 [A, B]/2).

Problem 28. Let H be a hermitian n × n matrix. Show that exp(H) is a positive definite matrix. Solution 28. If H is hermitian then H 2 , H 3 etc are hermitian and also exp(H). Let λj (j = 1, 2, . . . , n) be the real eigenvalues of H since H is hermitian. Then eλj (j = 1, 2, . . . , n) are the real eigenvalues of exp(H) and obviously eλj > 0 for (j = 1, 2, . . . , n). Thus exp(H) is a positive definite matrix. Problem 29. Let σ1 , σ2 , σ3 be the Pauli spin matrices. Does the set of 4 × 4 matrices { I2 ⊗ I2 , σ1 ⊗ σ1 , −σ2 ⊗ σ2 , σ3 ⊗ σ3 } form a group under matrix multiplication? Solution 29.

We have σ1 σ2 = iσ3 ,

σ2 σ1 = −iσ3 ,

σ3 σ2 = iσ1 ,

σ3 σ1 = iσ2 ,

σ2 σ3 = iσ1 , σ1 σ3 = −iσ2 .

Thus (σ1 ⊗ σ1 )(−σ2 ⊗ σ2 ) = −(σ1 σ2 ) ⊗ (σ1 σ2 ) = σ3 ⊗ σ3 (−σ2 ⊗ σ2 )(σ3 ⊗ σ3 ) = −(σ2 σ3 ) ⊗ (σ2 σ3 ) = σ1 ⊗ σ1 (σ3 ⊗ σ3 )(σ1 ⊗ σ1 ) = (σ3 σ1 ) ⊗ (σ3 σ1 ) = −σ2 ⊗ σ2 .

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92 Problems and Solutions The neutral element is I2 ⊗ I2 . Each element is its own inverse. Thus the set forms a group under matrix multiplication. Problem 30. The spin matrices for spin-2 particles (for example graviton) are given by 0 2  1 J1 =  0 2 0 0  0 2 i  J2 =  0 2 0 0  2 0 0 1  J3 =  0 0  0 0 0 0 

 2 √0 0 0 6 √0 0  √0  6 √0 6 0,  0 6 0 2 0 0 2 0  −2 0 0 0 √ 0 0  √0 − 6 √  6 √0 − 6 0 ,  0 6 0 −2 0 0 2 0  0 0 0 0 0 0   0 0 0 .  0 −1 0 0 0 −2

(i) Show that the matrices are hermitian. (ii) Find the eigenvalues and eigenvectors of these matrices. (iii) Calculate the commutation relations. (iv) Are the matrices unitary? Solution 30.

(i) Obviously the matrices are hermitian, i.e. J1∗ = J1 ,

J2∗ = J2 ,

J3∗ = J3 .

(ii) The eigenvalues of J1 are −2, 2, −1, 1, 0 with the corresponding normalized eigenvectors       1 1 1 −2  2  −1  1 1 1 √  √    u1 =  6  , u2 =  6  , u3 =  0  , 4 4 2    2 2 1 1 1 −1     1 1 √ 1  0  1 3   √ √   u4 =  0  , u5 = √  − 2/ 3  . 2   8 −1 0 −1 1

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The eigenvectors form an orthonormal basis in C5 . The eigenvalues of J2 are 2, 1, 0, −1, −2 with the corresponding normalized eigenvectors  1  −2i   1  1  u1 =  √  , 4 − 6   2i 1 

 1  2i √   1 − 6 u2 =  , 4  −2i    1 1 

 1 i  1   u4 =  0  , 2  i −1 

 1 −i  1   u3 =  0  , 2  −i −1 

  1 √  0  3 √ √  u5 = √  2/ 3  . 8 0  1

The eigenvalues of J3 are 2, 1, 0, −1, −2 with the corresponding eigenvectors (standard basis)   1 0   u1 =  0  ,   0 0

  0 1   u2 =  0  ,   0 0

  0 0   u3 =  1  ,   0 0

  0 0   u4 =  0  ,   1 0

  0 0   u5 =  0  .   0 1

(iii) The commutation relations are [J1 , J2 ] = iJ3 , [J2 , J3 ] = iJ1 , [J3 , J1 ] = iJ2 . (iv) No the matrices are not unitary. Note that det(J1 ) = det(J2 ) = det(J3 ) = 0 owing to the eigenvalue 0. Problem 31. Two orthonormal bases in an n-dimensional complex Hilbert space { |uj i : j = 1, 2, . . . , n }, { |vj i : j = 1, 2, . . . , n } are called mutually unbiased if the inner products (scalar products) between all possible √ pairs of vectors taken from distinct bases have the same magnitude 1/ n, i.e. 1 |huj |vk i| = √ n

for all

j, k ∈ { 1, 2, . . . , n }.

(i) Find such bases for the Hilbert space C2 . (ii) Find such bases for the Hilbert space C3 .

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94 Problems and Solutions Solution 31.

(i) As the first bases we select the standard basis     1 0 , . 0 1

For the second basis we could select    1 1 √ , 2 1 or    1 1 √ , 2 i

1 √ 2



1 −1



  1 1 √ . 2 −i Applying a unitary matrix to these two sets provide other such sets. (ii) An example is the standard basis       0 0   1 0, 1, 0   0 0 1 and    1  1 √ 1,  3 1

  −1 √ 1  √ (1 + 3)/2  , 3 (1 − √3)/2

√   (−1 + i√3)/2  1  √ (−1 − i 3)/2  .  3 1

Applying a unitary matrix to these two sets provide other such sets. Problem 32. Solution 32.

Find a 4 × 4 matrix A such that −A = A−1 = AT = A∗ . Let 02 be the 2 × 2 zero matrix.  0 0   02 −iσ2 0 0 A= ≡ 0 −1 −iσ2 02 1 0

We find  0 −1 1 0  . 0 0 0 0

This matrix plays a role for the charge conjugation in the Dirac equation. Problem 33. Consider the spin matrices for a spin-1 particle      0 1 0 0 −i 0 1 1 1 S1 = √  1 0 1  , S2 = √  i 0 −i  , S3 =  0 2 0 1 0 2 0 i 0 0 and the unit vector n = ( sin(θ) cos(φ) the scalar product

sin(θ) sin(φ)

n · S := n1 S1 + n2 S2 + n3 S3 .

0 0 0

 0 0  −1

cos(θ) ). We define

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Consider the Hamilton operator ˆ = ~ω(n · S). H ˆ Is the Hamilton operator H ˆ her(i) Calculate the Hamilton operator H. mitian? ˆ (ii) Calculate the trace of H. ˆ (iii) Find the eigenvalues and normalized eigenvectors of H. (iv) Do the eigenvectors form a basis in the Hilbert space C3 ? Solution 33.

(i) The Hamilton operator is given by √   cos(θ) √ sin(θ)e−iφ / 2 0 √ ˆ = ~ω  sin(θ)eiφ / 2 H 0 √ sin(θ)e−iφ / 2  . iφ − cos(θ) 0 sin(θ)e / 2

Since exp(iφ) = exp(−iφ) the Hamilton operator is hermitian. Thus the eigenvalues must be real. ˆ is 0. Thus the sum of the three eigenvalues of H ˆ must (ii) The trace of H be 0. ˆ are ~ω, 0, −~ω. The corresponding normalized (iii) The eigenvalues of H eigenvectors are √       −iφ −iφ (1 + cos(θ))e − sin(θ)e−iφ / 2 (1 − cos(θ))e√ /2 √ /2  ,  . sin(θ)/ 2 cos(θ) √  ,  − sin(θ)/ 2 sin(θ)eiφ / 2 (1 − cos(θ))eiφ /2 (1 + cos(θ))eiφ /2 (iv) The Hamilton operator is hermitian and the three eigenvalues are different. Thus the normalized eigenvectors form an orthonormal basis in the Hilbert space C3 . Problem 34. Consider a complex Hilbert space H and |φ1 i, |φ2 i ∈ H. Let c1 , c2 ∈ C. An antilinear operator K in this Hilbert space H is characterized by K(c1 |φ1 i + c2 |φ2 i) = c∗1 K|φ1 i + c∗2 K|φ2 i. A comb is an antilinear operator K with zero expectation value for all states |ψi of a certain complex Hilbert space H. This means hψ|K|ψi = hψ|LC|ψi = hψ|L|ψ ∗ i = 0 for all states |ψi ∈ H, where L is a linear operator and C is the complex conjugation. (i) Consider the two-dimensional Hilbert space H = C2 . Find a unitary 2 × 2 matrix such that hψ|U C|ψi = 0.

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96 Problems and Solutions (ii) Consider the Pauli spin matrices with σ0 = I2 , σ1 , σ2 , σ3 . Find 3 X 3 X

hψ|σµ C|ψig µ,ν hψ|σν C|ψi

µ=0 ν=0

where g µ,ν = diag(−1, 1, 0, 1). Solution 34.

(i) We find U = σ2 since ∗

hψ|σ2 C|ψi = hψ|σ2 |ψ i =

( ψ1∗

ψ2∗

 )

0 i

−i 0



ψ1∗ ψ2∗

 = 0.

(ii) We have 3 X 3 X

hψ|σµ C|ψig µ,ν hψ|σν C|ψi = −hψ|σ0 |ψ ∗ i2 + hψ|σ1 |ψ ∗ i2 + hψ|σ3 |ψ ∗ i2

µ=0 ν=0

= 0. Problem 35. Consider the Hilbert space Cd . Let |ji (j = 1, . . . , d) be an orthonormal basis in Cd . Then a d × d matrix A acting in Cd can be written as d X A= ajk |jihk| j,k=1

with ajk ∈ C. Obviously A depends on the underlying orthonormal basis. If we have the standard basis, then A reduces to the matrix A = (ajk ). We 2 can associate a vector |ψA i in the Hilbert space Cd with the matrix A via |ψA i =

d X

ajk |ji ⊗ |ki.

j,k=1

(i) Let d = 2 and consider the standard basis     1 0 , |2i = . |1i = 0 1 Find A and |ψA i. (ii) Let d = 2 and consider the Hadamard basis     1 1 1 1 , |2i = √ . |1i = √ 2 1 2 −1 Find A and |ψA i.

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97

(i) We have

A = a11 |1ih1| + a12 |1ih2| + a21 |2ih1| + a22 |2ih2|        a11 0 0 a12 0 0 0 = + + + 0 0 0 0 a21 0 0

0 a22



 =

a11 a21

a12 a22



and |ψA i = a11 |1i ⊗ |1i + a12 |1i ⊗ |2i + a21 |2i ⊗ |1i + a22 |2i ⊗ |2i           a11 0 0 0 a11  0   a12   0   0   a12  = + + + = . 0 0 a21 0 a21 0 0 0 a22 a22 (ii) We have A = a11 |1ih1| + a12 |1ih2| + a21 |2ih1| + a22 |2ih2|        1 1 1 1 1 1 1 −1 1 1 1 + a12 + a21 + a22 = a11 1 1 1 −1 −1 −1 −1 2 2 2 2   1 a11 + a12 + a21 + a22 a11 − a12 + a21 − a22 = 2 a11 + a12 − a21 − a22 a11 − a12 − a21 + a22

−1 1

and |ψA i = a11 |1i ⊗ |1i + a12 |1i ⊗ |2i + a21 |2i ⊗ |1i + a22 |2i ⊗ |2i         1 1 1 1 1 1 1  −1  1  1  1  −1  = a11   + a12   + a21   + a22   1 1 −1 −1 2 2 2 2 1 −1 −1 1   a11 + a12 + a21 + a22 1  a − a12 + a21 − a22  =  11 . 2 a11 + a12 − a21 − a22 a11 − a12 − a21 + a22 Extend to d = 3 and consider the orthonormal basis       1 0 1 1   1 0 , |2i =  1  , |3i = √  0  . |1i = √ 2 1 2 −1 0 Find A and |ψA i. Describe the connection of the map A 7→ |ψA i with the vec-operator. Problem 36. Let s be a spin with a fixed total angular momentum quantum number s ∈ {1/2, 1, 3/2, 2, . . .}.



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98 Problems and Solutions The (normalized) eigenstates of x3 -angular momentum |s, mi form a ladder with m = −s, −s + 1, . . . , s − 1, s. The eigenstates |s, mi form an orthonormal basis in a 2s + 1 dimensional Hilbert space. For example if s = 1/2 we have the two states |1/2, −1/2i, |1/2, 1/2i and can identify     1 0 |1/2, 1/2i 7→ , |1/2, −1/2i 7→ . 0 1 Thus we have the Hilbert space C2 . For s = 1 we have the three states |1, −1i, |1, 0i, |1, 1i and can identify       0 0 1 |1, −1i 7→  0  , |1, 0i 7→  1  , |1, 1i 7→  0  . 1 0 0 A spin coherent state |s, θ, φi for s = 1/2, 1, 3/2, . . . can be given by s m=s X (2s)! (cos(θ/2))s+m (sin(θ/2))s−m e−imφ |s, mi. |s, θ, φi = (s + m)!(s − m)! m=−s (i) Find |1/2, θ, φi and write it as a vector in C2 . (ii) Find |1, θ, φi and write it as a vector in C3 . (iii) For a given s find the scalar product hs, m|s, θ, φi. Solution 36.

(i) We obtain

|1/2, θ, φi = sin(θ/2)eiφ/2 |1/2, −1/2i + cos(θ/2)e−iφ/2 |1/2, 1/2i. Thus

 |1/2, θ, φi 7→

cos(θ/2)e−iφ/2 sin(θ/2)eiφ/2

 .

(ii) For s = 1 we obtain sin2 (θ/2)eiφ |1, −1i +

√

2 cos(θ/2) sin(θ/2)|1, 0i + cos2 (θ/2)e−iφ |1, 1i.

Thus we the state in C3  2 −iφ √ cos (θ/2)e |1, θ, φi 7→  2 cos(θ/2) sin(θ/2)  . sin2 (θ/2)eiφ 

(iii) Since hs, m0 |s, mi = δm0 ,m we obtain s (2s)! hs, m|s, θ, φi = (cos(θ/2))s+m (sin(θ/2))s−m e−imφ . (s + m)!(s − m)!

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99

Problem 37. Let n ≥ 1 and {|0ii, |1i, . . . , |ni} be an orthonormal basis in Cn+1 . Consider the linear operators ((n + 1) × (n + 1) matrices) an =

n X p j|j − 1ihj|,

a†n =

j=1

n √ X

k|kihk − 1|.

k=1

Find the commutator [an , a†n ]. Note that n X

|`ih`| = In+1 .

`=0

Solution 37. an a†n =

We have n n n X X X p √ k|k − 1ihk − 1| j k|j − 1ihj|kihk − 1| = j=1 k=1

k=1

n X n √ n X X p a†n an = k j|kihk − 1|j − 1ihj| = j|jihj|. j=1

k=1 j=1

Thus [an , a†n ] = an a†n − a†n an = In+1 − (n + 1)|nihn|. Problem 38.

Let z ∈ C. Consider the spin-1 matrix   0 −i 0 1  S2 = √ i 0 −i  . 2 0 i 0

Calculate exp(zS2 ). Then substitute z = −iωt. Since S23 = S2 , S24 = S22 etc we obtain    2  z3 z5 z z4 exp(zS2 ) = I3 + S2 z + + + · · · + S22 + + ··· 3! 5! 2! 4!

Solution 38.

