obtained thanks to https://t.me/HermitianSociety
443 146 984KB
English Pages [98] Year 2020
Table of contents :
Foreward
Acknowledgements
Notation and Useful Constants
Introduction
Abstract
The goal of physics
The connection between physics and mathematics
Paradigm shifts
The correspondence principle
Symmetry and Physics
Abstract
Learning outcomes
What is symmetry?
Role of symmetry in physics
Symmetry as a guiding principle
Symmetry and conserved quantities
Symmetry as a tool for simplifying problems
Symmetries were made to be broken
Spacetime symmetries
Parity violation
Spontaneously broken symmetries
Variational calculations: lifeguards and light rays
Formal Aspects of Symmetry
Abstract
Learning outcomes
Symmetries as operations
Definition of a symmetry operation
Rules obeyed by symmetry operations
Multiplication tables
Symmetry and group theory
Examples
The identity operation
Permutations of two identical objects
Permutations of three identical objects
Rotations of regular polygons
Continuous vs discrete symmetries
Noether's theorem
Supplementary: variational mechanics and the proof of Noether's theorem
Variational mechanics: principle of least action
EulerLagrange equations
Proof of Noether's theorem
Symmetries and Linear Transformations
Abstract
Learning outcomes
Review of vectors
Coordinate free definitions
Cartesian coordinates
Vector operations in component form
Position vector
Velocity and acceleration: differentiation of vectors
Linear transformations
Definition
Translations
Rotations
Reflections
Linear transformations and matrices
General discussion
Identity transformation and Inverse
Rotations
Reflections
Matrix representation of permutations of three objects
Pythagoras and geometry
Special Relativity I: The Basics
Abstract
Learning outcomes
Preliminaries
Frames of Reference
Spacetime diagrams
Newtonian relativity and Galilean transformations
Derivation of Special Relativity
The Fundamental Postulate
The problem with Galilean relativity
MichelsonMorley experiment
Maxwell's equations
Summary of consequences
Relativity of simultaneity
Surface of Simultaneity: what time did it happen?
Time dilation
Derivation:
Properties of time dilation
Proper time
Experimental confirmation of time dilation
Examples
Lorentz contraction
Derivation of Lorentz Contraction
Properties:
Proper length and proper distance.
Death Star Betrayal: an example
Special Relativity II: In Depth
Abstract
Learning outcomes
Lorentz transformations
Derivation of general form
Properties of Lorentz transformations
Lorentzian geometry
The light cone
Proper time revisited
Relativistic addition of velocities
Relativistic Doppler shift
Nonrelativistic Doppler shift review
Relativistic Doppler shift
Relativistic energy and momentum
Relativistic energymomentum conservation
Relativistic inertia
Relativistic energy
Relativistic threemomentum
Relationship between relativistic energy and momentum
Kinetic energy:
Massless particles
Spacetime vectors
Position fourvector:
Momentum fourvector:
Null fourvectors
Relativistic scattering
Relativistic units
Symmetry redux
Matrix form of Lorentz transformations
Lorentz transformations as a symmetry group
Supplementary: Fourvectors and tensors in covariant form
General Relativity
Abstract
Learning outcomes
Problems with Newtonian gravity
Review of Newtonian gravity
Perihelion precession of Mercury
Action at a distance
The Puzzle of inertial vs gravitational mass
Strong Equivalence Principle
Geometry of spacetime
Some consequences of general relativity:
Gravitational waves
Introduction
Detection
Early observations
Black holes
Properties of black holes:
Observational evidence for black holes
Cosmology
Introduction to the Quantum
Abstract
Learning outcomes
Light as particles
Review: light as waves
Photoelectric effect
Compton scattering
Blackbody radiation and the ultraviolet catastrophe
Blackbody radiation
Derivation of the RayleighJeans law
The ultraviolet catastrophe
Quantum resolution:
The early Universe: the ultimate blackbody
Particles as waves
Electron waves
de Broglie wavelength
Consequences
The Heisenberg uncertainty principle
The Wave Function
Abstract
Learning outcomes
Quantum vs Newtonian mechanics
Newtonian description of the state of a particle
Quantum description of the state of a particle
Measurements of position
Example: Gaussian wave function
Momentum in quantum mechanics
Pure waves
The momentum operator
Energy in quantum mechanics
Timeenergy uncertainty relation
The Schrödinger Equation
Abstract
Learning outcomes
The time independent Schrödinger equation
Stationary states
Examples of stationary states
Free particle in one dimension
Particle in a box with impenetrable walls
Simple harmonic oscillator
Absorption and emission
Tunnelling
Particle in a box with penetrable walls
Tunnelling through a potential barrier of finite width
Applications of tunnelling
The quantum correspondence principle
Recovering the everyday world
The Bohr correspondence principle
The time dependent Schrödinger equation
Heuristic derivation
Coherent states
Observables as linear operators
Symmetry in quantum mechanics
Supplementary: quantum mechanics and relativity
The Hydrogen Atom
Abstract
Learning outcomes
Newtonian dynamics
Supplementary: Symmetries of the hydrogen atom
Spherical Symmetry
Accidental symmetry of the hydrogen atom
The Bohr atom
Emission and absorption spectra
Three dimensional hydrogen atom
The Schrödinger equation
Solutions: symmetry to the rescue
Probability densities
Shells, orbitals and degeneracy
Fermions and the spin quantum number
Summary of the 3D hydrogen atom
The periodic table
Hydrogenlike atoms
Chemical properties and the periodic table
Nuclear Physics
Abstract
Learning outcomes
Properties of the nucleus
The constituents
Structure of nucleus
The nuclear force
Radioactivity
Isotopes
Neutrinos
Types of radioactive decay
Decay rates
Carbon dating
Fission and fusion
Binding energy
Binding energy per nucleon
Formation of elements
Mysteries of the Quantum World
Abstract
Learning outcomes
What is real?–a quantum conundrum
Bell's theorem and the nature of quantum reality
More on spin
Overview
Mathematical details
Summary
Experimental confirmation of quantum weirdness
Entanglement: the key to unlocking quantum weirdness
Quantum computation: entanglement as a resource
Quantum recap
Classical computers, in brief
What is a quantum computer?
