Mark Beck Quantum Mechanics Solution
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Quantum Mechanics: Theory and Experiment Solutions Manual
Mark Beck Whitman College
© 2012 This material is protected by copyright and cannot be reproduced without permission.
Chapter 1 Mathematical Preliminaries 1.1
(Sec. 1.1) Data A (xi): x
10, 13, 14, 14, 6, 8, 7, 9, 12, 14, 13, 11, 10, 7, 7
1 15 xi 10.33 15 i 1
x2
x 2
1 15 2 xi 114.60 15 i 1
x2 x
1.2
(Sec. 1.1)
1.3
(Sec. 1.1)
2
7.8
1 4 3 2 5 5.5 7.5 9.5 11.5 13.5 15 15 15 15 15 10.30
1.4
(Sec. 1.1) Using
x 2
2
x2 x :
1
2
Chapter 1 2 1 2 4 2 3 2 2 2 5 2 5.5 7.5 9.5 11.5 13.5 15 15 15 15 15 113.5
2 Using
2
x 2
2
2
113.5 10.3 7.4
x x
2
2
:
2 1 2 4 2 3 5.5 10.3 5.5 10.3 5.5 10.3 15 15 15 2 2 2 5 5.5 10.3 5.5 10.3 15 15 7.4
Both of these values agree, to within our precision; however, they do not agree with the value from Problem 1.1. This is because in this problem we used an estimate of the probability distribution, which is necessarily imprecise. The direct calculation of the variance from the data is more accurate. 1.5
(Sec. 1.1) Data B (xi,yi):(3,4), (5,8), (4,4), (8,5), (3,5), (4,5), (5,8), (8,5), (8,4), (3,4), (3,8), (4,8) 4 1 3 1 2 1 3 1 , P ( x 4) , P ( x 5) , P ( x 8) 12 3 12 4 12 6 12 4 4 1 4 1 4 1 P ( y 3) 0 , P( y 4) , P( y 5) , P ( y 8) 12 3 12 3 12 3 P ( x 3)
1.6*
(Sec. 1.1) P ( x 5, y 8)
2 1 12 6
P ( x 5 | y 8)
2 1 4 2
1 1 1 P ( x 5 | y 8) P ( y 8) P ( x 5, y 8) 2 3 6
Chapter 1 1.7
(Sec. 1.1) 10
10
1
p( x)dx c
10
5 3x
10
2
4
dx x 10
1 3x 5 c tan 1 6 2 x 10 1 c tan 1 12.5 tan 1 17.5 6 c 0.50 1 Therefore, c=2. 10
x
xp( x)dx
10 10
2
10
x
5 3x
2
4
dx x 10
1 5 2 5 3x log 5 3x 4 tan 1 9 9 2 x 10 1 5 1 5 log 629 tan 1 12.5 log 1229 tan 1 17.5 9 9 9 9 1.60 10
x
2
x 2 p ( x)dx
10 10
2
10
x2
5 3x
2
4
dx
10 2 7 5 3 x 10 2 log 5 3x 4 x tan 1 9 27 9 2 27
x 10
x 10
10 20 7 10 20 7 log 629 tan 1 12.5 log 1229 tan 1 17.5 27 9 9 27 9 9 6.53
x
x2 x
2
6.53 1.60 1.99 2
3
4
Chapter 1
1.8
(Sec. 1.1) ce x x 0 , p ( x) 0 x 0 where is a positive, real constant. Compute x and x for this distribution.
p ( x)dx c e x dx
0
c x e
0
x
x 0
c
1
Therefore, c
x
xp( x)dx
xe x dx 0
x 1 x e
x
x 0
1
x
2
x 2 p( x)dx
x 2 e x dx 0
x x 2 2 x e 2
x
x
x 0
2 2 x2 x
2
2
2 1 1 2
Chapter 1 1.9
5
(Sec. 1.1) This is a Gaussian, or normal, distribution. A standard integral table, or statistics textbook, will tell you that the mean of this distribution is , and the standard deviation is / 2 .
1.10
(Sec. 1.2) N
bi M ij a j j 1
b1 5 1 0 6 4 3 17 b2 31 1 6 0 3 9 b3 0 1 0 6 2 3 6
5 0 4 1 17 3 1 06 9 0 0 2 3 6
1.11
(Sec. 1.2) N
M ij A ik Bkj k 1
M11 5 0 0 8 4 3 12 M12 5 2 0 0 4 0 10 M13 5 7 0 1 4 0 35 M 21 3 0 1 8 0 3 8 M 22 3 2 1 0 0 0 6
M 23 3 7 11 0 0 22 M 31 0 0 0 8 2 3 6 M 32 0 2 0 0 2 0 0 M 33 0 7 0 1 2 0 0
5 0 4 0 2 7 12 10 35 3 1 0 8 0 1 8 6 22 0 0 23 0 0 6 0 0
6
Chapter 1
1.12* (Sec. 1.2) 5 0 2 1 10 5 a) 3 7 0 0 6 3 2 1 5 0 13 7 b) 0 03 7 0 0 Are they the same? Clearly not. These matrices do not commute. 1.13
(Sec. 1.2) 4 2 4 8 2 1 30 1 8
1.14
(Sec. 1.2) 5 0 4 1 0 3 0 3 1 0 4 3 1 0 5 0 2 2 2 2 0 2 0 2 5 2 0 4 2 2
1.15
(Sec. 1.2) a) 1 1 2 1 1 1 1 1 2 2 1 2 0 Eigenvalues are a 0, b 2 For a 0 : 1 xa 1 0 0 1 1 0 xb
xa xb 0 xb xa
Chapter 1 A normalized eigenvector is thus 1 1 . 2 1 For b 2 : 1 xa 1 2 0 1 1 2 xb
xa xb 0
xb xa A normalized eigenvector is thus 1 1 . 2 1 b) To show that they’re orthogonal, compute the inner product: 1 1 1 1 1 1 1 1 0 . 2 2 1 2 So, they’re orthogonal. 1.16* (Sec. 1.2) a) 0 i 2 1 0 0 i Eigenvalues are a 1, b 1 For a 1 : 0 1 i xa 0 0 1 xb i
ixa xb 0 xb ixa
7
8
Chapter 1 A normalized eigenvector is thus 1 1 . 2 i For b 1 : 0 1 i xa 0 0 1 xb i
ixa xb 0 xb ixa A normalized eigenvector is thus 1 1 . 2 i b) To show that they’re orthogonal, compute the inner product: 1 1 1 1 1 i 1 1 0 . 2 2 i 2 So, they’re orthogonal.
Chapter 2 Classical Description of Polarization 2.1
(Sec. 2.3) The magnitude of field reduced by εu , where u is the polarizer axis (a unit vector that makes an angle of w.r.t. horizontal). Intensity (power per unit area) is proportional to the square of 2 the field, and is thus reduced by ε u . 1 ux iuy ux cos uy sin 2 1 cos i sin 2 1 i e 2 1 2 1 2 ε L u 2 The amplitude of the electric field is reduced by a factor of 1/ 2 , and the intensity is reduced by 1/2. This result is independent of the angle . εL u
2.2 (Sec. 2.4) (a) Remember that the matrices are written right-to-left. 1 1 i i 0 1 1 1 1 1 i i i 2 i 1 0 1 2 1 1 2 i 1 1 1 1 0 2i 22 0 0 i 1 0 (b) 0 i 1 0 1 00 1 Vertically polarized light comes out.
9
10
Chapter 2
2.3* (Sec. 2.4) 1/ 2 1/ 2 J 45o 1/ 2 1/ 2 1/ 2 1/ 2 2 1/ 2 1/ 4 1/ 2 1/ 2 2 1 0 eigenvalues are a 1 , b 0 .
For a 1 1/ 2 1/ 2 0 1/ 2 1/ 2 0 1/ 2 0 normalized eigenvector: x a
1 1 2 1
For b 0 1/ 2 1/ 2 0 1/ 2 1/ 2 0 1/ 2 0 normalized eigenvector: xb
1 1 2 1
(b) 1 1 o ; this is a beam polarized at +45 , which is transmitted 100% ( a 1 ) 1 2 through the polarizer. 1 1 o b 0 , xb ; this is a beam polarized at -45 , which is completely blocked ( b 0 ) by 2 1 the polarizer. a 1 , xa
These make sense.
Chapter 2
11
2.4 (Sec. 2.4) In other words, we need to prove that J / 2 ε R cε L , where c is a complex number with c 1 . cos 2 sin 2 1 1 J / 2 ε R sin 2 cos 2 2 i 1 cos 2 i sin 2 2 sin 2 i cos 2 1 1 cos 2 i sin 2 sin 2 i cos 2 2 cos 2 i sin 2
cos 2 i sin 2 e i 2 sin 2 i cos 2 sin 2 i cos 2 cos 2 i sin 2 cos 2 i sin 2 cos 2 i sin 2 cos 2 i sin 2 sin 2 cos 2 i sin 2 2 i cos 2 2 cos 2 sin 2 cos 2 2 sin 2 2 i J / 2 ε R e i 2
1 1 2 i
cε L
2.5
(Sec. 2.4)
J /4 45o
1 1 2 1 i 2 1 i 1 1 i 1 1 i 2 2 /4 /4 i i e 1 e 2 e i/4 ei/4 ei/2 ei/4 1 2 e i/2 1 ei/4 1 i 2 i 1 ei/4 J /4 45
12
Chapter 2
Thus, they are the same to within an overall phase factor. 2.6* (Sec. 2.4) e i 0 1 e i i 1 i J ε H e e εH . 0 0 10 0 A horizontally polarized beam acquires a phase shift. e i 0 0 0 J εV εV 0 11 1 A vertically polarized beam is unaffected.
2.7 (Sec. 2.4) (a) cos 22 J / 2 2 J H J / 2 1 sin 22 cos 22 sin 22
sin 22 1 0 cos 21 sin 21 cos 22 0 0 sin 21 cos 21 sin 22 cos 21 sin 21 cos 22 0 0
cos 22 cos 21 cos 22 sin 21 sin 22 cos 21 sin 22 sin 21 (b) cos 2 cos sin J sin 2 cos sin The answer to part (a) is equal to this if 1 2 / 2 .
2.8 (Sec. 2.4) (a) cos 22 J / 2 2 JV J / 2 1 sin 22 cos 22 sin 22
sin 22 0 0 cos 21 sin 21 cos 22 0 1 sin 21 cos 21 sin 22 0 0 cos 22 sin 21 cos 21
sin 22 sin 21 cos 22 sin 21
sin 22 cos 21 cos 22 cos 21
Chapter 2
13
(b) The Jones matrix for a linear polarizer whose transmission axis makes an angle of from the vertical is the same as the Jones matrix for a linear polarizer whose transmission axis makes an angle of / 2 from the horizontal. cos 2 / 2 cos / 2 sin / 2 J / 2 sin 2 / 2 cos / 2 sin / 2 sin 2 sin cos cos 2 sin cos The answer to part (a) is equal to this if 1 2 / 2 . 2.9* (Sec. 2.4) Sandwich the PAHV between the two half-wave plates. There is more than one possible solution for the angles of the two wave plates. One possible solution is to set 1 2 22.5o . In this case the vertical axis of the PAHV will be equivalent to a linear polarizer oriented at
2 22.5o 45o from the vertical, which is the same as a polarizer at 45o from the
horizontal. The horizontal axis of the PAHV will be equivalent to a linear polarizer oriented at
2 22.5o 45o from the horizontal. This combination thus analyzes a beam into its 45o and 45o polarization components, so it is effectively a PA45. 2.10 (Sec. 2.4) (a) J J / 4 45 J H J / 4 45 1 1 i 1 0 1 1 i 2 i 1 0 0 2 i 1 1 1 i 1 i 2 i 1 0 0 1 1 i 2 i 1
14
Chapter 2
(b) ε Jε R 1 1 i 1 1 2 i 1 2 i 1 1 0 2 2 0 0 So the beam is completely blocked.
(c) ε Jε L 1 1 i 1 1 2 i 1 2 i 1 1 2 2 2 2i 1 1 εL 2 i So the beam is transmitted unattenuated with left-circular polarization – it is completely unaffected.
2.11 (Sec. 2.4) (a) J J / 4 45 JV J / 4 45 1 1 i 0 2 i 1 0 1 1 i 0 2 i 1 i
1 1 i 2 i 1
0 1 1 i 1 2 i 1 0 1
Chapter 2
15
(b) ε Jε R 1 1 i 1 1 2 i 1 2 i 1 1 2 2 2 2i
1 1 2 i
εR So the beam is transmitted unattenuated with right-circular polarization – it is completely unaffected.
(c) ε Jε L 1 1 i 1 1 2 i 1 2 i 1 1 0 2 2 0 0 So the beam is completely blocked.
2.12* (Sec. 2.4) Sandwich the PAHV between the two quarter-wave plates. Set the first quarter-wave plate so that its fast axis makes an angle of 45o from the horizontal, and the second quarter-wave plate so that its fast axis makes an angle of 45o from the horizontal. The horizontal output of the PAHV will pass the left-circular component unaffected, while the vertical output will pass the rightcircular component unaffected. 2.13 (Sec. 2.5) An arbitrary input unit polarization vector is a 2 2 ε i , where a b 1 b 0 e i a e i b ε Jε i 1 0 b a
16
Chapter 2 2 2 2 2 2 ε ei b a b a 1
2.14 (Sec. 2.5) Using the Jones matrix of Eq. (2.42), the Jones matrix for the -45o output port is 1 1 1 0 ei 1 1 ei J . 2 1 1 1 0 2 1 ei
If the input beam is polarized at +45o the polarization vector of the output beam is ε Jε 45 1 1 ei 1 1 2 1 e i 2 1
1 1 ei 2 2 1 e i
1 1 1 1 ei 2 2 1 1 1 ei ε 45 . 2
The intensity of this beam is 2
1 I I in 1 ei 2 I in 1 ei 1 e i 4 I in 2 ei e i 4 I in 1 cos . 2
The total output intensity is this, plus the intensity found in Eq. (2.45). I tot
I in I 1 cos in 1 cos Iin . 2 2
Energy is conserved.
Chapter 2
17
2.15*(Sec. 2.5)
The two PHHV’s essentially just split the beams apart and recombine them, so we can effectively ignore them (as long as we do include the relative phase shift). cos 23 J sin 23
sin 23 ei cos 23 0
0 0 1 cos 21 sin 21 1 1 0 sin 21 cos 21
cos 23 sin 23
sin 23 ei cos 23 0
0 sin 21 cos 21 1 cos 21 sin 21
cos 23 sin 23
sin 23 ei sin 21 ei cos 21 cos 23 cos 21 sin 21
ei cos 2 sin 2 sin 2 cos 2 ei cos 2 cos 2 sin 2 sin 2 3 1 3 1 3 1 3 1 i ei sin 2 sin 2 cos 2 cos 2 3 3 1 e sin 23 cos 21 cos 23 sin 21
The matrix needs to be real. Can be achieved for 0 or . sin 21 23 cos 21 23 J if 0 cos 21 23 sin 21 23 sin 23 21 cos 23 21 J if cos 23 21 sin 23 21 1 0 So, for 0 , want 3 1 45 to give J 0 1 almost: but vertical is phase shifted. 1 0 So, for , want 3 1 45 to give J 1 0 1 This is it: , 3 1 45 .
Complement 2.A Coherence and Interference 2.A.1* Eq. (2.A.4) says r1 r2 r
2
t1 t2 t
2
r t 1 .
Eqs. (2.A.5) and (2.A.6) tell us: E1 rEi E 2 tEi From Eqs. (2.A.2) and (2.A.3) we get
t 2 e i r 2 t 2 e i
E r tE1 rE 2ei Ei tr rtei
E d rE1 tE 2ei = Ei r 2 t 2ei
This means I d Ed Ei
2 2
r
2
I i r 4 t 4 2r 2t 2 cos I r Er
2
Ei
2
tr rte tr rte I r t r t 2r t cos i
2 2
2 2
i
2 2
i
The total intensity is then I tot I d I r
I r I r
I i r 4 t 4 2r 2t 2 cos I i 2r 2t 2 2r 2t 2 cos 4
t 4 2 r 2t 2
2
t2
i i
2
,
Ii
and energy is conserved (the energy output does not depend on the phase of the interferometer).
19
20
Complement 2.A
2.A.2 Assume r1 r2 r ,
2
t1 t2 t , and
2
r t 1
Eqs. (2.A.5) and (2.A.6) then tell us: E1 rEi E 2 tEi From Eqs. (2.A.2) and (2.A.3) we get
t 2 e i r 2 t 2 e i
E r tE1 rE 2ei Ei tr rtei
E d rE1 tE 2ei = Ei r 2 t 2ei
This means I d Ed Ei
2 2
r
2
I i r 4 t 4 2r 2t 2 cos I r Er
2
Ei
2
tr rte tr rte I r t r t 2r t cos i
2 2
2 2
i
2 2
i
The total intensity is then I tot I d I r
I r I r
I i r 4 t 4 2r 2t 2 cos I i 2r 2t 2 2r 2t 2 cos
i
4
t 4 2r 2t 2 4r 2t 2 I i cos
2
t2
i
2
,
4r 2t 2 I i cos
I i 4r 2t 2 I i cos
Which depends on the phase of the interferometer. For certain phases more energy would come out than is going in. For other phases less energy would be coming out than is going in. Clearly this does not conserve energy.
Complement 2.A
21
2.A.3 c
1 2f
lc cc
c 3 108 m / s lc 0.048 m 4.8 cm 2f 2 1 109 1/ s
2.A.4 f
c
c
2 1 2f 2c
df
c 2
d
f
530 109 m 2 lc cc 2 2 108 m
2.A.5 I max I 0 , I min 0.6 I 0 V
I 0 0.6 I 0
I 0 0.6 I 0
0.25
c 2
(only care about the spread, so take the absolute value)
2
4.5 106 m 4.5 m
Chapter 3 Quantum States 3.1* (Sec. 3.4) From Table 2.1 we know ε cos ε H sin εV , which means cos H sin V
3.2 (Sec. 3.4) P V | V
2
V V cos H V sin V sin
P | V
sin 2
3.3 (Sec. 3.4)
P 45 | 45
2
45 45 cos H 45 sin V cos
1 HV 2
H
sin
1 HV 2
V
1 cos sin 2 2 1 P 45 | cos sin 2
Note also: 45
1 1 cos sin 2 2
cos cos 45o sin sin 45o cos 45o
so, P 45 | cos 2 45o .
23
24
Chapter 3
3.4* (Sec. 3.4)
(a) PA: input state V , output state , probability to get through:
P | V
2
V
V
2
V V cos H V sin V sin
P | V
sin 2
PAHV: input state , output state H , probability to get through: PH | H
2
H H cos H H sin V cos
P H | cos 2 Ptot P | V
P H | sin cos 2
2
1 sin 2 2 4 (b) 1 Ptot sin 2 2 has a maximum of 1/4 at =/4, or =45o. 4 (c) If the PA is removed, we simply have vertical photons going into a horizontal polarizer. None of the photons will get through. PH | V
H V
2
0
Chapter 3
3.5 (Sec. 3.5) PR | R
2
R R cos H R sin V cos
1 H i V 2
H
sin
1 1 i cos i sin e 2 2
PR |
2
1 i 1 e 2 2
3.6* (Sec. 3.5) e1 cos H ei sin V
in general: e2 cH H cV V From orthogonality, we know e1 e2 cos cH e i sin cV 0 so: cos cH e i sin cV i sin cH e sin i cV cos e cos
one solution is thus: cH sin cV ei cos or e2 sin H ei cos V You can check that this is normalized.
1 H i V 2
V
25
26
Chapter 3
3.7 (Sec. 3.6) 1 1 H V H i V L 2 2 1 1 45 L H V H i V 2 2 1 1 i 2 1 1 1 45 L 1 1 HV 2 i HV 2 1 1 1 45 L 1 1 HV 2 2 i HV 45
3.8
1 1 i 2
(Sec. 3.6) e1 cos H ei sin V
1 1 45 e1 H V cos H ei sin V 2 2 1 cos ei sin 2 cos e1 i e sin HV cos 1 45 e1 1 1 HV i 2 sin e HV 1 cos ei sin 2
45
1 1 1 HV 2
Chapter 3
3.9
(Sec. 3.6) e1 cos H ei sin V
P R e1
R e1
2
1 1 H i V cos H ei sin V 2 2 2 1 cos iei sin 2 1 cos ie i sin cos iei sin 2 1 cos 2 sin 2 i ei e i cos sin 2 1 cos 2 sin 2 i 2i sin cos sin 2 1 cos 2 sin 2 2sin cos sin 2 1 1 sin sin 2 2
3.10 (Sec. 3.6) We want L 45 45 R 45 c
L 45 45 R 45 c
We know everything in terms of the hv-basis: 1 1 H i V H i V R L 2 2 1 1 H V H V 45 45 2 2 so
2
27
28
Chapter 3 1 H i V H V 2 1 1 i 2 1 1 1 i 2 2 1 i / 4 e 2
L 45
similarly, R 45
1 i / 4 e 2
L 45
1 i / 4 e 2
R 45
1 i / 4 e 2
so: 1 e i /4 e i /4 1 45 2 ei /4 c 2 i c 45
1 ei /4 e i /4 i 2 e i /4 c 2 1c
It doesn’t really matter what phase factor you pull out front, but it’s customary to pull out the same phase factor. 3.11* (Sec. 3.7) No. If the beam entered in state V , it would only take one path through the interferometer. After the half-wave plate it would be changed to state H , and would leave the PAHV in that state. It would split equally on the PA45, and there would be no phase dependence. 3.12
(Sec. 3.7) Yes. Just like the state 45 , the state R would also split equally on the first PAHV and take both paths through the interferometer. The only difference is the relative phase shift between the horizontal and vertical components on input. On output the beam would split on the PA45, and adjusting the phase of the interferometer would cause interference fringes.
Chapter 3
3.13* (Sec. 3.4) (a) P 1| H
H H
2
1 P 1| H
V H
2
0 (b) HV 1 P 1| H
1 P 1|
H
1 (c)
HV 2
HV
1/2
HV 2 HV
HV 2 1 P 1| H
H
(b) HV 1 P 1| 45 1 P 1| 45
2
12 P 1|
2 1/2
1 HV 1 1
1/2
0
3.14* (Sec. 3.4) (a) P 1| 45 H 45
2
1/ 2 P 1| 45 V 45
2
1/ 2
1 1 0 2 2
(c) HV
HV 2
1/2
HV 2 HV
2 1/2
29
30
Chapter 3 HV 2 1 P 1| 45 1 P 1| 45 2
1 1 1 2 2
HV 1 0
1/2
3.15
2
1
(Sec. 3.5) (a) P 1| e1 H e1
2
cos 2
P 1| e1 V e1
2
sin 2
(b) HV 1 P 1| e1 1 P 1| e1
cos 2 sin 2 cos 2 (c)
HV 2
HV
1/2
HV 2 HV
2 1/2
HV 2 1 P 1| e1 1 P 1| e1 2
2
cos 2 sin 2 1
HV 1 cos 2 2
1/2
sin 2
Chapter 4 Operators 4.1* (Sec. 4.2) The norm of is . The norm of Uˆ is the same: Uˆ †Uˆ , because Uˆ †Uˆ 1ˆ if Uˆ is unitary. 4.2 (Sec. 4.3) Pˆ
Pˆ 2
4.3 (Sec. 4.3)
PˆH H c*H V cV*
H
H
H H c*H H V H cV* H c*H H 4.4 (Sec. 4.4) cos sin Rˆ p . sin cos HV cos sin Rˆ †p Rˆ p . sin cos HV
31
32
Chapter 4 cos sin Rˆ †p Rˆ p sin cos HV
cos sin sin cos HV
cos sin sin cos cos 2 sin 2 2 2 sin cos cos sin sin cos HV 1 0 0 1 HV 1ˆ
4.5 (Sec. 4.6) cos sin Rˆ p . sin cos HV
Since O†ij O ji* [Eq. (4.75)], cos sin Rˆ †p . sin cos HV
Because cos cos and sin sin , cos sin Rˆ †p Rˆ p sin cos HV
4.6* (Sec. 4.4) cos sin Rˆ p sin cos HV
cos sin HV
cos cos sin sin cos sin sin cos HV cos sin HV
4.7 (Sec. 4.5) We want L H H R H C We know
LV V R V C
Chapter 4 L
1 H i V 2
R
1 H i V 2
so 1 H i V H 2 1 2 similarly, 1 i R H LV 2 2 so: 1 1 H 2 1 C L H
V
RV
i 2
i 1 2 1 C
4.8 (Sec. 4.5) 1 1 2 1 C i 1 V 2 1 C H
H V
i 1 i 1 1 1C (1 1) 0 2 2 1 C 2
4.9 (Sec. 4.5) e cos H ei sin V
We want 45 e e 45 e 45
33
34
Chapter 4 45 e 45 cos H 45 ei sin V cos 45 H ei sin 45 V 1 cos ei sin 2 45 e 45 cos H 45 ei sin V
cos 45 H ei sin 45 V 1 cos ei sin 2 i 1 cos e sin e i 2 cos e sin 45
4.10 (Sec. 4.5) PˆH L H H L 1 H 2 1 1 PˆH L L H 2 2 1 1 PˆH L R H 2 2 1 1 1 2 1 1C
L R PˆH
PˆH R H H R 1 H 2 1 1 L PˆH R L H 2 2 1 1 R PˆH R R H 2 2
PˆV L V V L
PˆV R V V R
i V 2 i 1 L PˆV L LV 2 2 i 1 R PˆV L R V 2 2 1 1 1 PˆV 2 1 1 C
i V 2 i 1 L PˆV R L V 2 2 i 1 R PˆV R RV 2 2
Chapter 4 1 1 PˆH 2 4 1 1 1 2 1 1 1 PˆH PˆV 4 1 1 0 4 0
35
1 1 1 1 1 PˆV 2 4 1 1 C 1 1 C 1 1 1 ˆ PV 2 1 1 C
1 1 1 1 C 1 1 C 1 ˆ PH 1 C 1 1 1 1 C 1 1 C 0 0 0 C
1 1 PˆV PˆH 4 1 1 0 4 0
1 1 1 1 C 1 1 C 0 0 0 C
4.11 (Sec. 4.5) In Problem 4.6 we showed that Rˆ p , so Rˆ p 45 45 cos 45 H sin 45 V Rˆ p 45 45 cos 45 H sin 45 V
45 Rˆ p 45 45 45
Rˆ p 45 Rˆ p 45
45 Rˆ p 45 45 Rˆ p 45
cos 45 H sin 45 V cos 45 H sin 45 V
45
45 cos 45 H sin 45 V 45 cos 45 H sin 45 V
1 cos 45 sin 45 cos 45 sin 45 . 2 cos 45 sin 45 cos 45 sin 45 45
Can simplify this some more
45
36
Chapter 4 1 Rˆ p 2
1 cos sin 2 1 cos sin 2
1 2 cos 2sin 2 2sin 2 cos 45
cos sin sin cos 45
1 sin cos 2 1 sin cos 2
1 cos sin 2 1 cos sin 2
1 sin cos 2 . 1 sin cos 2 45
Chapter 4
4.12 (Sec. 4.5)
1 1 1 1 Rˆ p 45o 45 V 2 1 1 45 0 45
1 1 2 1 45
1 2 1 2
45 45 1 H V 2
1 H V 2
V
4.13 (Sec. 4.5) 1 Rˆ p L Rˆ p H i V 2 1 ˆ R p H iRˆ p V 2 cos sin 1 Rˆ p H sin cos HV 0 HV
cos sin HV cos sin 0 Rˆ p V sin cos HV 1 HV sin cos HV 1 cos H sin V i sin H cos V Rˆ p L 2 1 cos i sin H sin i cos V 2 1 i e H ie i V 2 e i L
L is an eigenstate of Rˆ p , with eigenvalue e i .
37
38
Chapter 4 1 Rˆ p R Rˆ p H i V 2 1 ˆ R p H iRˆ p V 2 1 cos H sin V i sin H cos V 2 1 cos i sin H sin i cos V 2 1 i e H iei V 2 e i R
R is an eigenstate of Rˆ p , with eigenvalue ei .
L Rˆ p L ˆ Rp R Rˆ p L e i 0
L Rˆ p R ˆ R R p R
C
0 ei C
4.14* (Sec. 4.6) Physically this corresponds to a polarizer oriented along the +45° axis. From Table 2.2, its matrix representation is thus 1 1 1 Pˆ45 2 1 1 HV 1/ 2
1/ 2
1/ 2
1/ 2
1/ 2 1/ 4 2
2 1 0 eigenvalues are a 1 , b 0 .
For a 1 1/ 2 1/ 2 0 1/ 2 1/ 2 HV HV 0 HV
Chapter 4
39
1/ 2 0 normalized eigenstate:
1 1 45 2 1 HV
For b 0 1/ 2 1/ 2 0 1/ 2 1/ 2 HV HV 0 HV 1/ 2 0 normalized eigenstate:
1 1 45 2 1 HV
Physically, these make sense. The 45 state is transmitted with 100% probability (eigenvalue 1), while the 45 state is completely blocked (eigenvalue 0). 4.15 (Sec. 4.6) cos sin Rˆ p sin cos HV
cos sin 2 cos sin 2 sin cos
cos 2 2 cos 2 sin 2 2 2 cos 1
ei e i
2 ei e i 1
0 So, the eigenvalues are ei and e i . The eigenstate corresponding to ei is cos ei sin
sin cos ei HV
0 HV 0 HV
40
Chapter 4
1 i i 1 e e ei ei ei i sin 2 2 i 1 0 sin 1 i HV HV 0 HV i 0 i cos ei
The normalized eigenstate is thus 1 1 R 2 i HV This means that the eigenstate for e i must be L , because the eigenstates form an orthonormal basis. You can check that it is if you like. These results are consistent with Problem 4.13. 4.16* (Sec. 4.6) The eigenvalues and eigenstates of Oˆ are the solutions to the equation Oˆ , Oˆ . If this is true, then in general Oˆ † * . Oˆ † * * . If Oˆ is Hermitian, Oˆ † Oˆ , so Oˆ † Oˆ . The above then implies that * , and must be real. 4.17 (Sec. 4.6) We know that the matrix elements of the adjoint are given by O†ij O ji* [Eq. (4.75)], where the matrix elements are computed using arbitrary states. A ji j a1 a2 i
Chapter 4 A ji* j a1
*
a2 i
41
*
a1 j i a2 i a2 a1 j
a2
i
a1
j
A†ij i Aˆ † j
Comparing the last and the third to last lines yields Aˆ † a2 a1 . 4.18* (Sec. 4.6) Let 1
be eigenvectors of Oˆ , corresponding to distinct eigenvalues 1 and 2
and 2
( 1 2 ). Oˆ 1 1 1 and Oˆ 2 2 2 Oˆ † Oˆ , and the eigenvalues must be real (proved in Problem 4.17), so Oˆ 2 2 .
For Hermitian operators 1 Oˆ 1 1 and 2
1 Oˆ 2 1 Oˆ 2
1 2 2 2 1 2 ,
and 1 Oˆ 2 1 Oˆ 2
1 1 2 1 1 2 . 0 1 Oˆ 2 1 Oˆ 2 2 1 2 1 1 2 2 1 1 2 .
Since we’ve assumed 1 2 , then 1 2 0 , and 1 and 2 are orthogonal.
42
Chapter 4
4.19* (Sec. 4.6)
1 ˆn O n ! n 0
ˆ
eO
1 n n ! n 0
1 n n 0 n ! e Note that the second line is true ONLY if is an eigenstate of Oˆ . Otherwise, this problem is not solvable with the information we know. 4.20* (Sec. 4.6) First, express the function in terms of its power series:
cnOˆ n . n 0
f Oˆ
Then use the identity operator, written as a sum of the eigenstates of Oˆ ˆ ˆ f O 1 cnOˆ n i i n 0 i
i
cnOˆ n
n 0
i i
cn i n i i i n 0 f i i i . i
Complement 4.A Changing Bases Using Similarity Transformations 4.A.1 * We need to prove that S†S 1 . Sij ni o j [Eq. (4.75)] S †ij S ji* n j oi
*
oi n j
In Chapter 1 [Eq. (1.42)] we found that the matrix elements of the product S†S can be written as
S†S ij k S †ik Skj
oi nk
nk o j
k
oi nk nk o j k Since the states nk form a basis:
S†S ij oi 1ˆ o j oi o j ij .
Since the matrix elements of S†S are given by the Kronecker delta, S†S is equal to the identity matrix (as shown in Chapter 1).
43
44
Complement 4.A
4.A.2 The matrix that transforms the HV-basis to the circular polarization basis is C HV1 S 1 C2 HV1 L H R H
C1 HV2 C2 HV2 HV C LV RV HV C
1 1 i 2 1 i HV C
H C SH HV
1 1 1 i 2 1 i HV C 0 HV
1 1 2 1 C
V C SV HV
0 1 1 i 2 1 i HV C 1 HV
i 1 2 1 C
4.A.3 In the text we found [Eq. (4.A.11)] 1 1 1 S 2 1 1 HV 45 e 45 Se HV 1 1 1 cos i 2 1 1 HV 45 e sin HV i 1 cos e sin 2 cos ei sin
, 45
Complement 4.A
4.A.4 In the text we found [Eq. (4.A.11)] 1 1 1 S 2 1 1 HV 45 † R p45 SR HV p S
cos sin 1 1 1 1 1 1 2 1 1 HV 45 sin cos HV 2 1 1 HV 45
cos sin cos sin 1 1 1 2 1 1 HV 45 sin cos sin cos cos sin sin cos cos sin sin cos cos sin sin cos 45
cos sin sin cos cos sin sin cos 45
45
Chapter 5 Measurement 5.1*
(Sec. 5.1) In Problem 4.17 you showed that if Aˆ a1 a2 , then Aˆ † a2 a1 . This means PˆH H H PˆH † , and PˆV V V PˆV † ; both of these projection operators are Hermitian. ˆ HV 1 PˆH 1 PˆV , so We can write ˆ HV † 1 PˆH 1 PˆV † † 1 Pˆ 1 Pˆ H
†
V
1 PˆH 1 Pˆ ˆ HV . ˆ HV is Hermitian. So, 5.2*
(Sec. 5.1) ˆ HV H H 1 H H 1 V V H H 11 1 0 1 ˆ HV H V 1 H H 1 V V H V 1 0 1 0 0 ˆ HV V H 1 H H 1 V V V H 1 0 1 0 0 ˆ HV V V 1 H H 1 V V V V 1 0 11 1 1 0 ˆ HV 0 1 HV
47
48
Chapter 5
5.3
(Sec. 5.1) Work out the matrix elements: 1 ˆ HV L H i V ˆ HV H i V L 2 1 ˆ HV H i H ˆ HV V i V ˆ HV H V ˆ HV V H 2 1 1 0 0 1 2 0 1 ˆ HV R H i V ˆ HV H i V L 2 1 ˆ HV H i H ˆ HV V i V ˆ HV H V ˆ HV V H 2 1 1 0 0 1 2 1 ˆ HV is Hermitian ˆ ˆ HV R * 1 , because R HV L L 1 ˆ HV H i V H i V 2 1 ˆ HV H i H ˆ HV V i V ˆ HV H V ˆ HV V H 2 1 1 0 0 1 2 0
ˆ HV R R
0 1 ˆ HV 1 0 C 5.4
(Sec 5.1) ˆ HV H 1 H , ˆ HV V 1 V , so Eq. (5.3) says
ˆ HV cH ˆ HV H cV ˆ HV V cH H cV V . ˆ HV 1 H H 1 V V , so Eq. (5.4) says
Chapter 5 ˆ HV 1 H H 1 V V cH H cV V 1 H cH 1 V cV
49
cH H cV V .