= I3 + S2 sinh(z) + S22 (cosh(z) − 1). With z = −iωt we obtain exp(−iωtS2 ) = I3 − i sin(ωt)S2 + (cos(ωt) − 1)S22 . Since

 1 1 S22 =  0 2 −1

 0 −1 2 0  0 1

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100 Problems and Solutions we end up with √  − sin(ωt)/ 2 (− cos(ωt) +√ 1)/2 cos(ωt)√ − sin(ωt)/ 2  . sin(ωt)/ 2 1 + (cos(ωt) − 1)/2



1 + (cos(ωt) −√1)/2 exp(−iωtS2 ) =  − sin(ωt)/ 2 (− cos(ωt) + 1)/2

Problem 39. Consider the Hilbert space M2 (C) of all 2 × 2 matrices over C with scalar product hA, Bi := tr(AB ∗ ), The standard basis is    1 0 0 E11 = , E12 = 0 0 0

1 0

A, B ∈ M2 (C).



 ,

E21 =

A mutually unbiased basis is   1 1 1 0 µ0 = √ σ0 = √ , 2 2 0 1   1 1 0 −i µ2 = √ σ2 = √ , 2 2 i 0

0 1

0 0



 ,

E22 =

1 1 µ1 = √ σ1 = √ 2 2



1 1 µ3 = √ σ3 = √ 2 2



0 0

0 1





0 1

1 0

1 0

0 −1

,  .

(i) Express the Hadamard matrix 1 A= √ 2



1 1

1 −1



with this mutually unbiased basis. (ii) Express the Bell matrix 1 1 0 B=√  2 0 1 

0 1 1 0

0 1 −1 0

 1 0   0 −1

with the basis (sixteen dimensional) given by µj ⊗ µk , (j, k = 0, 1, 2, 3). Solution 39.

(i) We have the expansion A=

3 X

hA, µj iµj

j=0

with hA, µ0 i = 0,

hA, µ1 i = 1,

hA, µ2 i = 0,

hA, µ3 i = 1.

.

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Matrix Properties

101

Hence A = µ1 + µ3 . (ii) We have the expansion B=

3 X 3 X

hB, µj ⊗ µk i(µj ⊗ µk ).

j=0 k=0

The only nonzero expansion coefficients are √ √ hB, µ3 ⊗ µ0 i = 2, hB, µ1 ⊗ µ1 i = 2. Hence B=

√

2µ3 ⊗ µ0 +

√

2µ1 ⊗ µ1 .

Problem 40. Two orthonormal bases in an n-dimensional complex Hilbert space { |uj i : j = 1, 2, . . . , n }, { |vj i : j = 1, 2, . . . , n } are called mutually unbiased if inner products (scalar products) between all possible pairs of vectors taken from distinct bases have the same magnitude √ 1/ n, i.e. 1 |huj |vk i| = √ n

for all

j, k ∈ { 1, 2, . . . , n }.

(i) Find such bases for the Hilbert space C2 . Start of with the standard basis     1 0 u1 = , u2 = . 0 1 (ii) Find such bases for the Hilbert space C3 . Start of with the standard basis       1 0 0 u1 =  0  , u2 =  1  , u3 =  0  . 0 0 1 (iii) Find such bases for the Hilbert space C4 using the result from C2 and the Kronecker product. Solution 40.

(i) For the second basis we could select     1 1 1 1 , v2 = √ . v1 = √ 2 1 2 −1

Another selection would be 1 v1 = √ 2

  1 , i

1 v2 = √ 2



1 −i

 .

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102 Problems and Solutions (ii) A possible solution is √     1 1/ 3 √ 1 v1 = √  1  , v2 =  −i/2 − 1/(2√ 3)  , 3 1 i/2 − 1/(2 3)

√  1/ 3 √ v3 =  i/2 − 1/(2 √3)  . −i/2 − 1/(2 3) 

Problem 41. (i) Let A, B be n × n matrices over C such that A2 = In and B 2 = In . Furthermore assume that [A, B]+ ≡ AB + BA = 0n i.e. the anticommutator vanishes. Let α, β ∈ C. Calculate eαA+βB using eαA+βB =

∞ X (αA + βB)j j=0

j!

.

(ii) Consider the case that n = 2 and 

α = −iωt, β = −i∆t/~,

 1 0 A = σ3 = 0 −1   0 1 B = σ1 = . 1 0

(iii) Consider the case that n = 8 and α = −iωt, β = −i∆t/~, Solution 41.

A = σ3 ⊗ σ3 ⊗ σ3 B = σ1 ⊗ σ1 ⊗ σ1 .

(i) Since BA = −AB we have

(αA + βB)2 = (α2 + β)2 In ,

(αA + βB)3 = (α2 + β 2 )(αA + βB).

Thus in general we have for positive n (αA + βB)n = (α2 + β 2 )n/2 In

for n even

(αA + βB)n = (α2 + β 2 )n/2−1

for n odd

and Thus we have the expansion 1 1 1 2 (α + β 2 ) + (α2 + β 2 )2 + (α2 + β 2 )3 + · · ·) 2! 4! 6! 1 1 +(αA + βB)(1 + (α2 + β 2 ) + (α2 + β 2 )2 + · · ·). 3! 5!

eαA+βB = In (1 +

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103

This can be summed up to eαA+βB = In cosh(

p p αA + βB α2 + β 2 ) + p sinh( α2 + β 2 ). 2 2 α +β

(ii) We have p

α2 + β 2 =

p

−ω 2 t2 − ∆2 t2 /~2 =

it p 2 2 ~ ω + ∆2 . ~

√

~2 ω 2 + ∆2 . It follows that  αA + βB ~ωA + ∆B −~ω/E =− = 2 2 −∆/E α +β E

We set E :=

−∆/E ~ω

 .

Thus  p αA + βB −i sin(Et/~)~ω p sinh( α2 + β 2 ) = −i sin(Et/~)∆/E α2 + β 2

−i sin(Et/~)∆/E i sin(Et/~)~ω



and e

αA+βB

 =

cos(Et/~) − i sin(Et/~)~ω/E −i sin(Et/~)∆/E

−i sin(Et/~)∆/E cos(Et/~) + i sin(Et/~)~ω/E



Problem 42. Let H be an n × n hermitian matrix and λ1 , . . . , λn be the eigenvalues with the pairwise orthogonal normalized eigenvectors v1 , . . . , vn . Then we can write n X H= λ` v` v`∗ . `=1

Let P = In − vj vj∗ − vk vk∗ + vj vk∗ + vk vj∗ ,

j 6= k.

(i) What is the condition on the eigenvalues of H such that P HP ∗ = H. (ii) Find P 2 . Solution 42. (i) Note that P is hermitian. Utilizing v`∗ vj = δ`j we find by straightforward calculation P HP ∗ =

n X

λ` v` v`∗ + (λk − λj )vj vj∗ + (λj − λk )vk vk∗ .

`=1

Thus λj = λk . (ii) We obtain P 2 = In .

.

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104 Problems and Solutions Problem 43. Consider the orthogonal group O(n, R) ⊂ Rn×n with the linear product (v, w ∈ O(n, R)) hv, wi := tr(v T w). The orthogonal projection Π(g) : Rn×n → Tg O(n) is given by Π(g)v :=

1 (v − gv T g). 2

(i) Let n = 2 and  g=

0 1

1 0



 ,

v=

Find the orthogonal projection. (ii) Let n = 2 and   0 1 g= , 1 0

 v=

a c

b d



a c

b d



0 1

1 0



.

.

Find the orthogonal projection. Solution 43. (i) We have     1 a b 0 1 a − Π(g)v = c d 1 0 b 2

c d



1 = 2



a−d 0 0 d−a



(ii) We have    1 a b 0 − c d −1 2   1 a+d 0 . = 0 a+d 2

Note that

 tr

a−d 0 0 d−a

Consider the case that



1 g=√ 2





c d



a+d 0 0 a+d



1 0

Π(g)v =

1 1

1 −1

a b

0 −1

1 0



= 0.

 .

Problem 44. Let n ≥ 1 and m ≥ 1. Consider the T = (tj1 ...jm order-m tensor of size (n × · · · × n) (m-times), (j1 , . . . , jm = 1, . . . , n). One defines the operator on v ∈ Cn written as (T vm−1 )k :=

n X j2 =1

···

n X jm =1

tkj2 ...jm vj2 . . . vjm ,

k = 1, . . . , n.

.

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105

The (E−) eigenvector of T are the fixed points (up to scaling) of this operator T vm−1 = λv where v 6= 0. Let m = 3, n = 2 with t122 = 1, t211 = 1 and all other entries are 0. Solve the eigenvalue problem. Solution 44.

We obtain

(T v2 )1 = a122 v22 = v22 = λv1 ,

(T v2 )2 = a211 v12 = v12 = λv2

Now λ = 0 is not a solution, since v12 = v22 = 0 implies v1 = v2 = 0. If λ 6= 0, then v1 6= 0 and v2 6= 0. We also have v13 = v23 . Problem 45. curl(B) =

Starting from Maxwell’s equations in vacuum 1 ∂E , c2 ∂t

∂B , ∂t

curl(E) = −

div(E) = 0,

div(B) = 0

and Kramer’s vector F := E + icB, F∗ := E − icB show that the photon is a spin-1 particle. Solution 45. curl(F) =

Using Kramer’s vector we can write i ∂F , c ∂t

curl(F∗ ) = −

i ∂F∗ , c ∂t

div(F) = 0,

div(F∗ ) = 0.

Let jk`

  +1 if jk` are an even permutation of the integers 123 = −1 if jk` are an odd permutation of the integers 123 .  0 otherwise

Since (curlF)j =

3 3 X X

3

jk`

k=1 `=1

we can write

3 X 3  X k=1 `=1

3

XX ∂ ∂ F` = − kj` F` ∂xk ∂xk k=1 `=1

∂ −i ∂xk

 (−ikj` ) F` =

i ∂Fj . c ∂t

Introducing the differential operator pˆk := −i we find −

3 X 3 X j=1 k=1

∂ ∂xk

pˆk ikj` F` =

i ∂F` . c ∂t

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106 Problems and Solutions For fixed k, −ikj` is a 3 × 3 matrix, Sk(j,`) . The equation for F then takes the form ! 3 X i ∂F pˆk Sk F = (ˆ . p · S)F = c ∂t k=1

Using the definition of jk` , we obtain the representation for the matrices      0 0 0 0 0 1 0 −1 S1 = i  0 0 −1  , S2 = i  0 0 0  , S3 = i  1 0 0 1 0 −1 0 0 0 0

3×3  0 0. 0

The commutators are [S1 , S2 ] = iS3 , [S2 , S3 ] = iS1 , [S3 , S1 ] = iS2 . We have S × S = iS and S12 + S22 + S32 = 2I3 where I3 is the 3 × 3 identity matrix. Thus Maxwell’s equations describe a particle of spin-1.

Programming Problems Problem 1. Let σ1 , σ2 , σ3 be the Pauli spin matrices. Find the eigenvalues and eigenvectors of the 8 × 8 hermitian matrix H = (σ1 ⊗ σ1 + σ2 ⊗ σ2 + σ3 ⊗ σ3 ) ⊗ σ1 . The 8 × 8 hermitian matrix can be written as a direct sum   0 −1 0 2     0 1 2 0  0 1  −1 0 ⊕ . H= ⊕  1 0 1 0 0 2 0 −1 2 0 −1 0

Solution 1. Thus the eigenvalues can be calculated from the two 2 × 2 matrices and the 4 × 4 matrix. Applying the Maxima program /* directsum.mac */ sig1: matrix([0,1],[1,0]); sig2: matrix([0,-%i],[%i,0]); sig3: matrix([1,0],[0,-1]); sig11: kronecker_product(sig1,sig1); sig22: kronecker_product(sig2,sig2); sig33: kronecker_product(sig3,sig3); S: sig11 + sig22 + sig33; H: kronecker_product(S,sig1);

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Matrix Properties

107

eigenvectors(H); D: matrix([0,-1,0,2],[-1,0,2,0],[0,2,0,-1],[2,0,-1,0]); eigenvectors(D);

we find the eigenvalues −1 (3 times), +1 (3 times), +3 (1 times), −3 (1 times) of the matrix H. The eigenvalues of the 4 × 4 matrix are −1, 1, −3, 3 and the eigenvectors are  1  −1   , 1 −1 

  1 1  , 1 1

 1  −1   . −1 1

 1  1   , −1 −1





Problem 2. Let In be the n × n identity matrix. An invertible matrix 2 2 X ∈ Cn ×n satisfies the Yang-Baxter equation if (X ⊗ In )(In ⊗ X)(X ⊗ In ) = (In ⊗ X)(X ⊗ In )(In ⊗ X). If X satisfies the Yang-Baxter equation, then X ∗ satisfies the Yang-Baxter equation. If X satisfies the Yang-Baxter equation, then X −1 satisfies the Yang-Baxter equation. If X satisfies the Yang-Baxter equation and Q ∈ Cn×n is an arbitrary invertible matrix. Then e = (Q ⊗ Q)X(Q ⊗ Q)−1 X also satisfies the Yang-Baxter equation. Show that 1 (1 + i)  0 X=  0 2 −1 

0 1 −1 0

 0 1 1 0  1 0 0 1

satisfies the Yang-Baxter equation with n = 2. Solution 2.

The following Maxima program provides the proof

/* YBBell.mac */ I2: matrix([1,0],[0,1]); X: ((1+%i)/2)*matrix([1,0,0,1],[0,1,1,0],[0,-1,1,0],[-1,0,0,1]); T1: kronecker_product(X,I2); T2: kronecker_product(I2,X); F: (T1 . T2 . T1) - (T2 . T1 . T2);

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108 Problems and Solutions

3.3

Supplementary Problems

Problem 1. Let σ1 , σ2 , σ3 be the Pauli spin matrices. Consider the sixteen 4 × 4 matrices     σ1 0 2 σ2 02 I16 , Γ1 = , Γ2 = , 02 σ1 02 σ2  Γ3 =

02 σ3

σ3 02



 ,

Γ[µ,ν] =

Γ4 =

02 iσ3

−iσ3 02

1 i(Γµ Γν − Γν Γµ ), 2



 ,

Γ5 =

−σ3 02

02 σ3

 ,

µ, ν = 1, 2, 3, 4, 5.

Show that Γν Γµ + Γµ Γν = 2δνµ I4 [Γν , Γ[λ,µ] ] = 2iδνλ Γµ − 2iδνµ Γλ [Γ[ν,µ] , Γ[λ,σ] ] = 2iδµλ Γ[ν,σ] − 2iδνλ Γ[µ,σ] + 2iδνσ Γ[ν,λ] − 2iδµσ Γ[ν,λ] . Do the 16 matrices form an orthonormal basis in the Hilbert space of the 4 × 4 matrices? Problem 2.

Consider the 25 × 25 matrices A = I2 ⊗ I2 ⊗ σ1 ⊗ σ3 ⊗ σ3 B = I2 ⊗ I2 ⊗ σ2 ⊗ σ3 ⊗ σ3 C = σ3 ⊗ σ3 ⊗ σ3 ⊗ σ3 ⊗ σ3 .

Find A2 , B 2 , C 2 , [A, B], [B, C], [C, A], [A, B]+ , [B, C]+ , [C, A]+ . Problem 3. Let A be an n × n matrix over C and f : C → C be analytic in a region D containing the spectrum of A. Then the matrix f (A) can be defined as the Cauchy integral formula Z 1 (zIn − A)−1 f (z)dz. f (A) = 2πi ∂D Let 

0 A = 0 1

0 1 0

 1 0 0

with the spectrum +1 (twice) and −1. Find exp(A) applying the Cauchy integral formula.