Examples of quantum algorithms
Two important technical details
Interpretations of quantum mechanics: what does it all mean?
The Copenhagen interpretation
The ManyWorlds interpretation
Hidden variables and nonlocality
Conclusions
Abstract
Appendix: Mathematical Background
Abstract
Complex numbers
Probabilities and expectation values
Discrete distributions
Continuous probability distributions
Dirac delta function
Fourier series and transforms
Fourier series
Fourier transforms
The mathematical uncertainty principle
Dirac delta function revisited
Parseval's theorem
Waves
Moving pure waves
Complex waves
Group velocity and phase velocity
Wave packets
Wave number and momentum
List of Figures
List of Tables
Solutions to Exercises
Introduction
Symmetry and Physics
Formal Aspects of Symmetry
Symmetries and Linear Transformations
Special Relativity I: the basics
Special Relativity II: in depth
General Relativity
Introduction to the Quantum
The Wave Function
The Schrödinger Equation
The Hydrogen Atom
Nuclear Physics
Mysteries of the Quantum World
Conclusions
Appendix: Mathematical Background
Chapter 16 Solutions to Exercises 16.1
Introduction
No exercises
16.2
Symmetry and Physics
Exercise 1. Calculate the length D of roadway in Fig. 2.3 as a function of l and a, and verify that it is minimized by the value of l given in Eq.(2.1) above. Solution: Each diagonal segment on the right hand side of Fig. 2.3 has length: s a 2 a − l 2 + (16.1) d(l) = 2 2 The total length of pavement D(l) as a function of l is: s a 2 a − l 2 D(l) = l + 4d = l + 4 + 2 2
496
(16.2)
CHAPTER 16. SOLUTIONS TO EXERCISES
497
To find the value of lmin that minimizes the function D(l) we have to solve: dD(l) 2(a − l)(−1) = 1+ p dl lmin a2 + (l − a)2 = 0 (16.3) A bit of algebra yields: a2 3 Since l must be less than a the relevant solution is: 1 lmin = a 1 − √ 3 (a − lmin )2 =
Plugging this into the expression for D(l) we find: 1 D(lmin ) = a 1 + √ 3
(16.4)
(16.5)
(16.6)
which is less than the length of pavement required by building two diagonal roads directly through the center of the square, namely: √ √ (16.7) Ddiagonals = 2 a2 + a2 = 2 2a A straightforward calculation verifies that d2 D(l) 0 dl2 lmin
(16.14)
so that lmin is indeed a minimum as required.
Exercise 3. Derive Equation (2.2) using the diagram in Fig. 2.5. The distances a, b and c are given. You can use the fact that the total time Ttot taken along Path 2 is: p p c21 + b2 c22 + a2 Ttot = + (16.15) VG VS and find the equation for c1 that minimizes Ttot . Hint: in order to solve the resulting equation, you may need: c1 = tan(θG ) b c2 = tan(θS ) a
(16.16)
CHAPTER 16. SOLUTIONS TO EXERCISES
500
Solution: We wish to minimize the time take to get to swimmer with respect to c1 , so require: 1 2(−1)(c − c1 ) 1 2c dTtot p 1 + p = 2 2 dc1 2 VG c1 + b 2 (c1 − a)2 + a2 p c21 + b2 c1 VG p → = c − c1 VS (c − c1 )2 + a2
(16.17)
Now square both sides of the above, and divide both sides by a2 b2 to get: VG c22 c21 c21 c22 +1 = +1 (16.18) b 2 a2 V S a2 b 2 where we have replaced (c − c1 ) by c2 . Next use Eq. (16.16) to replace c1 /b and c2 /a by tan(VG ) and tan(VS ), respectively to obtain Snell’s law.
16.3
Formal Aspects of Symmetry
Exercise 1. 1. Verify that the entries in the multiplication table (3.3) follow from the actions of the symmetry operations listed in Eq.(16.99). 2. Verify that this multiplication table obeys the four axioms of group theory (i.e. properties of symmetry operations). Note: this should not require any calculations once you have the table completed. 3. Is this group of symmetry operations commutative? That is, does the order of operations matter in some cases? If so, which ones? Solution:
CHAPTER 16. SOLUTIONS TO EXERCISES
501
We consider the set of permutations S := {I, A, B, C, G, F } of three identical objects a b c where the symmetry operations corresponding to the permutations fined as: a a a b I : b → b ; A: b → a c c c c a a a c B : b → c ; C: b → b c b c a a c a b F : b → a ; G: b → c c b c a
are de
(16.19)
(16.20)
(16.21)
1. Fill in the rest of the multiplication table below for this group of symmetry operations: • Top horizontal row and left vertical row trivial since I · X = X · I for any X. • A · A = I since A just permutes the top two components a and b with each other twice so that this is the same as doing nothing at all. • Similarly B permutes 2nd and 3rd, and C permutes 1st and 3rd, so that B · B = I, and C · C = I. • For A · C, remember you act a b →C c
with C and then with A: c b b →A c a a
(16.22)
Note that the operations move the components based on their locations, not their letter names. The net effect is the same as G. Similarly F followed by G is:
CHAPTER 16. SOLUTIONS TO EXERCISES
502
a c a b →F a →G b c b c
(16.23)
The net effect is the identity. Continuing as above you get:
I A B C G F
I A B C G F I A B C G F A I F G C B B G I F A A C F G I B A G B C A F I F C A B I G
2. Verify that this multiplication table obeys the four axioms of group theory (i.e. properties of symmetry operations). Note: this should not require any calculations once you have the table completed. For associativity, just check on or two cases. –It is complete because the product of any two symmetry operations yields another symmetry operation. That is, all the entries in the table are one of the operations along the top row. –There is an identity operator, I –Every element has an inverse. A, B and C are their own inverses. The inverse of F is G and vice versa. –Verifying associativity is a bit of a pain, but for example: (A · C) · G = G · G = F A · (C · G) = A · B = F
(16.24)
So associativity holds in this case. 3. Is this group of symmetry operations commutative? That is, does the order of operations matter in some cases? If so, which ones? No it is not commutative. For example A · G = C whereas G · A = B. However, F · G = G · F = I. That is about it for commuting operations.