These are equal to each other. 5.5
(Sec 5.2) The possible results are the eigenvalues: +1 for horizontal, and –1 for vertical. For a measurement yielding +1, the probability is
P 1| H
2
1 2 i/3 H H e V 3 3 1 3
2
2
1 . 3
After such a measurement, the beam of photons is left in state H . For a measurement yielding –1, the probability is: P 1| V
2
1 2 i/3 H e V V 3 3
2 3
2
2
2 . 3
After such a measurement, the beam of photons is left in state V .
5.6*
(Sec 5.2)
Pˆ45 45 45 Pˆ45† , so the operator is Hermitian. This means it is an observable, as long as it can be physically measured. A physical implementation of this operator would simply be a PA45, set up to transmit the +45° output, and block the –45° output.
50
Chapter 5 We found the eigenvalues and eigenstates of Pˆ45 in Problem 4.14; they are Eigenvalue 1: 45 Eigenvalue 0: 45 . So, the possible outcomes from a measurement are either 1 or 0. If the outcome is 1, the system is left in state 45 ; if the outcome is 0, the system is left in state 45 (although this state is blocked in our implementation).
5.7*
(Sec 5.2) Remember that Rˆ p † Rˆ p Rˆ p . Since the operator is not Hermitian, it cannot correspond to an observable.
5.8
(Sec 5.3) 1 1 2 i/3 2 i/3 ˆ HV ˆ HV V V 3 H 3e 3 H 3e 1 ˆ HV H 2ei/3 H ˆ HV V 2e i/3 V ˆ HV H 2 V ˆ HV V H 3 1 1 1 0 0 2 . 3 3 This is the average. ˆ HV 2 1ˆ , so ˆ HV 2 1 . The standard deviation is then In Example 5.2 we showed that
ˆ HV 2 ˆ HV HV 1 2 1 3 0.943 .
1/2
2 1/2
Chapter 5
5.9
51
(Sec 5.3)
ˆ HV cos H sin e i V ˆ HV cos H sin ei V
ˆ HV H sin cos ei H ˆ HV V cos 2 H ˆ HV H sin 2 V ˆ HV V sin cos ei V cos 2 sin 2 cos 2 ˆ HV 2 1ˆ , so ˆ HV 2 1 . The standard deviation is then In Example 5.2 we showed that
ˆ HV 2 ˆ HV HV 1 cos 2 2 sin 2 2
2 1/2
1/2
1/ 2
sin 2 . 5.10
(Sec 5.3)
Pˆ45 cos H sin V Pˆ45 cos H sin V
cos 2 H Pˆ45 H sin cos H Pˆ45 V sin cos V Pˆ45 H sin 2 V Pˆ45 V Physically Pˆ45 corresponds to a polarizer oriented along the +45° axis. From Table 2.2, its matrix representation is thus 1 1 1 Pˆ45 2 1 1 HV Using these matrix elements: 1 Pˆ45 cos 2 sin cos sin cos sin 2 2 1 1 2sin cos 2 1 1 sin 2 2
52
Chapter 5
Pˆ45 2 45 45 45 45 45 45 Pˆ45 , so Pˆ45 2 Pˆ45 . The standard deviation is then P45
Pˆ45 2 Pˆ45
2 1/ 2
1/ 2
2 1 1 1 sin 2 1 sin 2 4 2
1/ 2
1 1 1 sin 2 1 2sin 2 sin 2 2 4 2 1/ 2
1 1 sin 2 2 4 1 cos 2 . 2 5.11
(Sec 5.5) ˆ 45 H H 1 45 45 1 45 45 H H 1 1 1 1 0 2 2 ˆ 45 V H 1 45 45 1 45 45 V H
ˆ 45 V
1 1 1 1 1 2 2 H V 1 45 45 1 45 45 H
1 1 1 1 1 2 2 ˆ 45 V V 1 45 45 1 45 45 V V 1 1 1 1 0 2 2
0 1 ˆ 45 1 0 HV
Chapter 5
5.12* (Sec 5.5) ˆ C 1 L L 1 R R ˆ C H H 1 L L 1 R R H H 1 1 1 1 0 2 2 ˆ C V H 1 L L 1 R R V H i i 1 1 i 2 2 ˆ C H V 1 L L 1 R R H V i i 1 1 i 2 2 ˆ C V V 1 L L 1 R R V V 1 1 1 1 0 2 2
0 i ˆC i 0 HV 5.13
(Sec 5.5) Example 5.4 shows that 0 2 ˆ HV , ˆ 45 2 0 HV ˆC. Problem 5.12 determines the matrix for 0 i ˆ C 2i 2i i 0 HV 0 2 2 0 HV ˆ HV , ˆ 45 2i ˆC. Thus,
53
54
Chapter 5
5.14
(Sec 5.5) The matrix representations are given in Problems 5.2 and 5.12. ˆ HV , ˆ C ˆ HV ˆ C ˆ C ˆ HV 1 0 0 i 0 i 1 0 0 1 HV i 0 HV i 0 HV 0 1 HV 0 1 0 1 i i 1 0 HV 1 0 HV 0 1 2i 1 0 HV ˆ 45 2i ˆ 45 was determined in Problem 5.11. The matrix representation of
5.15
(Sec 5.5) Pˆ45 , PˆV 45 45 V V V V 45 45 1 45 V V 45 2 1 H V V V H V 2 1 H V V H 2
5.16
(Sec 5.5) The possible results are the eigenvalues: +1 for +45°, and –1 for –45°. For a measurement yielding +1, the probability is
Chapter 5 P 1| 45
55
2
1 2 i/3 H e V 45 3 3
2
2
1 1 2 1 i/3 e 3 2 3 2 2 1 1 2 cos / 3 i 2 sin / 3 6 1 2 6 1 i 6 2 2
2
2 1 2 3 1 6 2 2 0.74 .
After such a measurement, the beam of photons is left in state 45 . For a measurement yielding –1, the probability is: P 1| 1 P 1| 0.26 . After such a measurement, the beam of photons is left in state 45 . 5.17* (Sec 5.5) 1 . 2
The probability of the first measurement is P HV 1| L P H | L H L
2
The probability of the second measurement is P C 1| L P R | L R L
2
0.
2
The joint probability of +1 and –1 is then P HV 1| L P C 1| L 0 . 5.18* (Sec 5.5) The probability of the first measurement is P HV 1| L P H | L H L
1 . 2
After the measurement, the photon would be left in state H . The probability of the second measurement is P C 1| H
PR |
H
R H
2
1 . 2
56
Chapter 5 The joint probability of +1 and –1 is then P HV 1| L P C 1| H
5.19
(Sec 5.5) In Problem 5.8 we found HV 0.943 for this state. ˆ 45 is given in Problem 5.11. The matrix representation of
ˆ 45 ˆ 45
1 1 3
2 e i /3
1 1 3
1 3
2 2 cos / 3 3
2 e i /3
0 1 1 1 HV 1 0 HV 3 2 ei /3 HV
2 ei /3 HV 1 HV
2 ei /3 2 e i /3
2 . 3
ˆ 45 2 1ˆ from Example 5.4, so
ˆ 45 2 ˆ 45 45
2 1/ 2
1/ 2
7 2 1 0.882 . 3 9 The commutator is given in Example 5.4 as
0 2 ˆ HV , ˆ 45 . 2 0 HV The expectation value of the commentator is
12 12 14 .
Chapter 5
1 1 3
ˆ HV , ˆ 45
2 e i /3
2 2 ei /3 HV 2 HV
1 1 3
2 3
2 2 2i sin / 3 3
i
2 e i /3
0 2 1 1 HV 2 0 HV 3 2 ei /3 HV
2 ei /3 2 e i /3
2 2 i1.633. 3
The indeterminacy relation then says HV 45
1 ˆ HV , ˆ 45 2
1 i1.633 2 0.832 0.817 , so the relation is satisfied.
0.943 0.882
5.20
(Sec 5.5) The matrix representations are given in Problems 5.2 and 5.12. In Problem 5.9 we found HV sin 2 for this state. ˆC ˆC cos sin e i
0 i cos HV i 0 HV sin ei HV
i cos sin e i
sin ei HV cos HV
i cos sin ei cos sin e i i cos sin i 2sin sin 2 sin .
0 i 0 i 1 0 ˆ ˆ C2 1 i 0 i 0 0 1 HV HV HV
57
58
Chapter 5
ˆ C2 ˆC C
2 1/2
1 sin 2 2 sin 2
1/ 2
ˆ HV , ˆ C 2i ˆ 45 . In Problem 5.14 you showed that ˆ HV , ˆ C 2i ˆ 45 2i cos sin e i
0 1 cos HV 1 0 HV sin ei HV
2i cos sin e i
sin ei HV cos HV
2i cos sin ei cos sin e i 2i cos sin 2 cos 2i sin 2 cos .
The indeterminacy relation is HV C
1 ˆ HV , ˆ C . Substituting in 2
1 2i sin 2 cos 2 sin 2 2 1 sin 2 2 sin 2 sin 2 2 cos 2 sin 2 1 sin 2 2 sin 2
1/2
1 sin 2 sin cos 1 sin 2 sin 1 sin 2
2
2
2
2
2
sin 2 2 sin 2 sin 2 This is true, so the indeterminacy relation is satisfied.
Chapter 5
5.21
(Sec 5.5) The matrix representations are given in Problems 5.11 and 5.12. ˆ 45 ˆ 45 cos sin e i
0 1 cos HV 1 0 HV sin ei HV
cos sin e i
sin ei HV cos HV
cos sin ei cos sin e i cos sin 2 cos sin 2 cos ˆ 45 2 1ˆ from Example 5.4, so
ˆ 45 2 ˆ 45 45
2 1/2
1 sin 2 2 cos 2
1/ 2
.
ˆC ˆC cos sin e i
0 i cos HV i 0 HV sin ei HV
i cos sin e i
sin ei HV cos HV
i cos sin ei cos sin e i i cos sin i 2sin sin 2 sin .
0 i 0 i 1 0 ˆ ˆ C2 1 i 0 i 0 0 1 HV HV HV
ˆ C2 ˆC C
2 1/2
1 sin 2 2 sin 2
1/ 2
59
60
Chapter 5
ˆ 45 , ˆ C ˆ 45 ˆ C ˆ C ˆ 45 0 1 0 i 0 i 0 1 1 0 HV i 0 HV i 0 HV 1 0 HV 1 0 1 0 i i 0 1 HV 0 1 HV 1 0 2i 0 1 HV ˆ HV 2i ˆ HV cos 2 for this state, so In Problem 5.9 we found
ˆ 45 , ˆ C i 2 ˆ HV i 2 cos 2 The indeterminacy relation is 45 C
1 ˆ 45 , ˆ C . Substituting in 2
1 sin 2 cos 1 sin 2 sin 12 i2 cos 2 1 sin 2 cos 1 sin 2 sin cos 2 1 sin 2 cos sin sin 2 sin cos cos 2 2
2
2
2
2
2
1/2
2
2
2
2
1/2
2
4
2
2
2
2
1 sin 2 2 sin 4 2 sin 2 cos 2 cos 2 2 cos 2 2 sin 4 2 sin 2 cos 2 cos 2 2 sin 4 2 sin 2 cos 2 0
This is true, so the indeterminacy relation is satisfied. 5.22
(Sec 5.5) The matrix representations are given in Problems 5.11 and 5.12. Use the 45 state as the eigenstate. ˆ 45 45 ˆ 45 45 1 45 45 1 ˆ 45 2 1ˆ from Example 5.4, so
Chapter 5
ˆ 45 2 ˆ 45 45 1 1
1/ 2
2 1/2
0.
ˆC ˆC 0 i 1 1 1 1 1 HV 2 i 0 HV 2 1 HV 1 1 i 1 1 HV 2 1 1 i 1 1 2 0.
0 i 0 i 1 0 ˆ ˆ C2 1 0 0 0 1 i i HV HV HV
ˆ C2 ˆC C 1 0
1/2
2 1/ 2
1
ˆ 45 , ˆ C ˆ 45 ˆ C ˆ C ˆ 45 0 1 0 i 0 i 0 1 1 0 HV i 0 HV i 0 HV 1 0 HV 1 0 1 0 i i 0 1 HV 0 1 HV 1 0 2i 0 1 HV ˆ HV 2i
61
62
Chapter 5 ˆ 45 , ˆ C i 2 ˆ HV 1 0 1 1 1 1 1 HV 2 0 1 HV 2 1 HV 1 i 1 1 HV 1
i2
i 1 1 0.
The indeterminacy relation is 45 C
1 ˆ 45 , ˆ C . Substituting in 2
1 0 2 00
0 1
This is true, so the indeterminacy relation is satisfied. 5.23* (Sec 5.5) a
ba bb
b
0 ba ba ab ba b a b a b a b a bb bb bb bb ba ab ab ba ab a a bb bb
a a 2 a a 0
a a bb ab
2
a a bb ab
2
0
ab bb ab bb
2
2
ab bb
2
2
bb
bb
2
Chapter 5 5.24
(Sec 5.5) ˆ ˆ BA ˆˆ Aˆ , Bˆ AB
Aˆ , Bˆ
ˆ ˆ BA ˆ ˆ AB †
†
†
ˆ ˆ BA ˆˆ AB
†
Bˆ † Aˆ † Aˆ † Bˆ † ˆˆ ˆ ˆ AB BA Aˆ , Bˆ We’ve used the fact that the operators are Hermitian. 5.25
(Sec 5.5) Cˆ i Aˆ , Bˆ
Cˆ † i Aˆ , Bˆ
†
ˆ ˆ BA ˆ ˆ i AB ˆ ˆ BA ˆˆ i AB
†
†
†
i Bˆ † Aˆ † Aˆ † Bˆ † ˆ ˆ ˆ ˆ AB i BA i Aˆ , Bˆ Cˆ
63
Complement 5.A Measuring a Quantum State 5.A.1 We'll express the state in the HV-basis, using H 1 ei V . . First we determine , H
2
P H | 0.5 1/ 2 .
Substituting this into Eq. (5.A.6), we find P (45 | )
1 1 1 1 2 1 cos 2 2 2 1 1 cos . 2
Solving for cos yields cos 2 P(45 | ) 1 2 0.75 1 0.50 . The two solutions for are then cos 1 0.5 / 3. As in Example 5.A.1, if 0 , P ( L | ) 0.5 , and if 0 , P ( L | ) 0.5 . Our data then indicate that we should choose the positive solution, which is = / 3 . The final solution is thus
1 H ei/3 V 2
.
65
66
Complement 5.A
5.A.2 We'll express the state in the HV-basis, using H 1 ei V . . First we determine , P( H | ) 0.67 2 / 3 (to within our precision).
Substituting this into Eq. (5.A.6), we find P (45 | )
1 2 2 1 2 1 cos 2 3 3 1 2 2 cos . 1 2 3
Solving for cos yields 2 2 cos 2 P(45 | ) 1 3 2 0.5 1 0. The two solutions for are then cos 1 0 0.54 rad = / 2. As in Example 5.A.1, if 0 , P ( L | ) 0.5 , and if 0 , P ( L | ) 0.5 . Our data then indicate that we should choose the positive solution, which is = / 2 . The final solution is thus
2 1 H i V . 3 3
5.A.3 The state of the photons is very nearly H . Since the coefficients in the state are given by the square root of the probabilities, the error in the coefficients will be of order 5%.
0.003 0.05 , or
Complement 5.A
67
While a good experimentalist always makes independent measurements (if possible) to verify his or her answer, making measurements at other device settings will probably not improve this answer very much. 5.A.4 A general linear polarization state can be written as cos H sin V . They key here is that the coefficients are real. We have been representing the H 1 ei V . Since the coefficients are real, we know that 0, .
state
as
So, we only need to perform measurements at two different device settings. Measurements with a PAHV determine , and measurements with a PA45 can distinguish between the two possible values of .
Chapter 6 Spin-1/2 6.1
(Sec. 6.1) F μB z
Bz Bz e z 2 z
The atoms are traveling in the y-direction at v=500 m/s, entering the magnetic field at y=0, and leaving it at y=0.035m. Assume no transverse velocity. The amount of time the atoms spend in the field is t y/v. The
acceleration
of
m (107.9u) 1.66 10
27
the
atoms
kg/u 1.79 10
in 25
the
z-direction
is
a F /m,
where
kg is the mass of a silver atom. The amount of the
deflection is then 2
1 1 F y d at 2 2 2 m v
e
Bz 2 2 2 z y e y Bz 2m v 4mv 2 z
1.76 10 s T 1.06 10 Js 0.035m 10 T/m 1.3 10 d 4 1.79 10 kg 500m/s 2
34
11 -1 -1
3
2
25
4
m 0.13mm
Negative sign is because atoms with spin-up (dipole moment down, because dipole moment comes from unpaired electrons) in an upward field gradient would be deflected downward. 6.2*
(Sec. 6.1)
31 11 -1 -1 2m e 2 9.1110 kg 1.76 10 s T ge 2.00 e 1.60 1019 C
In Dirac’s relativistic theory the g-factor is exactly 2. However, quantum electrodynamics yields corrections on the order of 0.1% to this value.
69
70
Chapter 6
6.3
(Sec. 6.2) A Hermitian operator can be written as the sum of the projection operators onto its eigenstates, weighted by the corresponding eigenvalues. Sˆ z z z z z . 2
6.4
(Sec. 6.3) 1 1 z z , x z z 2 2 z 1ˆ z x x x x z
We know: x
1 x x 2 1 z x z x x z x x x 2 z x z x x z x
6.5
(Sec. 6.3) z Sˆ x z
Sˆ x z Sˆ x z
z Sˆ x z z Sˆ x z z
Using the states found in Problem 6.4:
1 ˆ 1 Sˆ x z S x x Sˆ x x x x z 2 2 2 2 1 ˆ 1 Sˆ x z S x x Sˆ x x x x z 2 2 2 2 0 1 Sˆ x σx 2 1 0 z 2 6.6
(Sec. 6.3) 0 1 Sˆ x 2 1 0 z
Chapter 6 0
/2
/2
0
/ 2 2
2
0 eigenvalues are / 2 , / 2 . For / 2 / 2 / 2 0 / 2 / 2 z z 0 z / 2 0
normalized eigenstate:
1 1 x 2 1 z
For / 2 / 2 / 2 0 / 2 / 2 z z 0 z / 2 0
normalized eigenstate: 6.7
1 1 x 2 1 z
(Sec. 6.3) 1 1 z i z , y z i z 2 2 z 1ˆ z y y y y z
We know: y
1 y y 2 1 i z y z y y z y i y i y y y 2 2
z y z y y z y
6.8*
(Sec. 6.3) z Sˆ y z
Sˆ y z Sˆ y z
z Sˆ y z ˆ z S y z
z
71
72
Chapter 6 Using the result of Problem 6.7:
1 ˆ 1 Sˆ y z S y y Sˆ y y y y i z 2 2 2 2 i ˆ i Sˆ y z S y y Sˆ y y y y i z 2 2 2 2 0 i Sˆ y σy 2 i 0 z 2 6.9
(Sec 6.3) This is most easily done use the appropriate Pauli spin matrix: 0 i 1 0 Sˆ y z i z 2 i 0 z 0 z 2 i z 2 0 i 0 i Sˆ y z i z 2 i 0 z 1 z 2 0 z 2
6.10
(Sec. 6.3) 0 i Sˆ y 2 i 0 z
0 i / 2 i / 2
0
/ 2 2
2
0 eigenvalues are / 2 , / 2 . For / 2 / 2 i / 2 0 i / 2 / 2 z z 0 z / 2 i 0
i
normalized eigenstate: For / 2
1 1 y 2 i z
Chapter 6 / 2 i / 2 0 i / 2 / 2 z z 0 z / 2 i 0 i
normalized eigenstate: 6.11
1 1 y 2 i z
(Sec. 6.3) 1 1 z i z z z P S y / 2 | x y x 2 2 1 1 1 2 P S y / 2 | x 1 i 12 12 4 4 2
6.12
2
2
(Sec. 6.3) 1 1 z z z i z P S x / 2 | y x y 2 2 1 1 1 2 P S x / 2 | y 1 i 12 12 4 4 2 2
6.13* (Sec. 6.3)
un corresponds to angles and . Therefore
2
73
74
Chapter 6 n cos / 2 z ei () sin / 2 z cos / 2 / 2 z ei () sin / 2 / 2 z sin / 2 z ei ( ) cos / 2 z sin / 2 z ei cos / 2 z
6.14
(Sec. 6.3) Need the result of Problem 6.13 for this.
n n cos / 2 z ei sin / 2 z
sin / 2 z e
i
cos / 2 z
cos / 2 sin / 2 sin / 2 cos / 2 0 6.15
(Sec. 6.3) Need the result of Problem 6.13 for this.
Sˆn n n n n 2 z Sˆn z z Sˆn z Sˆn ˆ z z Sˆ z z S n n z 2 z Sˆn z z n n z z n n z z n z n 2 2 cos 2 / 2 sin 2 / 2 cos 2 2 z Sˆn z z n n z z n n z 2 cos / 2 e i sin / 2 sin / 2 e i cos / 2 2 e i 2 cos / 2 sin / 2 e i sin 2 2 * z Sˆn z z Sˆn z ei sin (the operator is Hermitian) 2 2 z Sˆn z z n n z z n n z z n z n 2 2 sin 2 / 2 cos 2 / 2 cos 2 2
2
2
Chapter 6
75
i cos e sin ˆ Sn 2 ei sin cos z
6.16
(Sec. 6.3) P Sn / 2 | z n z
6.17
2
cos / 2 z ei sin / 2 z z
2
sin 2 / 2
(Sec. 6.3) Need the result of Problem 6.13 for this. P Sn / 2 | x n x
2
1 sin / 2 z e i cos / 2 z z z 2
2 1 sin / 2 e i cos / 2 2 1 sin 2 / 2 cos 2 / 2 sin / 2 cos / 2 e i ei 2 1 1 sin cos 2
6.18
(Sec. 6.3) Sˆ z n Sˆ z n
cos / 2 e i sin / 2
z
cos / 2 ei sin / 2 2
1 0 cos / 2 2 0 1 z ei sin / 2 z
cos / 2 z ei sin / 2 z
2 cos / 2 sin 2 / 2 2 cos 2
2 1 0 1 0 2 1 0 2 ˆ Sˆ z 2 1 4 0 1 z 0 1 z 4 0 1 z 4
2 Sˆ z 2 4
2
76
Chapter 6 The standard deviation is then S z Sˆ z 2 Sˆ z
2 1/2
1/2
2 2 cos 2 4 4
1/2 1 1 sin 2 sin . 2 2
6.19* (Sec. 6.3) Need the result of Problem 6.13 for this. Sˆ x n Sˆ x n
sin / 2 e i cos / 2
z
sin / 2 e i cos / 2 2
0 1 sin / 2 2 1 0 z ei cos / 2 z
ei cos / 2 z sin / 2 z
i e sin / 2 cos / 2 ei sin / 2 cos / 2 2 sin / 2 cos / 2 ei e i 2 sin cos 2
2 0 1 0 1 2 1 0 2 ˆ 2 ˆ Sx 1 4 1 0 z 1 0 z 4 0 1 z 4
2 Sˆ x 2 4 The standard deviation is then S x Sˆ x 2 Sˆ x
2 1/2
1/2
2 2 sin 2 cos 2 4 4
1/2 1 sin 2 cos 2 . 2
Chapter 6
6.20* (Sec. 6.3) 1 3 z i z 2 2
The probability of the first measurement is P S z / 2 | z
2
3 . 4
After the measurement, the electron would be left in state z . The probability of the second measurement is P S x / 2 | z x z The joint probability of / 2 and / 2 is then P S z / 2, S x / 2 | P S z / 2 | P S x / 2 | 3 1 3 4 2 8
6.21
(Sec. 6.4) 0 1 Sˆ x 2 1 0 z
0 i Sˆ y 2 i 0 z
1 0 Sˆ z 2 0 1 z
0 1 1 0 1 0 0 1 Sˆ x , Sˆ z 2 1 0 2 0 1 2 0 1 2 1 0 z z z z 0 1 0 1 2 0 1 1 0 1 0 z 2 1 0 z z 2 0 i ˆ i i S y 2 i 0 z
6.22
2 4
(Sec. 6.4)
1 3 z i z 2 2
Sˆ x , Sˆ z iSˆ y
S x S z
1 Sˆ x Sˆ x 1 i 3 2
1 i 3 8
z
z i 13
ˆ Sy 2 0 1 1 1 0 z 2 i 3 z
z 2 1 i 8
3 3 0
2
1 . 2
77
78
Chapter 6 2 0 1 0 1 2 1 0 2 ˆ 1 Sˆ x 2 4 1 0 z 1 0 z 4 0 1 z 4
2 Sˆ x 2 4 The standard deviation is then S x Sˆ x 2 Sˆ x
2 1/2
1/2
2 0 4
1 Sˆ z Sˆ z 1 i 3 2
0 1 1 1 z 2 i 3 z
z 2 0 1
1 1 3 4 3 z 8 2 1 0 1 0 2 1 0 2 ˆ 1 4 0 1 z 0 1 z 4 0 1 z 4
Sˆ z 2
. 2
1 i 3 8
z i
2 Sˆ z 2 4 The standard deviation is then S z Sˆ z 2 Sˆ z
2 1/2
z 2 i
1 Sˆ y Sˆ y 1 i 3 2
1 i 3 8
1/2
2 2 4 16
3 i z 8
z
3 . 4
0 i 1 1 0 z 2 i 3 z
3 3
3 4
3 ˆ S x S z Sy 2 4 2 So, the indeterminacy relation is satisfied with an equality.
Chapter 6
79
6.23* (Sec. 6.5)
The SAz splits the incoming beam into two beams – spin-up z and spin-down z . For the top beam, it essentially acts as a projection operator z z , while for the bottom beam it acts as z z . The SA-z recombines the beams, without changing them, so the operator for the system is the sum of the top and bottom beam operators: Oˆ z z z z 1ˆ . The sum of the projection operators onto a complete set of basis states is the identity operator.
Chapter 7 Angular Momentum and Rotation 7.1*
(Sec. 7.2) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ AB, C ABC CAB ˆˆ ˆ ˆ CB Bˆ , Cˆ BC ˆˆ ˆ ˆ Bˆ , Cˆ CB BC
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ AB , C A B, C CB CAB ˆ ˆ ˆ CAB ˆˆ ˆ Aˆ Bˆ , Cˆ ACB ˆ ˆ CA ˆˆ Aˆ , Cˆ AC ˆ ˆ Aˆ , Cˆ CA ˆˆ AC
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ AB , C A B, C A, C CA B CAB Aˆ Bˆ , Cˆ Aˆ , Cˆ Bˆ 7.2*
(Sec. 7.2)
Jˆ Jˆ Jˆ Jˆ Jˆ Jˆ Jˆ 2
†
2
2
x
2
x
2
y
†
z
†
2
†
y
2
†
z
Jˆ x 2 Jˆ y 2 Jˆ z 2 Jˆ 2
81
82
Chapter 7
7.3
(Sec. 7.2) Jˆ 2 , Jˆ x Jˆ x 2 Jˆ y 2 Jˆ z 2 , Jˆ x Jˆ x 2 , Jˆ x Jˆ y 2 , Jˆ x Jˆ z 2 , Jˆ x 0
Using Eq. (7.8): Jˆ 2 , Jˆ x Jˆ y Jˆ y , Jˆ x Jˆ y , Jˆ x Jˆ y Jˆ z Jˆ z , Jˆ x Jˆ z , Jˆ x Jˆ z iJˆ Jˆ iJˆ Jˆ iJˆ Jˆ iJˆ Jˆ y z
z y
z y
y z
0. Jˆ 2 , Jˆ y Jˆ x 2 Jˆ y 2 Jˆ z 2 , Jˆ y Jˆ x 2 , Jˆ y Jˆ y 2 , Jˆ y Jˆ z 2 , Jˆ y 0
Jˆ x Jˆ x , Jˆ y Jˆ x , Jˆ y Jˆ x Jˆ z Jˆ z , Jˆ y Jˆ z , Jˆ y Jˆ z iJˆ x Jˆ z iJˆ z Jˆ x iJˆ z Jˆ x iJˆ x Jˆ z 0. 7.4
(Sec. 7.3) Jˆ Jˆ x iJˆ y σx i σ y 2 2 0 1 0 i i 2 1 0 i 0 0 1 . 0 0
z z z z
z z
z z z z
z z z
z
z z z z
0 1 . 0 0 z
Chapter 7
7.5
(Sec. 7.3) Jˆ Jˆ x iJˆ y
Jˆ Jˆ x iJˆ y
Invert these equations: 1 1 Jˆ x Jˆ Jˆ Jˆ y i Jˆ Jˆ 2 2
7.6
(Sec. 7.4) 1/2 Jˆ 1,1 1(1 1) 1(1 1) 1,1 1
0 1/2 Jˆ 1, 0 1(1 1) 0(0 1) 1, 0 1
2 1,1 1/2 Jˆ 1, 1 1(1 1) (1)(1 1) 1, 1 1
2 1, 0
1,1 Jˆ 1,1 Jˆ 1, 0 Jˆ 1,1 ˆ 1, 1 J 1,1
1,1 Jˆ 1, 0 1, 0 Jˆ 1, 0
1, 1 Jˆ 1, 0
1,1 Jˆ 1, 1 0 1 0 ˆ 1, 0 J 1, 1 2 0 0 1 . 1, 1 Jˆ 1, 1 0 0 0
1/2 Jˆ 1,1 1(1 1) 1(1 1) 1,1 1
2 1, 0 1/2 Jˆ 1, 0 1(1 1) 0(0 1) 1, 0 1
2 1, 1 1/2 Jˆ 1, 1 1(1 1) (1)(1 1) 1, 1 1
0
1,1 Jˆ 1,1 Jˆ 1, 0 Jˆ 1,1 ˆ 1, 1 J 1,1
1,1 Jˆ 1, 0 1, 0 Jˆ 1, 0 1, 1 Jˆ 1, 0
1,1 Jˆ 1, 1 1, 0 Jˆ 1, 1 1, 1 Jˆ 1, 1
0 0 0 2 1 0 0 . 0 1 0
83
84
Chapter 7
7.7*
(Sec. 7.4) From Problem 7.5
1 Jˆ x Jˆ Jˆ 2
1 Jˆ y i Jˆ Jˆ 2
Use the matrix representations from Problem 7.6
Sˆx
Sˆ y
7.8*
0 2 0 2 0 0 i 2 1 2 0
1 0 0 2 0 1 1 2 0 0 0 0 0 0 i 2 0 0 0 2 1 0 0
0 0 0 0 0 1 2 1 0 0 1 0 0 0 1 i 2 0 0 0
1 0 0 1 1 0 i 0 0 i i 0
(Sec. 7.4)
Sˆ x k k k
k
a b c z
Using the Matrix representation from Problem 7.7: 0 0 2 a a 0 b k b 2 c 2 c z z 0 0 2 z k 2 0
2
k 2
0 a b 0 2 c z k z
Set determinant of matrix equal to 0 to find the eigenvalues:
Chapter 7
k
2
2
k
0
2
k
k
2
k
0
0 2
2 2 k 0
2
k k 2
2 00 k
2 k 0 2 2 2
2 2 2 k k k 0 2 2
k 2 k 2 0 Eigenvalues are (not surprisingly): k , 0, Eigenstate for k 0 is 1, 0 x .
0
2
2
0
0
2
0 a b 0 2 c z 0 z
b0 2 a c0 2 2
Only use 2 equations (third is always redundant) Solution: a=–c, b=0
85
86
Chapter 7 1 1, 0 x a 0 1 z Normalize it:
x
1, 0 1, 0
1 a 1 0 1 z 0 1 z 2
x
2a
2
1 Therefore 1 a 2 1, 0
x
1 1 1 0 1,1 1, 1 2 2 1 z
Other eigenvalues k
2
2
0
2
a
0 a b 0 2 c z z
b0 2
b c 0 2
b 2a 1 c ba 2 1 1, 1 x a 2 1 z Normalize them:
Chapter 7 1 1 1 1, 1 x 2 1,1 2 1, 0 1, 1 2 2 1 z
7.9*
(Sec. 7.4)
Sˆ y k k k i
2
a a i b k b 2 c c z z 0 z
2
i
0
k
0
0 i
2
i
2
0
k i 2 0
a i b 0 2 c z k z 0
k 2
i
a b c z
Set determinant of matrix equal to 0 to find the eigenvalues: k i
2
i
2
k
0
i
k
i
2
k i
0
2
0 2
k
i
2
k
i i 2 2 0
2 00 k
i
87
88
Chapter 7 2 2 k k i k i 0 2 2 2 2
2 k 2 k 0 2 2
k
k 2 k 2 0
Eigenvalues are (not surprisingly): k , 0, Eigenstate for k 0 is 1, 0 y . i
0 2
0
2
i
0 i
2
a i b 0 2 c z 0 z 0
b0 2 i a i c0 2 2 i
Only use 2 equations (third is always redundant) Solution: a=c, b=0 1 1, 0 y a 0 1 z Normalize it:
y
1, 0 1, 0
1 a 1 0 1 z 0 1 z 2
y
2a 1 Therefore
2
Chapter 7 a
1, 0
1 2
y
1 1 1 0 1,1 1, 1 2 2 1 z
Other eigenvalues k i 2 0 a i i
i
a i b 0 2 c z z
2
0
i
2
b0 2
b c 0 2
b i 2 a i c b a 2 1 1, 1 y a i 2 1
z
Normalize them: 1 1 1 1, 1 y i 2 1,1 i 2 1, 0 1, 1 2 2 1 z
7.10
(Sec. 7.4) The first measurement prepares the particle in state 1,1 y .