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Matrix Properties Problem 4.

109

Let α > 0 and A be an n × n matrix over C. Show that Z i −αA e e−αλ (A − λIn )−1 dλ = 2π C

where C is the contour in the complex λ plane which encloses all eigenvalues of the matrix A. Let   0 1 A= . 0 0 Calculate the right-hand side. Problem 5.

Show that the vector is normalized in R4   cos(θ1 ) sin(θ1 ) cos(θ2 )    . sin(θ1 ) sin(θ2 ) cos(θ3 ) sin(θ1 ) sin(θ2 ) sin(θ3 )

Problem 6. Show that the equation of a hyperplane passing through the points x1 , x2 , . . . , xn in Rn can be given in the form   1 1 1 ··· 1 det = 0. x x1 x2 · · · xn Apply it to n = 4 with (Bell basis)         1 1 0 0 1  0  1 1 1  1  1 0 x1 = √   , x2 = √   , x3 = √   , x4 = √  . 0 2 0 2 2 1 2 −1 1 −1 0 0

Problem 7. Let σ1 , σ2 , σ3 be the Pauli spin matrices. (i) Show that the matrices 1 (I2 + σj ), 2

1 (I2 − σj ), 2

j = 1, 2, 3

are projection matrices. (ii) Show that the matrices 1 (I4 + σj ⊗ σk ), 2 are projection matrices.

1 (I4 − σj ⊗ σk ), 2

j, k = 1, 2, 3

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110 Problems and Solutions (iii) Show that the matrices 1 (I8 + σj ⊗ σk ⊗ σ` ), 2

1 (I8 − σj ⊗ σk ⊗ σ` ), 2

j, k, ` = 1, 2, 3

are projection matrices. (iv) Let U be an n × n unitary and hermitian matrix. Show that Π=

1 (In − U ) 2

is a projection matrix. Show that Π ⊗ Π is a projection matrix. Problem 8.

Consider the matrices    1 0 1 0 A = 1 0 1, B = 0 1 0 1 0

0 1 0

 1 0. 1

Write down all six 3 × 3 permutation matrices with     1 0 0 0 0 1 P0 =  0 1 0  , P5 =  0 1 0  . 0 0 1 1 0 0 Find the permutation matrices in this set such that Pj APjT = A. Find the permutation matrices in this set such that Pj BPjT = B. Problem 9.

Consider the 4 × 4 matrix  −1/2 1/2 1/2  1/2 −1/2 1/2 A= 1/2 1/2 −1/2 1/2 1/2 1/2

 1/2 1/2  . 1/2 −1/2

Find the eigenvalues of A without calculating the eigenvalues. Utilize the information from A2 , tr(A) and that the matrix A is symmetric over R. Then find the eigenvalues of A ⊗ A and A ⊗ I4 + I4 ⊗ A. Problem 10. Find the eigenvalues and normalized eigenvectors of the Hamilton operator (16 × 16 hermitian matrix) ˆ = ~ω1 (σ3 ⊗ σ3 ⊗ I2 ⊗ I2 + I2 ⊗ I2 ⊗ σ3 ⊗ σ3 ) + ~ω2 (σ1 ⊗ σ1 ⊗ σ1 ⊗ σ1 ). H Problem 11.

Consider the Hamilton operator

ˆ = ~ω1 (σ3 ⊗ I2 + I2 ⊗ σ3 ) + ~ω2 σ1 ⊗ σ1 + ~ω3 σ2 ⊗ σ2 . H

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111

ˆ Find the eigenvalues and normalized eigenvectors of H. Let n ≥ 2 and ω = e2πi/n . Consider the n × n matrices   0 0 ... 1 1 0 ... 0  D = diag(1 ω · · · ω n−1 ), Γ =   ... . . . ... 0  . 0 ... 1 0

Problem 12.

So Γ is a permutation matrix with Γn = In . Furthermore Dn = In . Find the commutator [D, Γ]. Problem 13. Let v0 , v1 , v2 , v3 be an orthonormal basis in the Hilbert space C4 . Show that the vectors u0 =

1 (v0 + v1 + v2 + v3 ), 2

u1 =

1 (v0 − v1 + v2 − v3 ), 2

u2 =

1 (v0 + v1 − v2 − v3 ), 2

u3 =

1 (v0 − v1 − v2 + v3 ) 2

also form an orthonormal basis in C4 . Problem 14. Let x ∈ R. Are the vectors         0 0 0 1 0  0   1  x v1 =  2  , v2 =   , v3 =   , v4 =   0 2 2x x 6 6x 3x2 x3 linearly independent? Find Gram’s matrix G = (vjT vk ),

j, k = 1, 2, 3, 4

and its determinant. Problem 15.

Let

 v(θ) =

cos(θ) sin(θ)

 .

Find all 2 × 2 matrices A, B such that v∗ ABv = (v∗ Av)(v∗ Bv). Problem 16.

Consider the  0 1 A = 1 0 0 1

symmetric binary matrices    0 1 0 1 1, B = 0 1 0. 0 1 0 1

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112 Problems and Solutions Find the eigenvalues and eigenvectors of A and B. Find the eigenvalues and eigenvectors of the anti-commutator [A, B]+ . Discuss. Problem 17.

Find all n × n matrices A and B such that eA ⊗ eB = eA⊗B .

Problem 18. Consider the 2 × 2 elementary matrices       1 0 0 0 0 1 E11 = , E22 = , E12 = , 0 0 0 1 0 0

 E21 =

0 1

0 0



with the commutator [E12 , E21 ] = E11 − E22 . Let θ ∈ R. Show that exp (−θ(E12 + E21 )) = exp(− tanh(θ)E12 ) exp(ln(cosh(θ))(E22 − E11 )) exp(− tanh(θ)E21 ). Problem 19. Let µ ∈ C, S the spin (S = 0, 1/2, 1, 3/2, 2, . . .) and |0i, |1i, . . . , |2Si be the standard basis in C2S+1 . The Bloch coherent states |µi are defined by 1/2 2S  X (2S)! 1 µp |pi. |µi = (1 + |µ|2 )S p=0 p!(2S − p)! Show that 1/2 2S  X 1 (2S)! (µ∗ )p hp|. hµ| = (1 + |µ|2 )S p=0 p!(2S − p)! Show that (completeness relation) Z 1 + 2S d2 µ |µihµ| = I2S+1 2 2 π C (1 + |µ| ) where d2 µ = d( 0. Consider the map AA∗ A→ρ= . tr(AA∗ ) (i) Show that ρ is a density matrix. (ii) Show that ρ is invariant under the map A → AU , where U is an n × n unitary matrix. (iii) Is AA∗ = A∗ A in general? A matrix is called a normal matrix if AA∗ = A∗ A. (iv) Consider the map A∗ A A→σ= . tr(A∗ A) Is σ = ρ? Solution 5. (i) Since A is a nonzero matrix we find that AA∗ is nonzero and tr(AA∗ ) 6= 0. The matrix ρ is positive-semidefinite and tr(ρ) = 1. (ii) We set A0 = AU . Thus we have 0

A0 A ∗ (AU )(AU )∗ AU U ∗ A∗ AA∗ ρ→ρ = = = = 0∗ tr(A0 A ) tr((AU )(AU )∗ ) tr(AU U ∗ A∗ ) tr(AA∗ ) =ρ 0

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119

where we used that U U ∗ = I. (iii) In general we have AA∗ 6= A∗ A. For example, let     0 1 0 0 ∗ A= , A = . 0 0 1 0 Then AA∗ =



1 0

0 0

 ,

A∗ A =



0 0

0 1

 .

Thus AA∗ 6= A∗ A. However, we have tr(AA∗ ) = tr(A∗ A). (iv) From (iii) it follows that in general we have ρ 6= σ. Problem 6. Find a normalized state |φi in the Hilbert space C2 such that we have the density matrix   1 1 I2 + √ (σ1 + σ3 ) . |φihφ| = 2 2

Solution 6.

We have to solve √ √   1 1 + 1/ 1/ 2√ √ 2 |φihφ| = . 1/ 2 1 − 1/ 2 2

We obtain the normalized state q √ √  ( 2 + 1)/(2 2) |φi =  q √ √ . ( 2 − 1)/(2 2)

Problem 7.

Consider the 2 × 2 matrix √ −iφ   3/4 2e /4 √ ρ= . 2eiφ /4 1/4

(i) Is the matrix a density matrix? (ii) If so do we have a pure state or a mixed state? (iii) Find the eigenvalues of ρ. (iv) Find tr(σ1 ρ), where σ1 is the first Pauli spin matrix. Solution 7. (i) We have tr(ρ) = 1 and the matrix is hermitian. Furthermore the eigenvalues are nonnegative. Thus we have a density matrix. (ii) Since ρ2 6= ρ we have a mixed state. √ √ (iii) The eigenvalues are λ1 = (2 + 3)/4, λ2 = (2 − 3)/4 which also indicate the state is mixed.

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120 Problems and Solutions (iv) We obtain 1 tr(σ1 ρ) = √ cos(φ). 2 Problem 8. (i) The Hilbert-Schmidt distance between any two density operators ρ1 and ρ2 is given by the Frobenius-Hilbert-Schmidt norm of their differences p DHS (ρ1 , ρ2 ) := tr((ρ1 − ρ2 )2 ). Let

 ρ1 =

1 0

0 0



 ,

ρ2 =

0 0

0 1

 .

Calculate DHS (ρ1 , ρ2 ). (ii) The Bures distance in the space of mixed quantum states described by the density matrices ρ1 and ρ2 is defined as q 1/2 1/2 DB (ρ1 , ρ2 ) := 2(1 − tr((ρ1 ρ2 ρ1 )1/2 )). Let

 ρ1 =

1/2 0

0 1/2



 ,

ρ2 =

3/4 0

0 1/4

 .

Calculate the Bures distance DB (ρ1 , ρ2 ). (iii) Let ρ, σ be two density operators acting in the same finite dimensional Hilbert space. The trace distance between ρ and σ is defined as D(ρ, σ) := Let

1/2 0 0 0 0  0 ρ= 0 0 0 1/2 0 0 

1 p ∗ tr( (ρ − σ ∗ )(ρ − σ)). 2

 1/2 0  , 0 1/2

1/2  0 σ= 0 0 

Find the trace distance. Solution 8.

(i) Since  ρ1 − ρ2 =



−1/4 0

0 1/4



0 1/8

√ we find DHS (ρ1 , ρ2 ) = 1/ 8. (ii) Since 1/2

1/2

ρ1 ρ2 ρ1

=

3/8 0



0 0 0 0 0 0 0 0

 0 0  . 0 1/2

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121

s √ √ 2 2− 3−1 √ DB (ρ1 , ρ2 ) = . 2

(iii) We have  0 0 0 1/2  0 0 0 0  ρ−σ = . 0 0 0 0 1/2 0 0 0 

Thus

1/4 0  (ρ∗ − σ ∗ )(ρ − σ) =  0 0 

0 0 0 0 0 0 0 0

 0 0  . 0 1/4

It follows that D(ρ, σ) = 1/2. Problem 9. space C4

Consider the linear operator (4 × 4 matrix) in the Hilbert

1 (1 − )I4 + (|0i ⊗ |0i)(h0| ⊗ h0|) 4 where  is a real parameter with  ∈ [0, 1] and the state   1 |0i = . 0 ρ=

Does ρ define a density matrix? Solution 9.

We find the diagonal matrix for ρ  (1 − )/4 +  0 0 0 (1 − )/4 0  ρ= 0 0 (1 − )/4 0 0 0

 0 0  . 0 (1 − )/4

Thus ρ = ρ∗ , tr(ρ) = 1, and hx|ρ|xi ≥ 0, for all x ∈ C4 . The last property follows since all entries on the diagonal are non-negative. Thus ρ defines a density matrix. Problem 10. A mixed state is a statistical mixture of pure states, i.e. the state is described by pairs of probabilities and pure states. Given a mixture { (p1 , |ψ1 i), . . . , (pn , |ψn i) } we define its density matrix to be the positive hermitian matrix ρ=

n X j=1

pj |ψj ihψj |

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122 Problems and Solutions where the pure states |ψj i are normalized (i.e. hψj |ψj i = 1), and pj ≥ 0 for j = 1, 2, . . . , n with p1 + p2 + · · · + pn = 1. (i) Find the probability that measurement in the orthonormal basis { |k1 i, . . . , |kn i} will yield |kj i. (ii) Find the density matrix ρU when the mixture is transformed according to the unitary matrix U . Solution 10. (i) From the probability distribution of states in the mixture we have for the probability P (kj ) of measuring the state |kj i (j = 1, 2, . . . , n) P (kj ) =

n X

pl |hkj |ψl i|2 =

l=1

n X

pl hkj |ψl ihψl |kj i = hkj |ρ|kj i.

l=1

(ii) After applying the transform U to the states in the mixture we have the new mixture { (p1 , U |ψ1 i), . . . , (pn , U |ψn i) }, with the density matrix   n n X X ∗ ρU = pj U |ψj ihψj |U = U  pj |ψj ihψj | U ∗ = U ρU ∗ . j=1

Problem 11.

j=1

(i) The Bell state 1 |ψi = √ (|0i ⊗ |0i + |1i ⊗ |1i) 2

has the density matrix 1 1 0 ρ=  2 0 1 

0 0 0 0

0 0 0 0

 1 0 . 0 1

Show that ρ can be written as linear combination of the matrices Λ00 = 1 1 1 1 2 (I2 ⊗ I2 ), Λ11 = 2 (σ1 ⊗ σ1 ), Λ22 = 2 (σ2 ⊗ σ2 ) and Λ33 = 2 (σ3 ⊗ σ3 ). (ii) The Werner state is described by the density matrix (1 − x)/4 0  = 0 0 

ρW

0 0 (1 + x)/4 −x/2 −x/2 (1 + x)/4 0 0

 0 0   0 (1 − x)/4

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123

where x ∈ [0, 1]. Show that ρW can also be written as linear combination of the operators given in (i). Solution 11.

(i) We find the linear combination ρ=

1 1 1 1 Λ00 + Λ11 − Λ22 + Λ33 . 2 2 2 2

(ii) We obtain the linear combination ρW =

1 x x x Λ00 − Λ11 − Λ22 − Λ33 . 2 2 2 2

Problem 12. Suppose we expand a density matrix for N qubits in terms of Kronecker products of Pauli spin matrices ρ=

3 3 3 X 1 X X cj0 j1 ...jN −1 σj0 ⊗ σj1 ⊗ · · · ⊗ σjN −1 . . . 2N j =0 j =0 j =0 0

N −1

1

where σ0 = I2 . (i) What is condition on the expansion coefficients if we impose ρ∗ = ρ? (ii) What is the condition on the expansion coefficients if we impose tr(ρ) = 1? (iii) Calculate tr(ρσk0 ⊗ σk1 ⊗ · · · ⊗ σkN −1 ). Solution 12. (i) Since σ1 = σ1∗ , σ2 = σ2∗ , σ3 = σ3∗ and I2 = I2∗ we find that the expansion coefficients are real. (ii) Since tr(A ⊗ B) = tr(A)tr(B) for square matrices A and B and tr(σ1 ) = tr(σ2 ) = tr(σ3 ) = 0,

tr(I2 ) = 2

we find c00...0 = 1. (iii) Since tr(σ1 σ2 ) = 0, tr(σ2 σ3 ) = 0, tr(σ3 σ1 ) = 0 we find tr(ρσk0 ⊗ σk1 ⊗ · · · ⊗ σkN −1 ) = ck0 k1 ...kN −1 . Problem 13. Let A and B be a pair of qubits and let the density matrix of the pair be ρAB , which may be pure or mixed. We define the spin flipped density matrix to be ρeAB := (σ2 ⊗ σ2 )ρ∗AB (σ2 ⊗ σ2 ) where the asterisk denotes complex conjugation and transpose in the standard basis { |0i ⊗ |0i,

|0i ⊗ |1i,

|1i ⊗ |0i,

|1i ⊗ |1i }

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124 Problems and Solutions and

 σ2 =

0 i

−i 0

 .