CHAPTER 16. SOLUTIONS TO EXERCISES
503
Exercise 2. Construct the multiplication table for the set of rotations that leave a hexagon invariant. Solutions As can be seen in Fig. 16.2 directly below, rotations by multiples60◦ take the diagonal lines onto diagonal lines and leave the hexagon invariant.
Figure 16.2: Hexagon
0 60 120 180 240 300
Exercise 3. Show that
0 0 60 120 180 240 300
dE dt
60 60 120 180 240 300 0
120 120 180 240 300 0 60
180 180 240 300 0 60 120
240 240 300 0 60 120 180
300 300 0 60 120 180 240
= 0 if x(t) satisfies Eq. (3.13)
CHAPTER 16. SOLUTIONS TO EXERCISES
504
Solution: Start with the expression for energy: 1 2 E := K + V = m[x(t)] ˙ + V (x(t)) = constant 2
(16.25)
Now calculate its time derivative, noting that all the time dependence is through x(t): dE 1 dV (x(t)) dx = m2x(t)¨ ˙ x(t) + dt 2 dx dt dV (x(t)) = v(t) x¨(t) + dx
(16.26)
which vanishes if x(t) obeys Newton’s equations. Note that if V := V (x(t), t)depended on time explicitly, there would be an extra term on the right hand side, so that energy would not be conserved..
Exercise 4. Derive the equations of motion for a bead moving on a circular frictionless wire of radius R (see Fig. 3.3) using the principle of least action.
Figure 16.3: Bead moving along frictionless circular wire of radius R.
CHAPTER 16. SOLUTIONS TO EXERCISES
505
Solutions: The action is the difference between the kinetic energy and potential energy. We neglect the transverse forces (i.e. perpendicular to the wire) that keep the bead moving in a circle since they do not contribute to the motion along the wire. The Lagrangian is then just equal to the kinetic energy of the bead, namely: L = K 1 2 mv = 2 t 1 = mR2 ω 2 2
(16.27)
where vt is the transverse speed of the bead along the wire and perpendicular to its position vector and ω := θ˙ (16.28) is its angular velocity. The only relevant coordinate in this case is θ. Applying Eq. (3.30) with xp = θ we find: d ∂L d ∂L ∂L − = 0− ∂xp dt ∂ x˙ p dt ∂ θ˙ = −mR2 θ¨ = 0 (16.29) Thus the Euler Lagrange equations tell us, as expected, that the angular acceleration of the bead is zero.
16.4
Symmetries and Linear Transformations
Exercise 1. Draw the three position vectors A, magnitude 2 m, direction due East (E), B magnitude 4 m, direction NorthNorthEast (NNE) and C, magnitude 3 m direction SouthWest (NW). (SW points between South and West, whereas NNE points in the direction midway between North and NorthEast. NorthEast points midway between North an East)
CHAPTER 16. SOLUTIONS TO EXERCISES
506
1. Show by using diagrams, not calculating components, that (A + B) − C = A + (B − C) = A−C+B
(16.30)
2. Calculate: A · B and the angle between A and B. Solutions:
1. In the Fig. 16.4, one can see geometrically that adding the vectors in the two different orders leads to the same point.
Figure 16.4: The red arrow represents the vector A, the blue arrow represents B, and the green arrow is C while the dark gree arrow is −C. The orange vector D is B − C. The diagram shows that A + D = A + B − C = F.
2. Fig. 16.5, one can see geometrically that adding B to A − C also yields the same vector F. 3. Choosing East as the positive x− axis and North as the positive y−axis: A = 2my B = 4m cos(67.5)y + 4m sin(67.5)ˆ y = 1.53my + 3.70ˆ y
(16.31) (16.32)
CHAPTER 16. SOLUTIONS TO EXERCISES
507
Figure 16.5: The red arrow represents the vector A, the blue arrow represents B, and the green arrow is C while the dark gree arrow is −C. The orange vector D is B − C. The diagram shows that A − C + B = (A + B) − C = F.
2 ˙ A · B = 21.53m +0 2 = 3.06m
(16.33)
The cosine of the angle between them is given by: cos(θ) =
A·B AB
3.06m2 p 4m2 (1.532 + 3.702 )m2 = 0.38 = √
(16.34)
The angle is therefore: θ = 67.5◦
(16.35)
Exercise 2. For the vectors A and B in the previous exercise, give the magnitude and direction of the following: 1. A × B,
CHAPTER 16. SOLUTIONS TO EXERCISES
508
2. B × A, 3. (A − B) × B, Solutions:
1. The magnitude of A × B = AB sin(θ). Using the results of Exercise 1: √ A = 2m, B = 1.532 + 7.302 m = 7.5m, θ = sin(67.5) = 0.92 (16.36) we get: A × B = 2m × 7.5m × 0.92 = 13.9m2
(16.37)
Using the right hand rule, the direction is out of the page, or straight up. 2. B × A has the same magnitude, but opposite direction. It points straight down, or into the page. 3. Since B × B = 0 for any vector: (A − B) × B = A × B
(16.38)
which is given above.