P S z 0 | 1,1
y
1, 0 1,1
2 y
89
90
Chapter 7
Using, 1,1
y
1 1,1 i 2 1, 0 1, 1 2
P S z 0 | 1,1
from Problem 7.9 yields
y
y
1 1,1 1,1 2i 1,1 1, 0 1,1 1, 1 2
2
1 1, 0 1,1 2i 1, 0 1, 0 1, 0 1, 1 2
2
1 2
Similarly:
P S z | 1,1
P S z | 1,1 7.11
y
1 4
1 1, 1 1,1 2i 1, 1 1, 0 1, 1 1, 1 2
2
1 4
(Sec. 7.4) First measurement prepares particle in state 1, 0 y .
1, 1 1, 0 1 1, 1 x 1,1 2 1, 0 1, 1 2
P S x | 1, 0 Using
2
y
x
y
and 1, 0
y
1 1,1 1, 1 2
from Problems 7.8
and 7.9 yields
P S x | 1, 0
y
2
1 1 1 1 1 . 2 2 2 2 2
7.12* (Sec. 7.4) 1/ 3 2 1,1 i 1, 0 2 1, 1 P S y | Using 1, 1
y
y
1, 1
2
1 1,1 i 2 1, 0 1, 1 2
from Problem 7.9 yields 2
2
2 2 2 2 1 2 2 1 1 2 P S y | i 0.0556 i 6 6 6 36 2 3 2 3 2 3 7.13* (Sec. 7.4)
1/ 3 2 1,1 i 1, 0 2 1, 1
Chapter 7 1 Sˆ 2 1,1 2 1, 0 i 1, 1 2 Sˆ 2 2 1,1 i 1, 0 2 1, 1 9 1 2 (1)(1 1) 1,1 2 1, 0 i 1, 1 2 2 1,1 i 1, 0 2 1, 1 9 2 2 4 1 4 2 2 9 1 Sˆz 1,1 2 1, 0 i 1, 1 2 Sˆz 2 1,1 i 1, 0 2 1, 1 9 1 1,1 2 1, 0 i 1, 1 2 2 1,1 2 1, 1 9 2 2 4 4 0 9 Using the matrix for Sˆ y found pin Problem 7.7: 0 i 0 2 ˆ Sy 2 i 2 i 0 i i 9 2 0 i 0 2 1 2 i 2 0 9 2 1 2 2 0 9 2
7.14
(Sec. 7.5) ˆ i /2 / Rˆ , uz z e iJ z / z e z e i/2 z ˆ i /2 / Rˆ , uz z e iJ z / z e z ei/2 z
z Rˆ , uz z
Rˆ , uz z Rˆ , uz z
e i/2 0
0 ei/2
z Rˆ , uz z z Rˆ , uz z
91
92
Chapter 7
7.15
(Sec. 7.5)
ˆ
iJ / Rˆ , u y z e y z
Using the result of Problem 6.7, we can write this as: iJˆ / 1 Rˆ , u y z e y y y 2 1 i( /2)/ e y e i( /2)/ y 2 1 i/2 1 1 z i z ei/2 z i z e 2 2 2 1 e i/2 ei/2 z i e i/2 ei/2 z 2 cos / 2 z sin / 2 z
ˆ
iJ / Rˆ , u y z e y z
Again using the result of Problem 6.7, we can write this as: iJˆ / i Rˆ , u y z e y y y 2 i ei( /2)/ y e i( /2)/ y 2 1 i i/2 1 z i z e i/2 z i z e 2 2 2 i ei/2 e i/2 z i ei/2 ei/2 z 2
sin / 2 z cos / 2 z
z Rˆ , u y z z Rˆ , u y z ˆ R , u y z Rˆ , u z ˆ z R , u y z y cos / 2 sin / 2 sin / 2 cos / 2
Chapter 7
7.16* (Sec. 7.5)
First apply Rˆ , u y to z . Using the result of Problem 7.15 we get cos / 2 sin / 2 1 cos / 2 Rˆ , u y z sin / 2 cos / 2 0 sin / 2
Apply Rˆ , uz to this. Using the result of Problem 7.14: e i/2 Rˆ , uz Rˆ , u y z 0
0 cos / 2 ei/2 sin / 2
e i/2 cos / 2 cos / 2 e i/2 i e sin / 2 ei/2 sin / 2 Apart from the overall phase, this is n cos / 2 z ei sin / 2 z
93
94
Chapter 7
7.17
(Sec. 7.5) ˆ Rˆ , ux z e iJ x / z ˆ 1 e iJ x / x x 2 1 i( /2)/ x ei( /2)/ x e 2 1 i/2 1 1 z z ei/2 z z e 2 2 2 1 i i z i i z 2 i z
Makes sense–should be spin down along z-axis. 7.18
(Sec. 7.5) ˆ Rˆ , uz 1,1 e iJ z / 1,1 ei / 1,1 e i 1,1 ˆ Rˆ , uz 1, 0 e iJ z / 1, 0 e i0/ 1, 0 1, 0 ˆ Rˆ , uz 1, 1 e iJ z / 1, 1 e i( )/ 1, 1 ei 1, 1
7.19
(Sec. 7.5) First we need to write 1,1 in the x-basis. Using the results of Problem 7.8: 1 1,1 1, 1 x 2 1 1, 1 x 1,1 2 1, 0 1, 1 2 1, 0
1,1 x 1, 1 x 1,1 1, 1 1,1 x 2 1, 0 1,1
x
1, 1 x 2 1,1
1 1,1 x 2 1, 0 2
x
1, 1
x
Chapter 7
ˆ 1 1,1 x 2 1, 0 x 1, 1 x Rˆ / 2, ux 1,1 e iJ x /2 2 1 e i/2 1,1 x 2 1, 0 x ei/2 1, 1 2 1 i 1,1 x 2 1, 0 x i 1, 1 x 2 i 1,1 x i 2 1, 0 x 1, 1 x 2
x
i 1 1 Rˆ / 2, ux 1,1 1,1 2 1, 0 1, 1 i 2 1,1 1, 1 2 2 2 1 1,1 2 1, 0 1, 1 2 i 2 1, 0 i 1,1 1, 1 2 1 1,1 i 2 1, 0 1, 1 2 1, 1 y
Where I’ve used the result of Problem 7.9. 7.20
(Sec. 7.5) First we need to write 1,1 in the x-basis. Using the results of Problem 7.8: 1 1,1 1, 1 2 1 1, 1 x 1,1 2 1, 0 1, 1 2 1, 0
x
1,1 x 1, 1 x 1,1 1, 1 1,1 x 2 1, 0 1,1
x
1, 1 x 2 1,1
1 1,1 x 2 1, 0 2
x
1, 1
x
95
96
Chapter 7
ˆ 1 1,1 x 2 1, 0 x 1, 1 x Rˆ , ux 1,1 e iJ x / 2 1 e i 1,1 x 2 1, 0 x ei 1, 1 x 2 1 1,1 x 2 1, 0 x 1, 1 x 2 1 1 1 Rˆ , ux 1,1 1,1 2 1, 0 1, 1 2 1,1 1, 1 2 2 2 1 1,1 2 1, 0 1, 1 2 1 1,1 1, 1 1,1 1, 1 2 1, 1
Complement 7.B Eigenvalues and Eigenstates of Angular Momentum 7.B.1
Jˆ† Jˆ x iJˆ y
†
Jˆ x† iJˆ †y Jˆ x iJˆ y Jˆ ,
where I’ve used the fact that the Jˆ x and Jˆ y are Hermitian.
Jˆ† Jˆ x iJˆ y
†
Jˆ x† iJˆ †y Jˆ x iJˆ y Jˆ .
7.B.2 Jˆ , Jˆ Jˆ x iJˆ y , Jˆ x iJˆ y Jˆ x , Jˆ x i Jˆ y , Jˆ x i Jˆ x , Jˆ y Jˆ y , Jˆ y 0
0
Jˆ z Jˆ z 2Jˆ z
7.B.3* j , m j Jˆ Jˆ j , m j j , m j Jˆ† Jˆ j , m j
Using Eq. (7.B.19): j , m j Jˆ Jˆ j, m j c * c 2 j , m j 1 j , m j 1 2
c 2 Using Eq. (7.B.16):
97
98
Complement 7.B
j , m j Jˆ Jˆ j , m j j , m j Jˆ 2 Jˆ z2 Jˆ z
j, m
j
2 j ( j 1) m j 2 m j j , m j j , m j 2 j ( j 1) m j (m j 1) c
2
j ( j 1) m j (m j 1) 1/2
c j ( j 1) m j (m j 1)
1/2 Jˆ j , m j j ( j 1) m j (m j 1) j, m j 1
7.B.4* Using Eq. (7.B.14):
Jˆ z Jˆ j , m j
Jˆ Jˆz
j, m j
Jˆ m j j , m j
m j 1 Jˆ j , m j
Jˆ j , m j is an eigenstate of Jˆ z with eigenvalue m j 1 , so Jˆ j , m j c j , m j 1 . j , m j Jˆ Jˆ j , m j j , m j Jˆ† Jˆ j , m j c * c 2 j , m j 1 j , m j 1 2
c 2
Using Eq. (7.B.15): j , m j Jˆ Jˆ j , m j j , m j Jˆ 2 Jˆ z2 Jˆ z
j, m
j
2 j ( j 1) m j 2 m j j , m j j , m j 2 j ( j 1) m j (m j 1) 2 c j ( j 1) m j (m j 1) 1/2
c j ( j 1) m j (m j 1)
Chapter 8 Two Particle Systems and Entanglement 8.1
(Sec. 8.1) ˆ sHV V , L ˆ sHV V , L V , L (1) V , L 1
ˆ iHV V , L ˆ iHV V , L s V
i
s V
i
s V
i
1 ˆ iHV V L H i i V s 2 1 L V s H i i V i 2 L V
s
i
R i 0
ˆ siHV V , L ˆ sHV ˆ iHV V , L
s
ˆ sHV V V
s i
1 ˆ iHV L H i i V 2
1 (1) i L H i i V 2 (1) i L R i 0
8.2
i
i
(Sec. 8.1)
1 2 ˆ sHV , 45 H H , 45 3 3
s 1 2 ˆ HV , 45 H H , 45 3 3
1 2 2 2 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 3 3 3 3 1 2 1 3 3 1 1 ˆ iHV 45 i H i V i H i V i 45 i 2 2 1 1 ˆ iHV 45 i H i V i H i V i 45 i 2 2
ˆ iHV ˆ iHV
99
100
Chapter 8 Therefore: 1 2 ˆ iHV 3 H , 45 3 H , 45
i 1 2 ˆ HV 3 H , 45 3 H , 45
1 2 2 2 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 3 3 3 3 2 2 2 2 3 3 3 1 si 1 2 2 ˆ HV , 45 , 45 H , 45 H , 45 H H 3 3 3 3
ˆ siHV
1 2 2 2 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 H , 45 3 3 3 3
8.3
2 2 2 2 ˆ iHV 3 3 3
(Sec. 8.1) P 45s , 45i Pˆ45S ,45i 1 2 H , 45 H , 45 3 3
45, 45 45, 45
1 2 H , 45 45, 45 H , 45 45, 45 3 3
1 1 1 0 3 2 1 6
2
2
1 2 H , 45 H , 45 3 3
Chapter 8
8.4
101
(Sec. 8.1) There are two possibilities: Vs , H i and H s ,Vi . Calculate the probabilities and add them. P Vs , H i PˆVs , Hi R, 45 V , H V , H R, 45 2
R, 45 V , H
s
RV
s i
45 H
2 i
2
i 1 1 . 4 2 2
P H s , Vi PˆH s ,Vi R, 45 H , V
H , V R, 45
R, 45 H , V
2
s
R H
s i
45 V
2 i
2
1 1 1 . 4 2 2
P P Vs , H i P H s , Vi 8.5*
1 . 2
(Sec. 8.2) a H , H b V ,V
Assume: s s cHs H s
s
i
s
cVs V
i
cHs H
s
, i
s
i
cHi H i cVi V
cVs V
s
c
i H
i
H i cVi V
i
cHs cHi H , H cHs cVi H ,V cVs cHi V , H cVs cVi V , V . In order for this to be equal to a H , H b V , V , the coefficients of the H , V and V , H terms must be zero. Starting with the H , V term, this means cHs cVi 0 . This implies that either cHs 0 , or cVi 0 . If the first is true then the H , H
term will be absent, while the second is
102
Chapter 8 true then the V , V term will be absent. This gives a contradiction, so our assumption is wrong, and the state cannot be factorized.
8.6*
(Sec. 8.2) Remember that if the sum of the projection operators onto a set of states is equal to the identity operator, then the states form a basis.
1 H , H V ,V 1 H , H V ,V 2 2 1 H , H V ,V 1 H , H V ,V 2 2 1 H , H H , H H , H V ,V V ,V H , H V , V V ,V 2 1 H , H H , H H , H V , V V ,V H , H V ,V V ,V 2 H , H H , H V ,V V ,V
1 H ,V V , H 1 H ,V V , H 2 2 1 H ,V V , H 1 H ,V V , H 2 2 1 H , V H ,V H ,V V , H V , H H ,V V , H V , H 2 1 H ,V H ,V H , V V , H V , H H , V V , H V , H 2 H ,V H ,V V , H V , H
H , H H , H V ,V V ,V H ,V
H ,V V , H V , H
1ˆ
This is the identity, because we know that the states H , H , V , V , H , V basis, and hence the projection operators onto those states add to the identity.
and V , H are a
Chapter 8
8.7
103
(Sec 8.2) (a) P 45s Pˆ45s 1 45, 45 45, 45 45 s s 45 1 45, 45 45, 45 2 2 1 s 45 i 45 s 45 i 45 45 s s 45 45 s 45 i 45 s 45 i 2 1 s 45 45 s i 45 s 45 45 s i 45 s 45 45 s 45 i s 45 45 s 45 i 2 1 i 45 45 i 2 1 . 2
(b) Probabilities must be normalized. P 45s P 45s 1
P 45s 1 P 45s 1
1 1 . 2 2
(c) P 45i Pˆ45i 1 45, 45 2 1 s 45 i 45 2 1 i 45 45 i s 2 1 s 45 45 2 1 . 2
45, 45 45 s 45
i
45
1 45, 45 45, 45 2
45 45 i i 45 45
45 i 45 45 s
i i
i s
45 i 45
s
45 i 45 45
i
s
45
45 s i 45 45
i
i
45 s
104
Chapter 8 P 45s , 45i Pˆ45S , 45i 1 45, 45 45, 45 45, 45 45, 45 1 45, 45 45, 45 2 2 2 1 45, 45 45, 45 45, 45 45, 45 2 1 . 2 P 45s , 45i 1/ 2 P 45s | 45i 1 . 1/ 2 P 45i
8.8
(Sec. 8.2) (a) P H s PˆH s 1 45, 45 45, 45 H s s H 1 45, 45 45, 45 2 2 1 s 45 i 45 s 45 i 45 H s s H 45 s 45 i 45 s 45 i 2 1 s 45 H s i 45 s 45 H s i 45 s H 45 s 45 i s H 45 s 45 i 2 1 1 1 1 1 45 45 45 i 45 i i i 2 2 2 2 2
1 1 1 1 2 2 2 2
(b) Probabilities must be normalized. P H s P Vs 1
P Vs 1 P H s 1
1 1 . 2 2
Chapter 8
105
(c) P H i PˆHi 1 45, 45 45, 45 H i i H 1 45, 45 45, 45 2 2 1 s 45 i 45 s 45 i 45 H i i H 45 s 45 i 45 s 45 i 2 1 i 45 H i s 45 i 45 H i s 45 i H 45 i 45 s i H 45 i 45 s 2 1 1 1 1 1 45 s 45 45 45 s s s 2 2 2 2 2
1 1 1 1 2 2 2 2
P H s , H i PˆH S , Hi 1 45, 45 45, 45 H , H H , H 2 2 1 45, 45 H , H 45, 45 H , H 2
1 1 1 1 1 2 2 2 2 2 1 2 P H s | Hi
P H s , H i 1/ 2 1 . 1/ 2 P Hi
2
1 45, 45 45, 45 2
106
Chapter 8
8.9*
(Sec. 8.2) es cos s H eis sin s V
1 H , H V ,V 2
(a)
e ,e
P es , ei
s
i
es , ei
ei cos i H eii sin i V
1 H , H V ,V es , ei es , ei 1 H , H V ,V 2 2 1 H , H V , V es , ei es , ei H , H V , V 2 2 1 H es H ei V es V ei s i i s s i i 2 s 2 1 i cos s cos i e s i sin s sin i 2
(b)
P ei
e
i
ei
12 H , H
1 H , H V ,V ei ei 2 1 H , H V , V ei ei 2 1 s H H ei s V V ei i i i 2
H , H V , V i
ei H i
i
H
1 s H cos i s V eii sin i cos i H 2 1 1 cos 2 i sin 2 i 2 2
(c) Using the results from a) and b): P (es , ei ) P (es | ei ) P(ei ) 2 1 i cos s cos i e s i sin s sin i 2 1/ 2
cos s cos i e
i s i
V ,V
sin s sin i
2
s
s
i
e
ei V
ii
i
V s
sin i V s
Chapter 8
107
(d) Using the result from (c), for s i and s i P (es | ei ) cos cos ei sin sin cos 2 sin 2
2
2
1 For a measurement that finds the idler photon to be in an arbitrary elliptical polarization state, there is always a measurement that can be performed on the signal beam that will find the polarization of the signal to be perfectly correlated with the idler polarization. (e) If s i 0 , then (d) means that measurements of linear polarizations along the angles s i are perfectly correlated between the two beams. The fact that every possible linear polarization in one beam is perfectly correlated with the same linear polarization in the other is a fairly remarkable (and highly nonclassical) result. 8.10* (Sec. 8.3)
P Hs P Hs H , H
P H, H PH
s
V ,V
1 1 H , H PˆH s H , H V ,V PˆH s V ,V 2 2 1 1 H , H H s s H H , H V ,V H s 2 2 1 . 2
P Hi P Hi H , H
P H, H PH
i
V ,V
P V ,V
s
H V ,V
P V ,V
1 1 H , H PˆHi H , H V ,V PˆH i V , V 2 2 1 1 H , H H i i H H , H V ,V H i i H V ,V 2 2 1 . 2
108
Chapter 8
P H s , Hi P H s , Hi H , H
P H, H PH , H s
i
V ,V
P V ,V
1 1 H , H PˆH s , Hi H , H V ,V PˆH s , Hi V ,V 2 2 1 1 H , H H , H H , H H , H V ,V H , H H , H V ,V 2 2 1 . 2 P H s | Hi
P H s , H i 1/ 2 1 . 1/ 2 P Hi
These are the same probabilities determined in Example 8.5. 8.11
(Sec. 8.3) (a) A horizontally polarized signal photon has a 50% probability of passing through a linear polarizer oriented at +45°. (b) A horizontally polarized idler photon has a 50% probability of passing through a linear polarizer oriented at +45°. (c) There are four possibilities: both photons are detected, the signal is detected, the idler is detected, neither photon is detected. All of these are equally likely, so the probability of detecting both is 25%. (d) If the idler is detected, there are two possibilities: the signal is detected, the signal is not detected. Both of these are equally likely, so the probability of detecting the signal, given that the idler is detected, is 50%. (e) A vertically polarized signal photon has a 50% probability of passing through a linear polarizer oriented at +45°. (f) A vertically polarized idler photon has a 50% probability of passing through a linear polarizer oriented at +45°. (g) There are four possibilities: both photons are detected, the signal is detected, the idler is detected, neither photon is detected. All of these are equally likely, so the probability of detecting both is 25%.
Chapter 8
109
(h) If the idler is detected, there are two possibilities: the signal is detected, the signal is not detected. Both of these are equally likely, so the probability of detecting the signal, given that the idler is detected, is 50%. (i)-(l) Notice that the measurement results are independent of whether the photons are prepared in the state H , H or the state V , V . Thus, the answers for the mixture will be the same as those above for either state preparation. These answers agree with those in Example 8.7. 8.12
(Sec. 8.3) (a) P 45s P 45s 45, 45 P 45, 45 P 45s 45, 45 P 45, 45
1 1 45, 45 Pˆ45s 45, 45 45, 45 Pˆ45s 45, 45 2 2 1 1 45, 45 45 s s 45 45, 45 45, 45 45 s s 45 45, 45 2 2 1 i 45 45 i 2 1 . 2
(b) Probabilities must be normalized. P 45s P 45s 1
P 45s 1 P 45s 1
1 1 . 2 2
110
Chapter 8
(c) P 45i P 45i 45, 45 P 45, 45 P 45i 45, 45 P 45, 45
1 1 45, 45 Pˆ45i 45, 45 45, 45 Pˆ45i 45, 45 2 2 1 1 45, 45 45 i i 45 45, 45 45, 45 45 i i 45 45, 45 2 2 1 s 45 45 s 2 1 . 2
P 45s , 45i P 45s , 45i 45, 45 P 45, 45 P 45s , 45i 45, 45 P 45, 45
1 1 45, 45 Pˆ45s ,45i 45, 45 45, 45 Pˆ45s ,45i 45, 45 2 2 1 1 45, 45 45, 45 45, 45 45, 45 45, 45 45, 45 45, 45 45, 45 2 2 1 . 2
P 45s | 45i
P 45s , 45i 1/ 2 1 ‘ 1/ 2 P 45i
These all agree with Problem 8.7. 8.13
(Sec. 8.3) (a) P H s P H s 45, 45 P 45, 45 P H s 45, 45 P 45, 45
1 1 45, 45 PˆH s 45, 45 45, 45 PˆH s 45, 45 2 2 1 1 45, 45 H s s H 45, 45 45, 45 H s s H 45, 45 2 2 1 1 1 1 1 45 45 i 45 45 i i i 2 2 2 2 2 1 . 2
Chapter 8
111
(b) Probabilities must be normalized. P H s P Vs 1
P Vs 1 P H s 1
1 1 . 2 2
(c) P H i P H i 45, 45 P 45, 45 P H i 45, 45 P 45, 45
1 1 45, 45 PˆHi 45, 45 45, 45 PˆHi 45, 45 2 2 1 1 45, 45 H i i H 45, 45 45, 45 H i i H 45, 45 2 2 1 1 1 1 1 45 45 s 45 45 s s s 2 2 2 2 2 1 . 2
P H s , H i P H s , H i 45, 45 P 45, 45 P H s , H i 45, 45 P 45, 45
1 1 45, 45 PˆH s , Hi 45, 45 45, 45 PˆH s , Hi 45, 45 2 2 1 1 45, 45 H , H H , H 45, 45 45, 45 H , H H , H 45, 45 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 1 . 4
P H s | Hi
P H s , H i 1/ 4 1 1/ 2 2 P Hi
Parts (a) and (b) agree with Problem 8.8, but not part (c), which is OK, because the states are not the same.
112
Chapter 8
8.14* (Sec. 8.4) 1 0.2 H , H 0.8 V ,V (a) P (θ A ,θ B )
A
θA
B
θ A
θB
1
A
θB
B
A
P (θ A ,θ B )
2
1
θA
B
θB
0.2 H
H
A
B
A A
0.2cosθ A cos θ B 0.8 sinθ A sin θ B
θ A 1
0.2 H , H θ A 0.2
B
0.8 V
A
V
0.2 θ A H θ B H 0.8 θ A V θ B V
0.2cosθ A cos θ B 0.8 sinθ A sin θ B
(b) P (θ A ) 1 θ A
A
V sin θ
0.8 V ,V θ A
H cos θ A 0.8
B
A
A
B
2
0.2
A
θ A H , H 0.8
0.2 cos θ A H
B
A
θ A V ,V
0.8 sin θ A V
B
0.2 cos 2 θ A 0.8sin 2 θ A P(θ B ) 1 θ B
B B
θ B 1
0.2 H , H θ B 0.2
A
B
V sin θ
0.8 V , V θ B
H cos θ B 0.8
A
B
B
0.2
B
θ B H , H 0.8
0.2 cos θ B H
A
B
θ B V ,V
0.8 sin θ B V
A
0.2 cos θ B 0.8sin θ B 2
P (θ A |θ B )
2
P (θ A ,θ B ) P(θ B )
P (θ B |θ A )
P (θ A ,θ B ) P(θ A )
Plug the angles given into a spreadsheet, using the formulas above. The numbers that are bold and underlined are the ones we are interested in. A 19 19 -35 -35
B -19 35 -19 35
P(A,B) 0.0930 0.2636 0.2636 0.0000
P(A) 0.2636 0.2636 0.3974 0.3974
P(B) 0.2636 0.3974 0.2636 0.3974
P(A|B) 0.3529 0.6633 0.9999 0.0001
P(B|A) 0.3529 0.9999 0.6633 0.0001
Chapter 8
113
(c) These results are consistent with Observations 1-4, but are not consistent with local realism. 8.15* (Sec. 8.4) 1 0.2 H , H 0.8 V ,V (a) P (θ A ) 1 θ A
A A
θ A 1
0.2 H , H θ A 0.2
B
A
V sin θ
0.8 V ,V θ A
H cos θ A 0.8
B
A
A
0.2
A
θ A H , H 0.8
0.2 cos θ A H
B
A
θ A V ,V
0.8 sin θ A V
B
0.2 cos θ A 0.8sin θ A 2
P(θ B ) 1 θ B
2
B B
θ B 1
0.2 H , H θ B 0.2
A
B
V sin θ
0.8 V , V θ B
H cos θ B 0.8
A
B
B
0.2
B
θ B H , H 0.8
0.2 cos θ B H
A
B
θ B V ,V
0.8 sin θ B V
A
0.2 cos θ B 0.8sin θ B 2
2
(b) Bob's measured probabilities depend only on θ B . This parameter is not under Alice's control, so Alice can in no way send a signal to Bob using her measurement apparatus. Note that we have only proved that Alice cannot send a signal to Bob by performing a measurement in the linear polarization basis. This does not prove that there is nothing she can do to send a message. However, more general proofs do show that there is no local operation that Alice can perform to send a message to Bob.
Complement 8.A The Density Operator 8.A.1 The density operator corresponding to this mixed state is 1 1 ˆ 45, 45 45, 45 45, 45 45, 45 2 2 (a) P H s PˆH s Tr PˆH s ˆ Tr H
s s
1 H 45, 45 45, 45 2
1 Tr H s s H 45, 45 45, 45 2 1 1 45, 45 H s s H 45, 45 45, 45 H s s H 45, 45 2 2 1 1 1 1 1 45 45 i 45 45 i i i 2 2 2 2 2 1 . 2
(b) Probabilities must be normalized. P H s P Vs 1
P Vs 1 P H s 1
1 1 . 2 2
115
116
Complement 8.A
(c) P H i PˆHi Tr PˆHi ˆ Tr H
i i
1 H 45, 45 45, 45 2
1 Tr H i i H 45, 45 45, 45 2 1 1 45, 45 H i i H 45, 45 45, 45 H i i H 45, 45 2 2 1 1 1 1 1 45 45 s 45 45 s s s 2 2 2 2 2 1 . 2 P H s , H i PˆH s , Hi Tr PˆH s , Hi ˆ 1 Tr H , H H , H 45, 45 45, 45 2
1 Tr H , H H , H 45, 45 45, 45 2 1 1 45, 45 H , H H , H 45, 45 45, 45 H , H H , H 45, 45 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 1 . 4
P H s | Hi
P H s , H i 1/ 4 1 . 1/ 2 2 P Hi
The results are the same as in Problem 8.13.
Complement 8.A
117
8.A.2
1/ 2
H , H
V ,V
The density matrix is ˆ
1 3
2 V ,V V ,V 3
1 Tr ˆ Tr 3 1 2 3 3
2 Tr V ,V V ,V 3 1 2 V ,V V ,V 1 3 3
2 2 1 1 ˆ 2 V , V V , V V , V V , V 3 3 3 3 1 2 2 4 V , V V , V V , V V , V V , V V , V 9 9 9 9 1 2 2 4 V , V V , V V , V V , V 9 9 9 2 9 2
2 2 1 4 V , V Tr V , V V , V V , V Tr Tr ˆ 2 Tr +Tr 9 9 9 2 9 2 1 2 2 4 V , V V , V 9 9 2 9 9 2 1 1 1 4 7 1 9 9 9 9 9 Since Tr ˆ 1 and Tr ˆ 2 1 , these are what we would expect.
Complement 8.B The Bell-Clauser-Horne Inequality 8.B.1* (a) P (θ A ,θ B )
A
θA
B
θ A
θB
1
A
θB
B
A
P (θ A ,θ B )
2
1
θA
B
θB
0.2 H
A
H
B
0.8 V
A
V
0.2 θ A H θ B H 0.8 θ A V θ B V 0.2cosθ A cos θ B 0.8 sinθ A sin θ B
0.2cosθ A cos θ B 0.8 sinθ A sin θ B
B
2
(b) Plug the angles given into a spreadsheet, using the formula from part (a) above. A 19 19 -35 -35
B -19 35 -19 35
A_p
B_p
109 109 55 55
71 125 71 125
P(A,B) 0.0930 0.2636 0.2636 0.0000
P(A,B_p) 0.1706 0.0000 0.1338 0.3974
P(A_p,B) 0.1706 0.1338 0.0000 0.3974
In this table A_p and B_p refer to A and B . The numbers that are bold and underlined are the ones we are interested in. (c) The Bell-Clauser-Horne inequality is
P A1 , B1 P A 2 , B 2 P A1 , B 2 P A 2 , B1 .
Plugging in values from the table gives 0.093 0 0 0 , which is clearly false, so the inequality is violated. The results are not consistent with local realism.
119
120
Complement 8.B
8.B.2 I'll solve this problem using conditional probabilities, as discussed in Sec. 8.3. You could also solve it using the density operator (Complement 8.A). (a) P θ A ,θ B P θ A ,θ B H , H
P H , H P θ
A
,θ B V ,V
P V ,V
H , H Pˆθ A ,θ B H , H 0.2 V ,V Pˆθ A ,θ B V ,V 0.8 0.2 H , H θ A ,θ B θ A ,θ B H , H 0.8 V ,V θ A ,θ B θ A ,θ B V ,V cos H sin V
P θ A ,θ B 0.2 cos θ A cos θ B cos θ A cos θ B 0.8 sinθ A sin θ B sinθ A sin θ B 0.2 cos 2 θ A cos 2 θ B 0.8 sin 2θ A sin 2 θ B (b) Plug the angles given into a spreadsheet, using the formula from part (a) above. A 19 19 -35 -35
B -19 35 -19 35
A_p
B_p
109 109 55 55
71 125 71 125
P(A,B) 0.1688 0.1479 0.1479 0.1766
P(A,B_p) 0.0948 0.1157 0.2495 0.2208
P(A_p,B) 0.0948 0.2495 0.1157 0.2208
In this table A_p and B_p refer to A and B . The numbers that are bold and underlined are the ones we are interested in. (c) The Bell-Clauser-Horne inequality is
P A1 , B1 P A2 , B 2 P A1 , B 2 P A2 , B1 .
Plugging in values from the table gives 0.1688 0.1157 0.1157 0.1766 0.4080 ,
which is true, so the inequality is satisfied. The results for the mixed state are consistent with local realism.
Complement 8.C Two Spin-1/2 Particles 8.C.1* Using Eqs. (8.C.14) and (8.C.15): Sˆ (1) Sˆ (2) z , z Sˆx(1) Sˆx(2) z , z Sˆ y(1) Sˆ y(2) z , z Sˆ z(1) Sˆz(2) z , z 1 2 1 1 z, z 2 z, z 2 z, z 4 4 4 1 2 z, z 4
, and the other 2-particle basis states]:
Using Eq. (8.C.7) [which also holds for Sˆ (1)
Sˆ 2 z , z Sˆ (1)
2
z , z Sˆ (2)
2
2
z , z 2 Sˆ (1) Sˆ (2) z , z
3 2 3 2 2 2 z, z 4 4 4 2 2 z , z 11 1 2 z , z
So s 1 .
Sˆz z , z Sˆz(1) Sˆz(2) z , z
z, z 2 2 z, z
So ms 1 .
121
122
Complement 8.C
8.C.2* Using Eqs. (8.C.14) and (8.C.15): Sˆ (1) Sˆ (2) z , z Sˆx(1) Sˆx(2) z , z Sˆ y(1) Sˆ y(2) z , z Sˆz(1) Sˆz(2) z , z 1 2 1 1 z , z 2 i i z , z 2 z , z 4 4 4 1 1 2 z, z 2 z, z 2 4
Sˆ (1) Sˆ (2) z , z Sˆx(1) Sˆx(2) z , z Sˆ y(1) Sˆ y(2) z , z Sˆz(1) Sˆz(2) z , z 1 2 1 1 z , z 2 i i z , z 2 z , z 4 4 4 1 1 2 z, z 2 z, z 2 4
, and the other 2-particle basis states]:
Using Eq. (8.C.7) [which also holds for Sˆ (1)
Sˆ 2 z , z Sˆ (1)
2
z , z Sˆ (2)
2
2
z , z 2 Sˆ (1) Sˆ (2) z , z
3 1 3 2 2 z, z 2 z, z 2 z, z 4 2 4 2 z, z 2 z, z
Sˆ 2 z , z Sˆ (1)
2
z , z Sˆ (2)
2
z , z 2 Sˆ (1) Sˆ (2) z , z
3 1 3 2 2 z, z 2 z, z 2 z, z 4 2 4 2 z, z 2 z, z
2 2 1 ˆ S z, z z, z z, z z, z z, z z, z 2 2 1 2 2 z, z z, z 2 1 11 1 2 z, z z, z 2 So s 1 .