Since both ρAB and ρeAB are positive operators, it follows that the product ρAB ρeAB , though non-hermitian, also has only real and non-negative eigenvalues. Consider the Bell state 1 |ψi := √ (|0i ⊗ |0i + |1i ⊗ |1i) 2 and ρ := |ψihψ|. Find the eigenvalues of ρAB ρeAB . Solution 13.

Since 1 1 0 ρ = |ψihψ| =  2 0 1 

0 0 0 0

0 0 0 0

 1 0  0 1

we have ρ∗ = ρ. Furthermore 0  0 σ2 ⊗ σ2 =  0 −1 

 0 0 −1 0 1 0  . 1 0 0 0 0 0

Thus ρe = ρ and ρe ρ = ρ with eigenvalues 1, 0, 0, 0. The tangle of the density matrix ρAB is defined as τAB := [max { µ1 − µ2 − µ3 − µ4 , 0}]2 where µj are the square root of the eigenvalues of ρAB ρeAB ordered in decreasing order. For the special case in which the state of AB is pure, the matrix ρAB ρeAB has only one non-zero eigenvalue. One can show that τAB = 4 det(ρA ) where ρA is the density matrix of qubit A, that is, the trace of ρAB over qubit B. Problem 14.

Consider the density matrix ρ=

1 (I2 + r · σ) 2

where r · σ := r1 σ1 + r2 σ2 + r3 σ3 and r2 ≤ 1. Consider the four normalized vectors a1 , a2 , a3 , a4 in R3         1 1 −1 −1 1   1  1 1 a1 = √ 1 , a2 = √ −1  , a3 = √  1  , a4 = √  −1  3 1 3 −1 3 −1 3 1

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Density Operators such that

4 1 δjk − = 3 3

aTj ak = We have

4 X



125

1 for j = k . −1/3 for j 6= k

4

3X aj aTj = I3 . 4 j=1

aj = 0,

j=1

Such a quartet of vectors consists of the vectors pointing from the center of a cube to nonadjacent corners. These four vectors can be viewed as the normal vectors for the faces of the tetrahedron that is defined by the other four corners of the cube. Owing to the conditions the four vectors are normalized. Each such quartets of aj ’s defines a positive operator-valued measure for minimal four-state tomography owing to 4 X

Pj = I 2 ,

Pj :=

j=1

1 (I2 + aj · σ) . 4

(i) Show that pj := hPj i = tr(Pj ρ) = 41 (1 + aj · r). (ii) Given pj for j = 1, 2, 3, 4 find the density matrix ρ. Solution 14.

(i) Since tr(σ1 ) = tr(σ2 ) = tr(σ3 ) = 0 we have

hPj i = tr(Pj ρ) =

1 tr((I2 + aj · σ)(I2 + r · σ)) 8

1 1 = tr(I2 + aj · σ + r · σ + (aj · σ)(r · σ)) = tr(I2 + (aj · σ)(r · σ)) 8 8 1 = tr(I2 + (aj · r)I2 ) 8 1 = (1 + aj · r). 4 (ii) Since 

 4 X 3 I3 r =  aj aTj  r 4 j=1 the vector r is obtained as r=

4 4 4 X X 3X 1 (aj · r)aj = 3 (1 + aj · r) aj = 3 pj aj 4 j=1 4 j=1 j=1

P4 where we used j=1 aj = 0. From aj · σ = 4Pj − I2 and substituting r from above yields the density matrix ρ=6

4 X j=1

pj Pj − I2 =

4 X j=1

hPj i(6Pj − I2 ) =

1 (I2 + r · σ). 2

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126 Problems and Solutions It follows that pj is restricted to the range 0 ≤ pj ≤ 1/2 and the probabilities pj obey the inequalities 4

1 X 2 1 3 + r2 ≤ ≤ . pj = 4 j=1 12 3 The upper bound is reached by all pure states, ρ = ρ2 and r2 = 1. The lower bound is reached for the completely mixed state, ρ = 13 I2 and r2 = 0. Problem 15. mixed states

Let |0i, |1i be the standard basis in C2 . Consider the     4 1 3 4 1 3 |0i + |1i + |0i − |1i 2 5 5 2 5 5

and

9 16 { |0i } + { |1i }. 25 25 Find the density matrices. Discuss. Solution 15. In the first case we have       1 1 9/25 12/25 9/25 −12/25 9/25 0 + = . ρ1 = 0 16/25 2 12/25 16/25 2 −12/25 16/25 In the second case we have    9 1 0 16 0 ρ2 = + 25 0 0 25 0

0 1



 =

9/25 0 0 16/25

 .

Thus these two different mixed states correspond to the same density matrix and thus they are indistinguishable. Problem 16. Let ρ1 and ρ2 be n × n density matrices. Let λj denote the eigenvalues of ρ1 − ρ2 with corresponding orthonormal eigenvectors |φj i where j = 1, 2, . . . , n. (i) Find the difference |D1 − D2 | between the probability distributions D1 and D2 for the measurement of the mixtures ρ1 and ρ2 in the basis { |φ1 i, . . . , |φn i }. (ii) Show that measurement in the basis { |φ1 i, . . . , |φn i } maximizes the difference |D1 − D2 |. Hint. Use Schur’s theorem. For any hermitian matrix A, let a1 ≥ a2 ≥ · · · ≥ an be the non increasing diagonal entries of A and µ1 ≥ µ2 ≥ · · · ≥ µn

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127

the non increasing eigenvalues of A. Then for 1 ≤ k ≤ n k X j=1

µj ≥

k X

aj

j=1

where equality holds for k = n. Solution 16. (i) We write ρ1 and ρ2 in the basis { |φ1 i, . . . , |φn i }. In this basis we have |D1 −D2 | =

n X

|hφj |ρ1 |φj i−hφj |ρ2 |φj i| =

j=1

n X

|hφj |(ρ1 −ρ2 )|φj i| =

j=1

n X

|λj |.

j=1

(ii) Let U be an arbitrary unitary transform (change of basis). We define P := U ρ1 U ∗ and Q := U ρ2 U ∗ . The matrix P − Q is hermitian. Let q1 ≥ q2 ≥ · · · ≥ qn denote the non decreasing diagonal entries of P − Q in the { |φ1 i, . . . , |φn i } basis and ν1 ≥ ν2 ≥ · · · ≥ νn be the non decreasing eigenvalues (i.e. λj ) of P − Q. Consider the difference |D10 − D20 | between the probability distributions D10 and D20 for the measurement of the mixtures ρ1 and ρ2 in the basis { U |φ1 i, . . . , U |φn i } |D10 − D20 | = =

n X j=1 n X

|hφj |U ∗ ρ1 U |φj i − hφj |U ∗ ρ2 U |φj i| |hφj |P |φj i − hφj |Q|φj i| =

j=1

n X

|qj |.

j=1

Since tr(P − Q) = tr(P ) − tr(Q) = 1 − 1 = 0 and tr(P ) − tr(Q) =

n X

(hφj |P |φj i − hφj |Q|φj i) =

j=1

we have for all 1 ≤ k ≤ n k n X X q = q j . j j=1 j=k+1 We conclude from the triangle inequality that X n X k |qj | ≥ 2 qj j=1 j=1

n X j=1

qj

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128 Problems and Solutions where equality holds for some 1 ≤ k0 ≤ n. Similarly k n X X νj . |νj | ≥ 2 j=1 j=1 From Schur’s theorem we have n X j=1

Thus

n X j=1

|νj | ≥

k0 X j=1

νj ≥

k0 X j=1

qj =

n X

|qj |.

j=1

|νj | = |D1 − D2 | ≥ |D10 − D20 | =

n X

|qj |.

j=1

Problem 17. Let σ1 , σ2 , σ3 be the Pauli spin matrices. (i) Is the 4 × 4 matrix ρ = 14 (I2 ⊗ I2 + σ1 ⊗ σ1 ) a density matrix? (ii) Is the 4 × 4 matrix ρ = 14 (I4 − σ1 ⊗ σ1 − σ2 ⊗ σ2 − σ3 ⊗ σ3 ) a density matrix? Solution 17. (i) We have tr(ρ) = 1 and ρ is hermitian. The eigenvalues of ρ are 1/2, 1/2, 0, 0. Thus the matrix ρ is a density matrix (mixed state). (ii) Obviously we have ρ∗ = ρ, tr(ρ) = 1. Furthermore the matrix is positive semidefinite. We have ρ2 = ρ. Thus ρ is a density matrix. We have a pure state. The density matrix is given by   0 0 0 0  0 1/2 −1/2 0  ρ= . 0 −1/2 1/2 0 0 0 0 0

Problem 18.  0 G1 =  1 0  0 G4 =  0 1

Consider the eight Gell-Mann matrices      1 0 0 −i 0 1 0 0 0 0  , G2 =  i 0 0  , G3 =  0 −1 0  , 0 0 0 0 0 0 0 0      0 1 0 0 −i 0 0 0 0 0  , G5 =  0 0 0  , G6 =  0 0 1  , 0 0 i 0 0 0 1 0     1 0 0 0 0 0 1 G7 =  0 0 −i  , G8 = √  0 1 0  . 3 0 0 −2 0 i 0

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129

They all have trace 0 and they are hermitian. (i) Find the anticommutation relations for these matrices. (ii) Consider the matrix   8 √ X 1 ρ= I3 + 3 nj Gj  3 j=1 where nj ∈ R. What is the condition on the vector n = (n1 n2 . . . n8 )T such that ρ is a density matrix for a pure state? Solution 18.

(i) We obtain for the anticommutators 8

[Gj , Gk ]+ =

X 4 δjk I3 + 2 djk` G` 3 `=1

where the nonzero components of the completely symmetric tensor djk` are 1 d118 = d228 = d338 = −d888 = √ , 3

1 d448 = d558 = d668 = d778 = − √ 2 3

d146 = d157 = −d247 = d256 = d344 = d355 = −d366 = −d377 =

1 . 2

(ii) The conditions for a density matrix ρ of a pure state are ρ∗ = ρ,

ρ2 = ρ,

tr(ρ) = 1.

Imposing these conditions we obtain for the vector n that n = n,

nT n = 1,

n?n=n

where (a ? b)j :=

8 √ X 3 djk` ak b` . k,`=1

Problem 19.

Let ρ denote the density matrix (mixed state)   1 1 0 ρ := 2 0 1

in C2 . Find a pure state |Ψi ∈ C2 ⊗ C2 such that the reduced density matrix found by taking the partial trace over the second system (C2 ) is ρ. In other words purify the density matrix ρ to obtain a pure state |Ψi.

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130 Problems and Solutions Solution 19. We begin with the Schmidt decomposition of |Ψi over the Hilbert space C2 ⊗ C2 Sch(|Ψi,C2 ,C2 )

|Ψi =

X

p

λj |ψj i ⊗ |φj i

j=1

where λ1 and λ2 are the eigenvalues of ρ and |ψ1 i and |ψ2 i are the corresponding orthonormal eigenvectors of ρ. The states |φ1 i and |φ2 i in C2 are also orthonormal. The eigenvalues and eigenvectors of ρ are given by λ1 = λ2 = 1/2 and     1 0 |ψ1 i = , |ψ2 i = . 0 1 Thus the spectral decomposition of ρ is given by     1 0 1 1 (1 0) + (0 ρ= 2 0 2 1

1).

Hence

    1 1 1 0 |Ψi = √ ⊗ |φ1 i + √ ⊗ |φ2 i 0 2 2 1 where hφ1 |φ1 i = hφ2 |φ2 i = 1 and hφ1 |φ2 i = hφ2 |φ1 i = 0. Thus we could take |Ψi as one of the Bell states         1 1 0 0 1  0  1 1 1  1  1 0 √  , √  , √  , √   0 2 0 2 2 1 2 −1 1 −1 0 0 but not a product state. Problem 20.

Let r ∈ [0, 1]. Consider the density matrix ρ = r|Φ+ ihΦ+ | + (1 − r)|00ih00|

where |Φ+ i is the Bell state 1 |Φ+ i = √ (|00i + |11i). 2 Calculate the eigenvalues of ρ. Solution 20.

The matrix representation of ρ is   1 − r/2 0 0 r/2 0 0 0 0   ρ= . 0 0 0 0 r/2 0 0 r/2

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131

The characteristic equation is   r  r r r 1− − λ (−λ)(−λ) − λ − (−λ)(−λ) = 0. 2 2 2 2 Thus two eigenvalues are 0 (λ2 = 0, λ3 = 0) with the corresponding eigenvectors     0 0 1 0 u2 =   , u3 =   . 0 1 0 0 The characteristic equation reduces to r   r2 r −λ 1− −λ − =0 2 2 4 with the eigenvalues 1 1p 1 + 2r(r − 1). λ1,4 = ± 2 2 If r = 0 the eigenvalues reduce to 1 and 0. If r = 1 the eigenvalues also reduce to 1 and 0. Problem 21.

(1)

Let ρj

(j = 1, 2, . . . , n) be density matrices in a finite(2)

dimensional Hilbert space H1 . Let ρj (j = 1, 2, . . . , n) be density matrices in a finite-dimensional Hilbert space H2 . Show that the convex combination ρ=

n X

(1)

(2)

λj ρj ⊗ ρj ,

λj ≥ 0,

j=1

n X

λj = 1

j=1

is also a density matrix. Solution 21.

We have   n n X X (1) (2) (1) (2) λj tr(ρj ⊗ ρj ) tr(ρ) = tr  λ j ρj ⊗ ρj  = j=1

=

n X

j=1 (1)

(2)

λj tr(ρj )tr(ρj ) =

j=1

n X

λj

j=1

= 1. (1)

Obviously ρ ≥ 0, since ρj Problem 22.

(2)

≥ 0, ρj

≥ 0 and λj ≥ 0 for j = 1, 2, . . . , n.

Given the Schr¨ odinger equation i~

∂ ˆ |ψi = H|ψi. ∂t

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132 Problems and Solutions Find the time-evolution of the density matrix ρ(t) :=

n X

|ψ (j) (t)ihψ (j) (t)|.

j=1

Solution 22.

From the Schr¨odinger equation we find −i~

∂ (j) ˆ hψ (t)| = hψ (j) (t)|H. ∂t

Thus n

∂ρ X = ∂t j=1



   ∂ (j) ∂ (j) |ψ (t)i hψ (j) (t)| + |ψ (j) (t)i hψ (t)| ∂t ∂t

n

=

   1 X  ˆ (j) ˆ H|ψ (t)i hψ (j) (t)| − |ψ (j) (t)i hψ (j) (t)|H i~ j=1

1 ˆ ˆ (Hρ(t) − ρ(t)H) i~ 1 ˆ = [H, ρ(t)]. i~ =

Note that the equation of motion for ρ(t) differs from the Heisenberg equation of motion by a minus sign. Since ρ(t) is constructed from state vectors it is not an observable like other hermitian operators, so there is no reason to expect that its time-evolution will be the same. The solution to the equation of motion is given by ˆ

ˆ

ρ(t) = e−iHt/~ ρ(0)eiHt/~ .