Exercise 3. 1. Consider the threevectors: A = 3i − 2j B = −i + j C = 2i + k Calculate the following in terms of components, not diagrams:
(16.39)
CHAPTER 16. SOLUTIONS TO EXERCISES
509
(a) A + B (b) A · B (c) A and B (d) A × B. (e) (A × B) · C (f) D = (A × B) × C and F = A × (B × C). Are D and F equal? Solutions:
(a) A + B = (3 − 1)i + (−2 + 1)j = 2i − j
(16.40)
A · B = 3(−1) + (−2)1 = −5
(16.41)
(b)
(c) A = =
p √
32 + (−2)2
2 p (−1)2 + 12 B = √ = 2
(16.42) (16.43)
(d) A × B = (Ax By − Bx Ay )k = [−3 − (−2)]k = −k
other components 0 (16.44)
(e) (A × B) · C = (−ˆ z ) × (2ˆ y + zˆ) = 0k + 0j + [0 − (−1)2]i
(16.45)
CHAPTER 16. SOLUTIONS TO EXERCISES
510
(f) Using A × B = −k, we have: D = (A × B) × C = −k × (2j + k = 2i + 0
(16.46)
where we have used the relation y × yˆ = zˆ + cyclic permutations, so that zˆ × yˆ = −y. Now calculate: B × C = (−i + j) × (2j + k) = −2ˆ z + yˆ + 0 + y
(16.47)
So that F = = = =
A × (B × C) (3i − 2j) × (i + j − 2k) (0 + 3k + 6j) + (+2k + 0 + 4i) 4i + 6j + 5k
(16.48)
They are not equal, so that surprisingly vector multiplication does not respect the distribution condition.
Exercise 4. by:
A ball of mass m = 10 gm moves in an elliptical path given r(t) = l cos(ωt)i + 2l sin(ωt)j
with l = 2 m and ω =
π 2
(16.49)
radians per second.
1. What is the period, T of the orbit? T =
2π = 4s ω
(16.50)
2. Sketch the orbit r(t) for one complete cycle, i.e 0 ≤ t < T Hint: Plot the points on the orbit at t = 0, T /4, T /2, 3T /4 and then fill in the rest as smoothly as you can.
CHAPTER 16. SOLUTIONS TO EXERCISES
511
3. Calculate the distance of the ball from the origin as a function of t, i.e the trajectory. √ r(t) := r·r q l2 cos2 (ωt) + 4l2 sin2 (ωt) = q = l 1 + 3 sin2 (ωt) (16.51) 4. Find the velocity v =
dr(t) dt
of the ball as a function of t.
dr(t) dt = l (−ω sin(ωt)i + 2ω cos(ωt)j)
v :=
a(t) := = = = =
dv(t) dt d l (−ω sin(ωt)i + 2ω cos(ωt)j) dt l −ω 2 cos(ωt)i − 2ω 2 sin(ωt)j = −ω 2 l (cos(ωt)i − 2 + sin(ωt)j) −ω 2~r(t)
(16.52)
(16.53)
5. Calculate r · p at t = 1 sec and t = 3/2 sec.
p = mv = mlω (− sin(ωt)i + 2 cos(ωt)j) r · p = xpx + ypy = ml2 ω (− cos(ωt) sin(ωt) + 4 cos(ωt) sin(ωt)) = 3ml2 ω cos(ωt) sin(ωt) (16.54) For l = 2, m = 0.01 kg , ω = π/2 rads/sec and t = 1: r · p = 3 × 0.01 × 22 × π/2 cos(π/2) sin(π/2) kg m2 /s = 0 (16.55) and for t = 3/2s: r · p = 0.03 × 22 × π/2 cos(3π/4) sin(3π/4) kg m2 /s = 0 = −0.03πkg m2 /s (16.56)
CHAPTER 16. SOLUTIONS TO EXERCISES
512
6. Calculate the angular momentum L := r × p at arbitrary time t.
r×p = = = =
(xpy − ypx )k (l cos(ωt)[2mlω cos(ωt)] − 2l sin(ωt)[−mlω sin(ωt)]) k 2ml2 ω(cos2 (ωt) + sin2 (ωt))k 2ml2 ωk (16.57)
Exercise 5. Verify that the transformation 0 x x 0 r= →r = y0 y cos(θ)x + sin(θ)y = − sin(θ)x + cos(θ)y
(16.58)
preserves the length of the vector, namely that: r0 2 := r0 · r0 = (x0 )2 + (y 0 )2 = r · r = x2 + y 2
(16.59) (16.60)
r · r = (x0 )2 + (y 0 )2 = (cos(θ)x + sin(θ)y)2 + (− sin(θ)x + cos(θ)y)2 = (cos2 (θ)x2 + 2 cos(θ) sin(θ)xy + sin2 (θ)y 2 ) +(sin2 (θ)y 2 − 2 sin(θ) cos(θ)xy + cos2 (θ)x2 ) = x2 + y 2
(16.61)
Solutions:
where the middle terms in each round bracket cancelled and we have used cos2 (θ) + sin2 (θ) = 1.
CHAPTER 16. SOLUTIONS TO EXERCISES
513
Exercise 6. To reflect through any other axis, say one that is rotated counterclockwise by an angle θ relative to the xaxis, one simply needs to first rotate the vector through θ in the clockwise direction in order to line up the reflection axis with the xaxis, then reflect through the xaxis, then rotate back (counterclockwise) by θ to get the reflection axis back to where it was. The net effect of these three transformations on a vector r = xi + yj is: cos2 (θ) − sin2 (θ) x + 2 cos(θ) sin(θ)y 0 r = 2 cos(θ) sin(θ)x + − cos2 (θ) + sin2 (θ) y cos(2θ)x + sin(2θ) = (16.62) sin(2θ)x − cos(2θ)y Check that the process outlined above produces Eq. (16.62) reflections through the x = y axis, i.e. θ = 45◦ . Solution: 1. The first step is to rotate r through 45◦ in counter clockwise (θ = 45◦ ) direction: cos(−45)x + sin(−45)y r1 := − sin(−45)x + cos(−45)y 1 x−y (16.63) = √ x+y 2 Now reflect r1 through the xaxis: 1 −(x − y) r2 = √ x+y 2
(16.64)
Finally rotate back, i.e. clockwise, through 45◦ : 1 cos(45)(−x + y) + sin(45)(x + y) 0 r = √ − sin(45)(−x + y) + cos(45)(x + y) 2 1 1 2y = √ √ 2x 2 2 y = (16.65) x
CHAPTER 16. SOLUTIONS TO EXERCISES
514
So reflection through the 45◦ axis interchanges the x and y components, as expected.