Complement 8.C
1 Sˆz z, z z, z Sˆz(1) Sˆz(2) 1 z, z z, z 2 2 1 2 2 z , z 2 2 z , z 2 0 So ms 0 . 8.C.3 0, 0 c1 z , z c2 z , z c3 z , z c4 z , z 1,1 0, 0 z , z 0, 0 c1 0 1, 1 0, 0 z , z 0, 0 c4 0
1, 0 0, 0
1 1 z , z 0, 0 z , z 0, 0 c2 c3 0 2 2
c3 c2 After using the normalization condition: 1 z , z z , z 0, 0 2 8.C.4* Sˆ (1) Sˆ (2) z , z Sˆx(1) Sˆx(2) z , z Sˆ y(1) Sˆ y(2) z , z Sˆz(1) Sˆz(2) z , z 1 2 1 1 z , z 2 i i z , z 2 z , z 4 4 4 1 2 1 2 z, z z, z 2 4
Sˆ (1) Sˆ (2) z , z Sˆx(1) Sˆx(2) z , z Sˆ y(1) Sˆ y(2) z , z Sˆz(1) Sˆz(2) z , z 1 2 1 1 z , z 2 i i z , z 2 z , z 4 4 4 1 2 1 2 z, z z, z 2 4
123
124
Complement 8.C
Sˆ 2 z , z Sˆ (1)
2
z , z Sˆ (2)
2
z , z 2 Sˆ (1) Sˆ (2) z , z
3 1 3 2 2 z, z 2 z, z 2 z, z 4 2 4 2 2 z, z z, z
Sˆ 2 z , z Sˆ (1)
2
z , z Sˆ (2)
2
z , z 2 Sˆ (1) Sˆ (2) z , z
3 1 3 2 2 z, z 2 z, z 2 z, z 4 2 4 2 2 z, z z, z 2 1 Sˆ 2 z, z z, z z, z z, z z, z z, z 2 2 0
So s 0 .
1 Sˆz z, z z, z Sˆz(1) Sˆz(2) 1 z, z z, z 2 2 1 2 2 z , z 2 2 z , z 2 0 So ms 0 . 8.C.5 1 1 x, x z 1 z 1 z 2 z 2 2 1 z, z z, z z, z z, z 2 1 1 1 1,1 0, 0 1, 1 2 2 2
1 1 ˆ2 1 Sˆ 2 x, x Sˆ 2 1,1 S 0, 0 Sˆ 2 1, 1 2 2 2 2 1,1 2 1, 1
2
Complement 8.C Sˆ 2 x, x Sˆ 2 x, x
1 1 1 1,1 0, 0 1, 1 2 1,1 2 1, 1 2 2 2 1 1 2 2 2 2
1 1 ˆ 1 Sˆ z x, x Sˆ z 1,1 S z 0, 0 Sˆ z 1, 1 2 2 2 1,1 1, 1 2 2 Sˆ z x, x Sˆ z x, x 1 1 1 0, 0 1, 1 1,1 1, 1 1,1 2 2 2 2 2 1 1 0 4 4
125
Chapter 9 Time Evolution and the Schrödinger Equation 9.1
(Sec. 9.3) d d (t ) Aˆ (t ) A dt dt d d d (t ) Aˆ (t ) (t ) Aˆ (t ) (t ) Aˆ (t ) dt dt dt i ˆ ˆ (t ) (t ) d Aˆ (t ) ˆ ˆ (t ) i (t ) AH (t ) HA dt
i d (t ) Hˆ , Aˆ (t ) (t ) Aˆ (t ) . dt
A term gets added. 9.2*
(Sec. 9.3) (a) (0) q12
is an eigenstate of the Hamiltonian. We know that if a system starts in an
eigenstate of the Hamiltonian, the state only picks up an overall phase factor as it evolves in time—the state itself doesn’t change. Therefore, in general, no expectation values will change in time. (b) 1 q12 q14 q16 is not an eigenstate of the Hamiltonian, it will evolve in 3 time, and expectation values will in general depend on time. Since (0)
(c) Operators that commute with the Hamiltonian have expectation values that do not depend on time. So, if Yˆ , Hˆ Yˆ , Qˆ 0 , Yˆ will be time independent.
127
128
Chapter 9
9.3
(Sec. 9.4) Hˆ Sˆ zB Sˆ ,
z ˆ / iHt
t e
x
ˆ
eiS zt / x
1 it /2 e z eit /2 z 2 1 eit /2 z e it z . 2
P x, t t Pˆ x t x t
2
1 1 z z eit /2 z e it z 2 2 2 1 1 e it 4 1 1 cos t . 2
2
P y, t t Pˆ y t y t
2
1 1 z i z eit /2 z e it z 2 2 2 1 1 ieit 4 1 1 sin t . 2
2
Chapter 9
129
P z , t t Pˆ z t 2
z t z eit /2
9.4
1 z e it z 2
2
1 . 2
(Sec. 9.4) Using Eqs. (9.19) and (9.22): d i S x (t ) Hˆ , Sˆ x (t ) dt i (t ) Sˆ z , Sˆ x (t ) i (t ) iSˆ y (t )
Sy
t .
Using Eq. (9.29) d S x sin t 2 dt S x cos t 2 This agrees with Eq. (9.28). 9.5
(Sec. 9.4) Pˆ z , Hˆ z z , Sˆ z z z Sˆ z Sˆ z z z
z z z z 2 2 0
By Eq. (9.19) Pˆ z should thus be independent of time. Since P z Pˆ z the probability of measuring spin up along the z-direction should be constant in time. This agrees with Problem 9.3.
130
Chapter 9
9.6*
(Sec. 9.4) ˆ
t eiHt / n
ˆ
eiHt / cos / 2 z ei sin / 2 z ˆ
ˆ
cos / 2 eiS zt / z ei sin / 2 eiS zt / z cos / 2 ei( /2)t / z ei sin / 2 ei( /2)t / z cos / 2 eit /2 z ei sin / 2 e it /2 z
eit /2 cos / 2 z ei sin / 2 e it z . S t S x t ux S y
t uy
S z t uz
Using Eqs. (6.14) and (6.15): S x t t Sˆ x t
cos / 2 z eit e i sin / 2 z Sˆ x cos / 2 z ei sin / 2 e it z
i t z cos / 2 z sin / 2 e z 2 2 i t i t cos / 2 sin / 2 e cos / 2 sin / 2 e 2 sin cos t , 2 cos / 2 z sin / 2 e
i t
Using Eq. (6.26): S y t t Sˆ y t
cos / 2 z eit e i sin / 2 z Sˆ y cos / 2 z ei sin / 2 e it z
cos / 2 z sin / 2 e
i t
i t z i cos / 2 z i sin / 2 e z 2 2
i t i t cos / 2 sin / 2 ie cos / 2 sin / 2 (i )e 2 sin sin t , 2
Chapter 9 S z t t Sˆ z t
131
cos / 2 z eit ei sin / 2 z Sˆ z cos / 2 z ei sin / 2 e it z
cos / 2 z sin / 2 e
i t
i t z cos / 2 z sin / 2 e z 2 2
cos 2 / 2 sin 2 / 2 2 cos 2 S t makes an angle of with the z-axis, and sweeps out a cone as it processes around the z-
axis at the Larmor frequency. 9.7
(Sec. 9.4) Hˆ Sˆ xB Sˆ , x
ˆ
t eiHt / z ˆ
eiHt /
1 x x 2
ˆ 1 iSˆxt / e x eiS xt / x 2 1 i( /2)t / e x ei( /2)t / x 2 1 eit /2 x eit x . 2
S t S x t ux S y
t uy
S z t uz
S x t t Sˆ x t 1 e it /2 2
x e
1 x eit 2
1 2 2 2 0
1 x Sˆ x eit /2 x eit x 2 x x e it x 2 2
it
132
Chapter 9
0 i 1 1 1 i Sˆ y x i x 2 i 0 z 2 1 z 2 2 i z 2 0 i 1 1 1 i Sˆ y x i x 2 i 0 z 2 1 z 2 2 i z 2 S y t t Sˆ y t 1 eit /2 2
x e
it
1 x Sˆ y eit /2 x e it x 2
1 x eit x i x ie it x 2 2 2 ieit ieit 4 sin t , 2
1 0 1 1 1 1 Sˆz x x 2 0 1 z 2 1 z 2 2 1 z 2 1 0 1 1 1 1 Sˆz x x 2 0 1 z 2 1 z 2 2 1 z 2
S z t t Sˆ z t 1 e it /2 2
x e
it it e e 4 cos t . 2
9.8
1 x eit 2
(Sec. 9.4) Hˆ Sˆ zB Sˆ z
1 x Sˆ z eit /2 x eit x 2 x x e it x 2 2
it
Chapter 9 ˆ
(t ) eiHt / 1, 1
x
ˆ ˆ 1 iSˆzt / e 1,1 2eiS zt / 1, 0 eiS zt / 1, 1 2 1 eit / 1,1 2 1, 0 ei( )t / 1, 1 2 1 eit 1,1 2 1, 0 e it 1, 1 2 eit 1 2 2 e it z
S t S x t ux S y
t uy
S z t uz
The matrices for the spin operators for spin-1 are found in Sec. 7.4. 0 1 0 ˆ Sx 1 0 1 2 0 1 0 z S x (t ) (t ) Sˆ x (t )
1 e it 2
4
e 2
it
2
2
eit
it 0 1 0 e 1 1 0 1 2 2 z 2 0 1 0 z e it z
eit
it 0 1 0 e 1 0 1 2 z 0 1 0 it z e z
2 eit e it z 2 z
e 2
2 e it eit e it eit 2
4
it
4 cos t
2
eit
133
134
Chapter 9 0 1 0 ˆ Sy i 1 0 1 2 0 1 0 z S y (t ) (t ) Sˆ y (t ) it 0 1 0 e 1 1 e it 2 eit i 1 0 1 2 z 2 2 0 1 0 2 e it z z it 0 1 0 e i e it 2 eit 1 0 1 2 z 4 2 0 1 0 it z e z
i e it 4 2
2
eit
2 eit e it z 2 z
i 2 e it eit e it eit 4 2 sin t
1 0 0 Sˆz 0 0 0 0 0 1
Chapter 9 S z (t ) (t ) Sˆ z (t )
1 e it 2
4
4
e 2 e 2
4 2 0 9.9
it
it
2
eit
it 1 0 0 e 1 0 0 0 2 z 2 0 0 1 2 eit z z
eit
2
eit
2
it 1 0 0 e 0 0 0 2 z 0 0 1 it z e z
eit 0 z it z e
1 1
(Sec. 9.4) Hˆ Sˆ zB Sˆ z
The initial state is an eigenstate of the Hamiltonian, with eigenvalue 0: Hˆ 1, 0 Sˆ z 1, 0 0 . As such, the state will not change in time, so P z1 , z2 , t P z1 , z2 , 0 z , z 1, 0
2
1 z, z z, z z, z 2 1 . 2
2
135
136
Chapter 9
9.10
(Sec. 9.4) Hˆ Sˆ zB Sˆ z
The initial state is an eigenstate of the Hamiltonian, with eigenvalue 0:
Hˆ z , z Sˆ z z , z Sˆ z(1) Sˆ z(2) z , z z , z 0 . 2 2 As such, the state will not change in time, so P z1 , z2 , t P z1 , z2 , 0 z, z z, z
2
0. 9.11
(Sec. 9.4)
q L× B 2m In the above picture the torque points out of the page (along u y ).
τ = μ× B
d q L L× B 2m dt Consider a small time interval t . Over this time interval the angular momentum will change by a small amount L . q q L L× B t LBt u y 2m 2m After t the angular momentum will be L ' L L . In a view from the top:
τ=
Chapter 9
137
Since we’re assuming L is huge and L is small, then L is nearly perpendicular to L ' , as well as being perpendicular to L . This means L ' L . The direction of L changes, but its magnitude does not. Note that τ will always be perpendicular to L . This means that the change in angular momentum L will always be perpendicular to L , so L will always change direction, but not magnitude. L will rotate about in a circle (clockwise in the above figure) and since μ is parallel to L , it will rotate as well. The dipole moment μ will process about the magnetic field. 9.12* (Sec. 9.5) Invert Eq. (9.31) to write the flavor eigenstates in terms of the mass eigenstates: e cos 1 sin 2 x sin 1 cos 2 The masses determine the energies: m 2c 4 m 2c 4 E1 E0 1 1 2 E2 E0 1 2 2 , 2 E0 2 E0 where E0 pc . The state at future times is ˆ
t e iHt / 0 ˆ
e iHt / e ˆ
ˆ
cos eiHt / 1 sin eiHt / 2 cos eiE1t / 1 sin e iE2t / 2 . The probability that this state will be detected as an electron neutrino is:
138
Chapter 9
P e e t
2
1 cos sin 2 cos eiE t / 1
cos 2 e iE1t / sin 2 e iE2t /
1 sin e iE2t / 2
2
2
2 i E E t / cos 2 sin 2 e 2 1
cos 4 sin 4 2 cos 2 sin 2 cos Et / , where we've defined E E2 E1
c4
m22 m12
2 E0
m 2c 4 . 2 E0 Applying a few trig identities: 1 2 1 2 1 P e 1 cos 2 1 cos 2 sin 2 2 cos Et / 4 4 2 1 1 2 2 cos 2 cos 2 2 2 cos 2 cos 2 2 sin 2 2 cos Et / 4 2 1 1 1 cos 2 2 sin 2 2 cos Et / 2 2 1 1 1 sin 2 2 sin 2 2 cos Et / 2 2 1 1 sin 2 2 1 cos Et / 2
1 sin 2 2 sin 2 Et / 2 , Substituting for E : m 2c 4t P e 1 sin 2 2 sin 2 4 E 0
m 2c 4 L 1 sin 2 2 sin 2 , 4 E c 0 where we've used the fact that L ct .
Chapter 9
139
9.13* (Sec. 9.5) P e 1 sin
2
2 sin
2 m
c L 4 E0 c 2 4
As a function of L / E0 this has peaks at m 2c 4 L n 4 E0 c
n 0,1, 2, .
The peak in Fig. 9.2 at L / E0 34 km/MeV should correspond to n 1 m 2c 4 L 4 E0 c 2
m
4E0 c Lc
4
4c
L / E0 c
4
4 6.6 1016 eV s 3.0 105 km / s 106 eV / MeV
34km / MeV c
4
7.3 105 eV 2 / c 4 Second peak at L / E0 70 km/MeV should correspond to n 2 m 2c 4 L 2 4 E0 c 2
m
8E0 c Lc
4
8c
L / E0 c
4
8 6.6 1016 eV s 3.0 105 km / s 106 eV / MeV
70km / MeV c
4
7.1 105 eV 2 / c 4 These correspond to an average of m 2 7.2 105 eV 2 / c 4 . The KamLAND collaboration performed a much more sophisticated analysis of all the data, and obtained m 2 7.6 105 eV 2 / c 4 . 9.14* (Sec. 9.2) (a) Need to find the eigenvalues of 1 0 Hˆ * 1 0 0 0 2
140
Chapter 9 1 * 1 0
0 0
0
2
0
1 1 2 * 2 2 2 1 1 2 2 2 21 12 0
one solution is 2 . Other solutions are given by 2 2 21 12 0 . Solve the quadratic equation
21 414 4 12
2
2
2 2 1
2
1
This gives two more solutions. (b) Before finding the time dependence, we need to find the eigenstates of the Hamiltonian. The eigenstate corresponding to the eigenvalue 2 is c , because 1 ˆ H c * 0 where I've set
1
0
0 0 0 0 0 2 0 2 c E3 E3 , 1 2 1
2 E3 , and E3 c .
Because the eigenstates of a Hermitian operator are orthogonal, the other states will have no c component. The other states are found by 1 1 a 0 0 * 1 1 b 0 0 0 2 1 0 Which yields (we know that the other equation will be linearly dependent, so we only need one equation).
Chapter 9 a b 0
Writing ei this becomes a eib 0
b a e i The properly normalized states are then, for E1 1 :
1 a e i b . 2 for E2 1 : E1
E2
1 a e i b . 2
Inverting: 1 a E1 E2 2
,
b
ei E1 E2 2
The time evolution of the state is given by ˆ
t e iHt / 0 ˆ
e iHt / a
ˆ ˆ 1 iHt e / E1 e iHt / E2 2 1 iE1t / e E1 eiE2t / E2 2 1 i E E t / e iE1t / E1 e 2 1 E2 2 1 i2 t / e iE1t / E1 e E2 2 The probability of a measurement yielding the result b is
141
142
Chapter 9 P b, t b t
2
b eiE1t /
1 i2 t/ E1 e E2 2
1 1 i 1 i i 2 t / e e e 2 2 2
2
2
2 1 i2 t / 1 e 4 1 2 2 cos 2 t / sin 2 t / . 4
Complement 9.A Magnetic Resonance 9.A.1
0 pB 2.68 108 s-1T -1 4T 1.07 109 rad/s
f 0 0 / 2 1.07 109 rad/s / 2 170 MHz
9.A.2 (a) In order to flip with 100% certainty, the field must be on resonance:
0 pBz 2.68 108 s-1T -1 8.5T 2.28 109 rad/s
f 0 0 / 2 2.28 109 rad/s / 2 363 MHz
(b) R
0 2 (1 / 2)2
On resonance:
R 1 / 2 pBx / 2 2.68 108 s-1T -1 106 T / 2 1.34 102 rad/s
A -pulse (which guarantees a spin flip) has a duration of:
t / R / 1.34 102 rad/s 23 ms .
143
Chapter 10 Position and Momentum 10.1
(Sec 10.1) (a) Let the normalization constant be c. We then have 2 x 2 dx x c dx rect a a /2
c
2
dx
a /2
c2 x
x a / 2 x a /2
c a 1. 2
So c 1/ a and x
1 x rect . a a
(b)
x
dx x x x
a/2
1 dx x a a/2
0 because the integrand is odd.
145
146
Chapter 10
x2
dx x x 2 x
a /2
1 dx x 2 a a/2
1 3 x 3a
x a /2
x a /2
1 3 3 a / 2 a / 2 3a 2 a 12
x 10.2
x
2
x
2 1/2
1/2
a2 0 12
a 2 3
(Sec 10.1) (a) Let the normalization constant be c. We then have
dx x c 2 dx e 2 x 2
0
c2 e 2 x 2
x
x 0
2
c 0 1 2
c2 1. 2
So c 2 and 0 x x 2 e
x0 x0
.
Chapter 10
(b) This probability is P 0 x 1/
1/
dx x
2
0
1/
2 dx e 2 x 0
e
2 x x 1/ x 0
2
e 1 0.865 . (c)
x
dx x x x
2 dx xe 2 x 0
x
1 e 2 x 2 x 1 2 x 0
1 2
x2
dx x x 2 x
2 dx x 2 e 2 x 0
x
1 2 e 2 x 2 2 x 2 2 x 1 2 x 0
x
1 2 2
x
2
x
2 1/2
1/2
1 1 2 2 2 4
1 x . 2
147
148
Chapter 10
10.3
(Sec 10.1) (a) Let the normalization constant be c. We then have 2 1 2 dx x c dx a 2 x 2 2 c2 2 2a
x
arctan x / a x 2 2 a a x x / 2 /2 0 0 a a
c2 2a 2
c2 1. 2a 3
So c 2a 3 / and 2a 3 1 2 a x2
x
(b) This probability is 2a 3 P a x a
a
dx
a
a
1
2
x2
2
x a
arctan x / a a x 2 2 a a x x a
a a / 4 a / 4 2 2 a 2a a 2a
a 1 1 1 0.818 . a 2a 2
(c)
x
dx x x x
2a 3
dx
a
x
2
x2
2
0 because the integrand is odd.
Chapter 10
x2
dx x x 2 x
2a 3
a3
x 10.4
dx
a
x2
2
x2
2 x
arctan x / a x 2 2 a a x x /2 / 2 0 0 a2 a a
a3
x2 x
a 0
2 1/2
1/2
2
a.
(Sec 10.1) (a) Let the normalization constant be c. We then have
dx x
2
c
2
dx sech
2
( x / a)
c 2 a tanh 2 ( x / a )
x x
2ac 1. 2
So c 1/ 2a and
x 1/ 2a sech x / a (b) This probability is P 0 x a
a
1 dx sech 2 ( x / a) 2a 0
1 x a tanh( x / a ) x 0 2 1 tanh(1) 0.381. 2
149
150
Chapter 10
(c)
x
dx x x x
1 2 dx x sech ( x / a) 2a
0 because the integrand is odd.
x2
dx x x 2 x
1 dx x 2 sech 2 ( x / a ) 2a 2 a 2 12 This integral was calculated using Maple.
x
x
2
x
2 1/2
a . 2 3
10.5* (Sec 10.2) Tˆ x dx Tˆ dx Tˆ x i 1ˆ pˆ x dx Tˆ x i Tˆ x pˆ xTˆ x dx ˆ ˆ T x dx T x i pˆ xTˆ x dx d ˆ i T x pˆ xTˆ x dx Tˆ x e ipˆ x x / , or Tˆ D e ipˆ x D /
Chapter 10
151
10.6* (Sec 10.2) Tˆ † D Tˆ D eipˆ x D / e ipˆ x D / 1ˆ , so the translation operator is unitary. 10.7* (Sec 10.2) ipˆ D / Tˆ † D eipˆ x D / e x Tˆ D
10.8
(Sec 10.2) ˆ ˆ , Cˆ Aˆ Bˆ , Cˆ Aˆ , Cˆ Bˆ . Equation (7.56) says: AB Rearranging: ˆ ˆ Aˆ Bˆ , Cˆ Aˆ , Cˆ Bˆ Cˆ , AB Aˆ Cˆ , Bˆ Cˆ , Aˆ Bˆ Therefore: Aˆ , Bˆ 2 Bˆ Aˆ , Bˆ Aˆ , Bˆ Bˆ c 2 Bˆ
Aˆ , Bˆ 3 Bˆ Aˆ , Bˆ 2 Aˆ , Bˆ Bˆ 2 c 2 Bˆ 2 cBˆ 2 c3Bˆ 2 The proof follows by induction. 10.9
(Sec 10.2) Eq. (10.71) says: if Aˆ , Bˆ c , Aˆ , Bˆ n cnBˆ n 1 .
Aˆ , f Bˆ Aˆ , f b f b Bˆ 1 f b Bˆ 2 ... 2 1 Aˆ , f b f b Aˆ , Bˆ f b Aˆ , Bˆ 2 ... 2 1 ˆ ... f b 2 Bc 2 c f b f b Bˆ ... Notice that the series in the square brackets is the derivative of the power series representation of the original function. Therefore, it is the derivative of the original function. 0 f b c
Aˆ , f Bˆ cf Bˆ .
152
Chapter 10
10.10 (Sec 10.2) Tˆ D e ipˆ x D / ˆ ipˆ x D / x xˆ eipˆ x D / xe
Eq. (10.85) says: if Aˆ , Bˆ c , Aˆ , f Bˆ cf Bˆ . Applying it: ipˆ x D / ˆ ipˆ x D / e ipˆ x D / xˆ xˆ , pˆ x i iD / e ipˆ x D / Deipˆ x D / xe e pˆ x ˆ ipˆ x D / e ipˆ x D / xˆ De ipˆ x D / xe
x eipˆ x D / e ipˆ x D / xˆ De ipˆ x D / xˆ D xD
So the average position shifts by D, as would be expected. ˆ ipˆ x D / eipˆ x D / e ipˆ x D / pˆ pˆ p p pˆ eipˆ x D / pe So the average momentum does not change, as would be expected. 10.11* (Sec 10.2) We'll start by rewriting x as x x 2 e x , where x is the step function 1 x 1/ 2 0
x0 x 0. x0
Differentiate x x 2 e x x 2 x e x x x x x x 2 e x 2 x e , where we've used the fact that the derivative of the step function is the delta function (in Complement 10.A). This yields
Chapter 10
p
dx x i x x
i 2 dx e 2 x x x x
i 2 e 0 i 2 0
2
dx e
2 x
0
i i 0
10.12
(Sec 10.2) For the particle in Problem 10.3: 2a 3 1 x , x a 2 a x2 2a 3 1 2a 3 2x x 2 2 x x a x a 2 x 2 2
p
dx x i x x
i
2a 3
dx a
2
1 2x 2 2 x a x 2 2
0
because the integrand is odd. 2 2a 3 x 2a 3 4 x 2 x x 2 2 x 2 x a 2 x 2 2 a 2 x 2 3 a 2 x 2 2 2
p2
dx x i x x
2
2 2a 2
4a 3
dx
1 4 x2 x a 2 x 2 a 2 x 2 3 a 2 x 2 2
153
154
Chapter 10
This integral was done using Maple. p
2 1/2
p2 p
2a
2a 2 2 so the indeterminacy relation is satisfied.
xp a
10.13
(Sec 10.2) For the particle in Problem 10.4:
x 1/ 2a sech x / a , x
a 2 3
1 1 x x x sech sech tanh x x 2a x a 2a a a a
p
dx x i x x
i
1 2a 2
dx sech
2
x x tanh a a
0 because the integrand is odd. 1 2 x x sech tanh x 2 x a 2a x a a 1 x x 2 x 2 x 2 sech tanh sech 1 tanh a 2a a a a a
1 a
2
x x sech 2 tanh 2 1 2a a a
Chapter 10 2
p2
155
dx x i x x
1 2a 3
2
dx sech
2
x 2 x 2 tanh 1 a a
2
3a 2
This integral was done using Maple.
3a a xp 2 3 3a 6 2 so the indeterminacy relation is satisfied.
p
p2 p
2 1/2
10.14 *
(Sec 10.3)
x x
dp
x p p
1 2
dp e
ipx /
p
where we've used Eq. (10.56). 10.15 *
(Sec 10.3)
For a position eigenstate x : p p x x p p x
1 e ipx / 2
The probability density corresponding to this wave function is 2 1 p , 2 which is constant over all space. This means that the uncertainty in the momentum is infinitely large. For a position eigenstate, the uncertainty in the position is 0. The only way to satisfy the indeterminacy principle with a position uncertainty of 0 is if the momentum uncertainty is infinitely large, so the state is consistent with the indeterminacy principle.
156
Chapter 10
10.16 *
(Sec 10.3)
x
1
p
2 2
x x /4 2 2
1/2
e
1
2
1/2
dx e
2 ipx / x x /4 2
e
1/4
2 p 22 / 2 ipx / e e This was integrated using Maple.
p
dp p p p
1/2
2
dp pe
2 p 2 2 / 2
0 because the integrand is odd.
p2
dp p p p
2
1/2
2
dp p e
2 2 p 2 2 / 2
2
4 2
This was integrated using Maple. p
p2 p
2 1/2
2
From Section 10.1.4 we know that x , so xp 2 2 The Gaussian wave packet is a minimum uncertainty wave packet, because the indeterminacy relation is satisfied with an equality. 10.17 (Sec 10.3)
Chapter 10 For the particle in Problem 10.3: 2a 3 1 x , x a 2 a x2
2a 3 1 dx e ipx / 2 a x2
1 2
p
a p a/ e
This was integrated using Wolfram Alpha (http://www.wolframalpha.com/).
p
dp p p p
a 2 p a / dp pe
0 because the integrand is odd.
p
2
dp p p p
2
a 2 p a / dp p 2 e
2 2 2a
This integral was done using Maple. p
p2 p
2 1/2
2a
2a 2 2 so the indeterminacy relation is satisfied.
xp a
157
Complement 10.A Useful Mathematics 10.A.1 1 p 2
dx e
1 x rect a a
ipx /
1 2a
a /2
dx e ipx /
a/2 x a / 2
1 ipx / e 2a ip x a /2
i ipa /2 ipa /2 e e 2a p
2sin pa / 2 p 2a
a sin pa / 2 2a pa / 2
a sinc pa / 2 2
10.A.2* We know: p
1 2
dx e
ipx /
x
Therefore: p
1 2
dx e
ipx /
x
And then
1 dx e ipx / x 2 Use the fact that for a real function: *( x) ( x) , so p
159
160
Complement 10.A
p
1 2
dx e
ipx /
p . x
Taking the magnitude of both sides: p p p p
10.A.3*
1 ipx / F x x0 dx e x x0 2 Let u x x0 , so du dx 1 F x x0 2
1 2
du e
ip u x0 /
u
du e
ipu / ipx0 /
e
u
1 e ipx0 / du e ipu / u 2
p . e ipx0 /
Chapter 11 Wave Mechanics and the Schrödinger Equation 11.1
(Sec. 11.1) (i) Substitute n x, t n x n t into Eq. (11.8): 2 2 V x n x, t i n x, t 2 t 2m x 2 2 V x n x n t i n x n t 2 t 2m x 2 2 V x n t n x n x i n t 2m x 2 t (ii) Divide the resulting equation by n x n t , and separate the temporal and spatial dependencies on opposite sides of the = sign: 2 2 1 1 V x x x i n t n t n n 2m x 2 x t t n x n t n n 2 2 1 V x i n t n x 2 n x 2m x n t t 1
(iii) Note that time changes cannot affect the spatial part of the equation, and vice versa. The only way for the temporal and spatial parts to be equal to each other is if they are both constants. Set both the temporal and spatial parts equal to the same constant, En , yielding two equations: 2 2 V x n x En n x 2m x 2 1 i n t En n t t 1
(v) Rearrange the spatial equation to obtain Eq. (11.19). 161
162
Chapter 11
2 2 V x n x En n x 2 2m x 2 2 d n x V x n x En n x 2m dx 2 (vi) Solve the temporal equation. Compare your solution to Eq. (11.20). E n t n n t t i The solution to this equation is n t e int , with En n . 11.2* (Sec. 11.3) Start with Eqs. (11.42) and (11.43) A1 B1 A2 . ik1 A1 ik1 B1 ik2 A2 . Substitute (11.42) into (11.43): ik1 A1 ik1B1 ik2 A1 B1
k1 k2 A1 k1 k2 B1 k k B1 1 2 A1 k1 k2
Substitute this back into (11.42) k k A1 1 2 A1 A2 k1 k2 A2
2k1 A k1 k2 1
(11.42) (11.43)
Chapter 11
11.3* (Sec. 11.3) (a) i k x t i k x t Im A2 e 2 2 A2e 2 2 x m k 2 2 A2 . m 2 4 E E V0 jxA2 x, t k2 A2 4k1k2 T A1 2 2 jx x, t k1 A1 k1 k2 E E V0
jxA2 x, t
2
(b) In part (a) we see that T
k2 A2 k1 A1
2 2
.
2
2
Since k2 k1 , it cannot be the case that T A2 / A1 . 11.4* (Sec. 11.3) Start with Eqs. (11.51) and (11.52) A1 B1 A2 ik1 A1 ik1 B1 A2 . Substitute (11.51) into (11.52) ik1 A1 ik1B1 A1 B1
ik1 A1 ik1 B1 ik A B1 1 ik1 1
Substitute this back into (11.51) ik A A A1 1 ik1 1 2 2ik1 A2 A ik1 1
(11.51) (11.52)
.
163
164
Chapter 11
11.5
(Sec. 11.3) jxA2 x, t T
11.6
Im A2 ex i1t A2 ex i1t x m 2 A2 Im e x e x 0 x m
jxA2 x, t 0 jxA1 x, t
(Sec. 11.3) We know from Eq. (11.53) that B1
ik1 A . ik1 1
This means ik A k12 2 A A . B1 1 ik1 1 k12 2 1 1 To find the phase shift, first note that ik1 ik1 ik1 k12 2 i 2k1 ik1 ik1 ik1 k12 2 So, 2 k B1 ei A1 , where tan 1 2 1 2 . k1
11.7* (Sec. 11.3) Since there is no right-traveling wave to the left of the cliff B1e ik1x x0 x ik x ik2 x 2 0 x A2 e B2 e Applying the continuity of the wave function and its derivative at x 0 yields B1 A2 B2 (A) ik1 B1 ik2 A2 ik2 B2 (B) Substituting (A) into (B)
Chapter 11
165
ik1 A2 B2 ik2 A2 ik2 B2 A2 k2 k1 k2 k1 B2 A2
k2 k1 B k2 k1 2
i k x t i k x t Im A2 e 2 2 A2e 2 2 x m k 2 2 A2 . m i k x t i k x t jxB2 x, t Im B2 e 2 2 B2e 2 2 x m k 2 2 B2 . m
jxA2 x, t
jxA2 x, t
R
jxB2 x, t
k2 A2 k2 B2
2 2
A2 B2
2 2
2 k1 k2 k1 k2 2
2 E V0
E E V0 E
2
Notice that this is exactly the same as Eq. (11.47), which describes a particle with E V0 incident on a barrier from the left. This means that the transmission coefficient T [Eq. (11.48)] must be the same as well, so
T
11.8
4 E E V0 E E V0
2
.
(Sec. 11.4) (a) We know that the wave function takes the form A1eikx B1e ikx x0 x 0 x L x A2 e B2 ex ikx A3e xL with 1/ 2
2mE k 2
2m V0 E 2
1/ 2
166
Chapter 11 Applying the boundary conditions (the wave function and its derivative are continuous) at x 0 yield A1 B1 A2 B2 ikA1 ikB1 A2 B2 .
(A) (B)
At x L the boundary conditions give A2 e L B2 eL A3eikL
(C)
A2 e L B2 eL ikA3eikL .