Problem 23.

Consider the Hamilton operator

γ γ ˆ H(t) = − σ · B(t) ≡ − (σ1 B1 (t) + σ2 B2 (t) + σ3 B3 (t)) 2 2 where γ denotes the gyromagnetic ratio and B(t) denotes the time-dependent magnetic induction. The time-evolution of the density matrix ρ(t) obeys the von Neumann equation i~

dρ(t) ˆ = [H(t), ρ(t)] dt

and the time-dependent expectation value of the spin vector is given by hσ(t)i := tr(σρ(t))

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133

or, written in components hσ1 (t)i = tr(σ1 ρ(t)),

hσ2 (t)i = tr(σ2 ρ(t)),

hσ3 (t)i = tr(σ3 ρ(t)).

It follows that the Bloch vector n(t) pertaining to ρ(t) is related to the spin vector as follows n(t) = hσ(t)i or, written in components n1 (t) = hσ1 (t)i,

n2 (t) = hσ2 (t)i,

n3 (t) = hσ3 (t)i.

Find the time-evolution of n(t). Solution 23.

We have dnj = dt



dσj (t) dt



  dρ(t) = tr σj dt

where j = 1, 2, 3. Inserting the right-hand side of the von Neumann equation, using the cyclic invariance of the trace tr(XY Z) = tr(ZXY ) = tr(Y ZX) and the properties σ1 σ2 = iσ3 , σ2 σ3 = iσ1 , σ3 σ1 = iσ2 , we obtain d γ n(t) = n(t) × B(t) dt ~ where × denotes the vector product. Problem 24.

Consider the state   cos(θ) |ψi = eiφ sin(θ)

in the Hilbert space C2 , where φ, θ ∈ R. Let ρ(t = 0) = ρ(0) = |ψihψ| be ˆ = ~ωσ1 . a density matrix at time t = 0. Given the Hamilton operator H Solve the von Neumann equation to find ρ(t). Solution 24.

We obtain for the density matrix at time t = 0   e−iφ cos(θ) sin(θ) cos2 (θ) ρ(0) = |ψihψ| = . eiφ cos(θ) sin(θ) sin2 (θ)

Now

ˆ

ˆ

ρ(t) = e−iHt/~ ρ(0)eiHt/~ = e−iωtσ1 ρ(0)eiωtσ1 .

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134 Problems and Solutions Since ˆ

eiHt/~ = eiωtσ1 =



cos(ωt) i sin(ωt) i sin(ωt) cos(ωt)



it follows that ∗

ρ(t) = U (t)



cos2 (θ) e−iφ cos(θ) sin(θ) iφ e cos(θ) sin(θ) sin2 (θ)

where

 U (t) =

cos(ωt) i sin(ωt) i sin(ωt) cos(ωt)

 U (t)

 .

Problem 25. Consider the Hilbert space Cn . Let ρ be a density matrix, i.e. ρ ≥ 0 and tr(ρ) = 1. The mean value of an observable A (hermitian n × n matrix) is given by hAi = tr(ρA). If the density ρ is unknown, then it may be determined using n2 mean values hA(k) i (k = 1, 2, . . . , n2 ) obtained from measurement if the set {A(k) } is a basis in the space of all hermitian n × n matrices. (i) Let n = 2,   0 −i A = σ2 = i 0 and tr(ρA) = 0, tr(ρA2 ) = 1, tr(ρA3 ) = 0, tr(ρA4 ) = 1. Find the density matrix. (ii) Let n = 2 and tr(ρI2 ) = 1, tr(ρσ1 ) = −1, tr(ρσ2 ) = 0, tr(ρσ3 ) = 0. Find ρ. Solution 25. matrix is

(i) Note that σ22 = I2 , σ23 = σ2 , σ24 = I2 . The density   1− 0 ρ= ,  ∈ [0, 1]. 0 

(ii) The 2×2 matrices I2 , σ1 , σ2 , σ3 form an orthogonal basis in the Hilbert space of the 2 × 2 matrices with scalar product hX, Y i = tr(XY ∗ ). The density matrix is   1/2 −1/2 ρ= . −1/2 1/2 Problem 26. Let |0i, |1i be the standard basis in C2 . Consider the entangled state 1 |ψi = √ (|0i ⊗ |1i − |1i ⊗ |0i) 2

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135

with the density matrix ρ = |ψihψ|. Find the reduced density matrix ρ1 . Discuss. Solution 26.

We obtain 1 1 ρ1 = (|0ih0| + |1ih1|) = 2 2



1 0

0 1

 .

Thus we have mixed state. Problem 27. Consider a mixture of 25% of the pure state (1, 0)T , 25% of the pure state (0, 1)T and 50% of the pure state √12 (1, 1)T described by the density matrix       1 0 1 1 1 1 1 1 √ (1 1). (1 0) + (0 1) + √ ρ= 4 0 4 1 2 2 1 2 Find the spectral representation of ρ. Use the spectral representation of ρ to find another mixture of pure states with the same (measurement) statistical properties as ρ. Solution 27.

The density matrix takes the form   1 2 1 ρ= 4 1 2

with the eigenvalues λ1 = 3/4 and λ2 = 1/4 with the corresponding normalized eigenvectors     1 1 1 1 , v2 = √ . v1 = √ 2 1 2 −1 Applying the spectral theorem ρ can be written as     3 1 1 1 1 1 1 1 √ (1 1) + √ √ ( 1 −1 ) . ρ= √ 4 2 1 4 2 −1 2 2 √ Consequently ρ can be realized √ by a mixture of 75% of the state (1, 1)T / 2 and 25% of the state (1, −1)T / 2. Problem 28. Consider the Hilbert space Cn . Let ρ be a density matrix in this Hilbert space and H and K be two hermitian n × n matrices. One defines hHi := tr(ρH), hH 2 i := tr(ρH 2 ) and analogously for K. Let p p ∆H := hH 2 i − hHi2 , ∆K := hK 2 i − hKi2 .

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136 Problems and Solutions Then we have the uncertainty relation (∆H)(∆K) ≥

1 |hi[H, K]i| . 2

Let 

1 1 ρ = 0 2 0

0 0 0

 0 0, 1



0 H = 1 0

1 2 0

 0 0, 0



 0 i 0 K =  −i 0 0  . 0 0 0

Show that the uncertainty relation becomes an equality for the given ρ, H and K. Solution 28. Note that ρ is a mixed state since ρ2 6= ρ. Straightforward calculations yield hHi = 0,

hKi = 0,

hH 2 i =

1 , 2

hK 2 i =

1 2

and the commutator of H and K is given by   −2i 2i 0 [H, K] =  2i 2i 0  . 0 0 0 Thus |hi[H, K]i| = 1/2 and the equality follows.

Programming Problems Problem 1.

Consider the matrix  3/4 ρ = √ iφ 2e /4

√

2e−iφ /4 1/4

 .

Check that the matrix is a density matrix. Is it a pure or mixed state? Apply computer algebra. Find tr(σ1 ρ). Solution 1.

The following Maxima program will do the job

/* densitycheck.mac */ load("nchrpl"); rho: matrix([3/4,sqrt(2)*exp(-%i*phi)/4],[sqrt(2)*exp(%i*phi)/4,1/4]); rhoT: transpose(rho); rhoTC: conjugate(rhoT); tr: mattrace(rho); E: eigenvalues(rho); e1: first(first(E)); e2: second(first(E));

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137

if(rho=rhoTC and tr=1 and e1>0 and e2>0) then print("matrix is density matrix") else print("matrix is not a density matrix"); rho2: rho . rho; if(rho2=rho and tr=1 and rho=rhoTC) then print("pure state") else print("not a pure state"); sig1: matrix([0,1],[1,0]); trsig1: mattrace(rho . sig1);

The matrix is a density matrix and a mixed state. The two eigenvalues are √ √ (2 + 3)/4, (2 − 3)/4. We find tr(σ1 ρ) = √12 cos(φ). Problem 2.

Consider the density matrix   1 1 −1 ρ= 2 −1 1

and let A be an 2 × 2 real symmetric matrix. Assume that tr(ρA) = −1, tr(ρA2 ) = 1. Reconstruct the matrix from this information. Solution 2.

Utilizing the Maxima program

/* density1.mac */ load("nchrpl"); rho: matrix([1/2,-1/2],[-1/2,1/2]); A: matrix([a11,a12],[a12,a22]); A2: A . A; r1: mattrace(rho . A); r1: ratsimp(r1); r2: mattrace(rho . A2); r2: ratsimp(r2); solve([r1+1=0,r2-1=0],[a11,a12,a22]);

we obtain

 A=

r 1+r

1+r r



with r an arbitrary real constant. Problem 3. Let S be the set of unit vectors in the Hilbert space Cn . Let u ∈ S. A function µ(u) from S to R is called a generalized probability measure if the following two conditions hold: (i) for u ∈ S, 0 ≤ µ(u) ≤ 1, n (ii) Pn if u1 , . . . , un form an orthonormal basis in the Hilbert space C , then j=1 µ(uj ) = 1. Let n ≥ 3. Then any generalized probability measure µ on Cn has the form µ(ρ) = tr(ρuu∗ ) for a uniquely defined density matrix ρ (Gleason 1957).

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138 Problems and Solutions (i) Consider the Hilbert space C3 , the orthonormal basis       1 0 1 1   1 u1 = √ 0 , u2 =  1  , u3 = √  0  2 1 2 −1 0 and the density matrix  1 1 1 ρ= 3 1

 1 1. 1

1 1 1

Find µ(u1 ), µ(u2 ), µ(u3 ). (ii) Consider the Hilbert space C4 , the orthonormal basis  eiφ 1  0  u1 = √  , 0 2 eiφ

 eiφ 1  0  u2 = √  , 0 2 −eiφ

 0 1  eiφ  u3 = √  iφ  , 2 e 0

 0 1  eiφ  u4 = √  iφ  2 −e 0









and the density matrix 1 1 1 ρ=  4 1 1 

1 1 1 1

1 1 1 1

 1 1 . 1 1

Find µ(u1 ), µ(u2 ), µ(u3 ), µ(u4 ). Solution 3.

Applying the Maxima program

/* Gleason.mac */ load("nchrpl"); u1: matrix([1],[0],[1])/sqrt(2); u1T: transpose(u1); u2: matrix([0],[1],[0]); u2T: transpose(u2); u3: matrix([1],[0],[-1])/sqrt(2); u3T: transpose(u3); rho: matrix([1,1,1],[1,1,1],[1,1,1])/3; muu1: mattrace(rho . u1 . u1T); muu2: mattrace(rho . u2 . u2T); muu3: mattrace(rho . u3 . u3T); v1: matrix([exp(%i*phi)],[0],[0],[exp(%i*phi)])/sqrt(2); v1T: transpose(v1); v1TC: conjugate(v1T); v2: matrix([exp(%i*phi)],[0],[0],[exp(%i*phi)])/sqrt(2);

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139

v2T: transpose(v2); v2TC: conjugate(v2T); v3: matrix([0],[exp(%i*phi)],[exp(%i*phi)],[0])/sqrt(2); v3T: transpose(v3); v3TC: conjugate(v3T); v4: matrix([0],[exp(%i*phi)],[exp(%i*phi)],[0])/sqrt(2); v4T: transpose(v4); v4TC: conjugate(v4T); rho4: matrix([1,1,1,1],[1,1,1,1],[1,1,1,1],[1,1,1,1])/4; muv1: mattrace(rho4 . v1 . v1TC); muv2: mattrace(rho4 . v2 . v2TC); muv3: mattrace(rho4 . v3 . v3TC); muv4: mattrace(rho4 . v4 . v4TC);

we find for (i) µ(u1 ) = 2/3, µ(u2 ) = 1/3, µ(u3 ) = 0 and for (ii) we find µ(u1 ) = 1/2, µ(u2 ) = 1/2, µ(u3 ) = 1/2, µ(u4 ) = 1/2.

4.3

Supplementary Problems

Problem 1. Consider a spin-1 system. Any pure state can be parametrized, with a suitable choice of its phase as  eiα sin(θ) cos(φ) |ψi =  eiβ sin(θ) sin(φ)  cos(θ) 

where 0 ≤ θ, φ ≤ π/2 and 0 ≤ α, β < 2π. Find the density matrix ρ = |ψihψ| and tr(ρS1 ), where S1 is the spin-1 matrix  1 1  S1 = √ 0 2 1

0 1 0

 1 0. 1

Note that the density matrix depends on all four parameters. Problem 2. Let σ0 , σ1 , σ2 , σ3 be the Pauli spin matrices, where σ0 = I2 is the 2 × 2 unit matrix and let v = ( v1

v2

T

v3 )

be a vector in R3 with kvk ≤ 1. (i) Show that ρv = 21 (σ0 + v1 σ1 + v2 σ2 + v3 σ3 ) is a density matrix. P3 (ii) Is ρ = 14 (σ0 ⊗ σ0 + j=1 vj σj ⊗ σj ) a density matrix? (iii) Is ρ=

3 X 1 (σ ⊗ σ ⊗ σ + vj σ j ⊗ σ j ⊗ σ j ) 0 0 0 23 j=1

a density matrix? Extend the result to n Kronecker products.