Exercise 7. Consider two linear transformations defined by: 1 1 3 1 T1 = ; T2 = 3 −2 2 1 2
(16.66)
1. Calculate the matrix T3 that represents the action of first T2 followed by T1 . 2. Calculate the matrix T4 that represents the action of first T1 followed by T2 . Do the two operations commute? Solution: Acting with T2 first requires putting it on the right, so that: 1.
1 −2
3 2
T3 = T1 T2 = ( 13 + 1) = (−2 × 13 + 1 × 2) 4 7 3 = 7 2 3
1 3
1
1 2
(1 + 2 × 3) (−2 × 1 + 2 × 2)
(16.67)
2.
1 3
1 2
T4 = T2 T1 = 1 1 − 2 (1 + 2) 3 = (1 − 4) (3 + 4) 5 3 3 = −3 7 T4 6= T3 , so the operations do not commute.
1 −2
3 2
(16.68)
CHAPTER 16. SOLUTIONS TO EXERCISES
515
Exercise 8. Calculate the inverses of T1 and T2 in the previous exercise. First check if they exist. Be sure to verify in both cases that you have obtained the correct inverse. Solution: A matrix has an inverse if its determinant is nonzero. det(T1 ) = 1 × 2 − (−2) × 3 = 8= 6 0 1 ×2−1×1 det(T2 ) = 3 1 = − 6= 0 3
(16.69)
(16.70) (16.71)
so that both have inverses. T1−1 T2−1
1 = det(T1 ) 1 = det(T2 )
1 2 −3 = 8 2 1 2 −1 2 −1 = −3 −1 13 −1 13 2 −3 2 1
(16.72) (16.73) (16.74)
CHAPTER 16. SOLUTIONS TO EXERCISES
516
Check that they are inverses: T1−1 T1
= = =
T2−1 T2 = = =
1 2 −3 1 3 −2 2 8 2 1 1 2+6 6−6 8 6+2 2−2 1 0 0 1 1 2 −1 1 3 −3 −1 13 1 2 1 −3 0 −3 0 − 13 1 0 0 1
(16.75)
(16.76)
For square matrices, if the left inverse exists then the right inverse also exists and is equal to the left inverse.
Exercise 9. form:
A general two dimensional linear transformation takes the T =
a b c d
(16.77)
1. Show that T preserves the length of an arbitrary vector (x, y) if and only if: a2 + c 2 = 1 b2 + d 2 = 1 ab + cd = 0
(16.78) (16.79) (16.80)
2. Show that Eq. (16.7816.80) imply that: a = ±d AND c = ∓b
(16.81)
Hint: Solve Eq.(16.80) for b and substitute this into Eq. (16.79).
CHAPTER 16. SOLUTIONS TO EXERCISES
517
Solution:
1. Consider the transformation T defined above acting on an arbitrary vector r = (x, y) to yield a new vector 4 = (x0 , y0 ): r0 = T r a b x = c d y ax + by = cx + dy
(16.82)
The norm squared of r0 is: 0
0
r ·r
= [ax + by, cx + dy]
ax + by cx + dy
= (ax + by)2 + (cx + dy)2 = (a2 + c2 )x2 + 2(ab + cd)xy + (b2 + d2 )y 2
(16.83)
The above will equal r·r = x2 +y2 for all possible x and y iff conditions 16.78, 16.79 and 16.80 are satisfied. 2. According to condition 16.80 b = −
cd a (16.84)
Putting this into condition 16.79 yields: c2 d 2 + d2 = 1 a2 d2 → 2 c 2 + a2 = 1 a
(16.85)
Using condition 16.78 yields d2 = 1 → d = ±a a2
(16.86)
Putting this back into condition 16.80 yields b = −(±)c = ∓c
(16.87)
CHAPTER 16. SOLUTIONS TO EXERCISES
518
Exercise 10. 1. Show by explicit calculations for θ = 30◦ , R(θ) as defined in Eq. (4.62) rotates the corners of a square with sides of length 2m centered on the origin by 30◦ clockwise. 2. Verify by matrix multiplication for arbitrary θ that R(−θ) is the matrix inverse of R(θ). That is: R(θ) · R(−θ) = R(−θ) · R(θ) = I
(16.88)
where I is the unit matrix defined in Equation (4.57). Solutions:
1. We place the corners of the square at points (x, y) = {(1, 1), (1, −1), (−1, −1), (−1, 1)} as shown in Fig. 16.6. Eq. (4.62) tells us that a clockwise rotation by
Figure 16.6: Square with sides of length 2m centered on origin.
CHAPTER 16. SOLUTIONS TO EXERCISES
519
30◦ can be represented by the action of the following matrix: cos(30) sin(30) R(30) = − sin(3) cos(30) " √ # =
3 2 − 12
1 √2 3 2
We now act with R(30) on the corners of the square: " √ # ! √ 3 3 1 1 1 + 2 2 2 √2 √ = ∼ 1 − 12 23 − 21 + 23 # ! " √ √ 3 3 1 1 1 − 2 2 2 √2 √ ∼ = −1 − 12 23 − 12 − 23 # ! " √ √ 3 − 3 1 1 − −1 2 2 √2 √2 ∼ = −1 − 12 23 + 12 − 23 " √ # ! √ 3 − 3 1 1 −1 + 2 2 √2 √2 ∼ = 1 − 12 23 + 12 + 23
(16.89)
1.35 0.35
(16.90)
0.35 −1.35
(16.91)
−1.35 −0.35
(16.92)
−0.35 1.35
(16.93)
Plotting the above transformed points on the same figure as before, we find:
Figure 16.7: Rotated square, in blue, with sides of length 2m centered on origin.