(D)
Multiply (A) by ik and add it to (B), to obtain 2ikA1 ik A2 ik B2
(E)
Multiply (C) by and add it to (D), to obtain 2B2 eL ik A3eikL B2
ik A eik L 2
(F)
3
Insert (F) into (D) to obtain ik A2 e L A3eik L eL ikA3eikL 2 A2 e L
ik A eikL ikA eikL 3
2 ik A eikL A2 e L 3 2 ik A eik L A2 3 2
3
Substitute (F) and (G) into (E) to obtain ik ik A3e ik L ik1 A3e ik L 2ik1 A1 ik1 2 2 1 2 2 ik e L ik eL A eikL 2 3 1 A3 2 k 2 i 2k eL 2 k 2 i 2k eL 2k A1 2 1 A3 2 k 2 2sinh L i 4k cosh L 2
(G)
Chapter 11 A1 A3
2 2
2 1 2 k 2 sinh 2 L 4 2 k 2 cosh 2 L 2 2 4 k
1 4 2 k 2
2
k2
4 k 2
2
4
2 2 k 2 k 4 sinh 2 L 4 2 k 2 1 sinh 2 L
2
sinh 2 L 1
V0 E E 1 sinh 2 L 4 V0 E E 2
1
V0 2 sinh 2 L 4 E V0 E
i kx t i kx t Im A3e A3e m x k 2 A3 . m 1 2 A3 jx x, t k A3 V0 2 2 T A1 sinh L 1 jx x, t k A1 2 4 E V0 E
jxA3 x, t
11.9
(Sec. 11.4) eL e L eL Lim sinh 2 L Lim L 1 L 1 2 2
Therefore: 1
16 E V0 E 2 L V0 2 e 2 L T 1 e 2 4 4 E V E V 0 0
167
168
Chapter 11
11.10 (Sec. 11.5)
dx m
L
2 mx nx x n x dx sin sin L0 L L xL
n m x n m x sin sin L L 2 nm nm x 0 sin n m sin n m nm nm 0 nm
2
Thus, the wave functions are orthogonal. The case of n m is the normalization integral given in the text [Eq. (11.69)]. 11.11* (Sec. 11.5) Assume an approximate equality for the indeterminacy relation: xp 2 The particle must be inside the well, which has width L. We also know that the wave function must be 0 at the edges of the well, so the particle is less likely to be near the edges. Assume that L the particle spends most of its time in the middle half of the well. This implies x , which 4 2 . means p L The particle is equally likely to travel to the left or right, so p 0 , which means p 2 p 2 p
2
p2
4 2 . L2
Inside the well the potential energy is 0, and the energy is all kinetic. E
p2 2m
4 2 . 2mL2
This is within factor of 3 of the actual ground state energy E1 .
Chapter 11 11.12* (Sec. 11.5)
dx n x x n x
x
L
2 nx dx x sin 2 L0 L xL
2nx 2nx cos 2 x sin 2 x L L 2 L 4 n n 4 8 L L x 0
1 L cos 2n 2 2 2 n n 4L 4L L L L 2
By symmetry, it should be fairly obvious that this is what we had to obtain. x
2
dx n x x
2
n x
L
2 nx dx x 2 sin 2 L0 L xL
2 2nx 2 n x 2 1 sin 2nx x cos L 2 x3 L L 2 3 L 6 n n 4 8 L L x 0
L2 cos 2n 2 3 n 2 L 1 1 L2 2 2 3 2n
x
x x 2
2 1/2
1/2
1 1 L2 L2 2 2 3 2n 4
1/ 2
1 1 L 2 2 12 2n
169
170
Chapter 11 We know
p 0 by symmetry (the particle is equally likely to be moving to the right or the
left), or because the average momentum is always 0 if the wave function is real (see Sec. 10.2). p2
2
dx n x i x n x
2 n L L
2L
0
n dx sin 2 x L x L
2n sin x 2 2 n x L 4n L L 2 L x 0 n L p
2
p2 p
2 1/2
n L 1/2
1 n n 1 xp L 2 2 , L 12 2 12 2n so the indeterminacy relation is satisfied. 11.13 (Sec 11.5) (a) Note that this wave function is equal to 1 2 2 x i 3 x 0 x L x, 0 3 3 elsewhere 0 Clearly this means that the coefficients in the expansion of x, 0 [Eq. (11.17)] are c2
1 2 , c3 i , and all other coefficients are 0. 3 3
As was the case in Example 11.1, the probability of measuring a particular energy is independent of time. Therefore, 1 2 P E2 , t , P E3 , t , all other probabilities are zero. 3 3
Chapter 11
171
(b) If the probability distribution is independent of time, the average will also be independent of time. E E2 P E2 E3 P E3
2 2 2 22 3 2 2mL2 3 3
112 2 3mL2
(c) x t
dx x, t x x, t
cm eimt cn e int m
x t
n
dx x x x m
n
2 mx nx i t cm cn e m n dx sin x sin L m,n L L 0 L
2 1 2 i 2 3 t i 3 2 t 2 a33 a e a e a22 i 23 32 3 L3 3 where L
mx nx amn anm dx sin x sin L L 0 xL
amn
n m x n m x x sin L cos L L L 2 2 nm 2 n m x 0
x L
n m x n m x x sin L cos L L L 2 2 nm 2 n m x 0 L2 cos n m 1 cos n m 1 2 2 22 n m n m note that this is NOT valid for n m
(A)
172
Chapter 11
a32
cos 5 1 L2 cos 1 2 25 2 L2 1 24 L2 1 2 25 252
For n m : L
nx ann dx x sin 2 L 0 x L
2nx 2nx cos 2 x sin x L L 2 4 n n 4 8 L L x 0
L2 cos 2n 1 2 2 4 n n 8 8 L L L2 4
Substituting back into (A): 1 2 24 2 x t 2 L i 2i sin 2 3 t 2 3 25 12 12
1 32 2 t sin L 2 3 2 2 25 1 L (0.18) sin 3 2 t 2
2
4 2 9 2 , 3 2mL2 2mL2
Chapter 11
(d) Using Ehrenfest's Thrm: d p t m x t dt d 1 32 2 mL sin 3 2 t 2 dt 2 25 32 2 3 2 cos 3 2 t 252 32 2 52 mL cos 3 2 t 252 2mL2 mL
16 2 cos 3 2 t 5L
11.14 (Sec. 11.5)
(a) First find the coefficients cn in the linear expansion of x, 0 :
173
174
Chapter 11
cn
dx n x x, 0
2 2 L
L /2
0
2x nx dx sin sin L L x L /2
n 2 x n 2 x sin sin L L 2 n2 n2 x 0 n 2 n 2 sin sin 2 2 2 n2 n2
2 1 n 1 sin 2 n 2 n 2 n sin 2 n 4 4 2
2
Note that this expression is not valid for n 2 , which we need to treat separately.
c2
dx 2 x x, 0
2 2 L
L /2
0
2x dx sin 2 L x L /2
4x sin 2 2 x L L 2 2 4 L x 0
2 2
It is possible to split the coefficients into three categories:
Chapter 11 4 2 n sin 2 n 4 2 cn 0 2 2
175
n odd n even, n 2 n2
x, t cn n x e int n
8 L
n 1,3,5,...
n nx i n sin sin e 2 2 L n 4 1
2 1 2x i 2 sin e L L
mL2 t
2 2
2 mL2 t
.
(b) As was the case in Example 11.1, the probability of measuring a particular energy is independent of time. Therefore,
P En , t cn
2
32 2 2 2 n 4 0 1 2
n odd n even, n 2 n2
11.15 (Sec. 11.5)
(a) First find the coefficients cn in the linear expansion of x, 0 :
176
Chapter 11
cn
dx n x x, 0
L
30 2 nx 5 dx x L x sin L L0 L L L 2 15 nx 3 L dx x sin L dx x 2 sin L 0 L 0 x L
2 15 L nx nx sin x cos Ln n L L x 0
x L
2 15 2 2 nx nx 2 cos 2 sin L n x n Lx 3 L L x 0 L2 n
2 15 2 15 4 15 2 cos n 2 n cos n , 3 n n n 3
For odd n we have 2 15 2 15 4 15 8 15 2 2 n (n odd) , cn 3 3 3 n n n n while for even n we have 2 15 2 15 4 15 2 2 n cn 0 (n even) . 3 n n n 3
The time-dependent wave function is then: x, t cn n x e int n
2 2 30 1 nx i n sin e L n1,3,5,... n3 L
8 3
2 mL2 t
.
(b) As was the case in Example 11.1, the probability of measuring a particular energy is independent of time. Therefore,
P En , t cn
2
960 6 n 0
n odd n even
.
Chapter 11
177
This probability decreases extremely rapidly with n. The most probable energy is E1 [with
P E1 0.9986 ]. We could probably have guessed this, because the initial wave function
x, 0 (see above, or Fig. 11.9) looks very similar to 1 x (Fig. 11.5). (c) Energy must be conserved, so the average energy must be independent of time. Since the potential energy is zero inside the well, the energy is all kinetic, and is given by E
1 p2 2m
2
1 dx x, 0 i x, 0 2m x
30 2 5
2mL
L
dx x
2
Lx 2
0
x L
30 2 2 3 x Lx 2 5 3 2mL x 0 30 2 2 3 3 L L 2mL5 3
10 2 2mL2
Notice E E1 , as we'd expect since P E1 0.9986 . 11.16 (Sec. 11.5) L . By symmetry, we would 2 expect this result to be time independent. Let's verify that this is the case.
By simply examining initial wave function, we can see that x 0
In general: x t
dx x, t x x, t
c e
im t m
m
cn e
in t
dx x x x m
n
n
2 mx nx i t x t cm cn e m n dx sin x sin L m ,n L L 0 L
(A)
178
Chapter 11 x L
n m x n m x x sin L cos L L L L mx nx dx sin L x sin L 22 2 nm n m 0 x 0
x L
n m x n m x x sin L cos L L L 2 2 nm 2 n m x 0
L2 cos n m 1 cos n m 1 2 2 22 n m n m
Not that this equation fails for n m , which we'll have to treat separately. In Problem 11.15, the time dependent wave function is x, t cn n x e int n
8 3
2 2 30 1 nx i n sin e L n1,3,5,... n3 L
2 mL2 t
.
So, the only contributions to the wave function occur for odd terms in the expansion. If m and n are both odd, then n m and n m are both even. In these cases this integral above is 0, because the cosine's are 1. L
mx nx x sin 0 m odd, n odd . L L
dx sin 0
This means that the only possible nonzero terms in the sum of Eq. (A) are when n m (we haven't worked out this integral yet). Calculating the integral for n m :
Chapter 11
179
x L
2nx 2nx cos 2 x sin L L L 2 nx x sin dx x 2 n L 4 n 0 4 8 L L x 0
1 L2 cos 2n 2 2 4 n n 8 8 L L L2 4
x t
2 2 2 L c n 4 L n 1,3,5,...
L P En 2 n 1,3,5,...
Here we've used the fact that the square magnitude of the coefficients in the expansion are equal to the probabilities of the energies. Since the probabilities must be normalized, the sum must add to 1, and L x t . 2 11.17 (Sec. 11.5) Ehrenfest's Theorem states that d p t m x t . dt From the previous problem we know that x t is constant, so its derivative is zero, and hence p t 0 .
180
Chapter 11
11.18 (Sec. 11.5)
(a) First find the coefficients cn in the linear expansion of x, 0 :
cn
dx n x x, 0
L /2 L L 24 nx nx nx 2 dx x sin L dx sin dx x sin L L L 0 L L /2 L /2 x L /2
24 L L nx nx 2 sin x cos L n n L L x 0
x L
24 L nx nx L nx L cos x cos sin 2 n L L x L /2 L n L
24 nL
L n L n n sin 2 2 cos 2
24 nL
L L n L n n sin n n sin 2 2 cos 2
24 n n 2sin sin n 2 2 2
n sin n 2 4 6
2 2
This can be simplified as
Chapter 11 4 6 ( n 1)/2 2 2 1 cn n 0
181
n odd n even
x, t cn n x e int n
8
2
2 2 3 1 ( n 1)/2 nx i n 1 sin e L n1,3,5,... n 2 L
2 mL2 t
.
(b) First, note that E t is independent of time, so we just need to find E t 0 . There are two ways to approach this problem. The first is to use a sum. We know P En cn
2
96 n 4 4 0
n odd n even
Therefore E En P En n
n 2 2 2 96 2 4 4 n 1,3,5,... 2mL n 48 2 1 2 2 mL n 1,3,5,... n 2
48 2 2 2 2 mL 8
6 2 mL2
The other way is to use: 1 E p2 2m
2
1 dx x, 0 i x, 0 2m x
182
Chapter 11 12 0 x L/2 3 L 12 x, 0 3 L / 2 x L x L 0 elsewhere 12 1 2 x L / 2 0 x L L3 elsewhere 0 where x is the step function (see Complement 10.A). 2 12 x , 0 2 3 x L / 2 2 x L
E
2 12 dx x, 0 x L / 2 m L3 2 12 12 L m L3 L3 2 6 2 mL2
11.19 (Sec. 11.5) Using EQ. (9.19): d i p (t ) Hˆ , pˆ (t ) dt pˆ 2 i (t ) V xˆ , pˆ (t ) 2m i (t ) V xˆ , pˆ (t ) i (t ) pˆ , V xˆ (t ) Using Eq. (10.72):
Chapter 11
183
d i p (t ) iV xˆ (t ) dt d V x dx F x 11.20 (Sec. 11.5) Expand F x in a power series about the point x x F x F x F x
x x 12 F x x x
F x F x F x F x F x F x
1 F x 2 1 x x F x 2
2
...
x x
x x
x x
2
...
2
...
1 F x x 2 ... 2
Here we have used the fact that the variance of x is x 2
x x
2
. F x F x
We have an exact correspondence
if the second and higher derivatives of
F x are all zero. This occurs if F x const , or F x x .
Note that the harmonic oscillator is one of the special cases where F x F x . Despite this fact, expectation values for the harmonic oscillator can still behave in ways that we would not expect classically, as we'll see in Chapter 12. The approximation F x F x
is valid if the uncertainty in the particle's position x is
small compared to the length scale over which F x varies.
184
Chapter 11
11.21 (Sec. 11.5)
If E V0 , the wave function solution looks like: A1e x B1ex x A2 eikx B2 e ikx A e x B ex 3 3
x0 0 xL xL
with 1/ 2
2mE k 2 ,
2m V0 E 2
1/ 2
The normalization condition requires that the solutions may not diverge, which reduces the allowed wave function to B1ex x A2 eikx B2 e ikx A3e x
x0 0 xL xL
The boundary conditions (wave function and its derivative continuous) at x 0 mean B1 A2 B2 B1 ikA2 ikB2 Substituting (A) into (B) yields A2 B2 ikA2 ikB2
(A) (B)
Chapter 11 A2
ik B2 ik
(C)
The boundary conditions at x L mean A2 eikL B2 e ikL A3e L ikA2 e
ikL
ikB2 e
ikL
A3e
(D) L
(E)
Multiplying (D) by and adding it to (E) yields
ik A2eikL ik B2eikL 0 Substituting (C) into this equation yields ik ikL ikL ik B2 e ik B2 e 0 ik
ik eikL ik eikL 0 2 2 ik eikL ik eikL 0 2 2 Im ik eikL 0 2
2
Im 2 k 2 i 2k cos kL i sin kL 0 2k cos kL 2 k 2 sin kL 0 2k cos kL k 2 2 sin kL tan kL
2k k 2 2
k and are both functions of E, and the solutions of this equation determine E.
185
Complement 11.A Wave Packet Propagation 11.A.1* Equation (11.A.1) is 1 x 2 /4 2 ip0 x / x, 0 e e 1/2 2
p
dx x, 0 i x x, 0
x, 0 x
1
x 2 /42 ip0 x / e e x
2
p0 x 2 /42 ip0 x / 2 x i e e 1/2 2 4 2
1/2
1
p 2 x 2 i 0 x, 0 4
p i
p 2 x dx x, 0 42 i 0 x, 0
2 2 2 i 2 dx x x, 0 p0 dx x, 0 4
The first integral is 0 because the integrand is an odd function. The second integral is 1 because it is just the normalization integral of the wave function. Therefore p p0 .
187
188
Complement 11.A
11.A.2 Equation (11.A.4) is 1/4
2 p p0 2 2 / 2 p . e
p
dp p p p *
2 1/4
2
dp pe
2 p p0 2 / 2 2
Substitute: u p p0 , du dp 2 1/4 p
2
du u e
2
2 u 2 2 / 2
2 1/ 4 2 u 2 2 / 2 p 0 du e
The first integral is 0 because the integrand is an odd function. The second integral is
p p0 du u p0 . 2
The integral is 1 because the wave function is normalized. 11.A.3*
E k 22 2m
vg
k p d 0 0 v0 dk k k0 m m
vp
k0 p0 v0 k0 2 m 2 m 2
Chapter 12 The Harmonic Oscillator 12.1* (Sec. 12.2) m i i † aˆ , aˆ 2 xˆ m pˆ , xˆ m m i pˆ , xˆ xˆ, pˆ 2 m 1 i 2i 2 1
12.2
pˆ
(Sec. 12.2) nˆ , aˆ † aˆ † aˆ , aˆ † aˆ † aˆ , aˆ † aˆ † , aˆ † aˆ aˆ † . Here we've used Eq. (7.56).
12.3* (Sec. 12.2) nˆ , aˆ † na ˆ ˆ † aˆ † nˆ aˆ † ˆ ˆ † aˆ † nˆ aˆ † na nˆ aˆ † n aˆ † nˆ aˆ † n aˆ † n aˆ † n n 1 aˆ † n
n 1 c n 1
because aˆ n is an eigenstate of nˆ with eigenvalue n 1 . †
aˆ † n c n 1 n aˆ n 1 c
190
Chapter 12
ˆ ˆ † n n 1 c c n 1 n aa
ˆ ˆ † aˆ † aˆ 1 aa ˆ ˆ † aˆ † aˆ 1 Note: aˆ , aˆ † aa n aˆ † aˆ 1 n c
2
n nˆ 1 n c
2
n 1 c
2
n 1 c aˆ † n n 1 n 1 12.4* (Sec. 12.2) In Example 12.1 we showed that for the state n , 1/2
1 x n 2 m
En . m2
For this same state we have p n pˆ n i 2 i 2 i 2 0
n aˆ aˆ † n n aˆ n n aˆ † n n n n 1 n 1 n n 1
Chapter 12
191
p 2 n pˆ 2 n 2 2 2 2 2 2 2 2 2
n aˆ aˆ † n 2
n aa ˆ ˆ n n aa ˆ ˆ † n n aˆ † aˆ n n aˆ † aˆ † n n n aˆ n 1 n 1 n aˆ n 1
n n aˆ † n 1 n 1 n aˆ † n 1 2 2 n(n 1) n n 2 (n 1) n n 2 n n n (n 1)(n 2) n n 2 m 2n 1 . 2
p
p p 2
2 1/ 2
1 m n 2
1 xp n 2 m 12.5
1/2
1/2
mEn 1/2
1 m n 2
1 n 2 2
(Sec. 12.2) aˆ n n n 1 amn m aˆ n n m n 1 n m ,n 1 this means that all of the elements are 0, except that in column n, the element in row n 1 is equal to n . (NOTE on numbering: since n 0,1, 2,... , I'm numbering the columns 0,1,2,…, not 1,2,3,…) In other words: 0 0 aˆ 0 0
1 0 0 0
0 2 0 0
0 3 0 0
192
Chapter 12
12.6
(Sec. 12.2) aˆ † n n 1 n 1 † amn m aˆ † n n 1 m n 1 n 1 m ,n 1
This means that all of the elements are 0, except that in column n, the element in row n 1 is equal to n 1 . (NOTE on numbering: since n 0,1, 2,... , I'm numbering the columns 0,1,2,…, not 1,2,3,…) In other words: † aˆ
12.7
0
0
1
0
0 0
2 0
0 0 0 0 3 0 0
0
(Sec. 12.2) 1 1 2 2 x xˆ
1
aˆ aˆ †
2 1 aˆ aˆ † 2 1 1 aˆ 1 1 aˆ 2 2 aˆ 1 2 aˆ 2 1 aˆ † 1 1 aˆ † 2 2 aˆ † 1 2 aˆ † 2 2 2 1 0 2 0 0 0 0 2 0 2 2 1 x 2 xˆ 2 2 1 aˆ aˆ † 2 2 1 ˆ ˆ aa ˆ ˆ † aˆ † aˆ aˆ † aˆ † 2 aa 2
Chapter 12
193
ˆ ˆ † aˆ † aˆ 1 , so Using aˆ , aˆ † 1 , aa 1 ˆ ˆ aˆ † aˆ † aˆ † aˆ 1 aˆ † aˆ x 2 2 aa 2 0 0 1 2 2nˆ 1 2 1 2 1 2nˆ 1 1 1 2nˆ 1 2 2 2nˆ 1 1 2 2nˆ 1 2 4 1 3 0 0 5 42 2 2
x
x
2
x
2 1/2
2
2 1 1 2 2
p pˆ aˆ aˆ † i 2 aˆ aˆ † i 2 1 aˆ 1 1 aˆ 2 2 aˆ 1 2 aˆ 2 1 aˆ † 1 1 aˆ † 2 2 aˆ † 1 2 aˆ † 2 i2 2 0 2 0 0 0 0 2 0 i2 2 0
194
Chapter 12 p 2 pˆ 2 2 2 2 aˆ aˆ † 2 2 2 ˆ ˆ aa ˆ ˆ † aˆ † aˆ aˆ † aˆ † aa 2 2 2 ˆ ˆ aˆ † aˆ † aˆ † aˆ 1 aˆ † aˆ aa 2 0 0 2 2 2nˆ 1 2 2 2 1 2nˆ 1 1 1 2nˆ 1 2 2 2nˆ 1 1 2 2nˆ 1 2 4
2 2 3 0 0 5 4 2 2 2
p
p2 p
1 xp 12.8
2 1/2
2
2 2
2
(Sec. 12.2) 1 1 2 2 x xˆ
1
aˆ aˆ † 2 1 2 aˆ 2 2 aˆ 4 4 aˆ 2 4 aˆ 4 2 aˆ † 2 2 aˆ † 4 4 aˆ † 2 4 aˆ † 4 2 2 1 0 0 0 0 0 0 0 0 2 2 0
Chapter 12
195
x 2 xˆ 2 2 1 aˆ aˆ † 2 2 1 ˆ ˆ aa ˆ ˆ † aˆ † aˆ aˆ † aˆ † 2 aa 2
ˆ ˆ † aˆ † aˆ 1 , so Using aˆ , aˆ † 1 , aa
x2
1 ˆ ˆ aˆ † aˆ † aˆ † aˆ 1 aˆ † aˆ aa 2 2
1 1 1 ˆ ˆ 4 4 aˆ † aˆ † 2 2nˆ 1 2 aa 2 2 2 2 1 2 4 3 4 3 2 2nˆ 1 2 2 2nˆ 1 4 4 2nˆ 1 2 4 2nˆ 1 4 4 1 2 4 3 5009 4 5.23 2
x
x
2
x
2 1/2
5.23 2.29
p pˆ aˆ aˆ † i 2 aˆ aˆ † i 2 2 aˆ 2 2 aˆ 4 4 aˆ 2 4 aˆ 4 2 aˆ † 2 2 aˆ † 4 4 aˆ † 2 4 aˆ † 4 i2 2 0 0 0 0 0 0 0 0 i2 2 0
196
Chapter 12 p 2 pˆ 2
2 2 2 2 2 2 2 2 2 2 2 2
aˆ aˆ † 2
ˆ ˆ aa ˆ ˆ † aˆ † aˆ aˆ † aˆ † aa aa ˆ ˆ aˆ † aˆ † aˆ † aˆ 1 aˆ † aˆ 1 1 ˆ ˆ 4 4 aˆ † aˆ † 2 2nˆ 1 2 2 aa 2
2 2 4 3 4 3 2 2nˆ 1 2 2 2nˆ 1 4 4 2nˆ 1 2 4 2nˆ 1 4 4 2 2 2 3 5009 4 2.63 22
p
p2 p
2 1/2
2.63 1.62
2.29 xp 1.62 3.71 2 12.9
(Sec. 12.2) From Section 11.5, we know that Ehrenfest's theorem says d x dt d d p V x dt dx p m
d 1 2 2 m x dx 2
m2 x
For a Fock state: x 0 from Example 12.1.
(A)
(B)
Chapter 12
197
p n pˆ n
i 2 i 2 i 2 0
n aˆ aˆ † n n aˆ n n aˆ † n n n n 1 n 1 n n 1
Since x 0 and p 0 , (A) and (B) are both satisfied. 12.10* (Sec. 12.3) The differential equation is 0 x 2 x 0 x x Substitute in the solution 1/ 4
2 2 x2 /2 , 0 x e
and we obtain 1/4
1/4
2 2 2 2 x2 /2 2 2 x e x /2 e x 2 x 0 x
So it is a solution to the differential equation. The function 0 x is a Gaussian wave packet, as given in Complement 10.A, with x 0 , and 2 1/ 22 . We know that such a wave packet is properly normalized.
12.11 (Sec. 12.3) The recursion formula 'Eq. (12.43)] is: H n 1 x 2 xH n x 2nH n 1 x
We know H 0 x 1 , H1 x 2 x .
198
Chapter 12 H 2 x 2 xH1 x 2 1 H 0 x 2 x 2 x 2 1 4 x 2 2
H 3 x 2 xH 2 x 2 2 H1 x 2 x 4 x 2 2 4 2 x 8 x 3 12 x
H 4 x 2 xH 3 x 2 3 H 2 x 2 x 8 x3 12 x 6 4 x 2 2 16 x 4 48 x 2 12 12.12 (Sec. 12.3) n
2 d H n x 1 e e x dx
n x2
H 0 x e x e x 1 2
2
2 d 2 2 2 H1 x 1 e x e x 1 e x 2 x e x 2 x dx
2
2 2 d 2 2 2 d H 2 x 1 e e x e x 2 x e x e x 2 4 x 2 e x 4 x 2 2 dx dx
2
x2
12.13 (Sec. 12.3) 1/4
2 n x
1 2 n! n
H n x e x
2 2
1/4
2 x 2 n x n n 1 x
/2
1 x 2 n H n x 2 n!
22 H n 1 x e x /2 2n 1 n 1 ! 1
n
1/4
2
1 x 2 n 1 H n x 2 n!
22 2n H n 1 x e x /2 n 1 2 n n 1 ! 1
1/4
2
1 2
n 1
2xH n x 2nH n 1 x e x
2 2
n!
/2
Using the recursion formula [Eq. (12.43)] H n 1 x 2 xH n x 2nH n 1 x :
Chapter 12 1/4
2 x 2 n x n n 1 x
1 2
n 1
H n 1 x e x
2 2
n!
1/4
2
n 1
1 2
n 1
n 1!
n 1 n 1 x 12.14* (Sec. 12.4) 1 1 e iE1t / 2 e iE2t / 2 1 1 e i 3t /2 2 e i 5t /2 2 1 1 e it 2 e i 2 t e it /2 2 x t t xˆ t
t
1
t aˆ aˆ † t
2 1 t aˆ t t aˆ † t 2 1 i 2 t i 2 t 1 aˆ 2 e 2 aˆ † 1 e 2 2 all other matrix elements are 0.
1 2 e it 2 e it 2 2 1 cos t
x t
Ehrenfest's theorem says d p m x , dt so p t m
d 1 m sin t cos t dt
/2
H n 1 x e x
2 2
/2
199
200
Chapter 12 H t t Hˆ t 1 t nˆ t 2 1 2 2 2 2
This is time-independent, as it should be. 12.15* (Sec. 12.4) 1 n e iEnt / n e iEnt / 2 1 n e int n e int e it /2 2 x t t xˆ t
t
1
t aˆ aˆ † t
2 1 n aˆ n n aˆ † n n aˆ n ei n n t n aˆ † n ei n n t 2 2
i n nt i n nt n aˆ n e n aˆ † n e n aˆ n n aˆ † n 1 i n nt i n nt n n ,n ' 1 n 1 n , n '1 e n n,n 1 n 1 n,n 1 e 2 2
Three possibilities: If n n 1 x t
1 n 1 e it n 1 eit 2 2 1 n 1 cos t 2
If n n 1 x t
1 n eit n e it 2 2 1 n cos t 2
Chapter 12
201
Otherwise, x t 0 . Thus, if n and n are adjacent states ( n n 1 ) the expectation value oscillates. Otherwise, it does not. 12.16 (Sec. 12.5) From Section 11.5, we know that Ehrenfest's theorem says d x dt d d p V x dt dx p m
d 1 2 2 m x dx 2
m2 x
For a coherent state, we learned in Sec 12.5 that x t A cos t 0 . p t mA sin t 0 .
m
d x t mA sin t 0 p t dt
d p t m2 A cos t 0 m2 x t dt Thus, Ehrenfest's theorem is satisfied.
(A)
(B)
202
Chapter 12
12.17* (Sec. 12.5) From Eq. (12.68) we know that: 1 X 2 P Pˆ 1 aˆ aˆ † 2i 1 aˆ aˆ † 2i 1 2i Im
P 2 Pˆ 2 1 aˆ aˆ † aˆ aˆ † 4 1 ˆ ˆ † aˆ † aˆ aˆ 2 aˆ †2 aa 4 ˆ ˆ † aˆ † aˆ 1 , so Using aˆ , aˆ † 1 , aa
1 aˆ 2 aˆ †2 2aˆ † aˆ 1 4 2 1 2 2 1 4
P2
2
1 1 2i 4 1 2 Im() . 4
P
P2 P
2
2 xp X
1 2
2P
. 2
Coherent states are minimum uncertainty states of position and momentum.
Chapter 12 12.18 (Sec. 12.5) 2
aˆ † , aˆ aˆ † , aˆ
2
Thus, Glauber's formula says: † † Dˆ eaˆ aˆ eaˆ e aˆ e
2
/2
/2 aˆ † aˆ 0 Dˆ 0 e e e 2
e
2
/2 aˆ †
e
0
This is true because 0 is an eigenstate of aˆ , with eigenvalue 0.
Dˆ 0 e
/2
e
/2
2
2
n 0
n 0
aˆ aˆ aˆ aˆ aˆ
aˆ
† n
n!
n † n aˆ 0 n!
† 0
0 0
† 1
0 11
† 2
0 aˆ † 1 1 2 1 2
† 3
0 aˆ †
† n
0 n! n
Dˆ 0 e
0
2 1 2 3 2 1 3
2
/2
n0
n n n!
12.19* (Sec. 12.5)
x aˆ t x t t m i xˆ 2 m
pˆ t x t t 1 d it x 2 x, t 0 e x, t dx 2 where we've used the fact that t 0 e it . Subsituting in for x, t :
x
203
204
Chapter 12 1/4
1/4
2 2 1 d 2 2 x x t 2 /2 i p t x / 2 x x t /2 i it x e e e e 0 e 2 dx 2 p t 1 2 it x 2 x x t i 0e 2
p t x /
p t x t i 0 ei0 e it 2 2
From Eqs. (12.72) and (12.73) we know 2 2 0 cos t 0 0 cos t 0 m
x t
p t 2m 0 sin t 0 2 0 sin t 0
Substituting these in, we see, 0 cos t 0 i 0 sin t 0 0 e 0 e
i t 0
0 e
i t 0 i t 0
So x, t is the solution. 12.20 (Sec. 12.5) 1/4
2 2 x x t 2 /2 i e x, t e
x t
p t x /
dx x x, t 2
1/2
2
2 2 x x t dx xe
This is exactly the integral for the expectation value for a Gaussian wavepacket, given in Complement 10.A (with x t x , and 2 1/ 22 ), so x t x t .
Chapter 12
p t
dx x, t i x, t x 1/2
2
p t 2 x x t 2 dx 2 x x t i e 2 2 dx x x, t x t dx x, t
i
i 2
p t dx x, t
2
We showed above that
dx x x, t x t , 2
and from normalization we know that
2
dx x, t 1 ,
so p t i 2 x t x t p t p t 12.21* (Sec. 12.5)
P n n e
2
2
/2
2n
n!
e
n n!
2
.
2
205
Complement 12.A Solving the Schrödinger Equation Directly 12.A.1 d2 d
2
2
h e
2
/2
d2 2 2 /2 2 /2 h e h e d 2
2 2 d 2 /2 d h h e /2 2 h e /2 e d d
e
2
/2
d2 d 2
h 2e
2 h e d2 d
2
h 2
2
2
n
hn x a j x j j 0
2 j 1 n a j 2 j 1 j 2 j 1 2n 1
j 2 j 1
For n 0
2 2 d h 2 h e /2 h e /2 d
/2
d h 1 h 0 . d
12.A.2
a j2
/2
aj
208
Complement 12.A h0 x a0 .
If a0 1 , h0 x 1 H 0 x . For n 1 a0 0 h1 x a1 x
If a1 2 , h1 x 2 x H1 x . For n 2 a1 0 4 2 h2 x a0 a2 x 2 a0 a0 x 2 1 If a0 2 , h2 x 4 x 2 2 H 2 x . For n 3 a0 0 26 3 h3 x a1 x a3 x3 a1 x a1 x 3 2 If a1 12 , h3 x 8 x 3 12 x H 3 x .
Chapter 13 Wave Mechanics in Three Dimensions 13.1
(Sec. 13.1) In the tables below, the energy is expressed in units of 2 2 / 2mL2 : nx ny nz E nx ny 1 1 1 3 2 2 1 1 2 6 2 3 1 2 1 6 3 2 2 1 1 6 1 1 1 2 2 9 1 4 2 1 2 9 4 1 2 2 1 9 1 3 1 1 3 11 3 1 1 3 1 11 3 3 3 1 1 11 1 2 2 2 2 12 1 4 1 2 3 14 2 1 1 3 2 14 2 4 2 1 3 14 4 1 2 3 1 14 4 2 3 1 2 14 3 2 1 14 The degeneracies of the levels are: Degen. E 3 1 6 3 9 3 11 3 12 1 14 6 17 3 18 3 19 3 21 6
nz 3 2 2 4 1 1 3 3 1 4 2 4 1 2 1
E 17 17 17 18 18 18 19 19 19 21 21 21 21 21 21
210
Chapter 13
13.2
(Sec. 13.1) The Schrödinger equation is 1 2 2 2 2 2 2 2 2 2 2 2 r m x y z r E r 2m x 2 y z
Substitute in a solution of the form r X x Y y Z z , and then divide through by r . The result is 1 1 2 1 2 2 1 2 X x Y y Z z m2 x 2 y 2 z 2 E 2 2 2 2 Y y y Z z z 2m X x x 2 2 1 2 1 1 2 1 2 2 X x m x Y y m2 y 2 2 2 2 2 2m X x x 2m Y y y 2 1 2 1 2 2 Z z m z E 2 2 2m Z z z
Each of the terms in brackets depends only on a single variable, and hence must be constant in order for the terms to add to a constant. We thus have Ex E y Ez E , and the differential equation for the x-direction is: 2 2 1 X x m2 x 2 X x E x X x . 2 2m x 2 This is exactly the equation for the one-dimensional harmonic oscillator [Eq. (12.A.1)], so the solutions must be the same. 1 Enx nx nx 0,1, 2, . . . 2 and the wave functions are given by the Hermite Gaussians, Eq. (12.42). Since the differential equations for the other directions are the same, they will also have the same solutions, so the final energies are 3 Enx ny nz nx n y nz nx , ny , nz 0,1, 2, . . . 2 The final wave functions are given by the product of Hermite Gaussians along three directions.