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140 Problems and Solutions Problem 3. the states

Let |0i, |1i, . . . , |ni be an orthonormal basis in Cn+1 . Are n

1 1 X |ψ0 i = √ |0i ⊗ |0i + √ |ji ⊗ |ji 2 2n j=1 n

1 1 X |ψ1 i = √ |0i ⊗ |0i − √ |ji ⊗ |ji 2 2n j=1 normalized? Are the state orthogonal to each other? Is ρ = (|ψ0 ihψ0 |) ⊗ (|ψ1 ihψ1 |) a density matrix? Problem 4. Consider the Pauli spin matrices σ1 and σ2 . Let ρ be a 2 × 2 density matrix with tr(ρσ1 ) = 1, tr(ρσ2 ) = 0. Reconstruct the density matrix ρ from this information. Show that   1 1 1 . ρ= 2 1 1

Problem 5. Let H1 and H2 be two Hilbert spaces and H1 ⊗ H2 be the product Hilbert space. Let ρ be a density operators of the Hilbert space H1 ⊗ H2 . Show that if one of the reduced density operators trH2 (ρ) = ρ1 or trH1 (ρ) = ρ2 is pure, then ρ = ρ1 ⊗ ρ2 . If both ρ1 and ρ2 are pure, then ρ is pure too. Problem 6. Find a normalized state |ψi in the Hilbert space C2 such that we have the density matrix   1 1 |ψihψ| = I2 + √ (σ1 + σ2 + σ3 ) . 2 3

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Chapter 5

Trace and Partial Trace

5.1

Introduction

Let H be the finite dimensional Hilbert space Cn with an orthonormal basis { |φj i : j = 1, 2, . . . , n }. Let A be a linear operator (n × n matrix) acting in this Hilbert space. Then the trace of A is defined as tr(A) :=

n X

hφj |A|φj i .

j=1

The trace is independent of the chosen orthonormal basis. For the trace we have cyclic invariance. Let A, B, C be n × n matrices over C. Then tr(AB) = tr(BA) and (cyclic invariance) tr(ABC) = tr(CAB) = tr(BCA). The trace of an n × n matrix A is the sum of the eigenvalues counting multiplicities. The eigenvalues of A can be reconstructed from tr(A) = λ1 + λ2 + · · · + λn tr(A2 ) = λ21 + λ22 + · · · + λ2n .. . tr(An ) = λn1 + λn2 + · · · + λnn . 141

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142 Problems and Solutions If |ψi is a normalized state in Cn , then tr(|ψihψ|) = 1. For any n × n matrix over C we have the identity det(exp(A)) = exp(tr(A)). Let A be an n × n matrix over C and B be an m × m matrix over C. Then tr(A ⊗ B) = tr(A)tr(B). The calculation of the partial trace plays a central role in quantum computing. Suppose a finite dimensional quantum system SAB is a system composed of two subsystems SA and SB . The finite dimensional Hilbert space H of SAB is given by the tensor product of the individual Hilbert spaces HA ⊗ HB . Let NA := dim(HA ) and NB := dim(HB ). Let ρAB be the density matrix of SAB . Using the partial trace we can define the density operators ρA and ρB in the subspaces HA and HB as follows ρA := trB (ρAB ) ≡

NB X

(IA ⊗ hφj |)ρAB (IA ⊗ |φj i)

j=1

and ρB := trA (ρAB ) ≡

NA X

(hψj | ⊗ IB )ρAB (|ψj i ⊗ IB i)

j=1

where IA is the identity operator in HA , IB is the identity operator in HB and |φj i, (j = 1, 2, . . . , NB ) is an orthonormal basis in HB and |ψj i, (j = 1, 2, . . . , NA ) is an orthonormal basis in HA . For example we could select the standard bases in the two finite dimensional Hilbert spaces HA and HB . The partial trace can also be calculated as follows. Consider a bipartite state |ψi =

n−1 X n−1 X

cjk |jki ≡

j=0 k=0

n−1 X n−1 X

cjk |ji ⊗ |ki ,

j=0 k=0

n−1 X n−1 X

cjk c∗jk = 1

j=0 k=0 n

in the finite-dimensional Hilbert space H = C ⊗ Cn . We can define the n × n matrix Λjk := cjk , j, k = 0, 1, . . . , n − 1. Then we have (prove it) ρA = trB (ρ) = trB (|ψihψ|) = ΛΛ† .

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Trace and Partial Trace

5.2

143

Solved Problems

Problem 1. Consider the Pauli spin matrix σ1 . Calculate the trace of σ1 with respect to the standard basis. Calculate the trace of σ1 with respect to the Hadamard basis      1 1 1 1 √ , √ . 2 1 2 −1 Solution 1. Obviously for the standard basis we find tr(σ1 ) = 0. Since         1 1 1 1 0 1 1 0 1 1 √ (1 1) √ √ = 1, √ (1 − 1) = −1 1 0 1 0 2 2 1 2 2 −1 we also obtain (as expected since the trace is independent of the chosen orthonormal basis) for the Hadamard basis the result 0. In the Hadamard basis we have the Pauli spin matrix σ3 . Problem 2.

Consider the hermitian matrix   1 1 1 . H=√ 2 1 −1

Find tr(H) and tr(H 2 ) and then the eigenvalues from this information. Solution 2. Since tr(H) = 0, tr(H 2 ) = 2 and λ1 + λ2 = 0, λ21 + λ22 = 2 we obtain the eigenvalues λ1 = 1, λ2 = −1. Problem 3.

Consider the entangled state in C4

  0         1 1  1  1 0 0 1 |ψi = √ ⊗ − ⊗ =√   0 1 1 0 2 2 −1 0 and the density matrix ρ = |ψihψ|. Find ρ1 = tr2 (ρ), ρ2 = tr1 (ρ) i.e. calculate the partial trace. Solution 3.

We have 1 hψ| = √ ( 0 2

Thus

0 1 0 ρ=  2 0 0 

1 −1

0 1 −1 0

0 −1 1 0

0).  0 0 . 0 0

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144 Problems and Solutions Using the basis    1 1 ⊗ 0 0

0 1

 ,

   0 1 ⊗ 1 0

0 1



and the basis 

1 0

0 1



  1 ⊗ , 0



we find the density matrix (mixed state)  1 1 ρ1 = ρ2 = 2 0 Problem 4.

1 0

0 1

0 1



  0 ⊗ 1

 .

Consider the 4 × 4 matrix (density matrix) |uihu| = (uj u ¯k ),

j, k = 1, . . . , 4

in the product Hilbert space HA ⊗ HB ≡ C4 , where HA = HB = C2 . (i) Calculate trA (|uihu|), where the basis is given by     1 0 ⊗ I2 , ⊗ I2 0 1 and I2 denotes the 2 × 2 unit matrix. (ii) Find the partial trace trB (|uihu|), where the basis is given by     1 0 I2 ⊗ , I2 ⊗ . 0 1

Solution 4.

(i) Since

 1   1 0 ⊗ I2 =  0 0 0

 0 1 , 0 0

 0   0 0 ⊗ I2 =  1 1 0

 0 0  0 1

we find, using the transpose of these matrices on the left-hand side of |uihu|, that    ¯1 u1 u ¯2 u1 u ¯3 u1 u ¯4 1 0   u1 u 1 0 0 0  u2 u ¯1 u2 u ¯2 u2 u ¯3 u2 u ¯4   0 1  trA (|uihu|) =    0 1 0 0 u3 u ¯1 u3 u ¯2 u3 u ¯3 u3 u ¯4 0 0 u4 u ¯1 u4 u ¯2 u4 u ¯3 u4 u ¯4 0 0    ¯1 u1 u ¯2 u1 u ¯3 u1 u ¯4 0 0   u1 u 0 0 1 0  u2 u ¯1 u2 u ¯2 u2 u ¯3 u2 u ¯4   0 0  +   . 0 0 0 1 u3 u ¯1 u3 u ¯2 u3 u ¯3 u3 u ¯4 1 0 u4 u ¯1 u4 u ¯2 u4 u ¯3 u4 u ¯4 0 1

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145

Using matrix multiplication and matrix addition we obtain  trA (|uihu|) =

u1 u ¯ 1 + u3 u ¯3 u2 u ¯ 1 + u4 u ¯3

u1 u ¯2 + u3 u ¯4 u2 u ¯2 + u4 u ¯4

 .

(ii) Since  1   1 0 I2 ⊗ = 0 0 0

 0 0 , 1 0

 0   0 1 I2 ⊗ = 1 0 0

 0 0  0 1

we find   u1 u ¯1 u1 u ¯2 u1 u ¯3 u1 u ¯4 1 0 1 0 0 0  u2 u ¯1 u2 u ¯2 u2 u ¯3 u2 u ¯4   0 0  trB (|uihu|) =    0 0 1 0 u3 u ¯1 u3 u ¯2 u3 u ¯3 u3 u ¯4 0 1 u4 u ¯1 u4 u ¯2 u4 u ¯ 3 u4 u ¯4 0 0    u u ¯ u u ¯ u u ¯ u u ¯ 0 0 1 1 1 2 1 3 1 4   0 1 0 0  u2 u ¯1 u2 u ¯2 u2 u ¯3 u2 u ¯4   1 0  +  .  0 0 0 1 0 0 u3 u ¯1 u3 u ¯2 u3 u ¯ 3 u3 u ¯4 0 1 u4 u ¯1 u4 u ¯2 u4 u ¯ 3 u4 u ¯4 





Using matrix multiplication and matrix addition yields  trB (|uihu|) =

u1 u ¯1 + u2 u ¯2 u3 u ¯1 + u4 u ¯2

u1 u ¯3 + u2 u ¯4 u3 u ¯3 + u4 u ¯4

 .

We see that trA (|uihu|) 6= trB (|uihu|). However tr (trA (|uihu|)) = tr (trB (|uihu|)) , det (trA (|uihu|)) = det (trB (|uihu|)) .

Problem 5.

Consider the 9 × 9 matrix (density matrix) |uihu| = (uj u ¯k ),

j, k = 1, . . . , 9.

Find the partial trace trC3 (|uihu|), where the basis is given by   1  0  ⊗ I3 , 0

  0  1  ⊗ I3 , 0

and I3 denotes the 3 × 3 unit matrix.

  0  0  ⊗ I3 1

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146 Problems and Solutions Solution 5.

We have 

1 0  0    1 0  0  ⊗ I3 =  0  0 0  0  0 0 and

0 1 0 0 0 0 0 0 0

0 0  0    0 1  1  ⊗ I3 =  0  0 0  0  0 0

 0 0  1  0  0,  0  0  0 0



0 0  0    0 0  0  ⊗ I3 =  0  1 0  1  0 0 

The respective transposes of the  1 (1 0 0) ⊗ I3 =  0 0  0 (0 1 0) ⊗ I3 =  0 0  0 (0 0 1) ⊗ I3 =  0 0

0 0 0 0 0 0 0 1 0

0 0 0 0 1 0 0 0 0

 0 0  0  0  0  1  0  0 0

 0 0  0  0  0.  0  0  0 1

above matrices are given by  0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0  0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0  0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0. 0 0 0 0 0 0 0 1

Taking this basis we find trA (|uihu|) = 

u1 u ¯ 1 + u4 u ¯4 + u7 u ¯7  u2 u ¯1 + u5 u ¯4 + u8 u ¯7 u3 u ¯1 + u6 u ¯4 + u9 u ¯7 Problem 6.

u1 u ¯ 2 + u4 u ¯5 + u7 u ¯8 u2 u ¯2 + u5 u ¯5 + u8 u ¯8 u3 u ¯2 + u6 u ¯5 + u9 u ¯8

(i) Consider the Bell state 1 |ψi = √ (|00i + |11i). 2

 u1 u ¯3 + u4 u ¯6 + u7 u ¯9 u2 u ¯3 + u5 u ¯6 + u8 u ¯9  . u3 u ¯3 + u6 u ¯6 + u9 u ¯9

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147

√ Hence c00 = c11 = 1/ 2, c01 = c10 = 0. Find ρA . (ii) Under a unitary transformation U , V (U and V are n×n unitary matrices) the matrix Λ is changed to Λ → U T ΛV , where T denotes the transpose. Apply the transformation to ΛΛ† . Calculate the Wigner function tr(ΛΛ† )2 . √ Solution 6. (i) Since c00 = c11 = 1/ 2, c01 = c10 = 0, we find the matrix ! √1 0 2 Λ= . √1 0 2 Thus we obtain the density matrix (mixed state)  1 0 † 2 . ρA = ΛΛ = 0 12 For the other three Bell states we find the same result. (ii) We have ΛΛ† → (U T ΛV )(V † Λ† U T † ) = U † ΛΛ† U T † since V † V = V V † = In . Furthermore, tr(ΛΛ† )2 stays invariant under the transformation since U T U T † = (U † U )T = In . Problem 7. Let { |0i, |1i, . . . , |d − 1i } be an orthonormal basis in the Hilbert space Cd . The discrete Wigner operator is defined as ˆ p) := A(q,

d−1 X d−1 X

  2π δ2q,r+s exp i p(r − s) |rihs| d r=0 s=0

where q and p take integer values from 0 to d − 1 and δm,u denotes the Kronecker delta. The arithmetic in the subscript is modulo N arithmetic, i.e. 2q mod d and (r + s) mod d. The (p, q) pairs constitute the discrete phase space. For a state described by the density matrix ρ the discrete Wigner function is defined as W (p, q) :=

1 ˆ tr(ρA). d

Let ρ = |0ih0|. Calculate W (p, q). Solution 7.

Since h0|ri = δ0r we obtain   ! d−1 X 1 2π W (p, q) = tr |0i δ2q,s exp −i ps hs| . d N s=0

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148 Problems and Solutions To calculate the trace we have !   d−1 d−1 X 1X 2π W (p, q) = hk|0i δ2q,s exp −i ps hs|ki . d N s=0 k=0

Using hk|0i = δk0 and hs|ki = δsk we arrive at 1 δ2q,0 . d

W (p, q) =

Problem 8. For a bipartite state with subsystems 1 and 2 described by the joint density matrix the joint Wigner function is given by W (q1 , q2 , p1 , p2 ) :=

1 tr(ρ(12) (Aˆ1 (q1 , p1 ) ⊗ Aˆ2 (q2 , p2 ))) d2

where the Wigner operators are given by Aˆ1 (q1 , p1 ) :=

d−1 X d−1 X

  2π δ2q1 ,r+s exp i p1 (r − s) |rihs| d r=0 s=0

and Aˆ2 (q2 , p2 ) :=

d−1 X d−1 X

  2π δ2q2 ,r+s exp i p2 (r − s) |rihs|. d r=0 s=0

Wigner functions describing a subsystem are obtained by summing the joint Wigner functions in the corresponding set of the respective variables, e.g. W (q1 , p1 ) =

d−1 X d−1 X

W (q1 , p1 , q2 , p2 )

q2 =0 p2 =0

W (q2 , p2 ) =

d−1 d−1 X X

W (q1 , p1 , q2 , p2 ).

q1 =0 p1 =0

Consider the EPR state d−1

1 X |ψi = √ |ki ⊗ |ki. d k=0 Let ρ = |ψihψ|. Find W (q1 , q2 , p1 , p2 ). Discuss. Solution 8.

Straightforward calculation yields the Wigner function W (q1 , q2 , p1 , p2 ) =

1 δq ,q δp ,−p . d2 1 2 1 2

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149

The Wigner function given above shows the connection with the EPR state for continuous-variable teleportation δ(q1 − q2 ) ⊗ δ(p1 + p2 ) where δ denotes the Dirac delta function. The trace of an m × m matrix A over C is defined as

Problem 9.

trm (A) =

m X

(b∗m,j A bm,j )

j=1

where { bm,1 , bm,2 , . . . , bm,m } is an orthonormal basis for Cm . The partial traces of an mn × mn matrix C over C are defined as m X

tr1m,n (C) =


b∗m,j ⊗ In C (bm,j ⊗ In )

j=1 n X

tr2m,n (C) =


Im ⊗ b∗n,k C (Im ⊗ bn,k )

k=1

where { bn,1 , bn,2 , . . . , bn,n } is an orthonormal basis for Cn and Im and In denote the m × m and n × n identity matrices respectively. Let B be an n × n matrix over C. (i) Show that the above definition of trm (A) is independent of the choice of orthonormal basis { bm,1 , bm,2 , . . . , bm,m }. (ii) Show that
trmn (C) = trn tr1m,n (C)


trmn (C) = trm tr2m,n (C) .

and

(iii) Use your favourite orthonormal basis for C2 to calculate 0 2 1 tr2,2  1 0 

tr12,2



0 1

1 0



 ⊗

1 3

2 4

 and

1 0 0 1

1 0 0 −1

 0 1  . −1 0

Solution 9. (i) Let { φ1 , φ2 , . . . , φm } be an orthonormal basis for Cn . Expanding each of the bm,k in terms of the φj and vice versa yields bm,j =

m X k=1

φ∗k bm,j φk ,

φj =

m X k=1

b∗m,k φj bm,k

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150 Problems and Solutions where we require m X

b∗m,j bm,k = =

m X

b∗m,j φu φ∗u φ∗v bm,k φv =

u,v=1 m X

b∗m,j φu φ∗v bm,k (φ∗u φv )

u,v=1

φ∗u bm,k b∗m,j φu = δj,k

u=1

so that orthonormality holds. It follows that m X

φ∗k Aφk =

m X m X m X

φ∗k bm,j b∗m,j Ab∗m,l φk bm,l

k=1 j=1 l=1

k=1

=

=

m X m m X X j=1 l=1 m X m X

! φ∗k bm,j b∗m,l φk

b∗m,j Abm,l

k=1

δj,l b∗m,j Abm,l

j=1 l=1

=

m X

b∗m,j Abm,j = trm (A).

j=1

(ii) We have trn tr1m,n (C)




=

=

n X

b∗n,j

j=1 m n X X

!

m X

(b∗m,k k=1

⊗ In )C(bm,k ⊗ In ) bn,j

(I1 ⊗ b∗n,j )(b∗m,k ⊗ In )C(bm,k ⊗ In )(I1 ⊗ bn,j )

j=1 k=1

=

n X m X

(b∗m,k ⊗ b∗n,j )C(bm,k ⊗ bn,j )

j=1 k=1

= trmn (C) where the last equivalence follows from taking the trace of C using the basis { bm,j ⊗ bn,k : j ∈ {1, 2, . . . , m} k ∈ {1, 2, . . . , n} }.
The result trmn (C) = trm tr2m,n (C) follows similarly. (iii) Using any basis, for example the standard basis     1 0 , 0 1 we find tr12,2



0 1

1 0



 ⊗

1 3

2 4



 =

0 0

0 0



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Trace and Partial Trace and

0 1 tr22,2  1 0 

Problem 10. C2 ⊗ C2 ⊗ C2 )

1 0 0 1

1 0 0 −1

 0  1  0 = −1 2 0

2 0

151

 .