CHAPTER 16. SOLUTIONS TO EXERCISES
520
2. "
√
1 0
R(−30)R(30) = =
" R(30)R(−30) = =
3 2 1 2
1 − √2 3 2
0 1
√
3 2 − 21
1 0
√
3 2 − 12
1 √2 3 2
#
(16.94)
1 √2 3 2
0 1
#"
#"
√
3 2 1 2
1 − √2
#
3 2
(16.95)
Exercise 11. 1. Check that the general reflection matrix given in Eq.(4.73) preserves lengths and that it has determinant 1. The matrix form of Eq. (4.46) as given in Eq. (4.73) is: cos(2θ) sin(2θ) R= (16.96) sin(2θ) − cos(2θ) It preserves lengths because the components satisfy conditions 16.7816.80 above. Its determinant is: − cos2 (2θ) − sin2 (2θ) = −1
(16.97)
2. Find the inverse transformation of the reflection Equation (4.46) for arbitrary θ. The inverse is: − cos(2θ) − sin(2θ) −1 R = −1 − sin(2θ) cos(2θ) cos(2θ) sin(2θ) = (16.98) sin(2θ) − cos(2θ) Thus R is its own inverse.
CHAPTER 16. SOLUTIONS TO EXERCISES
521
Exercise 12. Consider the group of permutations {I, A, B, C, G, F } of three identical objects as given in Eq. (16.99). 1. Find a 3x3 matrix that represents each operation. Hint: the matrix contains only ones and zeros. 2. Verify explicitly that matrix multiplication is consistent with the multiplication table 3.3 in the following cases: F · C, C · F and F · F . 1. The eight operations we a I: b → c a B: b → c a F : b → c
need to consider are: a a b ; b A: c c a a c ; b C: b c c a a ; b G: b c
The first is straightforward. It is the 1 0 I = 0
b → a c c → b (16.99) a b → c a
unit matrix: 0 0 1 0 0 1
(16.100)
The second operation interchanges the first and three entries, so that: 0 1 0 A = 1 0 0 (16.101) 0 0 1 Operations B and C are similar, since they interchange first and third, and second and third, respectively. 0 0 1 1 0 0 B = 0 1 0 ; C = 0 0 1 (16.102) 1 0 0 0 1 0
CHAPTER 16. SOLUTIONS TO EXERCISES
522
F and G are different in that the represent cyclic permutations in each of two directions: 0 0 1 0 1 0 F = 1 0 0 ; G = 0 0 1 (16.103) 0 1 0 1 0 0 2. Matrix multiplication reveals: 1 C ·F = 0 0 0 0 = 1 = B 0 1 F ·C = 0 0 1 = 0 = A 0 1 F ·F = 0 0 = 0 1 = B
16.5
0 0 1 0 1 0
0 0 1 1 0 0 1 0 0
0 0 1
1 0 0
0 0 1
0 1 0
0 0 1
(16.104) 0 0 1 1 0 0
1 1 0 0 0 0 0 0 1
(16.105) 0 0 1 1 0 0
1 0 0 1 0 0 0 1 0
1 0 0
Special Relativity I: the basics
(16.106)
CHAPTER 16. SOLUTIONS TO EXERCISES
523
Exercise 1. A Trip to the grocery store: You travel to a store 5 km away from your home in a car moving at 100km/hr = 30 m/s. 1. What is the speed of light in km/hr? 2. How long does your journey appear to take as measured by your friend sitting at home? 3. How much proper time elapses for you during the journey? Solutions: 1. c = 300 million m/s = 300 million m/s
100 km/hr 1 m/s
= 1 billion km/hr
(16.107)
2. In the rest frame of the store, the time elapsed is: ∆t =
5 km = 1/20 hr = 3 minutes 100 km/hr
(16.108)
3. In the rest frame of the car: ∆t0 =
δt γ
(16.109)
where 1 γ = p 1 − (v/c)2 1 = p 1 − (100/109 )2 = 2 min 59.999 sec
(16.110)
Thus p 1 − (100/109 )2 ∆t 1 −7 = 1 − 10 3 minutes 2
∆t0 =
(16.111)
CHAPTER 16. SOLUTIONS TO EXERCISES
524
Exercise 2. Two politicians travel to Mars at 0.001 c in order to negotiate a fee trade agreement with the Martians. On arrival they see only rocks so they had back immediately at the same speed. The distance to Mars is about 55 million km. How long would the return trip take from NASA’s point of view. How much would the politicians age during the return journey? (i.e. What is the proper time elapsed along the journey?) Solution: From NASA’s point of view the distance is d = 2 × 55 × 109 m, at speed v = 0.001 × 3 × 108 m/s, so the time required is: d v 110 × 109 m = 3 × 105 m/s = 370, 000s = 103 hours
∆tN ASA =
1 γ = q 1− = √
(16.112)
v2 c2
1 1 − 0.0012
(16.113)
The politicians’ proper time is therefore 1 ∆tN ASA γ √ 1 − 0.0012 103 hours = ∼ (1 − 5 × 10−5 ) hours
∆tP =
(16.114)
The politicians therefore age 5 × 10−5 hrs ∼ 0.18 seconds less than the NASA staff during the return trip.