Chapter 13
13.3
(Sec. 13.2) z z r x 2 y 2 z 2 , cos 1 cos 1 x2 y 2 z 2 r
y , tan 1 x
x r sin cos , y r sin sin , z r cos r x x r x x 2x 2 xz 1 1 y 2 2r 3 y 2 x 2 2r r z 1 1 x r sin cos
1 sin 1 sin cos cos 2 2 r r sin 1 tan r sin cos
sin cos
sin cos cos r r r sin
r y y r y y 2y 2 yz 1 1 1 2 2r 3 y 2 x 2r r 1 z 1 x r sin sin
1 1 1 sin sin cos 2 r r sin 1 tan r sin cos
sin sin
cos cos sin r r sin r
211
212
Chapter 13 r z z r z z 2z2 1 2z 1 0 3 2 2r r r 2 r 1 z r 1 cos 2 1 cos r sin r sin cos r r
13.4
(Sec. 13.2)
Yl Yl
ml
, 1
ml
ml
, 1
2l 1 l ml ! 4 l ml !
ml
1/ 2
Pl ml cos eiml
2l 1 l ml ! 4 l ml !
1/ 2
Pl ml cos e iml
From (13.48) we know m l ml ! m Pl ml x 1 l P l x l ml ! l 1/ 2
Yl
ml
, 1
ml
1
ml
2l 1 l ml ! ml l ml ! m Pl l x e iml 1 l ml ! 4 l ml ! 1/2 ml 2l 1 l ml ! im m 1 Pl l cos e l 4 l ml !
1 l Yl ml , m
13.5
(Sec. 13.2) Call the spherical harmonics Yl ml , clml Pl ml cos eiml , where
clml 1
ml
2l 1 l ml ! 4 l ml !
1/ 2
Chapter 13 2
0
d d sin Ylml , Ylm , 0
2
clml clml
0
iml l d d sin Plml cos eiml Plm cos e 0
clml clml d sin Pl
ml
cos
l Plm
0
2
cos
d eiml eiml
0
Use Eq. (13.42): 2
0
d d sin Ylml , Ylm , 0
l clml clml d sin Plml cos Plm cos 2 ml ml
0
l clml clml 2 ml ml d sin Plml cos Plm cos
0
Use Eq. (13.49): 2
0
13.6
d d sin Ylml , Ylm , clml clml 2 ml ml 0
(Sec. 13.3) In 3 dimensions: pˆ i , so Lˆ rˆ pˆ i r
2 l ml ! l l ml ml l l 2l 1 l ml !
213
214
Chapter 13
13.7* (Sec. 13.3)
x r sin cos , y r sin sin , z r cos ˆ ˆ z zp ˆˆ y Lˆx yp i y z y z cos sin cos sin i r sin sin cos r cos sin sin r r r r r sin i sin 2 sin cos 2 sin cot cos i sin cot cos ˆˆ z ˆ ˆ x xp Lˆ y zp
i z x z x sin cos cos sin i r cos sin cos r sin cos cos r r r r sin r i cos 2 cos sin 2 cos cot sin i cos cot sin ˆˆ y yp ˆˆx Lˆz xp i x y x y cos sin cos i r sin cos sin sin r r r sin sin cos cos r sin sin sin cos r r r sin i cos 2 sin 2 i
Chapter 13
215
13.8* (Sec. 13.3) Lˆ2x i sin cot cos i sin cot cos 2 2 sin 2 2 sin cos cot cot cos sin cot 2 cos
cos
1 2 2 sin 2 2 sin cos cot 2 cot cos sin cos sin 2 cot 2 cos cos 2 sin 2 2 2 2 2 2 sin 2 2 cot sin cos cot cos 2 cot cos 2
sin cos cot 2 cos sin 2 sin
Lˆ2y i cos cot sin i cos cot sin 2 2 cos 2 2 cos sin cot cot sin cos cot 2 sin
sin
1 2 2 cos 2 2 cos sin cot 2 cot sin cos sin sin 2 cot 2 sin sin 2 cos 2 2 2 2 2 2 cos 2 2 cot sin cos cot sin 2 cot sin 2 2 2 2 ˆ Lz 2
sin cos cot 2 cos sin 2 sin
216
Chapter 13 Lˆ2 Lˆ2x Lˆ2y Lˆ2z 2 2 2 2 2 cot 2 2 cot 2 1 1 2 sin 2 sin 2 2 sin
13.9
(Sec. 13.3) Starting from Eq. (13.114):
Lˆ i r 1 1 i rur ur rur u rur u r r sin r 1 i u u sin 13.10 (Sec. 13.3) 1 rˆ rˆ pˆ pˆ 2 r r 1 rˆ 1 rˆ r pˆ r r pˆ r pˆ 2 r 2 r ih ur r ur r 2
pˆ r
Using Eqs (13.116) and (13.118) ih 1 2 r pˆ r r 2 r r 2 r r r ih ih r 2 2r r r 2 r 2 r 2r r ih ih r r r r 1 i r r r
1 1 pˆ r i i r r r r r
Chapter 13 13.11* (Sec. 13.3) r Hˆ r E pˆ r2 Lˆ2 r V rˆ r E 2m 2 I
Using Eqs. (13.60) and (13.63) this becomes 2 1 2 2 r 2 2m r r r 2 I
1 1 2 sin V r r E r \ 2 2 sin sin
Using I mr 2 and assuming a central potential: 1 2 1 1 2 r 2 r r r r r 2 sin sin r r 2 sin 2 2 r V r r E r .
2 2m
13.12* (Sec. 13.3) From Eq. (7.19): Lˆ Lˆ x iLˆ y Using Eqs. (13.57) and (13.58) Lˆ i sin cot cos cos cot sin cos i sin i cot cos i sin ei i cot ei i cot Yl ml , e
i
1
ml
e
i
1
ml
2l 1 l ml ! 4 l ml !
1/2
2l 1 l ml ! 4 l ml !
1/2
ml iml i cot Pl cos e ml iml ml Pl cos i cot Pl cos iml e
217
218
Chapter 13 ei i cot Yl ml , 1
ml
2l 1 l ml ! 4 l ml !
1/ 2
ml i ml 1 ml Pl cos ml cot Pl cos e
From Eq. (13.45): Pl ml cos 1 cos 2
ml /2
m
l d Pl cos d cos
m
l d sin Pl cos d cos
ml
Pl
ml 1
cos sin
ml 1
sin
ml 1
d d cos
ml 1
Pl cos m
l d d Pl cos d cos d cos
Let x cos , then dx d cos sin d
Pl
ml 1
cos sin
ml 1
m
l d d Pl cos sin d d cos
m
l d d sin Pl cos d d cos d 1 sin ml Pl ml cos ml d sin cos d ml 1 sin ml ml Pl cos ml Pl ml cos ml 1 sin sin d
ml
d Pl ml cos ml cot Pl ml cos d Substitute this into (A):
(A)
Chapter 13
219
ei i cot Yl ml , 1
ml
2l 1 l ml ! 4 l ml !
1/ 2
Pl ml 1 cos ei ml 1 1/2
l ml 1 ! l ml 1 ! l ml 1 ! l ml 1 ! l ml ! l ml 1! l ml 1! l ml !
1
1/2
1
ml 1
ml 1
2l 1 l ml ! 4 l ml !
1/2
2l 1 l ml 1 ! 4 l ml 1 !
Pl ml 1 cos e
i ml 1
1/2
Pl ml 1 cos e
i ml 1
l ml l ml 1 Yl ml 1 , 1/2
l 2 l ml2 ml Yl ml 1 , 1/2
l l 1 ml ml 1 Yl ml 1 , 1/2
This is consistent with the position representation of Eq. (7.20). 13.13* (Sec. 13.4) Eq. (13.74) is: l l 1 1 2 2 R R R 0 2 2 4 Eq. (13.82) is: R l e /2 w , so 2 2 l e /2 w R 2 2
l 1 / 2 1 l e w l e /2 w l e /2 w 2
1 l l 1 l 2 w ll 1w ll 1 w 2 1 1 1 w ll 1w l w l 2 4 2 ll 1
1 2 w l w l 2 w e /2 2
2 1 l l 1 l l l 2 l 1 w e / 2 l l 1 l w 2l w 2 4
220
Chapter 13
2 2 1 R ll 1w l w l w e /2 2 w e /2 2ll 2 l 1 w 2l 1
Substituting these into Eq. (13.74), rearranging terms, and dividing by a common factor of e /2 yields: l
2 1 w 2ll 1 l w l l 1 l 2 ll 1 l w 2 4
2l 1
l
l l 1 1 l w 2ll 2 l 1 w w 0 2 4
2 w 2 l 1 l 1 l w 2
1 1 l l 1 l 2 ll 1 l 2ll 2 l 1 l 1 l l 1 l 2 l w 0 4 4 l
2 w 2 l 1 l 1 l w l 1 l 1 w 0 2
2 2 w 2 l 1 w l 1 w 0
This is Eq. (13.83). 13.14* (Sec. 13.4) ˆˆ y yp ˆˆx Lˆz xp Lˆz , xˆ xp ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ y ypx , x y px , x iy Lˆz , yˆ xp ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ y ypx , y x p y , y ix Lˆz , zˆ 0 Lˆz , pˆ x xp ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ y ypx , px x, px p y ip y Lˆz , pˆ y xp ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ y ypx , p y y, p y px ipx Lˆz , pˆ z 0
Chapter 13
221
2 pˆ 2 pˆ pˆ 2 Hˆ x y z V xˆ 2 yˆ 2 zˆ 2 f pˆ x2 pˆ y2 pˆ z2 V xˆ 2 yˆ 2 zˆ 2 2m 2m 2m
Lˆz , pˆ x2 pˆ y2 pˆ z2 Lˆz , pˆ x2 Lˆz , pˆ y2 Lˆz , pˆ z2 Using Eq. (7.56): Lˆz , pˆ x2 pˆ y2 pˆ z2 pˆ x Lˆz , pˆ x Lˆz , pˆ x pˆ x pˆ y Lˆz , pˆ y Lˆz , pˆ y pˆ y i pˆ x pˆ y pˆ y pˆ x pˆ y pˆ x pˆ x pˆ y 0
Therefore: Lˆz , f pˆ x2 pˆ y2 pˆ z2 0 Lˆz , xˆ 2 yˆ 2 zˆ 2 Lˆz , xˆ 2 Lˆz , yˆ 2 Lˆz , zˆ 2 xˆ Lˆz , xˆ Lˆz , xˆ xˆ yˆ Lˆz , yˆ Lˆz , yˆ yˆ ˆˆ yx ˆ ˆ yx ˆ ˆ xy ˆˆ i xy 0 Therefore: Lˆz ,V xˆ 2 yˆ 2 zˆ 2 0 This means Lˆz , Hˆ 0 . We could repeat the above for Lˆx and Lˆ y , or simply note that the Hamiltonian is radially symmetric, this means that there's nothing special about the z-axis, and we could just as well call it the x-axis or the y-axis. Therefore: Lˆx , Hˆ 0 Lˆ y , Hˆ 0 Therefore Lˆ2 , Hˆ Lˆ2x Lˆ2y Lˆ2z , Hˆ 0
222
Chapter 13
13.15 (Sec. 13.4) The final state for the Lyman series ends at n 1 . The shortest wavelength corresponds to the largest frequency, which is the largest energy. The largest energy transition has as its initial state E 0 . The energy is thus
E Ei E f E E1 13.6 eV The corresponding wavelength is 15 8 c hc 4.14 10 eV s 3.00 10 m/s 91.3 nm 13.6 eV f E which is in the ultraviolet.
13.16 (Sec. 13.4) The final state for the Paschen series ends at n 3 . The longest wavelength corresponds to the smallest frequency, which is the smallest energy. The smallest energy transition has as its initial state E4 . The energy is thus 1 1 E E4 E3 13.6 eV 2 2 0.661 eV 4 3 The corresponding wavelength is 15 8 c hc 4.14 10 eV s 3.00 10 m/s 1.88 m 0.661 eV f E which is in the infrared.
13.17 (Sec. 13.4) Ignoring constant multipliers, the radial wave function for l n 1 is: 2r n 1 r / na0 , Rn l n 1 r r n 1e r / na0 L20n 1 r e na0 because from Eq. (13.87) Lk0 x e k e k 1 . The radial probability density is then
pn l n 1 r r 2 r n 1e r / na0
Differentiating,
2
r 2 n e 2 r / na0 .
Chapter 13
223
d 2 2 n 2 r / na0 pn l n 1 r 2nr 2 n 1e 2 r / na0 r e dr na0 Solving for the root of this equation, we find that the max is located at: 2 2 n 2 r / na0 r e 0 2nr 2 n 1e 2 r / na0 na0 n
1 r na0
r n 2 a0 13.18* (Sec. 13.4) R21 r
1 1 r r /2 a0 e 2 6a03 a0 2
1 1 r r / a0 1 4 r / a0 p21 r r R21 r r r e e 3 4 6a0 a0 24a05 2
2
2
Integrate this to get the probability (use www.wolframalpha.com for the integral): 3 a0 3 a0 1 4 r /a 0 dr p21 r 24a05 0 dr r e 0
1 3a05 8 131e 3 5 24a0
131 8e3 0.18 1
13.19 (Sec. 13.4) For 3s the wave function is 2 1 300 r R30 r Y , 9 3a03 0 0
2 r 2 r 2 3 e r /3a0 a0 9 a0
1 4
For a small volume the wave function is approximately constant, so the probability is given by 2
P r 0 300 0
2
2 1 1 4 3 dV 3 R , 3 4 3 9 3a0
where R is the radius of the nucleus. This gives
224
Chapter 13 P r 0
3 1 1 4 1015 m 3.34 1016 3 9 3 0.529 1010 m 3
13.20* (Sec. 13.4) 100 r R10 r Y00 , 2
1 r / a0 e a03
1 4
1 r / a0 e a03 1 Y00 , 4 R10 r 2
4 2 r / a0 e a03
p10 r r 2 R10 r r 2 2
Use www.wolframalpha.com for the integrals.
4 dr rp10 r 3 a0
r 0
r 2 dr r 2 p10 r 0
dr r 3e 2 r / a0
4 a03
0
4 3a04 3 a0 a03 8 2
dr r 4 e 2 r / a0
0
4 3a05 2 3a0 3 a0 4
z r cos
2
0
0
0
z dr d d r 2 sin r cos r 2
2 0
2
dr r 2 R10 r r d d sin cos Y00 , 0
0 2
0
0
0
dr p10 r r d d sin cos
2
1 4
r
1 d sin cos 2 0
0 Could have noted that this integral is odd in z, so it will be 0.
Chapter 13
2
0
0
0
z 2 dr d d r 2 sin r cos r 2
2 2
dr r R10 r r 2
0
0 2
0
0
2
2
d d sin cos 2 Y00 ,
2
0
dr p10 r r 2 d d sin cos 2 0
225
1 4
r2
1 d sin cos 2 2 0
3a02
12 a02 23
13.21 (Sec. 13.4) Let 1 E1 / and 2 E2 / , then 1 i1t 1 i1t t e e 1, 0, 0 e it 2,1, 0 1, 0, 0 e i2t 2,1, 0 2 2
,
where 2 1 .
z t t zˆ t 1, 0, 0 zˆ 1, 0, 0
1 1, 0, 0 zˆ 1, 0, 0 2 Re e it 1, 0, 0 zˆ 2,1, 0 2,1, 0 zˆ 2,1, 0 2
dxdydz z 100 r
2
all space 1/2 1 x 2 y 2 z 2 / a0 dxdydz z 2 3 e a0 all space
This integral is odd in z, so it must be 0.
1 4
2
226
Chapter 13 2,1, 0 zˆ 2,1, 0
dxdydz z 210 r
2
all space
dxdydz z R21 r Y10 ,
2
all space
2 3 dxdydz z R r cos 2 21 4 all space
2 z 3 dxdydz z R21 r 4 all space r
R r 3 dxdydz z 3 21 4 all space r
2
2
Since r is even in z, this integral is odd in z, so it is 0 as well. 2 1 1, 0, 0 zˆ 2,1, 0 dr d d r 2 sin 2 3 e r / a0 a0 0 0 0
1 1 3 2 6a0
r r /2 a0 3 e cos 4 a0
1 4a03 a0 1
1 4a04
2 3 4 3r /2 a0 2 dr r e d sin cos d 3 24a0 0 0 0
256a05 2 2 2 81 3
256 a0 0.74a0 243 2
Used www.wolframalpha.com for the integrals. z t 0.74a0 cos t .
13.22 (Sec. 13.4)
V r
e2 4 0 r
p10 r r 2 R10 r r 2 2
1 r cos 4
4 2 r / a0 e a03
Chapter 13 V r
e2 4 0
e2 4 1 p10 r r 4 0 a03
dr
0
dr re 2 r / a0
0
227
e 2 4 a02 . 4 0 a03 4
Used www.wolframalpha.com for the integral. Simplify: 2
e2 e2 1 e2 me 2 m 2 E1 27.2 eV 40 a0 40 4 0 2 4 0
V r
This is twice the total energy of the ground state. The average kinetic energy is the difference of the total and the potential: K E1 V 13.6 eV 27.2 eV=13.6 eV 13.23 (Sec. 13.5)
a
1 1 1 3! 1
a
1 1 6 1 1 6
3
1
2
1
2
2
2
3
2 1
1
1
2
2 3
2
1
3
2
3 3
3
2
3
3
2
2 2
1
3
2
2
3
1
3
3 3 1 3
1
1
1
3
2
2
1
2
1
3
3
2 3
3
2
3
2
1
3
1
3
2
1
1
2
1
2 3
2 2
3 3 2
1
2 3
3
2
1
3
The symmetric combination is the same, with all the minus signs changed to plus signs: s
1 1 6
3
1
1
1
2 2
2
2
3 3 1 3
3
1
1
2
3 2
13.24 (Sec. 13.4)
e2 V r 4 0 r e2 r V r r V r 2 4 0 r
Using the virial theorem, this means
2
1
2 3
3
2
1
1
2
3 3 2
1
3
2
1
3
228
Chapter 13
K
1 V 2
E K V K
1 V 2
1 V E 2
Complement 13.A Quantum Dots and the Infinite Spherical Well 13.A.1 Light is produced when an electron and hole recombine. The photon energy is the energy of the electron-hole pair. The longest wavelength corresponds to the minimum energy, which we know is Eeh [Eq. (13.A.13)]: 15 8 hc 4.14 10 eV s 3.00 10 m/s 2.04 eV Eeh hf 610 109 m
Eeh Eg
2 2 2 2 2me R 2 2mh R 2
2 2 1 1 Eeh Eg 2 2 R me mh 1/2
1 1 2 2 R 2 Eeh Eg me mh
R 6.58 1016 6.58 1016 3.6 nm
1 1 1 eV s 3 2 2 2.04 eV 1.75 eV 511 10 eV / c 0.13 0.45 2 3.00 108 m/s 1 1 eV s 2 0.29 eV 511 103 eV 0.13 0.45
1/2
1/ 2
Complement 13.B Series Solution to the Radial Equation 13.B.1* This is a central potential, so we know that the angular solutions are given by the spherical harmonics. The radial equation is: 2l l 1 1 2 2m 1 2 2 r R r m r E R r 0 2 2 2 r r r 2mr 2 Using m / r , this becomes l l 1 2 m m 1 2 R 2 2 E R 0 2 2 2 2 2 2 l l 1 1 2 R E R 0 2 2 2 2
Using 2 E / : l l 1 1 2 R 2 R 0 . 2 2 l l 1 2 2 R R 2 R 0 2 2
In the limit , this becomes 2 R 2 R 2
R Ae
2
/2
Be
2
/2
Must have B 0 for a normalizable solution, so we get R e
2
/2
In the limit 0 , (A) becomes
(A)
232
Complement 13.B l l 1 2 2 R R R 0 2 2
This is the same equation we got for 0 for the hydrogen atom, so the solution is R l 0 The above suggests the substitution 2 R l e /2 w 2 2 l 2 /2 R w e 2 2
2 2 l 1 2 /2 l e w l 1e /2 w l e /2 w
w l l 1 l 2 w ll w ll 1 w l 1 l w l 2 w l 1 ll 1
2 2 w l 1 w l 2 w e /2
2 l l 1 l 2 2l 1 l l 2 w 2ll 1 2l 1 w l 2 w e /2 2 2 2 R ll 1w l 1w l w e /2 2 w e /2 2ll 2 2l w 2l 1
Putting all this into Eq. (A) yields: 2 w 2ll 1 2l 1 w l l 1 l 2 2l 1 l l 2 w 2 2l 1 w 2ll 2 2l w l 2 l l 1 l 2 l w 0
l
l
2 w 2 l 1 l 1 2l 1 w 2
2l 1 2 l l 1 2l l l 1 w 0 l
l 2
Complement 13.B
2 w 2 l 1 2 w 2l 3 w 0 2
233
(B)
Look for a series solution to this equation. w c j j j 0
c j j j 1 j 2 2 l 1 2 c j j j 1 2l 3 c j j 0 j 0
j 0
j 1 2 l 1 jc
j 1
j
j 0
j 0
2l 3 2 j c j j 1 0 j 0
c1 2 l 1 j 1 2 l 1 jc j j 1 2l 3 2 j c j j 1 0 j 2
j 0
Set j j 2 in the first sum: c1 2 l 1 j 1 2 l 1 j 2 c j 2 j1 2l 3 2 j c j j 1 0 j 0
j 0
Set j j and combine: c1 2 l 1 j 0
j 1 2 l 1 j 2 c
j2
2l 3 2 j c j j 1 0
(C)
Each term in the sum must be 0. This means:
c1 0 c j 2
2l 3 2 j 2 j 2l 3 cj c j 1 2 l 1 j 2 2l 3 j j 2 j
Because c1 0 , all of the values of c j for odd j are 0. This menas that the sum contains only odd powers of [even values of j in the series of Eq. (C)]. Limiting behavior: c j2 2j 2 2 j cj j j
234
Complement 13.B Compare this to Eq. (12.A.20), and this tells us that the series diverges as e , so R would 2
2
diverge as e /2 if the series doesn't truncate. This is not normalizable, so the series must truncate, which means:
2 jmax 2l 3 0 2E 2 jmax 2l 3 3 E jmax l 2 jmax must be an even integer, so set jmax 2n , where n 0,1, 2,... , then 3 Enl 2n l 2 We already know that for the spherical harmonics l 0,1, 2,... 13.B.2* Cartesian: 3 Enx ny nz nx n y nz 2
nx , n y , nz 0,1, 2, . . .
Spherical: 3 Enl 2n l , n, l 0,1, 2,... . 2 Define the principal quantum number N. For Cartesian N nx n y nz , while for spherical N 2n l . Energies are then 3 EN N 2
N 0,1, 2, . . . .
Allowed states for N=0 (degeneracy of 1): nx
ny
nz
n
l
ml
0
0
0
0
0
0
Complement 13.B
Allowed states for N=1 (degeneracy of 3): nx
ny
nz
n
l
ml
1 0 0
0 1 0
0 0 1
0 0 0
1 1 1
-1 0 1
Allowed states for N=2 (degeneracy of 6): nx
ny
nz
n
l
ml
1 1 0 2 0 0
1 0 1 0 2 0
0 1 1 0 0 2
0 0 0 0 0 1
2 2 2 2 2 0
-2 -1 0 1 2 0
235
Chapter 14 Time-Independent Perturbation Theory 14.1
(Sec. 14.1) En1 n0 V0 n0 V0 To first order, all levels are shifted by V0 . Note that this is also the exact result. m V0 n 0
n1
0
0
m0 V0
En Em mn No corrections to the wave functions. mn
m n 0
0
0
0
0
E n Em
m0 0
En 2 n0 Hˆ p n1 0 No second-order corrections, because there are no corrections to the wave functions. 14.2
(Sec. 14.1) (a) En1 n0 Hˆ p n0 L /2 a /2
2 n V0 dx sin 2 x L L /2 a /2 L 2 n na na 2 n 2 cos 2 sin 2 cos 2 L sin 2 L a V0 n L na cos n sin L a V0 n L n na a 1 sin L . V0 n L
238
Chapter 14
(b) For a L : n a a 1 En V0 1 L L n even 0 2a L V0 n odd 14.3
(Sec. 14.1) 1 0 0 En n Hˆ p n
2 n M dx sin x sin x L0 L L L
V0
0 V0
nM nM
It does seem reasonable that states with the same periodicity as the perturbation are the most strongly affected. 14.4
(Sec. 14.1) n n0 n1 2 n2 n n n0 n0 n1 n0 n0 n1 O 2
But we know n1 n0 0 , and n0 n0 1 , so n n 1 O 2 , Which is normalized to first order.
Chapter 14
14.5
(Sec. 14.1) 1
n
mn
m0 Hˆ p n0 0
m0
0
En Em
2 m n m0 Hˆ p n0 dx sin x qEx sin x L0 L L L
1 n m 4qEL nm 1 1 2 2 2 2 n m n m even 0 8qEL nm n m odd 2 n 2 m 2 2
If n is even: 8qEL 2mL2 nm 1 1 n 2 2 2 m0 2 2 2 2 2 mn n m n m m odd 16qEmL3 4 2 If n is odd: 16qEmL3 1 n 4 2
mn m odd
mn m even
n
nm 2
n
m
nm 2
2 3
m
2 3
m0
m 0
14.6* (Sec. 14.1) D ˆ / Tˆ D e ipD 1 ipˆ D Tˆ D n 1 ipˆ n n
m D aˆ aˆ † n 2
n
m D 2
n n 1 n 1 n 1
239
240
Chapter 14 In Example 14.2, the exact solution tells us that the wave functions are shifted by D qE / m2 m qE Tˆ D n n 2 m2 n qE
1 2m3
n n 1 n 1 n 1
n n 1 n 1 n 1
This is consistent with the first-order wave function correction in Example 14.2 [Eq. (14.33)]. 14.7
(Sec. 14.2) 2 Hˆ p 1/ 2 xˆ 2 aˆ x aˆ x† aˆ x2 aˆ x†2 aˆ x aˆ x† aˆ x† aˆ x aˆx2 aˆ x†2 2nˆx 1 4m 4m 4m The ground state is nondegenerate, so
1 E111 0, 0, 0 Hˆ p 0, 0, 0 4m The ground state is shifted by
. 4m
To find the corrections for the first excited state, we need to diagonalize: 1, 0, 0 Hˆ p 1, 0, 0 1, 0, 0 Hˆ p 0,1, 0 1, 0, 0 Hˆ p 0, 0,1 0,1, 0 Hˆ p 0,1, 0 0,1, 0 Hˆ p 0, 0,1 H p 0,1, 0 Hˆ p 1, 0, 0 0, 0,1 Hˆ p 1, 0, 0 0, 0,1 Hˆ p 0,1, 0 0, 0,1 Hˆ p 0, 0,1 Perturbation does not affect y or z, so matrix elements must be proportional to ny ny nz nz : 1, 0, 0 Hˆ p 1, 0, 0 0 Hp 0 3 0 0 0 1 0 4m 0 0 1
0 0,1, 0 Hˆ p 0,1, 0 0
0 0, 0,1 Hˆ p 0, 0,1 0
This is diagonal, so we know that the states 0,1, 0 and 0, 0,1 are shifted by the same amount as the ground state, and 1, 0, 0 is shifted by 3 times that amount.
Chapter 14
14.8
(Sec. 14.3) rˆ 1
40 V rˆ , e2
where V rˆ is the potential energy for the Hydrogen atom. The virial thrm says: 2 K r V r e2 r V r r V r 2 4 0 r
Applying the virial theorem: K
1 V 2
E K V
1 V 2 2
e2 1 V m 2 4 0 n 2 4 0 e 2 1 me 2 1 1 1 ˆr 2 m 2 2 2 e 40 n 4 0 n a0 n 2 14.9
(Sec. 14.3) If m and n are positive integers: 0 mn m m! mn n n ! m n ! 1 n0
241
242
Chapter 14
r
1
1
11
2n
1
na0 2
1
11
1 1! k 0
1
k
1 1 n l k n l k 1 k 1 1 1 1
0
2 0 k 0 n l k n l k 1 1 0 2n na0 k 0 k 0 1 0 n l n l 1 2 n a0 0 0 0
1 n a0 2
For v 2, 0 2
1 na r 2 0 2 2n 0 1 ! k 0 0
1
0 k
0 n l 1 k 0 k 2 l 0 1 k 0 1
0 n l 1 2 1 0 0 na0 2n 2l 1 1 2
2 1 2 n a0 2l 1
1 n a l 1/ 2
3
3
2 0
14.10 (Sec. 14.3) Hˆ SO Sˆ Lˆ Sˆ x Lˆ x Sˆ y Lˆ y Sˆ z Lˆ z We know: Lˆ2 , Lˆi 0 , Sˆ 2 , Sˆi 0 , Lˆi , Sˆ j 0 ,
i , j x, y , z
These imply Lˆ2 , Hˆ SO Lˆ2 , Sˆ Lˆ 0 and Sˆ 2 , Hˆ SO Sˆ 2 , Sˆ Lˆ 0 .
Chapter 14
243
Lˆ z , Hˆ SO Lˆ z , Sˆ x Lˆ x Sˆ y Lˆ y Sˆ z Lˆ z Lˆ z , Lˆ x Sˆ x Lˆ z , Lˆ y Sˆ y iLˆ Sˆ iLˆ Sˆ y x
x y
0 Sˆ z , Hˆ SO behaves similarly. 14.11* (Sec. 14.3) In Problem 13.14 we showed that Hˆ 0 commutes with Lˆ2 and Lˆ z . There is nothing special about the z-direction, so Hˆ 0 will commute with Lˆ x and Lˆ y as well. There is no spin dependence in Hˆ 0 , so it will commute with all the components of Sˆ , and with Sˆ 2 .
Jˆ 2 Jˆ Jˆ Sˆ + Lˆ Sˆ + Lˆ Sˆ 2 Lˆ2 2 Sˆ Lˆ
Jˆ z Sˆ z Lˆ z
Since Hˆ 0 commutes with Lˆ2 , Sˆ 2 , and all the components of Sˆ and Lˆ , it will commute with Jˆ 2 and Jˆ z . Hˆ SO Sˆ Lˆ Sˆ x Lˆ x Sˆ y Lˆ y Sˆ z Lˆ z We know: Lˆ2 , Lˆi 0 , Sˆ 2 , Sˆi 0 , Lˆi , Sˆ j 0 ,
i , j x, y , z
These imply Lˆ2 , Hˆ SO Lˆ2 , Sˆ Lˆ 0 and Sˆ 2 , Hˆ SO Sˆ 2 , Sˆ Lˆ 0 . An operator must commute with itself, so Sˆ Lˆ , Sˆ Lˆ 0 .
Jˆ 2 Jˆ Jˆ Sˆ + Lˆ Sˆ + Lˆ Sˆ 2 Lˆ2 2 Sˆ Lˆ
Hˆ SO commutes with each of the 3 terms, so Jˆ 2 , Hˆ SO 0 .
244
Chapter 14 Jˆ z , Hˆ SO Lˆ z , Hˆ SO Sˆ z , Hˆ SO Lˆ z , Sˆ x Lˆ x Sˆ y Lˆ y Sˆ z Lˆ z Sˆ z , Sˆ x Lˆ x Sˆ y Lˆ y Sˆ z Lˆ z Lˆ z , Lˆ x Sˆ x Lˆ z , Lˆ y Sˆ y Sˆ z , Sˆ x Lˆ x Sˆ z , Sˆ y Lˆ y iLˆ Sˆ iLˆ Sˆ iSˆ Lˆ iSˆ Lˆ y x
0
x y
y x
x y
Chapter 14
14.12 (Sec. 14.3) For v 3, 1 3
1 1 na r 3 0 2 2n 1 1 ! k 0
1
1 k
1 n l 1 k 1 k 2l 1 1 k 11
1 n l 1 1 n l 2 1 0 1 1 1 2l 2 2l 1 na0 4n 2 2 3
n l 1 nl (2l 2)(2l 1) (2l 1)(2l )
2 n a03
2 n l 1 nl 3 n a0 (2l 2)(2l 1) (2l 1)(2l )
(2l ) n l 1 (2l 2) n l 2 n a (2l 1) (2l 2)(2l ) (2l 2)(2l )
4
4
4
3 0
(2nl 2l 2 2l ) 2nl 2l 2 2n 2l 1 4 3 n a0 (l 1/ 2) (2l 2)(2l )
1 2n n a (l 1/ 2) (2l 2)(l )
1 n a (l 1)(l 1/ 2)(l )
4
3 0
3 3 0
245
246
Chapter 14
14.13* (Sec. 14.3) Must have j l 1/ 2 , then Eq. (14.79) becomes: 1
ESO
mc 2 4 l 1/ 2 l 3 / 2 l l 1 3 / 4 4n3 l l 1/ 2 l 1
2 2 mc 2 4 l 2l 3 / 4 l l 3 / 4 4n 3 l l 1/ 2 l 1 mc 2 4 l 3 4n l l 1/ 2 l 1 1 mc 2 4 3 4n l 1/ 2 l 1
l 0: 1 ESO
mc 2 4 2n 3
14.14 (Sec. 14.3) r1 x , etc.: 2V rirj ri rj V 1 3 2V rirj ri 2 i , j 1 ri rj ri
V 1 3 ri 2 i , j 1 ri
3
V r r V r i 1
3
V r r V r i 1
Compute averages: c
x
0 c
dV x
2
c
0
0
0
d d
dr r 2 sin r sin cos
dV
0
4 3 c 3
0
Can see that this integral is 0 in two ways: i) the integral yields 0, ii) x is positive over half the sphere, and negative over the other half. Similarly y and z will be 0. All the terms in the first sum for V r r are 0.