Consider the GHZ state in the Hilbert space C8 (C8 ∼ =

1 |GHZi = √ 2

            1 1 1 0 0 0 ⊗ ⊗ + ⊗ ⊗ . 0 0 0 1 1 1

Then the density matrix is given by the 8 × 8 matrix  1 0 0 0 0 0 0 0 0 0 0 0  0 0 0 0 0 0 1 0 0 0 0 0 0 ρ = |GHZihGHZ| =  2 0 0 0 0 0 0  0 0 0 0 0 0  0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0

 1 0  0  0 . 0  0  0 1

(i) Calculate the partial trace ρAB = trC (ρ) with the basis     1 0 I4 ⊗ , I4 ⊗ . 0 1 (ii) Calculate the partial trace ρA = trB (ρAB ) with the basis     1 0 I2 ⊗ , I2 ⊗ . 0 1

Solution 10. (i) We have   ∗      ∗    1 1 0 0 trC (ρ) = I4 ⊗ ρ I4 ⊗ + I4 ⊗ ρ I4 ⊗ 0 0 1 1     1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 =  +   2 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 1   1 0 0 0 1 0 0 0 0 =  . 2 0 0 0 0 0 0 0 1

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152 Problems and Solutions (ii) We find ρA = trB (ρAB )   ∗      ∗    1 1 0 0 = I2 ⊗ ρAB I2 ⊗ + I2 ⊗ ρAB I2 ⊗ 0 0 1 1     1 0 0 1 1 0 + = 2 0 0 2 0 1   1 1 0 = . 2 0 1

Programming Problems Problem 1.

Consider the hermitian  0 A= 0 −i

matrix  0 i 0 0. 0 0

Find the eigenvalues of A utilizing tr(A) = λ1 + λ2 + λ3 ,

Solution 1.

tr(A2 ) = λ21 + λ22 + λ23 ,

tr(A3 ) = λ31 + λ32 + λ33 .

The following Maxima program will do the job

/* traceeigen.mac */ load("nchrpl"); A: matrix([0,0,%i],[0,0,0],[-%i,0,0]); r1: mattrace(A); r2: mattrace(A . A); r3: mattrace(A . A . A); solve([x1+x2+x3-r1=0,x1*x1+x2*x2+x3*x3-r2=0, x1*x1*x1+x2*x2*x2+x3*x3*x3-r3=0],[x1,x2,x3]);

The eigenvalues are λ1 = −1, λ2 = 0, λ3 = +1. Problem 2. Consider the 6 × 6 matrix A = (ajk ). Calculate the partial trace with the basis     1 0 ⊗ I3 , ⊗ I3 . 1 0 Calculate the partial trace with the basis     1 0  0  ⊗ I2 ,  1  ⊗ I2 , 0 0

  0  0  ⊗ I2 . 1

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153

Apply computer algebra. Solution 2.

The following Maxima program will do the job.

/* partialtrace.mac */ load("nchrpl"); A: matrix([a11,a12,a13,a14,a15,a16],[a21,a22,a23,a24,a25,a26], [a31,a32,a33,a34,a35,a36],[a41,a42,a43,a44,a45,a46], [a51,a52,a53,a54,a55,a46],[a61,a62,a63,a64,a65,a66]); I2: matrix([1,0],[0,1]); I3: matrix([1,0,0],[0,1,0],[0,0,1]); v1: matrix([1],[0]); v2: matrix([0],[1]); u1: matrix([1],[0],[0]); u2: matrix([0],[1],[0]); u3: matrix([0],[0],[1]); b1: kronecker_product(v1,I3); b1T: transpose(b1); b2: kronecker_product(v2,I3); b2T: transpose(b2); ptrA1: b1T . A . b1 + b2T . A . b2; c1: kronecker_product(u1,I2); c1T: transpose(c1); c2: kronecker_product(u2,I2); c2T: transpose(c2); c3: kronecker_product(u3,I2); c3T: transpose(c3); ptrA2: c1T . A . c1 + c2T . A. c2 + c3T . A. c3;

The output is the 3 × 3 matrix for the first basis [ a44 + a11 [ a54 + a21 [ a64 + a31

a45 + a12 a55 + a22 a65 + a32

a46 + a13 ] a46 + a23 ] a66 + a33 ]

for the first two-dimensional basis and the 2 × 2 matrix [ a55 + a33 + a11 [ a65 + a43 + a21

a46 + a34 + a12 ] a66 + a44 + a22 ]

for the second three dimensional basis.

5.3

Supplementary Problems

Problem 1.

Let c1 , c2 , c3 ∈ C and |c1 |2 + |c2 |2 + |c3 |2 = 1. Show that √ √   2 c1 c∗2 / 2 c1 c∗2 / 2 |c1 |√ c1 c∗3√ ∗ 2 2 ∗ |c2 | /2 c2 c3 /√2   c c / 2 |c2 | /2 ρ =  1∗ 2 √  c1 c2 / 2 |c2 |2 /2 |c2 |2 /2 c2 c∗3 / 2 √ √ ∗ ∗ ∗ 2 c1 c3 c2 c3 / 2 c2 c3 / 2 |c3 |

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154 Problems and Solutions is a density matrix. Taking the partial trace we obtain √   |c1 |2 + |c2 |2 /2 (c1 c∗2 + c2 c∗3 )/ 2 √ ρp = . (c∗1 c2 + c∗2 c3 )/ 2 |c3 |2 + |c2 |2 /2 Show that ρp is a density matrix. Discuss. Problem 2. Consider the finite dimensional Hilbert spaces H1 = Cn1 ˆ 1 be an n1 × n1 matrix and Q ˆ 2 be an n2 × n2 matrix. and H2 = Cn2 . Let Q n1 Let ρ1 be a density matrix in C and ρ2 be a density matrix. Show that ˆ1 ⊗ Q ˆ 2 )(ρ1 ⊗ ρ2 )) = tr(Q ˆ 1 ρ1 )tr(Q2 ρ2 ). tr((Q

Problem 3. Consider the finite-dimensional Hilbert spaces H1 = Cn1 and H2 = Cn2 . Let H1 ⊗ H2 be the product Hilbert space. Let |ψi and |φi be states in the product Hilbert space H1 ⊗ H2 . Show that if trH2 (|ψihψ|) = trH2 (|φihφ|) then there exists a unitary matrix U acting in the Hilbert space H2 such that |ψi = (In1 ⊗ U )|φi where In1 is the identity matrix in the Hilbert space H1 . Problem 4. Consider the finite dimensional Hilbert spaces H1 = Cn1 and H2 = Cn2 . Let v, u be normalized vectors in the product Hilbert space Cn1 ⊗ Cn2 . Show that if tr1 (vv∗ ) = tr1 (uu∗ ) then there exists a n2 × n2 unitary matrix U such that v = (In1 ⊗ U )u.

Problem 5. Consider the product Hilbert space `2 (N0 ) ⊗ C2s+1 , where s = 1/2, 1, 3/2, 2, . . . is the spin. Find the partial trace over C2s+1 .

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Chapter 6

Boolean Functions and Quantum Gates

6.1

Introduction

A truth table (or function table) is a tabular description of a combinational circuit (such as an AND gate, OR gate, NAND gate) listing all possible states of the input variables together with a statement of the output variable(s) for each of those possible states. The truth table for the AND gate, OR gate, XOR gate and NOT gate are

0 0 1 1

AND 0 1 0 1

0 0 0 1

0 0 1 1

OR 0 1 0 1

0 1 1 1

0 0 1 1

XOR 0 1 0 1

0 1 1 0

NOT 0 1 1 0

The NAND gate is an AND gate followed by a NOT gate. The NOR gate is an OR gate followed by a NOT gate. Both are universal gates, i.e. all other gates can be built from these gates. A boolean function f on n variables is a mapping {0, 1}n into {0, 1}. Let xj ∈ {0, 1} for j = 1, . . . , n. We set x = (x1 , x2 , . . . , xn ). In the following · denotes the AND operation, + denotes the OR operation, ⊕ the XOR 155

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156 Problems and Solutions operation and functions

is the NOT operation. For n = 1 we have the four boolean

f1 (x) = 0,

f2 (x) = 1,

f3 (x) = x,

f4 (x) = x ¯.

The last two are of course reversible. Let X = {0, 1} and xj ∈ X. A boolean function f with n input variables, x1 , . . . , xn and n output variables, y1 , . . . , yn is a function f : X n → X n obeying f (x1 , . . . , xn ) 7→ (y1 , . . . , yn ). Here (x1 , . . . , xn ) ∈ X n is called the input vector and (y1 , . . . , yn ) ∈ X n is called the output vector. An n-input and n-output boolean function f is reversible if it maps each input vector to a unique output vector, i.e. the map is a bijection. Quantum gates are described by unitary operators. In the finite dimensional Hilbert space Cd we have d × d unitary matrices. We describe how 2n+1 × 2n+1 unitary matrices can be associated with a non-reversible boolean function f and how 2n × 2n unitary matrices can be associated with reversible boolean functions f . Finally the associated Hamilton operator has to be constructed. Reversible gates are gates that function in both directions. CMOS implementations of such gates have been designed. A special pass transistor logic family has been applied: reversible MOS. Many different reversible logic gates are candidates as universal building blocks. The Feynman gate, the controlled NOT gate, the Fredkin gate can be implemented. They dissipate very little energy. Owing to their use of reversible truth tables, they are even candidates for zero-power computing. Circuit synthesis takes advantage of mathematical group theory. Algorithms have been developed for the synthesis of arbitrary reversible circuits. A reversible logic gate has a corresponding quantum version, whose properties are completely defined by the truth table for the classical version. Reversible circuits are applicable to nanotechnolgy, quantum and optical computing as well as reducing power in CMOS implementations. In adiabatic circuits, current is restricted to flow across devices with low voltage drop and the energy stored on their capacitors is recycled. One uses reversible energy recovery gates capable of realizing the functions { AN D, OR } or { N AN D, N OR }.

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Boolean Functions and Quantum Gates

6.2

157

Solved Problems

Problem 1.

The Feynman gate is a 2 input/2 output gate given by x01 = x1 ,

x02 = x1 ⊕ x2 .

(i) Give the truth table for the Feynman gate. (ii) Show that copying can be implemented using the Feynman gate. (iii) Show that the complement can be implemented using the Feynman gate. (iv) Is the Feynman gate invertible? Solution 1.

(i) The truth table is x1 0 0 1 1

x2 0 1 0 1

x01 0 0 1 1

x02 0 1 1 0

(ii) Setting x2 = 0, we have x02 = x1 ⊕ 0 = x1 . Thus we have a copy. (iii) Setting x2 = 1, we have x02 = x1 ⊕ 1 = x ¯1 . Thus we generated the complement. (iv) From the truth table we see that the transformation is invertible. The inverse transformation can be found as follows. Since x1 ⊕ x1 = 0 we have x01 ⊕ x02 = x1 ⊕ x1 ⊕ x2 = 0 ⊕ x2 = x2 . Thus x1 = x01 , x2 = x01 ⊕ x02 . Problem 2.

Consider the 3-input/3-output gate given by x01 = x1 ,

x02 = x1 ⊕ x2 ,

x03 = x1 ⊕ x2 ⊕ x3 .

Give the truth table. Is the transformation invertible? Solution 2.

The truth table is given by x1 0 0 0 0 1 1 1 1

x2 0 0 1 1 0 0 1 1

x3 0 1 0 1 0 1 0 1

x01 0 0 0 0 1 1 1 1

x02 0 0 1 1 1 1 0 0

x03 0 1 1 0 1 0 0 1

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158 Problems and Solutions From the truth table we see that the transformation is invertible, i.e. we have a 1 − 1 map. The inverse transformation is given by x1 = x01 ,

Problem 3.

x2 = x01 ⊕ x02 ,

x3 = x01 ⊕ x02 ⊕ x03 .

Consider the 3-input/3-output gate given by x01 = x1 ,

x02 = x1 ⊕ x2 ,

x03 = x3 ⊕ (x1 · x2 ).

Give the truth table. Is the gate invertible? Solution 3.

The truth table is given by x1 0 0 0 0 1 1 1 1

x2 0 0 1 1 0 0 1 1

x3 0 1 0 1 0 1 0 1

x01 0 0 0 0 1 1 1 1

x02 0 0 1 1 1 1 0 0

x03 0 1 0 1 0 1 1 0

From the truth table we see that the map is invertible, i.e. we have a 1-1 map. The inverse transformation is given by x1 = x01 x2 = x01 ⊕ x02 x3 = x03 ⊕ (x01 · (x01 ⊕ x02 )) where we used that x ⊕ x = 0. Problem 4. For reversible gates the following boolean expression plays an important role (a11 · a22 ) ⊕ (a12 · a21 ) where a11 , a12 , a21 , a22 ∈ { 0, 1 }. It could be considered as the determinant of the 2 × 2 binary matrix   a11 a12 . a21 a22 Discuss. Find the inverse of the matrix when it exists.

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Boolean Functions and Quantum Gates Solution 4. is given by

The inverse exists iff (a11 · a22 ) ⊕ (a12 · a21 ) = 1. The inverse 

since





Problem 5. given by

159

a11 a21

a22 a21

a12 a22



a12 a11



a22 a21

a12 a11

 =

(a11 · a22 ) ⊕ (a12 · a21 ) (a11 · a12 ) ⊕ (a12 · a11 ) (a21 · a22 ) ⊕ (a22 · a21 ) (a21 · a12 ) ⊕ (a11 · a22 )

 .

Consider a two input gate (x, y) / two output gate (x0 , y 0 ) x0 = a · x ⊕ b · y ⊕ c,

y 0 = a0 · x ⊕ b0 · y ⊕ c0

where a, b, a0 , b0 , c, c0 ∈ { 0, 1 }. (i) Let a = 0, b = 1, a0 = 1, b0 = 0 and c = c0 = 0. Find the output (x0 , y 0 ) for all possible inputs (x, y). Is the transformation invertible? (ii) Let a = 1, b = 1, a0 = 1, b0 = 1 and c = c0 = 0. Find the output (x0 , y 0 ) for all possible inputs (x, y). Is the transformation invertible? Solution 5.