CHAPTER 16. SOLUTIONS TO EXERCISES
16.6
525
Special Relativity II: in depth
Exercise 1. Use the Galilean transformations in Eq. (5.3) to substitute for (t0 , x0 ) on the right hand side of Eq. (6.8), thereby proving explicitly that Eq. (5.3) do not satisfy Eq. (6.8) in general. Solutions: We substitute the one dimensional Galilean transformations: dt0 = dt dx0 = dx + vdt
(16.115) (16.116)
into c2 dt2 − dx2 = c2 dt02 − dx02
(16.117)
to get c2 dt2 − dx2 = c2 dt2 − (dx + vdt)2 = c2 dt2 − dx2 − 2dxvdt − v 2 dt2
(16.118)
collecting terms we find: 2
2
c dt − dx
2
= c 1− 2 = c 1− 2 = c 1+ 2
v2 dt2 − dx2 − 2vdxdt c2 dx v2 dt2 − dx2 − 2v dt2 2 c dt 2 v dt2 − dx2 c2
(16.119)
where we have used dx/dt = v. Note that this is almost invariant, except for the v 2 /c2 correction to 1 in front of the dt2 . This shows that for v/c 1, Galilean invariance preserves c2 dt2 − dx2 to a high degree of accuracy.
Exercise 2. Verify by direct substitution of dt0 and dx0 in terms of dx and dt as given by Equations (6.9) and (6.10) that Equation (6.8) is satisfied.
CHAPTER 16. SOLUTIONS TO EXERCISES
526
Eq. (6.8) in general. Solutions: The 1D Lorentz transformations are: v cdt0 = γ(v)cdt − γ(v)dx c v 0 dx = − γ(v)cdt + γ(v)dx c
(16.120) (16.121)
Substituting into Eq. (6.8) gives: v v c2 dt2 − dx2 = (γ(v)cdt − γ(v)dx)2 − (− γ(v)cdt + γ(v)dx)2 c 2 c v v2 2 2 2 = c γ (v) 1 − 2 − γ (v) 1 − 2 dx2 c c 2 2 2 = c dt − dx (16.122) where thecross terms dxdt have cancelled and we have used the definition v2 of γ = 1/ 1 − c2 .
Exercise 3. Verify explicitly that a Lorentz transformation with speed v followed by a Lorentz transformation with speed −v takes one back to the original coordinates. Solutions: Start with Lorentz transformation either in infinitesimal or finite form: v cdt0 = γ(v)cdt − γ(v)dx c v 0 dx = − γ(v)cdt + γ(v)dx c
(16.123) (16.124)
Now do another Lorentz transformation on dt0 , dx0 but with v replaced by
CHAPTER 16. SOLUTIONS TO EXERCISES
527
−v, to get dt00 , dx00 : v cdt00 = γ(v)cdt0 − γ(v)dx0 c −v −v −v = γ(−v)c γ(−v)cdt − γ(−v)dx − γ(v) − − γ(−v)cdt + γ(−v)dx c c c = cdt v 00 dx = − γ(v)cdt0 + γ(v)dx0 c v v v = − γ(v)c γ(v)cdt − γ(v)dx + γ(v) − γ(v)cdt + γ(v) c c c = dx (16.125)
Exercise 4. Taylor expand Equations (6.9) and (6.10) to obtain (6.16). By keeping the next term in the Taylor expansion calculation the (v/c)2 corrections to the Galilean transformations. Solution: We need to expand γ(v): −1/2 v2 γ(v) = 1− 2 c 1 v2 = 1 + 2 + O(v 4 /c4 ) 2c
(16.126)
Putting this into Equations (6.9) and (6.10) yields v 1 v2 1 v2 4 4 4 4 0 ct = 1 + 2 + O(v /c ) ct − 1 + 2 + O(v /c ) x 2c c 2c (16.127) 2 2 1v v 1v x0 = − 1 + 2 + O(v 4 /c4 ) ct + 1 + 2 + O(v 4 /c4 ) x c 2c 2c (16.128)
CHAPTER 16. SOLUTIONS TO EXERCISES
528
Exercise 5. Prove that for speeds much less than the speed of light the relativistic and nonrelativistic expressions for Doppler shift agree. Solution: The relativistic expression for the Doppler shift in frequency is: p 1 + v/c 0 f = p f 1 − v/c
(16.129)
Doing a Taylor expansion of the square roots gives: 1 + 12 v/c 0 f = f 1 − 12 v/c Using the Taylor expansion (1 − )−1 ∼ 1 + gives: 1 1 f = 1 + v/c 1 + v/c f 0 2 2 2 2 ∼ 1 + v/c + O(v /c ) f 0
(16.130)
(16.131)
which is the same as the nonrelativistic expression Eq. (6.43) up to order v 2 /c2 .