Chapter 14 c
xy
dV xy
0
c
2
c
0
0
0
d d
247
dr r 2 sin r sin cos r sin sin 4 3 c 3
dV
0
0
Once again, the integral yields 0; also, the product xy is positive over half the sphere, and negative over the other half. All the other cross terms are 0 as well. Only terms left are the squares: c
x 2
dV x2
0
c
dV
0
2
c
0
0
0
2 d d dr r sin r sin cos
4 3 c 3 2
c
0
0
0
2 3 4 d cos d sin dr r
4 3 c 3
4 1 5c 3 5 4 3 c 3 1 2c 5 y2 and z 2 are the same. This means: 1 3 2V 2 ri 2 i 1 ri 2 1 1 3 2V V r 2c 2 2 5 i 1 ri
V r r V r
V r
2 2V r 2 2 10m c
2
248
Chapter 14
14.15 (Sec. 14.3) For l 0 : Hˆ D
2 2e2 n 00 0 2 2 8m c 0 2 2e2 Rn 0 0 Y00 , 2 2 8m c 0 2
2e2 1 Rn 0 0 2 2 8m c 0 4 Using Eq. (13.94): 2 4 n 1 ! 1 2e2 ˆ HD L 0 1 n 8 4 0 m 2 c 2 a03 n 4 n !
Using Eqs. (13.91) and (14.103): 3
Hˆ D
me 2 4 n ! 2e2 8 4 0 m 2 c 2 4 0 2 n5 n 1 ! 4
mc 2 e 2 4 2 n 8 40 c n5
mc 2 4 2n 3
For l 0 Rnl 0 0 , so 1 ED Hˆ D
mc 2 4 2n3 0
l 0 l0
14.16 (Sec. 14.3) 0
En
mc 2 2 2n 2
[Eq. (14.65)]
Use Eqs. (14.66), (14.80), and (14.86). First, for l 0 , j 1/ 2 :
2
Chapter 14 1 1 1 1 EF ER ESO ED
mc 2 4 mc 2 4 4n 3 0 8n 4 0 1/ 2 2n3 0 0 En 2 En 2 8n 3 n 4n 2 0 En 2 8n 3 n n 2 4 0 En 2 3 n 2 4 n
3 2 0 0 1 Enj En EF En 1 2 n 4 n This is consistent with Eq. (14.87), for l 0 , j 1/ 2 . l 0 , j l 1/ 2 : l j 1/ 2 1 1 1 1 EF ER ESO ED
mc 2 4 j j 1 l l 1 3 / 4 mc 2 4 4n 3 0 4 n3l l 1 l 1/ 2 8n 4 l 1/ 2 0 0 En 2 j j 1 j 1/ 2 j 1/ 2 3 / 4 En 2 4n 3 2n j 1/ 2 j 1/ 2 j 4n 2 j 0 En 2 n 3 n j 2 j j 2 1/ 4 3 / 4 n 2 j 4 2 j 1/ 2 j 1/ 2 j 0 3 En 2 n n 1 n 2 j 2 j 1/ 2 j 4 0 3 En 2 n 1 1 n 2 j 2 j 1/ 2 4 0 En 2 n 2 j 1 1 3 n 2 j 2 j 1/ 2 4 0 En 2 n 3 n 2 j 1/ 2 4
249
250
Chapter 14 3 2 n 0 0 1 Enj En EF En 1 2 n j 1/ 2 4 This is consistent with Eq. (14.87). l 0 , j l 1/ 2 : l j 1/ 2 1 1 1 1 EF ER ESO ED
mc 2 4 j j 1 l l 1 3 / 4 mc 2 4 4n 3 0 4 n3l l 1 l 1/ 2 8n 4 l 1/ 2 0 0 E 2 j j 1 j 1/ 2 j 3 / 2 3 / 4 E 2 4n n 2 3 n 2n j 1/ 2 j 3 / 2 j 1 4n j 1 0 E 2 n 3 n j 2 j j 2 2 j 3 / 4 3 / 4 n 2 n j 1 4 2 j 1/ 2 j 3 / 2 j 1 0 3 En 2 n j 6/ 4 n n 2 j 1 2 j 1/ 2 j 3 / 2 j 1 4 0 3 En 2 n j6/4 1 n 2 j 1 2 j 1/ 2 j 3 / 2 4 2 0 En 2 n 2 j 2 j 3 / 4 j 6 / 4 3 n 2 j 1 2 j 1/ 2 j 3 / 2 4 0 3 E 2 n 2 j2 5 j 3 n 2 j 1 2 j 1/ 2 j 3 / 2 4 n 0 E 2 n 2 j 1 j 3 / 2 3 n 2 j 1 2 j 1/ 2 j 3 / 2 4 n 0 E 2 n 3 n 2 n j 1/ 2 4 3 2 n 0 0 1 Enj En EF En 1 2 n j 1/ 2 4
This is consistent with Eq. (14.87).
Chapter 14
14.17 (Sec. 14.3) 2 n 3 0 1 EF En 2 n j 1/ 2 4 1 137
n 2 , j 1/ 2 :
3 1 13.6 eV 2 5.66 105 eV EF 2 4 24 137 n 2, j 3/ 2: 1 13.6 eV 3 1 1.13 105 eV EF 2 4 2 137 4
Separation of n 2 and n 3 levels 13.6 eV 13.6 eV 1.89 eV E3 E2 32 22 The fine structure shifts are much smaller than the separation of the electronic levels. 14.18* (Sec. 14.4) Hˆ DD
3 Sˆ e ur
Sˆ u Sˆ Sˆ r
p
r
e
p
3
Need to average this, assuming a spherically symmetric wave function. First note:
3 Sˆ e ur
Sˆ
p
r
ur Sˆ e Sˆ p 3
3 Sˆ p r
Sˆ r
r
e
5
Sˆ Sˆ
Start with the first term: Sˆ e r Sˆ p r Sex x Sey y Sez z S px x S py y S pz z 5 r r5
e
p
r
3
(A)
251
252
Chapter 14 Sex S py xy
Notice that this has terms such as
r5
. For the moment, we are only interested in the spatial
contributions. In other words, for this term we need to compute
xy r5
assuming a spherically
symmetric wave function [ r r ].
all space
dV
2 xy r 0 , 5 r
because the product xy is positive half the time, and negative the other half. Similarly, all other cross terms will also be 0. The only nonzero contributions will come from terms that involve squares of the position variables. Furthermore: 2 2 2 2 x2 y2 z2 1 r2 dV r dV r dV r dV r 5 5 5 5 r r r 3 all r all all all space
space
3 Sˆ p r
Sˆ r
r
e
5
space
3
S dV
ex
all space
space
S px x 2 Sey S py y 2 Sez S pz z 2 r
5
r
2
2 2 1 1 r2 r2 3 Sex S px dV 5 r Sey S py dV 5 r 3 all 3 all r r space space 2 1 r2 Sez S pz dV 5 r r 3 all space 2 1 Sˆ e Sˆ p dV 3 r r all
space
Sˆ Sˆ e
p
r
3
Substituting this back into (A), we see that the dipole-dipole term is 0 for spherically symmetric wavefunctions (wavefunctions with l 0 are spherically symmetric).
Chapter 14
253
14.19* (Sec. 14.4) 1 EHF
2 g p me2 c 2 4 6 s s 1 3m p 4
The energy difference between the s 1 and s 0 states is E
4 g p me2 c 2 4 3m p
4 g p me c 2 4 2
3m p c
2
4 5.59 511103 eV 3 938 10 eV 137 6
2
4
5.89 106 eV
f E / h 5.89 106 eV / 4.14 1015 eV s 1.42 109 Hz
c / f 3.00 108 m/s / 1.42 109 Hz 21.1 cm 14.20 (Sec. 14.5) (a)
eB ˆ Hˆ Ze Lz 2Sˆz 2m
If the magnetic field is stronger than the spin-orbit field, the external field is perturbing the original Hamiltonian, Hˆ 0 (the Coulomb Hamiltonian), not the fine structure. The fine structure will be a perturbation on the Zeeman effect, not vice versa. Thus, we're looking for the eigenstates of operators that commute with Hˆ 0 and Hˆ Ze . We already know that Lˆ2 and Lˆz commute with Hˆ 0 , and since the unperturbed Hamiltonian has no spin dependence, then Sˆ 2 and Sˆ must commute with it as well. z
We know: Lˆ2 , Lˆi 0 , Sˆ 2 , Sˆi 0 , Lˆi , Sˆ j 0 , i, j x, y, z , so Lˆ2 , Sˆ 2 , Lˆz and Sˆz all ˆ commute with H Ze . This means that we can use the eigenstates of those operators, l , ml , s, ms , for our perturbation calculation, since they are non-degenerate.
254
Chapter 14
(b) Using these states, we find that the first-order energy corrections are 1 EZe Hˆ Ze
eB ml 2ms 2m eB ml 1 . 2m
14.21 (Sec. 14.5) (a) The perturbing Hamiltonian is: H p z qEz eEz The ground state is nondegenerate. For the ground state: 1 ES H s
eE
2
dV z 100 r 0 ,
all space
because the z integral is odd. So there is no first-order shift. (b) For degenerate theory of the n 2 level, we need to diagonalize:
2, 0, 0 Hˆ s 2, 0, 0 2,1, 0 Hˆ 2, 0, 0
2, 0, 0 Hˆ s 2,1, 0 2,1, 0 Hˆ 2,1, 0
2, 0, 0 Hˆ s 2,1,1 2,1, 0 Hˆ 2,1,1
2,1,1 Hˆ s 2, 0, 0 2,1, 1 Hˆ s 2, 0, 0
2,1,1 Hˆ s 2,1, 0 2,1, 1 Hˆ s 2,1, 0
2,1,1 Hˆ s 2,1,1 2,1, 1 Hˆ s 2,1,1
s
s
s
We need to calculate the matrix elements 2, l , ml Hˆ s 2, l , m eE
dV 2lml r z 2lml r
all space
We know z r cos , so the angular integrals are:
2, 0, 0 Hˆ s 2,1, 1 2,1, 0 Hˆ s 2,1, 1 2,1,1 Hˆ s 2,1, 1 2,1, 1 Hˆ s 2,1, 1
Chapter 14
0
255
2
d d sin cos Yl ml* , Yl ml , 0
cl ml clml
0
2
d sin cos Pl ml cos Pl ml cos d e
i ml ml
0
This is 0 unless ml ml , so the matrix simplifies to: 2, 0, 0 Hˆ s 2, 0, 0 2,1, 0 Hˆ s 2, 0, 0 0 0
2, 0, 0 Hˆ s 2,1, 0 2,1, 0 Hˆ 2,1, 0
0
0 0
0
0 2,1,1 Hˆ s 2,1,1
0
0
s
0 2,1, 1 Hˆ s
2,1, 1
Look at integrals:
d sin cos P cos P cos d sin cos 0
d sin cos P10 cos P10 cos d sin cos cos 2 0
d sin cos P11 cos P11 cos d sin cos sin 2 0
0
0
0
0 0
0 0
0
0
0
Therefore, are the diagonal terms are 0: 2,1, 0
0 Hˆ s 2, 0, 0 0 0
2, 0, 0 Hˆ s 2,1, 0 0 0 0
0 0 0 0 0 0 0 0
Therefore, we only need to calculate 2 matrix elements, and they are complex conjugates of each other, so we only need to do 1 integral:
256
Chapter 14 2, 0, 0 Hˆ s 2,1, 0 eE
dV 200 r r cos 210 r
all space
eE
0
2
0
0
drr 3 R20 r R21 r d sin cos d Y00 , Y10 ,
1 1 eE 2 2a3 0
1 1 2 6a3 0
r r drr 3 2 e r /2a0 e r /2a0 a0 a0 0
2
1 3 d sin cos d 4 cos 4 0 0 3 r /2 a0 r r /2 a0 r 3 d sin cos 2 eE 3 drr 2 e e 8a0 3 2 0 a0 a0 0 1 4 2 eE a 72 0 16a03 3 1
3eEa0 0 3eEa0 0 0
3eEa0 0 0 0
0 0 0 0 0 0 0 0
Find eigenvalues and eigenvectors: So two of the levels, 2,1, 1 are unaffected by the electric field (eigenvalues are 0). To find the other eigenvalues and eigenvectors, diagonalize 0 3eEa 0
3eEa0 0
You'll find that: 1 2, 0, 0 2,1, 0 . 2 1 Eigenvalue 3eEa0 (energy decreases), corresponding eigenstate 2, 0, 0 2,1, 0 . 2
Eigenvalue 3eEa0 (energy increases), corresponding eigenstate
Chapter 14
257
14.22 (Sec. 14.2) 0 0 Aˆ n, j a j n, j
, and the values of a j are nondegenerate ( a j a j if j j .) Aˆ is
Hermitian, because it corresponds to an observable. We know Hˆ p , Aˆ 0 , therefore 0 0 0 0 0 ˆ ˆ 0 n, j Hˆ p , Aˆ n, j n, j Hˆ p Aˆ n, j n, j AH p n, j
a j a j
0 0 n, j Hˆ p n, j
0 0 0 Since a j a j if j j , n, j Hˆ p n, j 0 if j j , so the matrix is diagonal.
Chapter 15 Time-Dependent Perturbation Theory 15.1
(Sec. 15.1) P12 t c2 t
2
t
i c2 t dt ei21t H p 21 t 0 H p 21 t 2 Vˆ 1 V21 t
i c2 t V21 dt ei21t 0 t t
1 i21t i V21 e i21 t 0
V21 i21t e 1 21
V P12 t 21 ei21t 1 21 V 21 21
2
2
2 2 cos 21t 2
V 4 21 sin 2 21t / 2 21 15.2
(Sec. 15.1) (a) P12 t c2 t c2 t
t
2
i dt ei21t H p 21 t 0
260
Chapter 15 0t T V H p 21 t 2 Hˆ p t 1 21 t T 0 For t > T: T
i c2 t V21 dt ei21t 0 t T
1 i21t i V21 e i 21 t 0
V21 i21T 1 e 21
V P12 t 21 ei21T 1 21 V 21 21
2
2
2 2 cos 21T 2
V 4 21 sin 2 21T / 2 21 (b) V21 / T 2
P12 t 4 sin 2 21T / 2 21T
2 2 sin 21T / 2 21T / 2 2
sinc2 21T / 2
2
Lim T 0 : P12 t
2
Chapter 15
15.3
(Sec. 15.1) V xˆ , t qExˆ 0 x L, 0 t T during the pulse.
H p 21 t 2 V xˆ , t 1 L
2 2 x qEx sin x dx sin L0 L L
16qEL
0 t T
92 T
i 16qEL c2 t dt ei21t 2 9 0 t T
i 16qEL 1 i21t e 92 i21 t 0
16qEL 2
9 21
P12 t
e
16qEL 92 21
16qEL 16qEL 2
e
1
i21T
2
1
2
92 21
4
i21T
9 21
2 2 cos 21T 2
sin 2 21T / 2
15.4* (Sec. 15.1) i x
2 sin x L L
f x
2 2 sin x 1.1L 1.1L
0 xL 0 x 1.1L
261
262
Chapter 15 Pif t f i
2
L
2 2 dx sin x sin x L 1.1 0 L 1.1L
2
2
0.142 0.020 The integral was calculated using Maple. 15.5* (Sec. 15.2) Hˆ p t Vˆ cos t Follow the derivation in Sec. 15.2, just swap 1 and 2: 2 1 P21 t c1 t
2
t
i c1 t dt ei12t H p 12 t 0 t
i V12 dt e i0t cos t 0
iV i t i t 21 dt e 0 e 0 2 0 t
V 21 2
ei 0 t 1 e i 0 t 1 0 0
V12 . Now make resonance Here we've used the fact that the matrix elements satisfy V21 approximation: i t V e 0 1 P21 t 21 2 4 0 2
2
2 V21 2 2 cos 0 t 2 4 2 0 2 V sin 0 t / 2 212 2 0 2
P12 t
where P12 t is given in Eq. (15.29).
Chapter 15 15.6
(Sec. 15.3) Hˆ ED t dˆ zE t d 2 dˆ 1 z
z
P12 t c2 t
2
t
i c2 t dt ei0t H p 21 t 0 t
i d zE dt ei0t 1 e t/ 0
t t
ei0t e i0 1/ t i d zE i0 i0 1/ t 0 t t
ei0t i e t / d zE 1 i0 1 i / 0 t0 ei0t i d zE i0
1 e t / 1 1 1 1 i / 0 i0 1 i / 0
This expression is valid in both limits. (a) 1/ 0 , 0 1 ei0t i c2 t d zE i0
e t / 1 1 1 1 1 i 1 0
ei0t i 1 e t / d zE i0
P12 t c2 t 2
2
2 d E2 z 2 2 1 e t / 0
263
264
Chapter 15
(b) 1/ 0
, 1 0
ei0t i c2 t d zE i0
e t / 1 1 1 1 i / 0 i0 i / 0
ei0t i 1 d zE 1 i0 e t / 1 i0 i0 i0
ei0t 1 i d zE i0 P12 t c2 t
2
2
dz E 2 2 2 2 2 cos 0t 0 2
dz E 2 4 2 2 sin 2 0t / 2 0 Note that this result agrees with that of Problem 15.1, as it should. 15.7
(Sec. 15.3) 1/ A21
3 0 c 3 2
d 30 2
Need to calculate d . 2
2
2
d dx d y dz
2
d z 2,1, 0 ezˆ 1, 0, 0 e dV 210 r z 100 r all space
analogous integrals for d y and d x . 8 / 3 1 Y1 , Y11 , 2 i 8 / 3 1 Y1 , Y11 , y r sin sin r 2 z r cos r 4 / 3Y10 ,
x r sin cos r
Chapter 15
265
Angular integral for d x : 8 / 3 1 1 0 1 d d sin Y , Y , Y , 1 1 1 0 0 4 0 , 2 because the spherical harmonics are orthogonal. So d x 0 , and d y 0 for the same reason. 2
Angular integral for d z : 4 0 1 1 d d sin Y10 , Y1 , , 3 0 3 4
2
0
because the spherical harmonics are normalized. Radial integral for d z :
0
drr R21 r rR10 r drr 3 R21 r R10 r 2
0
1 1 drr 3 3 0 2 6a0
r r /2 a0 1 r /a 2 3 e 0 e a0 a0
128 6 a0 243 This integral was done using Maple.
2
2
d dz
2
1 128 6 e a0 0.55 e 2 a02 3 243 2
1 1 13.6 eV 2 2 E E 1 2 1 1.55 1016 s 1 0 2 6.58 1016 eV s
3 0 c3 2
d 30 3 8.85 1012 C2 / J m 3.00 108 m / s 1.06 1034 J s 3
0.55 1.60 1019 C
1.6 ns
2
0.529 10
10
m 1.55 1016 s 1 2
3
266
Chapter 15
15.8* (Sec. 15.3) Start with exact equations, Eq. (15.9), and assume 1 : d i i t cn t e nj H p nj t c j t j dt Hˆ ED t dˆ zE cos t 0 2 1 d 2 dˆ 1 z
z
Write 2-level equations. Start with level 1: d i c1 t H p 11 t c1 t ei12t H p 12 t c2 t dt i d*zEei0t cos t c2 t Ignore the first term, because selection rules tell us that l2 l1 l 1 ; matrix elements of the
dipole moment are 0 otherwise. So, the diagonal matrix elements H p 11 t and H p 22 t must be 0. Expand the cosine, and only keep the near resonant term: d i * i0t it it c1 t e e c2 t d zEe dt 2 i * i 0 t c2 t d zEe 2
(A)
Level 2: d i c2 t H p 22 t c2 t ei21t H p 21 t c1 t dt i d zEei0t cos t c1 t i i t d zEe 0 c1 t 2 c1 t
2 id zE
e
i 0 t
d c2 t dt
Take the second derivative of the level-2 equation:
(B)
(C)
Chapter 15
d2 dt
c 2 2
t
1 i i t i t d d zE 0 e 0 c1 t d zEe 0 c1 t 2 2 dt
Substitute (A) and (C) into this, yielding: d2 dt
c t 2 2
1 i t 2 i t d d zE 0 e 0 e 0 c2 t 2 dt id zE i i t i i t d zEe 0 d*zEe 0 c2 t 2 2 2
dz E 2 d c2 t i 0 c2 t dt 4 2 2
dz E 2 d c t i c t c2 t 0 2 0 2 dt dt 2 4 2 d2
Assume c2 t eit , then: 2 0
2
dz E 2 4 2
0
Solve for :
0 R 2
,
where R
0
2
2
dz E 2 2
General solution is: c2 t aeit beit Substitute this into (C): 2 i 0 t c1 t e a eit b eit dE z
Use the initial conditions: c2 0 a b 0 b a
267
268
Chapter 15 c1 0 a
2
a 1
d zE
d zE dE z 2 2 R
c2 t
d zE eit eit 2 R
d zE i 0 t /2 i Rt /2 i Rt /2 e e e 2 R
d zE i 0 t /2 2i sin R t / 2 e 2 R
P12 t c2 t
2
2
dz E 2
2
2R 2 2
sin 2 R t / 2
dz E 1 1 cos R t 2 2R 2
In the perturbative (weak field) limit R 0 , and P12 t
2
dz E 2
2
0
2
sin 2 0 t / 2
This agrees with Eq. (15.34).
Complement 15.A Einstein's A and B Coefficients 15.A.1 Let Rsp 1/ sp and Rc 1/ c be the rates of the decay. Ignoring absorption and stimulated emission: d N 2 Rsp N 2 Rc N 2 Rsp Rc N 2 dt The total rate is the sum of the rates, so the effective lifetime is the inverse of that:
1 1 Rsp Rc 1/ sp 1/ c
Chapter 16 Quantum Fields 16.1
(Sec. 16.1) pˆ 2 1 Hˆ m2 xˆ 2 2m 2
d i pˆ 2 1 m2 xˆ 2 , pˆ pˆ 2m 2 dt
im2 2 xˆ , pˆ 2
im2 xˆ xˆ, pˆ xˆ, pˆ xˆ 2 m2 xˆ .
For a classical harmonic oscillator, we know that m2 , where is the spring constant. Hooke's law says F x m2 x From Newton's second law: F
d p m2 x dt
This is equivalent to the Heisenberg equation of motion.
272
Chapter 16
16.2
(Sec. 16.1) pˆ 2 Hˆ 2m d i pˆ 2 , xˆ xˆ 2m dt i 2 pˆ , xˆ 2m i pˆ pˆ , xˆ pˆ , xˆ pˆ 2m 1 pˆ . m Classically, we know: d 1 xv p , dt m which is equivalent to the Heisenberg equation of motion. d i pˆ 2 , pˆ pˆ dt 2m 0
The momentum of a free particle is conserved, which is also true classically. 16.3
(Sec. 16.1) Hˆ μˆ B Sˆ B z
Sˆ z .
The initial conditions are:
S x t 0 x Sˆ x x 2 S y t 0 x Sˆ y x 0 S z t 0 x Sˆ z x 0
Chapter 16
273
The Heisenberg equations of motion for the x- and y-directions are: d ˆ i S x Sˆ z , Sˆ x dt Sˆ y d Sx S y dt d ˆ i S y Sˆ z , Sˆ y dt Sˆ x d S y S x dt
(A)
(B)
(A) and (B) represent coupled differential equations for the expectation values. To solve, we first uncouple them by differentiating (A): d2 dt
2
Sx
d Sy dt
Substitute (B) into this: d2 S x 2 S x 2 dt The solution to this is: S x t C cos t D sin t
From (A) we know that Sy
1 d S x C sin t D cos t dt
We can solve for C and D using the initial conditions. The final solution is: S x t cos t 2 S y t sin t . 2 These agree with Eqs. (9.28) and (9.29). The Heisenberg equations of motion for the z-directions is:
274
Chapter 16 d ˆ i S z Sˆ z , Sˆ z dt 0 d Sz 0 dt Therefore S z t constant , and using the initial condition we learn that Sz t 0 .
This agrees with Eq. (9.30). 16.4* (Sec. 16.1) 1 Hˆ nˆ 2 d i 1 aˆ t nˆ , aˆ t dt 2 iaˆ t
where we used Eq. (12.12). The solution to this equation is: aˆ t aˆ 0 e it aˆ e it .
16.5
(Sec. 16.1) If there is no explicit time dependence, the Heisenberg equation of motion is: d ˆ i H t Hˆ t , Hˆ t 0 dt This means that Hˆ t is constant in time, so Hˆ t Hˆ Hˆ . H
S
16.6* (Sec. 16.2) A0 eik r ik x A0 x eik r ik y A0 y eik r ik z A0 z eik r ik A0 eik r
Maxwell's equations for free fields say:
Chapter 16
275
E r , t E t eik r E t e ik r ik E t eik r ik E t e ik r 2 Re ik E t eik r 0
Which means that k E t 0 . Thus, k must be perpendicular to E r , t , and E r , t is transverse. B r , t is also transverse because B r , t 0 . 16.7
(Sec. 16.2)
E r , t E t eikx E t e ikx uz (16.15)
E t E 0 e it
(16.18)
Combine these: E r , t E 0 ei kx t E t e i kx t uz
(A)
Wave equation is: 1 2 2E 2 2 E 0 c t Substitute (A) into this, and it becomes 2 uz 12 2 E 0 ei kx t E t e i kx t uz c t 2 i kx t i kx t uz 2 E 0 ei kx t E t e i kx t uz k 2 E 0 e E t e c 2 k 2E r , t 2 E r , t c 0
2 E 0 e
i kx t
E t e
i kx t
This is true, as long as ck . 16.8
(Sec. 16.2) The electric and magnetic fields are related by E B t E r , t E t eikx E t e ikx uz
E 0 e it eikx E 0 eit e ikx uz
276
Chapter 16
Here we've used Eq. (16.18). Take the curl of this E r , t E 0 e it eikx uz E ks 0 eit e ikx uz E 0 e it ikeikx uy E ks 0 eit ike ikx uy
1 iE 0 e it eikx iE ks 0 eit e ikx uy , c where we've used ck . Integrate this to get the magnetic field
B r , t dt E
1 E 0 e it eikx E 0 eit e ikx uy c 1 E t eikx E t e ikx uy c 16.9
(Sec. 16.2) E
1 X iP 2 0V 1 X 0V
E E X
0V
E E
P i 0V E E
E E
V d X 0 dt V 0 0V
d E t E t dt d E 0 e it E 0 eit dt
iE 0 e iE 0 e i V E t E t
it
0
P
i P 0V
it
Chapter 16
16.10 (Sec. 16.4) Eˆ r , t
i k r t i k r t n aˆ n e ε n aˆ † n e ε 20V n n n 1 ei k r t ε n 1 n n 1 e i k r t ε 20V
0.
E
2
E 2 r, t E r, t
2
0
Eˆ r , t Eˆ r , t
2 2 i 2 k r t ˆe ˆ † e i 2 k r t ε ε aa ˆ ˆ † aˆ † aˆ a a ε ε 20V
2 ˆ 2 n ei 2 k r t εε n aˆ † n e i 2 k r t ε ε n a 20V
ˆ ˆ † n n aˆ † aˆ n n aa 0 0 (n 1) n 20V
2n 1 . 20V
16.11 (Sec. 16.4) Eˆ r , t
i k r t i k r t ε aˆ † e ε aˆ e 2 0V
i k r t i k r t ε *e ε e 2 0V
i k r t 2 Re e ε . 2 0V
277
278
Chapter 16
16.12 (Sec. 16.4)
1 ˆ i aˆ †ei Xˆ ae 2
[Eq. (16.37)]
X Xˆ
1 ˆ i aˆ †ei ae 2 1 aˆ ei aˆ † ei 2 1 e i ei 2
Re ei . This is just a phase-shifted version of X Re , so changing effectively shifts the phase of the coherent state. X 2 Xˆ 2
1 ˆ i aˆ †ei ae ˆ i aˆ †ei ae 4 1 ˆ ˆ † aˆ † aˆ aˆ 2 ei 2 aˆ †2ei 2 aa 4 1 aˆ 2 ei 2 aˆ †2ei 2 2aˆ † aˆ 1 4 2 2 1 e i ei 2 1 4 2 1 1 e i ei 4 4 2 1 Re e i 4 2 1 X . 4
X
X 2 X
2
1 2
This is independent of theta.
Chapter 16
16.13 (Sec. 16.4) From Eq. (16.53), we know that
P n
2n
e
n!
2
n nP n n 0
e
2
n
n!
n0
e
2
n
e
2
2n
n!
n 1
2
2n
2 n 1
n 1! n 1
let k n 1 n e
2
2
2
k!
k 0
e
2
e
2 k
2
2
n2 n2 P n n 0
e
2
n
2
n 0
e
2
2
2n
n! n
2 n 1
n 1! n 1
e
2
2
n 1
1 1 n
2 n 1
n 1!
2 n 1 2 n 1 2 2 2 2 e n 1 e n 1! n 1! n 1 n 1 Above we showed that the second term in [ ] is equal to n , so:
279
280
Chapter 16
n
2 2 e
n n 2 ! 2 n2
2
2
n2
Let k n 2 2 2 e
n 2 k
n2
e
2
2
e 2 2
k!
k 0
2
n
2
n n
n
n2 n
2
2
n n n
n
1/2
2
16.14 (Sec. 16.4)
1 1 ˆ ˆ r exp r aˆ †2 aˆ 2 aˆ exp r aˆ 2 aˆ †2 Sˆ † r aS 2 2 Using Eq. (16.119): 1 1 1 1 ˆ ˆ r aˆ aˆ , r aˆ 2 aˆ †2 aˆ , r aˆ 2 aˆ †2 , r aˆ 2 aˆ †2 Sˆ † r aS 2 2! 2 2
2
1 1 1 aˆ r aˆ , aˆ 2 aˆ †2 r aˆ , aˆ 2 aˆ †2 , aˆ 2 aˆ †2 2 2! 2
aˆ , aˆ 2 aˆ †2 aˆ , aˆ †2
aˆ † aˆ , aˆ † aˆ , aˆ † aˆ †
2aˆ † 1 1 1 ˆ ˆ r aˆ r 2 aˆ † r Sˆ † r aS 2! 2 2
2
2 aˆ † , aˆ 2 aˆ †2
Chapter 16
aˆ † , aˆ 2 aˆ †2 aˆ † , aˆ 2
aˆ aˆ † , aˆ aˆ † , aˆ aˆ 2aˆ
2
1 1 2 1 ˆ ˆ r aˆ r 2 aˆ † r 2 aˆ Sˆ † r aS 2! 2 2 1 1 aˆ 1 r 2 aˆ † r r 3 2! 3! aˆ cosh r aˆ † sinh r †
Sˆ † r aS ˆ ˆ r Sˆ † r aˆ † Sˆ † r † † Sˆ r aˆ Sˆ r
†
aˆ † cosh r aˆ sinh r 16.15 (Sec. 16.4)
1 Xˆ Xˆ 0 aˆ aˆ † 2 ˆ X r X r
1 Pˆ Xˆ /2 aˆ aˆ † 2i
ˆ ˆ r 0 0 Sˆ † r XS
1 ˆ ˆ r 0 0 Sˆ † r aˆ † Sˆ r 0 0 Sˆ † r aS 2 1 0 aˆ 0 cosh r 0 aˆ † 0 sinh r 0 aˆ † 0 cosh r 0 aˆ 0 sinh r 2 1 0 0 0 0 2 0
Similarly, P 0 .
281
282
Chapter 16 X 2 r Xˆ 2 r 0 Sˆ † r Xˆ 2 Sˆ r 0 1 ˆ ˆ † aˆ † aˆ Sˆ r 0 0 Sˆ † r aˆ 2 aˆ †2 aa 4 1 0 Sˆ † r aˆ 2 aˆ †2 2aˆ † aˆ 1 Sˆ r 0 4 1 ˆ ˆ r Sˆ † r aS ˆ ˆ r Sˆ † r aˆ † Sˆ r Sˆ † r aˆ † Sˆ r 0 Sˆ † r aS 4 ˆ ˆ r 1 0 , 2 Sˆ † r aˆ † Sˆ r Sˆ † r aS
where we've used the fact that Sˆ r is unitary. X2
1 0 aˆ cosh r aˆ † sinh r aˆ cosh r aˆ † sinh r aˆ † cosh r aˆ sinh r aˆ † cosh r aˆ sinh r 4 2 aˆ † cosh r aˆ sinh r aˆ cosh r aˆ † sinh r 1 0
We know that expectation values of terms containing aˆ 2 or aˆ †2 will be 0. Eliminating them, we find: 1 ˆ ˆ † aˆ † aˆ 2 cosh 2 r aˆ † aˆ 2sinh 2 r aa ˆ ˆ † 1 0 0 2 cosh r sinh r aa 4 1 0 2 cosh r sinh r 2nˆ 1 2 cosh 2 r nˆ 2sinh 2 r nˆ 1 1 0 4 1 2 cosh r sinh r 2sinh 2 r 1 4 1 1 1 e r e r e r e r er e r er e r 1 4 2 2
X2
1 1 2 r 2 r 1 2 r e e e e 2 r 2 1 4 2 2 1 e 2 r 4
X
X2 X
2
1 r e 2
Chapter 16
283
P 2 r Pˆ 2 r 0 Sˆ † r Pˆ 2 Sˆ r 0 1 ˆ ˆ † aˆ † aˆ Sˆ r 0 0 Sˆ † r aˆ 2 aˆ †2 aa 4 1 0 Sˆ † r aˆ 2 aˆ †2 2aˆ † aˆ 1 Sˆ r 0 4 1 ˆ ˆ r Sˆ † r aS ˆ ˆ r Sˆ † r aˆ † Sˆ r Sˆ † r aˆ † Sˆ r 0 Sˆ † r aS 4 ˆ ˆ r 1 0 , 2 Sˆ † r aˆ † Sˆ r Sˆ † r aS
where we've used the fact that Sˆ r is unitary. P2
1 0 aˆ cosh r aˆ † sinh r aˆ cosh r aˆ † sinh r aˆ † cosh r aˆ sinh r aˆ † cosh r aˆ sinh r 4 2 aˆ † cosh r aˆ sinh r aˆ cosh r aˆ † sinh r 1 0
We know that expectation values of terms containing aˆ 2 or aˆ †2 will be 0. Eliminating them, we find: 1 ˆ ˆ † aˆ † aˆ 2 cosh 2 r aˆ † aˆ 2sinh 2 r aa ˆ ˆ † 1 0 0 2 cosh r sinh r aa 4 1 0 2 cosh r sinh r 2nˆ 1 2 cosh 2 r nˆ 2sinh 2 r nˆ 1 1 0 4 1 2 cosh r sinh r 2sinh 2 r 1 4 1 1 1 e r e r e r e r er e r er e r 1 4 2 2
P2
1 1 2 r 2 r 1 2 r e e e e 2 r 2 1 4 2 2 1 e2 r 4
P
P2 P
2
1 r e 2
284
Chapter 16
16.16 (Sec. 16.4)
1 Sˆ r exp r aˆ 2 aˆ †2 2
n 0
1 1 r aˆ 2 aˆ †2 n! 2
n
The squeezing operator Sˆ r contains only even powers of aˆ †2 and aˆ 2 . Thus Sˆ r 0
n 0
cn n will contain only terms with even n .