(i) We have x0 = 0 · x ⊕ 1 · y ⊕ 0,

y 0 = 1 · x ⊕ 0 · y ⊕ 0.

Thus x0 = 0 ⊕ y ⊕ 0 = y,

y 0 = x ⊕ 0 ⊕ 0 = x.

The truth table follows as x 0 0 1 1

y 0 1 0 1

x0 0 1 0 1

y0 0 0 1 1

Therefore the transformation is invertible. (ii) We have x0 = 1 · x ⊕ 1 · y ⊕ 0,

y 0 = 1 · x ⊕ 1 · y ⊕ 0.

Thus x0 = x ⊕ y ⊕ 0 = x ⊕ y, The truth table follows as

y 0 = x ⊕ y ⊕ 0 = x ⊕ y.

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160 Problems and Solutions x 0 0 1 1

y 0 1 0 1

x0 0 1 1 0

y0 0 1 1 0

Therefore the transformation is not invertible. Problem 6.

Consider the Toffoli gate

T : {0, 1}3 → {0, 1}3 ,

T (a, b, c) := (a, b, (a · b) ⊕ c)

and the Fredkin gate F : {0, 1}3 → {0, 1}3 ,

F (a, b, c) := (a, a · b + a ¯ · c, a · c + a ¯ · b)

where a ¯ is the NOT operation, + is the OR operation, · is the AND operation and ⊕ is the XOR operation. 1. Express N OT (a) exclusively in terms of the TOFFOLI gate. 2. Express N OT (a) exclusively in terms of the FREDKIN gate. 3. Express AN D(a, b) exclusively in terms of the TOFFOLI gate. 4. Express AN D(a, b) exclusively in terms of the FREDKIN gate. 5. Express OR(a, b) exclusively in terms of the TOFFOLI gate. 6. Express OR(a, b) exclusively in terms of the FREDKIN gate. 7. Show that the TOFFOLI gate is invertible. 8. Show that the FREDKIN gate is invertible. Thus the TOFFOLI and FREDKIN gates are each universal and reversible (invertible). Solution 6. (1) We have N OT (a) = BIT 3(T (a, a, 1)), where BIT 3(a, b, c) = c. This follows from T (a, a, 1) = (a, a, a ⊕ 1). (2) N OT (a) = BIT 3(F (a, 1, 0)). (3) AN D(a, b) = BIT 3(T (a, b, 0)). (4) AN D(a, b) = BIT 3(F (a, 0, b)).

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(5) We can use OR(a, b) = N OT (AN D(N OT (a), N OT (b))) and simply expand N OT and AN D in terms of TOFFOLI gate. Noting that ¯ · ¯b = (¯ a, ¯b, a + b) T (¯ a, ¯b, 1) = (¯ a, ¯b, (¯ a · ¯b) ⊕ 1) = (¯ a, ¯b, a we implement the OR operation as OR(a, b) = BIT 3(T (N OT (a), N OT (b), 1)) = BIT 3(T (BIT 3(T (a, a, 1)), BIT 3(T (b, b, 1)), 1)). (6) We can use OR(a, b) = N OT (AN D(N OT (a), N OT (b))) and simply expand N OT and AN D in terms of the FREDKIN gate. Noting that F (a, b, 1) = (a, a · b + a ¯ · 1, a · 1 + a ¯ · b) = (a, b + a ¯, a + b) we implement the OR operation as OR(a, b) = BIT 3(F (a, b, 1)). (7) We have T −1 (a, b, c) = T (a, b, c) since the XOR (⊕) is invertible if we remember at least one of the arguments. In other words T (T (a, b, c)) = T (a, b, (a · b) ⊕ c) = (a, b, (a · b) ⊕ ((a · b) ⊕ c)) = (a, b, ((a · b) ⊕ (a · b)) ⊕ c) = (a, b, c). (8) We have F −1 (a, b, c) = F (a, b, c) since the swap operation on two bits is its own inverse, and the FREDKIN gate swaps the last two bits whenever the first argument is 0. In other words F (F (a, b, c)) = F (a, a · b + a ¯ · c, a · c + a ¯ · b) = (a, a · (a · b + a ¯ · c) + a ¯ · (a · c + a ¯ · b), a · (a · c + a ¯ · b) +¯ a · (a · b + a ¯ · c)) = (a, a · b + a ¯ · b, a · c + a ¯ · c) = (a, b, c).

Problem 7. A generalized Toffoli gate T(x1 , x2 , . . . , xn , xn+1 ) is a gate that maps a boolean pattern (x1 , x2 , . . . , xn , xn+1 ) to (x1 , x2 , . . . , xn , xn+1 ⊕ (x1 · x2 · . . . · xn ))

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162 Problems and Solutions where ⊕ is the XOR operation and · the AND operation. Show that the generalized Toffoli gate includes the NOT gate, CNOT gate and the original Toffoli gate. Solution 7. The NOT gate is given by T(1, a), where a ∈ { 0, 1 }. The CNOT-gate is given by T(a, b), where a, b ∈ { 0, 1 }. The original Toffoli gate is given by T(a, b, c). Problem 8. The Fredkin gate F(x1 , x2 , x3 ) has 3 inputs (x1 , x2 , x3 ) and three outputs (y1 , y2 , y3 ). It maps boolean patterns (x1 , x2 , x3 ) → (x1 , x3 , x2 ) if and only if x1 = 1, otherwise it passes the boolean pattern unchanged. Give the truth table. Solution 8.

We have x1 0 0 0 0 1 1 1 1

x2 0 0 1 1 0 0 1 1

x3 0 1 0 1 0 1 0 1

y1 0 0 0 0 1 1 1 1

y2 0 0 1 1 0 1 0 1

y3 0 1 0 1 0 0 1 1

Problem 9. The generalized Fredkin gate F(x1 , x2 , . . . , xn , xn+1 , xn+2 ) is the mapping of the boolean pattern (x1 , x2 , . . . , xn , xn+1 , xn+2 ) → (x1 , x2 , . . . , xn , xn+2 , xn+1 ) if and only if the boolean product x1 · x2 · . . . · xn = 1 (· is the bitwise AND operation), otherwise the boolean pattern passes unchanged. Let n = 2 and (x1 , x2 , x3 , x4 ) = (1, 1, 0, 1). Find the output. Solution 9.

Since x1 · x2 = 1, we find x1 1

Problem 10.

x2 1

x3 0

x4 1

y1 1

y2 1

y3 1

Is the gate (a, b, c ∈ { 0, 1 }) (a, b, c) → (a, a · b ⊕ c, a · c ⊕ b)

y4 0

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reversible? Solution 10. truth table

We set x = a, y = a · b ⊕ c and z = a · c ⊕ b. We find the a 0 0 0 0 1 1 1 1

b 0 0 1 1 0 0 1 1

c 0 1 0 1 0 1 0 1

x 0 0 0 0 1 1 1 1

y 0 1 0 1 0 1 1 0

z 0 1 1 0 1 1 0 0

Thus from the truth table we see that the gate is reversible. Problem 11.

Show that one Fredkin gate (a, b, c) → (a, a · b + a · c, a · c + a · b)

is sufficient to implement the XOR gate. Assume that either b or c are available. Solution 11. Choosing b = c (equivalently c = b) we find that the Fredkin gate yields (a, b, c) → (a, a · b + a · b, a · c + a · c) ≡ (a, a ⊕ b, a ⊕ c). Thus we can apply the Fredkin gate to (a, b, b) and use the second bit to obtain a ⊕ b or equivalently apply the Fredkin gate to (a, c, c) and use the third bit to obtain a ⊕ c. Problem 12. Consider the 2 × 2 identity matrix and the Pauli spin matrices σ1 , σ2 , σ3 using the following notation   1 0 σ00 = τ00 = I2 = , 0 1   0 1 σ01 = τ01 = σ1 = , 1 0   1 0 σ10 = τ10 = σ3 = , 0 −1   0 −i σ11 = σ2 = , i 0   0 1 τ11 = iσ2 = . −1 0

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164 Problems and Solutions Let n be a positive integer. If v, w ∈ Zn2 with  v1 . v =  ..  ,

 w1 . w =  .. 





vn and

wn 

b :=

v w



∈ Z2n 2

we define σb := σv1 w1 ⊗ σv2 w2 ⊗ · · · ⊗ σvn wn and τb := τv1 w1 ⊗ τv2 w2 ⊗ · · · ⊗ τvn wn . Thus we can associate a bit string b with each σb and vice versa. (i) Let n = 3 and σb1 = σ1 ⊗ σ3 ⊗ σ1 ,

σb2 = I2 ⊗ σ3 ⊗ σ1 .

Find the corresponding bit strings b1 and b2 for the given σb1 and σb2 . Then XOR the two bit strings and find the corresponding σb3 . Calculate the matrix product σb1 σb2 . Discuss. Solution 12.

(i) We obtain the vectors   0 1   0 b1 =   , 1   0 1

  0 1   0 b2 =   . 0   0 1

The XOR operation provides the bit string   0 0   0 b3 =   1   0 0 with the corresponding 8 × 8 matrix σb3 = σ1 ⊗ I2 ⊗ I2 which is σb3 = σb1 σb2 .

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Problem 13. Let x ∈ {0, 1} and |0i, |1i be the standard basis in C2 . Consider the boolean function f (x) = x. Find a 4 × 4 permutation matrix P such that P |xi ⊗ |0i → |xi ⊗ |f (x)i ≡ |xi ⊗ |xi. Solution 13.

We have to satisfy     1 0 0   1 P (|0i ⊗ |0i) = |0i ⊗ |1i → P   =   0 0 0 0     0 0 0 0 P (|1i ⊗ |0i) = |1i ⊗ |1i → P   =   . 1 0 0 1

A solution is

0 1 P = 0 0 

Problem 14.

0 0 1 0

0 0 0 1

 1 0 . 0 0

Consider the reversible gate (Feynman gate) x1 → x1 ,

x2 → x1 ⊕ x2 .

The inverse function is given by (x1 , x2 ) → (x1 , x1 ⊕ x2 ). Let |0i, |1i be the standard basis in the Hilbert space C2 . Find the unitary matrix which implements (x1 , x2 ∈ {0, 1}) |x1 i ⊗ |x2 i 7→ |x1 i ⊗ |x1 ⊕ x2 i.

Solution 14.

We have |0i ⊗ |0i 7→ |0i ⊗ |0i,

|0i ⊗ |1i 7→ |0i ⊗ |1i

|1i ⊗ |0i 7→ |1i ⊗ |1i,

|1i ⊗ |1i 7→ |1i ⊗ |0i.

This provides the 4 × 4 permutation matrix 1 0 U = 0 0 

0 1 0 0

0 0 0 1

 0  0 1 ≡  1 0 0

0 1



 ⊕

0 1

1 0



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166 Problems and Solutions which is the CNOT-gate and ⊕ denotes the direct sum. Problem 15.

Let x1 , x2 ∈ {0, 1} and ⊕ be the XOR operation. Then (x1 , x2 ) 7→ (x1 ⊕ 1, x1 ⊕ x2 )

is a 2-bit reversible gate since (0, 0) 7→ (1, 0),

(0, 1) 7→ (1, 1),

(1, 0) 7→ (0, 1),

(1, 1) 7→ (0, 0).

Let |0i, |1i be the standard basis in C2 . Find the 4 × 4 permutation matrix P such that P (|x1 i ⊗ |x2 i) = |x1 ⊕ 1i ⊗ |x1 ⊕ x2 i. Solution 15. We calculate the Kronecker products of the vectors. This provides the four equations for P     0 1 0 0 P   =  , 1 0 0 0

    0 0 1 0 P   =  , 0 0 1 0

    0 0 0 1 P   =  , 0 1 0 0

    1 0 0 0 P   =  . 0 0 0 1

Consequently we obtain the 4 × 4 permutation matrix 0 0 P = 1 0 

0 0 0 1

0 1 0 0

 1 0  0 0

with the eigenvalues +1, −1, +i, −i. Problem 16.

Given the 4 × 4 permutation matrix 0 0 U = 1 0 

1 0 0 0

0 0 0 1

 0 1  0 0

with the eigenvalues +1, −1, +i, −i. Find the corresponding boolean function.

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Solution 16. Since the matrix has 4 = 22 rows, the function f : {0, 1}2 → {0, 1}2 has two arguments. The first column (i.e. the column numbered 0) has a 1 in the third row (the row numbered 2) for which b−1 (0) = (0, 0) and b−1 (2) = (1, 0). Thus (0, 0) → (1, 0). From the second column (0, 1) → (0, 0). The third column provides (1, 0) → (1, 1) and the fourth column (1, 1) → (0, 1). Thus we have the map (0, 0) 7→ (1, 0),

(0, 1) 7→ (0, 0),

(1, 0) 7→ (1, 1),

(1, 1) 7→ (0, 1).

The right hand side provides the boolean expression f (x1 , x2 ) = (x1 · x2 + x1 · x2 , x1 · x2 + x1 · x2 ) = (x2 , x1 ).

Problem 17. Given the boolean function f (x1 , x2 ) = x1 · x ¯2 . Thus the map is (x1 , x2 , y ∈ {0, 1}) |x1 i ⊗ |x2 i ⊗ |yi 7→ |x1 i ⊗ |x2 i ⊗ |y ⊕ (x1 · x ¯2 )i with |0i ⊗ |0i ⊗ |0i 7→ |0i ⊗ |0i ⊗ |0i,

|0i ⊗ |0i ⊗ |1i → |0i ⊗ |0i ⊗ |1i

|0i ⊗ |1i ⊗ |0i 7→ |0i ⊗ |1i ⊗ |0i,

|0i ⊗ |1i ⊗ |1i 7→ |0i ⊗ |1i ⊗ |1i

|1i ⊗ |0i ⊗ |0i 7→ |1i ⊗ |0i ⊗ |1i,

|1i ⊗ |0i ⊗ |1i 7→ |1i ⊗ |0i ⊗ |0i

|1i ⊗ |1i ⊗ |0i 7→ |1i ⊗ |1i ⊗ |0i,

|1i ⊗ |1i ⊗ |1i 7→ |1i ⊗ |1i ⊗ |1i.

Find the 8 × 8 permutation matrix for this map. Solution 17.

The map leads to the 8 × 8 permutation matrix   0 1 Uf = I4 ⊕ ⊕ I2 1 0

where ⊕ denotes the direct sum and In is the n × n identity matrix.

Programming Problems Problem 1.

Consider the reversible 3-input/3-output gate given by x01 = x1 ⊕ x3 x02 = x1 ⊕ x2 x03 = (x1 · x2 ) ⊕ (x1 · x3 ) ⊕ (x2 · x3 ).

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168 Problems and Solutions The inverse is given by x1 = x01 · x02 · x03 + x01 · x03 + x02 · x03 x2 = x01 · x02 · x03 + x01 · x03 + x02 · x03 x3 = x01 · x02 · x03 + x01 · x03 + x02 · x03 . Give an implementation in C++ utilizing the bitset class. Solution 1. In the bitset class & is the AND operation, | the OR operation, ^ the XOR operation and ~ the NOT operation. // reversiblegate.cpp #include #include #include using namespace std; int main(void) { bitset x1(string("1")); bitset x2(string("0")); bitset x3(string("1")); cout