16.17 (Sec. 16.5) (a) Problem 16.6 proves that the wave vector must be orthogonal to the polarization vector. Clearly the vectors uk uz , k1 ux and k 2 u y are all mutually orthogonal. (b) Generalizing Eq. (4.51) to rotations in three dimensions, rotation by about the y-axis is: cos 0 sin 1 0 Ry 0 sin 0 cos Rotation by about the z-axis: cos sin 0 R z sin cos 0 0 0 1 cos sin 0 cos 0 sin cos cos sin sin cos 1 0 cos sin cos sin sin R z R y sin cos 0 0 0 0 1 sin 0 cos sin 0 cos cos cos sin sin cos 0 sin cos uk R z R y uz cos sin cos sin sin 0 sin sin sin 0 cos 1 cos uk sin cos ux sin sin u y cos uz
Chapter 16
285
cos cos sin sin cos 1 cos cos k1 R z R y ux cos sin cos sin sin 0 cos sin sin 0 cos 0 sin k1 cos cos ux cos sin u y sin uz
k 2
k 2
cos cos sin sin cos 0 sin R z R y u y cos sin cos sin sin 1 cos sin 0 cos 0 0 sin ux cos uz
16.18* (Sec. 16.6) Iˆ r , t Eˆ r , t Eˆ r , t
20V
Iˆ r , t 0
k ,s
20V
k ,s
k aˆk†s t e ik r εk s k ,s k aˆk†s t eik r εk s k ,s
k aˆks t eik r ε ks k aˆks t eik r ε 0 0 ks
0 Iˆ r , t 0 0 0 Iˆ2 r , t 0 0
I 2 0 The expectation value and variance of the field for a vacuum state were calculated in Example 16.3. The expectation value was 0, but the variance was not. The vacuum field fluctuations that lead to a nonzero field variance do not affect the variance the intensity. Intensity measurements are not sensitive to the vacuum.
286
Chapter 16
16.19 (Sec. 16.6) Iˆ r , t Eˆ r , t Eˆ r , t
I
2
aˆ † aˆ 2 0V
n nˆ n 2 0V
n . 2 0V
Iˆ 2 r , t Iˆ r , t 2 0V
2
n nˆ
2
2
n n2
2
2 2 n n 2 0V 0. 16.20 (Sec. 16.6) Iˆ r , t Eˆ r , t Eˆ r , t
aˆ † aˆ 2 0V
2 2 0V
n . 2 0V
Chapter 16
I
2
Iˆ 2 r , t Iˆ r , t 2 0V
2
2 0V
2
2 0V
2
287
2
aˆ aaˆ ˆ aˆ †
4
†
aˆ aˆ aˆ 1 aˆ †
4
4
†
2
4
2
2 2 0V 2
n . 2 0V 16.21 (Sec. 16.6) Referring to Fig. 15.5, the atom makes a transition from the lower energy ( E1 ) to the upper
energy ( E ) state, so fi E E1 / 0 , and c f t H ED
fi
t
i dt ei0t H ED 0
t
fi
t
.
E , F f dˆ Eˆ r , t 1, Fi
E dˆ 1 F f Eˆ r , t Fi E dˆ 1 F f Eˆ r , t Fi F f Eˆ r , t Fi Eˆ r , t
Eˆ r , t
20V
k aˆks t eik r ε ks
2 0V
k aˆk†s e ik r eik t εks
i c f t 20V
k ,s
k aˆks eik r e ik t ε ks
k ,s
k ,s
k F f aˆks Fi e
k ,s
i 20V
20V
k ,s
ik r
E dˆ 1 ε ks
t
dt ei0t eik t
0
k
F f aˆk†s Fi e ik r
E dˆ 1 εks
t
dt e 0
i0t ik t
e
288
Chapter 16 Near resonance approximation: k 0 . The second term averages to 0 because the integrand oscillates rapidly, so it can be dropped. Left with just the first term, which is the positive frequency part, and it can be written as c f t
t
i dt ei0t E dˆ 1 F f Eˆ r , t Fi . 0
16.22 (Sec. 16.7) if r1 r2 r ,
t1 t2 t ,
r 2 t 2 1,
then aˆ3 , aˆ3† r1aˆ1 t2 aˆ2 , r1aˆ1† t2 aˆ2† r12 t22 r 2 t 2 1 aˆ4 , aˆ4† t1aˆ1 r2 aˆ2 , t1aˆ1† r2 aˆ2† t12 r2 2 t 2 r 2 1 aˆ3 , aˆ4† r1aˆ1 t2 aˆ2 , t1aˆ1† r2 aˆ2† r1t1 r2t2 r t r t 0 16.23* (Sec. 16.7) aˆ3 , aˆ3† r aˆ1 , r aˆ1† r 2 1 aˆ4 , aˆ4† taˆ1 , taˆ1† t 2 1 aˆ3 , aˆ4† raˆ1 , taˆ1† r t 0 16.24 (Sec. 16.7) Create 2 photons in mode 1, and use Eq. (16.110):
aˆ
† 2 1
2 2 2
1
1
0 raˆ3† taˆ4† 0 2
3 2 0 2 r 2 aˆ3† t 2 aˆ4† rt aˆ3† aˆ4† aˆ4† aˆ3† 0 1 02 2 r 2 2 3 0 4 2 t 2 0 3 2 4 2rt 1 3 1 4 . 2
Chapter 16
289
16.25* (Sec. 16.7) Create 1 photons in each input mode, and use Eqs. (16.110) and (16.111): aˆ2† aˆ1† 0 taˆ3† raˆ4† raˆ3† taˆ4† 0 3 2 1 1 1 2 rt aˆ3† rt aˆ4† t 2 aˆ3† aˆ4† r 2 aˆ4† aˆ3† 0
1 1 1 2 rt 2 2
3
0 4 rt 2 0
3
2 4 t 2 r 2 1 3 1 4 .
For a 50/50 beamsplitter r t 1/ 2 : 11 12
1 2 2
3
0 4 0
3
2
4
.
There is 0 probability for one photon to leave from each port. Both photons will be measured to come out the same port (either 3 or 4). 16.26* (Sec. 16.7)
The operator for the difference of the number of photons striking each detector is nˆ12 nˆ1 nˆ2 aˆ1†aˆ1 aˆ2†aˆ2
1 † 1 aˆ S aˆ R† aˆS aˆ R aˆS† aˆ R† aˆS aˆ R 2 2 1 aˆ S† aˆS aˆ R† aˆ R aˆ R† aˆS aˆS† aˆ R aˆS† aˆ S aˆ R† aˆ R aˆ R† aˆ S aˆ S† aˆ R 2 aˆ R† aˆS aˆS† aˆ R .
Taking the expectation value of this in state operator becomes
R
ei
R
, the photon-difference-number
290
Chapter 16 nˆ12 aˆS aˆS†
e i aˆ S ei aˆ S†
2 Xˆ where Xˆ is the quadrature amplitude operator of the signal field [Eq. (16.37)].
Complement 16.A Second-Order Coherence and the Grangier Experiment 16.A.1 From Eq. (16.A.14): 2 g 0
aˆ I†aˆ I†aˆ I aˆ I aˆ I†aˆ I
2
ˆ ˆ † aˆ † aˆ 1 , so aˆ †aˆ aa ˆ ˆ † 1 , and Using the commutation relationship aˆ , aˆ † aa 2 g 0
n aˆ I† aˆ I aˆ I† 1 aˆ I n
n nˆ I n
2
n nˆI 2 nˆ I n n
2
2
n n
n2 1 1 n 2 For n 1 this is 0, which agrees with what we already know. As n increases, g 0 increases. 2 In the limit of large n g 0 approaches 1, but it is always less than 1. It always violates the
classical inequality.
Chapter 17 Quantum Information 17.1* (Sec. 17.3)
AB
1 0 2
0
A
B
1
A
1
B
0
1 0 1 2
1
0
1 0 1 2
1
AB
AB
1 0 1 2
1 0 1 2
1 0 A 1 A 0 B 1 B 0 A 1 A 0 2 2 1 0 A 0 B 1 A 1 B 0 A 1 B 1 A 0 B 2 2 1 0 A 0 B 1 A 1 B 0 A 1 B 1 A 0 B 2 2
1 0 2
A
0
B
1
A
1
B
B
0
B
1
B
17.2* (Sec. 17.4)
a
1 0 2
AB a
1 a 0
A
0
a
1 a 1
A
1 B.
Write the identity as a sum of projection operators onto the Bell basis: 1ˆ
aA aA
aA aA
aA aA
aA aA
294
Chapter 17 1 1ˆ 2 1 2 1 2 1 2
aA aA
1 0 0 1 0 0 1 0 0 1 0
0
a
A
a
a
A
aA aA
aA aA
1 2
a
a
A
a
a
A
aA aA
1 0 0 B aA 2 1 0 0 B aA 2 1 0 0 B aA 2
1 1ˆ 2
0
a
0
B
0 0
a
1 a 1
a
1 a 1
0
B
0
B
A
1 B
A
1 B
a
1 a 1
a
1 a 1
0
B
0
A
1 B
A
1 B
B
1
1 0 1 B 2 1 0 1 B 2
1 0 1 B 2 1 1 1 0 1 B 0 0 B aA 2 2 2
1 2
1 0 B 1 B 1ˆ aA 2 1 0 B 1 B aA 2 1 0 B 1 B aA 2 1 0 B 1 B aA 2
17.3
(Sec. 17.4) If Alice measures 0
B
aA
. Bob's qubit is projected into the state 0
B
. Bob must turn
into 1 B , and vice versa. The matrix representation of the unitary transformation that will
turn this into Alice's state is 0 1 Uˆ 1 0 Note that this is the X-gate (or NOT gate) of Sec. 17.5.
Chapter 17 If Alice measures turn 0
B
aA
Bob's qubit is projected into the state 0
into 1 B , and 1
B
into 0
B
B
1
0 1 Uˆ 1 0 Note that this is the Y-gate (apart from an overall phase factor of -i) of Sec. 17.5. (Sec. 17.4) The state of the three-particle system is
a
1 0 2
AB
a
1 a 0
A
1
B
0
a
1 a 1
A
0
.
B
Write the identity as a sum of projection operators onto the Bell basis: 1ˆ
aA aA
1 1ˆ 2 1 2 1 2 1 2
aA aA
aA aA
aA aA
a
a
A
B
a
a
A
B
a
a
A
B
a
a
A
B
aA aA
aA aA
a
a
A
B
a
a
A
B
a
a
A
B
a
a
A
B
aA aA
1 0 1 B aA 2 1 0 1 B aA 2 1 0 1 B aA 2
1 1ˆ 2
aA aA
1 0 1 0 1 1 0 0 1 0 1 0 1 1 0 0 1 0 1 0 1 1 0 0 1 0 1 0 1 1 0
0
B
. Bob must
. The matrix representation of the unitary transformation
that will turn this into Alice's state is
17.4
295
1 0 0 2
B
1 1 0 0 B 2 2 1 1 0 0 B 2 2 1 1 1 0 1 B 0 0 B aA 2 2 2
296
Chapter 17
1 1 B 0 B 1ˆ aA 2 1 1 B 0 B aA 2 1 1 B 0 B aA 2 1 1 B 0 B aA 2
From this, we can see that if Alice measures
aA
, Bob's qubit is projected into the state of
Alice's original qubit. If Alice measures any other Bell state, Bob's qubit can be transformed into the state of Alice's original qubit with a unitary transformation. 17.5
(Sec. 17.5)
A 0 0 1 1 17.6
B 0 1 0 1
C 1 1 1 0
D 1 1 0 1
(Sec. 17.5) 1 1 1 0 1 0 1 2 2 ˆ 0 1 X 1 0 ˆ 0 0 1 1 1 1 1 0 X 2 1 1 0 2 1 ˆ 1 0 1 1 1 1 1 1 X 2 1 1 0 2 1 0
E 1 0 1 1
Q 0 1 1 0
Chapter 17
297
This does not look like the behavior of a NOT gate in the prime-basis, as in the prime-basis we'd like a NOT gate to turn 0 into 1 , and vice versa. 17.7
(Sec. 17.5) ˆ X B
AB
A
A
1 0 2 1 0 2
ˆ Y B
AB
ˆ 0 X B
B
1
A
1
B
1
A
0
B
ˆ 1 X B
AB
1 ˆ 0 1 Y ˆ 1 0 A Y B B B A 2 1 i 0 A 1 B i 1 A 0 B 2
B
i
B
AB
The phase factor in front is not physically significant. Zˆ B
AB
A
A
1 0 2 1 0 2
Zˆ B 0 0
B
1
B
A
1
A
1
B
Zˆ B 1
B
AB
It is also possible to operate on qubit A, and achieve essentially the same results. 17.8* (Sec. 17.5) (a)
Cin
Tin
Cout
Tout
0
0
0
0
0
1
0
1
1
0
1
1
1
1
1
0
298
Chapter 17 (b) From Eq. (17.17): 1 0 CNOT 0 0
0
0
1
1
C
C
C
C
0
1
0
1
T
T
T
T
0 0 0 1 0 0 0 0 1 0 1 0 01
1 0 2
C
C
1 0 2
1 0 2
C
C
1 0 2
1 0 2
C
C
1 0 2
1 0 2
C
1 0 2
C
1C
0
0
T
1C
0
T
T
0
T
1T 1C 0
T
1 1 1 2 1 101
1 1 1 2 1 101
1 1 1 2 1 1 01
1C 1T
C
1T 1C 0
T
1C 1T
0 T 1 T
0
1C
C
1 1 1 2 1 101
0 T 1 T
0
1C
0
0 T 1 T
C
1T 1C 0
T
1C 1T
0 T 1 T
0
C
1T 1C 0
T
1C 1T
Chapter 17
0 CNOT
0 CNOT
1 CNOT
1 CNOT
C
C
C
C
0
1
0
1
T
T
T
T
1 0 0 0
0 0 0 1 1 1 0 0 1 1 1 1 0 0 0 1 2 1 2 1 0 1 0 01 101 101
1 0 0 0
0 0 0 1 1 1 0 0 1 1 1 1 1 0 0 1 2 1 2 1 0 1 0 01 101 1 01
1 0 0 0
0 0 0 1 1 1 0 0 1 1 1 1 1 0 0 1 2 1 2 1 0 1 0 01 101 101
1 0 0 0
0 0 0 1 1 1 0 0 1 1 1 1 0 0 0 1 2 1 2 1 0 1 0 01 1 01 101
0
C
C
C
C
Cin
Tin
Cout
Tout
0
0
0
0
0
1
1
1
1
0
1
0
1
1
0
1
299
T
1
0
1
T
T
T
(c) We see from the truth table in part (b) that in the 0'1'-basis a CNOT data has no effect on the target qubit, but flips the control qubit if the target qubit is 1 . Clearly the control qubit can change in a CNOT gate. In the 0'1'-basis it appears that the action of the control and target qubits is opposite from what it is in the 01-basis.
300
Chapter 17
17.9
(Sec. 17.5)
(a) The truth table for a CZ gate is Cin Tin
Cout
Tout
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
1
Therefore, its matrix representation is 1 0 0 0 0 1 0 0 CZ 0 0 1 0 0 0 0 101
(b) The truth table for a Hadamard gate operating on the "target" qubit is Cin Tin Cout Tout 0
0
0
1 0 1 2
0
1
0
1 0 1 2
1
0
1
1 0 1 2
1
1
1
1 0 1 2
So the matrix representation is
Chapter 17 1 1 ˆ 1 1 1 H T 2 0 0 0 0
301
0 0 0 1 1 1 101 0
1 1 H ˆ CNOT ˆ 1 1 1 H T T 2 0 0 0 0 1 1 1 1 1 2 0 0 0 0 1 0 0 0
0 1 0 0 0 1 1 0 1 101 0
0 0 0 1 1 1 0 0 1 1 0 0 1 0 0 0 1 0 01 0 0
0
0 1 1 0 0 1 1 1 1 0 0 1 101 0 0
0 0 0 1 1 1 101 0
0 0 0 1 1 1 1 01
0
0
0 1 0 0 0 1 0 0 0 101 0 0
CZ
17.10 (Sec. 17.5) We know that the input to the U f -gate is:
1 1 0 1 11 0 212 2 2 2 1 0 1 0 2 0 1 1 2 11 0 2 11 2
12
Qubit 1 is x, and is unchanged. Qubit 2 is y, and gets replaced with y f x . The functions are defined in Table 17.3. In the text we showed that for f1 x : f1 0 0,
1 0 1 0 2 0 1 1 2 11 0 2 11 2 1 1 0 1 11 0 212 2 2
3
12
.
f1 1 0 , and the output state is:
302
Chapter 17 For f 2 x : f 2 0 1,
f 2 1 1 , and the output state is:
1 0 1 1 2 0 1 0 2 11 1 2 11 0 2 2 1 1 0 1 11 0 212 2 2
3
For f3 x : f3 0 0,
f3 1 1 , and the output state is:
1 0 1 0 2 0 1 1 2 11 1 2 11 2 1 1 0 1 11 0 212 2 2
3
For f 4 x : f 4 0 1,
0
2
f 4 1 0 , and the output state is:
Therefore: 3
1 0 1 1 2 0 1 0 2 11 0 2 11 1 2 2 1 1 0 1 11 0 212 2 2
3
1 1 0 1 11 0 212 2 2 1 1 0 1 11 0 212 2 2
f f1 , f 2
f f3 , f 4
17.11* (Sec. 17.5) The input state is: 1 1 0 1 11 0 212 2 2 2 1 0 1 0 2 0 1 1 2 11 0 2 11 2
12
The output states are worked out in the previous problem. For f 2 x : 3
1 0 1 1 2 0 1 0 2 11 1 2 11 0 2
2
ˆ to qubit 2: This is obtained from 2 by applying X
Chapter 17
303
ˆ Uˆ f 2 X 2 For f3 x : 3
1 0 1 0 2 0 1 1 2 11 1 2 11 0 2
2
with 1 as the control, and 2 as the target: This is obtained from 2 by applying CNOT Uˆ CNOT f3
For f 4 x :: 3
1 0 1 1 2 0 1 0 2 11 0 2 11 1 2 2
to , ˆ to the state for f x , so first we apply CNOT This is obtained by applying X 2 3 3 2 ˆ : then we apply X 2
ˆ CNOT Uˆ f 4 X 2
(b) Using the operations from part (a):
Lab 1 Spontaneous Parametric Downconversion Lab Ticket
We know that k p ks ki . z-components: k p k s z ki z cos k s ki
n n2 c Since the signal and idler have the same wavelength and index of refraction, they have the same magnitude of wave vector. k s ki k k
k p 2k cos n p 2
p np
p
n 4 cos
n2 cos
306
Lab 1 We know that p / 2 , so cos
np n
1.659 o cos 1 2.8 1.661
Q1: Should be around 3 degrees (or whatever angle your crystal is cut for). Q2: Should be around 3 degrees (or whatever angle your crystal is cut for). Q3: Depends on the thickness of the crystal, the alignment of the focusing optics, etc. Hopefully it's smaller than one degree. Q4: The polarizer should be oriented parallel to the pump polarization. The A detector counts should be minimized because the downconverted photons are perpendicular to the pump beam. Q5: This beam should behave exactly the same as the other beam. Q6: Rotating the wave plate 45° (which rotates the pump polarization by 90°) should extinguish the downconversion counts. The polarization of the downconversion should not change as the pumpbeam polarization changes (can verify this with the polarizer). Q7: Hopefully about 1ns or less. Q8: Typically a few nanoseconds. Q9: Students should see a peak on top of a constant background. The peak represents true coincidences, while the background represents accidentals. They should determine the total number of accidentals in an 8ns window (assuming constant background), and compare it to the total number of coincidences in that window. This is easy enough to do by importing the data into a spreadsheet.
Lab 2 “Proof” of the Existence of Photons Lab Ticket Two-detector measurements use only the two detectors monitoring the outputs of the beam splitters. Three-detector measurements further condition the results on the detection of an idler photon. We should measure g (2) 0 1 for three-detector measurements, because the signal beam is only prepared in a true single-photon state conditionally upon the detection of an idler photon. Measuring g (2) 0 1 is a sign that a quantum mechanical description of the field is necessary to explain the results, because classical waves must satisfy g (2) 0 1 . For the two-detector measurements we expect to measure g (2) 0 1 , because the signal beam is not prepared in a single-photon state without conditioning on the third detector. Q1: Depends on how the crystal is aligned--whether it produces vertical or horizontal pairs. Horizontals are transmitted to B, verticals are reflected to B’. Rotate the wave plate 45° to maximize other count rate, because a half-wave plate rotates the polarization by twice its rotation angle. Q2: Depends on crystal orientation. Q3: There are always accidental coincidences. Q4: This number is calculated using the formula in Thorn et al. (Ref. [L2.2]) for the expected g ( 2) 0 , given the measured rates and the coincidence window: N t N g (2) 0 N A B B T N AB N AB
If the coincidence window t is set properly in the program, the measured value should agree reasonably well.
308
Lab 2
Q5: Because the measurements involve different quantum states for two-detector and three-detector measurements. Two-detector measurements correspond to the field in a thermal-like (classical) state, while three-detector measurements correspond to the field in a single-photon state because of the conditional state preparation.
Lab 3: Single Photon Interference Lab Ticket
The Jones matrix of the combination of the two BDPs, wave-plate 2, and the phase shift is given by J J J /2 45o JV J /2 45o J H ei 0
00 10 0 0 11 0 11 00 1 1 00 0
0 e i 0 0 0 e i . 0 0 1 0 1 0
The Jones matrix for the whole interferometer, corresponding to the beam transmitted through the PBS to detector-B is:
310
Lab 3 J PI J H J /2 22.5o J J /2 1
1 0 1 1 1 0 ei cos 21 sin 21 0 0 2 1 1 1 0 sin 21 cos 21 1 1 0 1 1 ei sin 21 ei cos 21 2 0 0 1 1 cos 21 sin 21 i i 1 1 0 e sin 21 cos 21 e cos 21 sin 21 2 0 0 ei sin 21 cos 21 ei cos 21 sin 21 1 ei sin 21 cos 21 ei cos 21 sin 21 2 0 0 For vertically polarized input: 0 1 ei sin 21 cos 21 ei cos 21 sin 21 0 J PI 2 0 0 1 1
1 ei cos 21 sin 21 2 0
So, the intensity of the horizontally polarized output is: 2 1 i e cos 21 sin 21 2 1 sin 2 21 cos 2 21 2sin 21 cos 21 cos 2 1 1 2sin 21 cos 21 cos 2 1 1 sin 41 cos 2
I
Calculate the visibility: 1 1 1 sin 41 1 sin 41 2 V2 1 1 1 sin 41 1 sin 41 2 2 sin 41
Lab 3
311
Q1: Beams get alternately brighter and dimmer--when one is bright the other is dim. The input to wave-plate 3 is linearly polarized (by the PBS), and the wave plate rotates the input polarization to the BDP; the beams split differently on the BDP for different input linear polarizations. Q2: The intensity of each of the beams (horizontal and vertical) coming out of interferometer is constant. The intensity doesn’t get brighter or dimmer in front of the polarizer because there is no interference between these two beams, as they are orthogonally polarized. The total intensity is constant. However, the relative phase of the horizontal and vertical beams does change. Inserting the polarizer projects each of the beams emerging from the interferometer onto the polarizer axis. After the polarizer the beams are polarized along the same direction, and will interfere. Only after the polarizer is the interference apparent. Q3: Use the formula from the ticket to calculate the visibility and compare it to the data. The measured visibility will almost certainly be less than theory, because it's hard to get perfect interference. Q4: The pattern with 1 22.5o should have higher visibility. With 1 0o the photons strike the first BDP with vertical polarization (referring to Fig. L3.1), so we know they will take the top path through the interferometer. With 1 22.5o the path information is erased; we don't know which path the photon takes, because it will split at the first BDP. With 1 0o we have the path information, so we cannot see interference, while with 1 22.5o we don't have path information, so we should see interference. Q5: The pattern with 3 22.5o should have higher visibility. With 3 0o the photons detected at detector-B (referring to Fig. L3.1) have emerged from the second BDP with horizontal polarization, so we know they have taken the top path through the interferometer. With 3 22.5o the path information is erased; we don't know which path the photon has taken, because either path will have equal probability of being transmitted through the PBS. With 3 0o we have the path information, so we cannot see interference, while with 3 22.5o we don't have path information, so we should see interference. Q6: The rates should be roughly equal--only one polarization is incident on the polarizing beam splitter, so the beam splits equally. Q7: Same as previous question.
312
Lab 3
Q8: AB maximum, AB' minimum. This is because the two beams interfere, and the phase of the interferometer has been adjusted to send all of the signal photons to detector-B.
Q9: Should be less than 1.
Lab 4 Quantum State Measurement Lab Ticket The beam goes through QWP then HWP, then Horizontal polarizer: Both waveplates set at 0 deg: a 1 J H J /2 0o J /4 0o i be 0 1 0 1 0 a 0 P H a2
0 1 0 1 0 a 0 0 1 0 i bei 0 1 0 a 0 0 i bei 0 a 0 bei
[Eq. (L4.2)]
QWP @ 45, HWP @ 22.5 a 1 0 1 1 1 1 1 i a J H J /2 22.5o J /4 45o i i be 0 0 2 1 1 2 i 1 be 1 1 1 1 i a 2 0 0 i 1 bei 1 1 i i 1 a 0 bei 2 0 1 1 a 1 1 i i 2 0 0 be a bei i /4 1 e 2 0
314
Lab 4
PH
a b i e 2 2
2
1 2 a 2ab cos b 2 2 1 1 2ab cos 2 This is the same as P 45 [Eq. (L4.5)].
QWP @ 45, HWP @ 0 a 1 0 1 0 J H J /2 0o J /4 45o i be 0 0 0 1 1 1 0 1 2 0 0 i
1 1 i a i 2 i 1 be i a 1 bei
1 1 i a i 2 0 0 be i 1 a ibe 2 0
2
a b i PH e i 2 2 1 a 2 ab iei iei b 2 2 1 1 2ab sin 2 This is the same as P L [Eq. (L4.6)].
Q1:
It should be! Should have a 1 b 0 , or a 0 b 1, depending on source polarization. The phase is insignificant (see Q2).
Q2:
a H bei V If a is very small, then b 1 , and bei V V ; the phase is an overall phase, which is not physically significant. If b is very small, then a 1 , and a H H , so the phase doesn't matter.
Lab 4
315
Q3:
45
1 H V 2 1 ab , 0 2
1 H V 2 1 ab , 2
Q4:
45
30 cos 30 H sin 30 V a cos 30 0.866
b sin 30 0.5 , 0
Q5:
1 H i V 2 1 ab , /2 2
L
Q6:
1 H i V 2 1 ab , / 2 2
R
I'll assume that the photons coming from the downconversion crystal are vertical. A quarter wave plate at an arbitrary angle then produces the state
316
Lab 4 cos 2 i sin 2 1 i sin cos 0 ˆJ /4 V 1 i sin cos sin 2 i cos 2 1 1 i sin cos 2 sin i cos 2 e i/4 2 sin cos 1 1 i cos 2 e i/4 2 sin cos 1 ei/4 2 cos 2 2 sin cos e i/4 e i/4 2 cos 2 1 sin 2 2 e i/4 1 2 1 i 2 cos 2 1 sin 2 2 e i/4 1 1 2 1 2 cos i 2 2
a H bei V a
1 sin 2 2
1/2
2 1 1 b 1 2 cos 2 2 2 1 1 tan 1 2 cos 2
1 1 2 cos 2 2
For 15o : 1 1 sin 30 a 0.35 2 2 2 b
1 2
2 1 2 cos 15
2
1/2
1
0.935
2
1/2
1
Lab 4 1 tan 1 1 2 cos 2 15
tan 1 1 2.28rad 0.727 130 deg 0.866
317
Lab 5 Testing Local Realism Lab Ticket From Carlson et al. the probabilities are P (θ A ,θ B )
A
θA
B
θ A
θB
1
A
θB
B
A
P (θ A ,θ B )
θA
2
1
B
θB
0.2 H
A
H
B
0.8 V
A
V
0.2 θ A H θ B H 0.8 θ A V θ B V 0.2cosθ A cos θ B 0.8 sinθ A sin θ B
0.2cosθ A cos θ B 0.8 sinθ A sin θ B
B
2
Can use this in a spreadsheet to generate all the plots. Plots of P, and P , are identical (apart from negative signs):
This is 0 at approximately 19o ,
320
Lab 5
H
H (alpha=35 deg.) 0.2 0.15 0.1 0.05 0 -0.05 0 -0.1 -0.15 -0.2 -0.25 -0.3
20
40
60
80
100
H
Beta
with 35o and 19o , P , , P , and P , should all be
For 1
approximately 0. Notice in the plot of H, as decreases from 19o , H will increase. [This is because P, increases faster than P, and P , in this region]. So, decreasing
is the key to maximizing H for 1 . P , and P , are both less than 0.01 as long
as 120 . Q1: We know that an expectation value (or average) can be calculated by multiplying the probability of a measurement by the measured value, and then summing over all possible measurements. If Alice measures her photon polarization to be along A she gets value of +1, while if she measures perpendicular to that she gets a value of -1. Bob's values are the same, whether he measures along B or perpendicular to it. The measured value for the joint polarization operator is the product of Alice and Bob's measurements. Therefore: ˆ AB E A , B A B
1 1 P A , B 1 1 P A , B 1 1 P A , B 1 1 P A , B P A , B P A , B P A , B P A , B
Lab 5
Q2: P (θ A ,θ B )
A
θA
B
θB
A
θA
θB
B
θA
A
B
321
2
1 θB H 2
H
A
B
V
A
V
1 θ A H θB H θ A V θB V 2 1 cosθ A cos θ B sinθ A sin θ B 2
B
1 2 cosθ A cos θ B sinθ A sin θ B 2 1 cos 2 θ A θ B 2
P (θ A ,θ B )
S E A1 , B1 E A 2 , B1 E A 2 , B 2 E A1 , B 2
For the given angles, we can easily calculate the probabilities, expectation values and S using a spreadsheet: A
B
P(A,B) 0 22.5 0.426777 45 22.5 0.426777 45 ‐22.5 0.073223 0 ‐22.5 0.426777 Here Ap=A+90, and Bp=B+90
P(Ap,Bp) 0.426777 0.426777 0.073223 0.426777
P(Ap,B) 0.073223 0.073223 0.426777 0.073223
P(A,Bp) 0.073223 0.073223 0.426777 0.073223 S=
E(A,B) 0.707107 0.707107 ‐0.70711 0.707107 2.828427
Q3: Probability of an A’B coincidence is given by: PA' B
A V
B
A
V H
B
H
2
a
A V
B
H
0
Likewise for an AB’ coincidence.
H
A
H
B
1 aei
A V
B
H
V
A
V
B
322
Lab 5
Q4: The A beam is unchanged, but the half-wave plate in the B beam transforms the state. We can determine the transformation by using the operator corresponding to a half-waveplate at 45o on the B beam: out Jˆ B / 2 45o out a H
A
a H
A
Jˆ B /2 45o H V
B
B
1 a e i V
1 a e i V
A
H
A
Jˆ B /245o V
B
B
This state has H A V B photons which make AB’ coincidences, and V A H B photons which make A’B coincidences, but no other terms. You should see AB’ and A’B coincidences, but no AB or A’B’ coincidences. Q5: Equal probabilities of AB or A’B’ coincidences mean a and 1 a are equal. We don't know anything about the phase, so with proper normalization the source state is 1 s H A H B ei V A V B . 2
Q6: The state from the source is:
1 H A H B ei V A V B 2 Again, the half-wave plates act as operators to transform the state: s
Jˆ A / 2 22.5o H
A
45
, Jˆ A / 2 22.5o V
A
A
45 A ,
A
and likewise for B. So, the state after the wave plates is:
1 out Jˆ A /2 22.5o Jˆ B /2 22.5o s 45 2
45
Probability of an A’B coincidence is then given by: PA' B
A
V
B
H out
2
B
ei 45
A
45
B
Lab 5
A V
out
1 V B H 45 A 45 B ei A 2 1 1 1 1 1 ei 2 2 2 2 2 For this to be 0, must have 0 . The source state is thus 1 s H A H BV AV B . 2 B
H
Q7: P (θ A ,θ B )
A
θA
B
θB
A V
B
A
θA
s
θB
B
θA
A
2 s
1 θB H 2
B
H
A
B
V
A
1 θ A H θB H θ A V θB V 2 1 cosθ A cos θ B sinθ A sin θ B 2
V
B
1 2 cosθ A cos θ B sinθ A sin θ B 2 1 cos 2 θ A θ B 2 2 1 P A , B cos A cos B sin A sin B 2 1 cos 2 A B 2 P (θ A ,θ B )
Q8: P A , B H A , H B
Q9: P A , B VA , VB
A
A
θA
θA
B
B
θB
θB
H
V
A
A
V
H
B
B
2
2
cos 2 A cos 2 B
sin 2 A sin 2 B
Q10: Pmix A , B
1 cos 2 A cos 2 B sin 2 A sin 2 B 2
For the graphs, want plots of: 1 Pmix 0, B cos 2 B 2
H
323
45 A 45 B
324
Lab 5 Pmix 45o , B
1 1 cos 2 B sin 2 B 4 4
Q11: No Q12: There's a symmetry: the difference between the states 1 and 2 is that horizontal and vertical have been flipped, which corresponds to a reflection about a line that makes an angle of 45°. For 1 is 10° less than 45°, so for 2 will be 10° more than 45°. can be found in a similar manner by reflecting it about 45°. This means 55o 71o You can verify that these angles yield the proper probabilities. Redoing the lab ticket: for 2 with 55o and 71o , P , , P , and P ,
should all be 0. As increases from 71o , H will increase. [This is because P, increases faster than P , and P , in this region]. So, increasing is the key to
maximizing H for 2 .