Exercise 6. Calculate the energy equivalent in Joules of 1 gram of fissionable material (assumed to be at rest). How many tons of TNT are required to produce this much energy in an explosion? (1 ton of TNT can produce 4 × 109 Joules. Solution:
E = mc2 = 0.001 kg × (3 × 108 m/s)2 = 9 × 1013 Joules
(16.132)
CHAPTER 16. SOLUTIONS TO EXERCISES
529
Number of tons required to produce same energy is: 9 × 1013 Joules = 2.3 × 104 tons of TNT 9 4 × 10 Joules per Ton
(16.133)
Exercise 7. 1. Taylor expand the expression for T (u) in Eq. (6.63) in a power series in u/c to order (v/c)6 . 2. For a 1 kg soccer ball moving at u = 50m/s, what is the percentage error made by calculating the kinetic energy using the last line of Eq. (6.64) and neglecting the extra terms you calculated in the first part of this exercise. Solutions:
1. Eq. (6.63) can be written as: T (u) = mc2 γ(u) − mc2 r !−1/2 2 u 1− 2 − 1 = mc2 c := mc2 f ()
(16.134) (16.135)
where v2 = 2 c f () = mc2 (1 − )−1/2 − 1
(16.136)
We need to expand this in to order 3 , which is O(v 2 /c2 ). This requires
CHAPTER 16. SOLUTIONS TO EXERCISES
530
the following derivatives: f () = 0 at = 0 −1 1 f 0 () = (1 − )−3/2 (−1) = at = 0 2 2 1 −3 3 00 −5/2 f () = (1 − ) (−1) = at = 0 2 2 4 3 −5 15 000 −7/2 f () = (1 − ) (−1) = at = 0 4 2 8
(16.137) (16.138)
The Taylor expansion of T (U ) is therefore: 1 000 1 00 2 2 3 4 0 T (U ) = mc f (0) + f (0) + f (0) + f (0) + O( ) 2 3·2 1 3 2 1 15 3 1 2 + + ... (16.139) = mc 0 + + 2 24 6 8 1 v 2 1 3 v 4 1 15 v 6 2 = mc 0 + 2 + + + ... (16.140) 2c 2 4 c2 6 8 c4 1 2 3m v 4 15m v 6 = mv + + + ... (16.141) 2 8 c2 48 c4 (16.142) 2. The kinetic energy as given by the first (Newtonian term) is: 1 TN = mv 2 = 0.5 × 1 kg × (0.05km/s)2 ∼ 2.5 × 10−2 Joules (16.143) 2 The approximated difference between the above and the first term in Eq. (6.64) is: 1 2 1 2 3m v 4 15m v 6 + + ... ∆T (u) :=: T (U ) − TN = mv − mv + 2 2 8 c2 48 c4 3mv 2 v 2 15m v 6 = + + ... (16.144) 8 c2 48 c4 Since v/c = (.05/3 × 108 ) ∼ 2 × 10−9 and v 2 /c2 ∼ 4 × 10−18 we can keep just the lowest order term, so that: 3 × 1 ‘kg × 0.052 km2 /sec2 × 4 × 10−18 8 ∼ 0.4 × 10−20 Joules (16.145)
∆T (u) ∼
CHAPTER 16. SOLUTIONS TO EXERCISES
531
The fractional difference is: ∆T (u) 0.4 × 10−20 ∼ ∼ 6 × 10−22 −2 TN 2.5 × 10
(16.146)
which is too small to be measured.
Exercise 8. You are in a frame O in which you see a particle of mass m, moving to the right with speed v, so that its momentum fourvector is given by Eq.(6.81). Now boost to a Lorentz frame, O0 , moving with speed v in the same direction as the particle. O0 is the rest frame of the particle. Do the appropriate Lorentz transformation to verify that E 0 = mc2 and p0 = 0 in the boosted frame O0 , as expected. Solution: Start with the four momentum of a particle moving with speed u with respect to a frame O: E c P := p mγc = (16.147) mγ(u)u lf u is positive, the particle is moving to the right, and if u is negative it is moving to the left. Now act on it with a Lorentz transformation which is also moving with speed u to the right (assuming u is positive for the moment) relative to O. γ(u) −γ(u)u/c mγc 0 P = L(u)P = −γ(u)u/c γ(u) mγ(u)u ! 2 mγ 2 (u) 1 − uc2 = mγ 2 (u)(u − u) 0 mc2 E /c = = (16.148) 0 0 This is the fourmomentum in the particles rest frame as expected.
CHAPTER 16. SOLUTIONS TO EXERCISES
532
Exercise 9. A pion, rest mass mπ = 270me , decays from rest into a muon (mµ = 205me ) and an antineutrino, ν, rest mass mν ∼ 0). The mass of an electron is approximately 10−30 gms. 1. Show that after the collision the speed vµ of the muon is given by: m2π − m2µ vµ = 2 c mπ + m2µ
(16.149)
2. What is energy of the muon after the collision, expressed in terms of mπ and mµ ? What is it in Joules? 3. What is the momentum of the neutrino after the collusion? Note that working with MKS units to analyze the collisions of elementary particles is somewhat tedious. In Sec. 6.10 we will introduce units that are much more convenient for this purpose. Solution:
1. We need to conserve energy and momentum in the collision. momentum conservation tells us that the neutrino and muon must be emitted in opposite directions as shown in Fig. 6.7, so the problem is essentially one dimensional. Without loss of generality we take the muon to travel along the positive xaxis (positive momentum), so the neutrino travels along the negative xaxis and has negative momentum. Ei = mπ c2 Pi = 0 Ef = Eµ + Eν = mµ γ(vµ )c2 + Eν Eν Pf = mµ γ(vµ )vµ − c
(16.150) (16.151) (16.152) (16.153)
CHAPTER 16. SOLUTIONS TO EXERCISES
533
Energy conservation Ef = Ei and momentum conservation Pf = Pi imply mµ γ(vµ )c2 + Eν = mπ c2 Eν mµ γ(vµ )vµ − = 0 c
(16.154) (16.155)
This is a system of two equations in two unknowns (vµ and Eν ). Solving for Eµ in Eq. (16.154) and putting it into Eq. (16.155) yields after multiplication by c: mµ γ(vµ )vµ c − mπ c2 − mµ γ(vµ )c2 = 0 → mµ γ(vµ )vµ c + mµ γ(vµ )c2 = mπ c2 v mπ +1 = → γ(vµ ) c mµ s v 1+ c mπ → = v 1− c mµ →
m2π − m2µ v = c m2π + m2µ
(16.156)
where the last line was obtained by squaring both sides of the previous line, collecting terms in v/c and then solving. 2. To determine the energy of the muon we need to first calculate γ(vµ ): v2 γ(vµ ) = (1 − 2 )−1/2 c −1/2 m2π − m2µ = 1− 2 mπ + m2µ −1/2 2m2µ = m2π + m2µ 2 1/2 mπ + m2µ = 2m2µ
(16.157)
CHAPTER 16. SOLUTIONS TO EXERCISES
534
The energy of the muon is: Eµ = mµ γ(vµ )c2 2 1/2 mπ + m2µ = mµ c2 2m2µ r m2π + m2µ 2 = c 2
(16.158)
3. The momentum of the neutrino is: Eν c vµ = mµ gamma(vµ ) c c 2 2 1/2 mπ + mµ m2π − m2µ = mµ 2m2µ m2π + m2µ
pν =
= √
m2π − m2µ 2(m2π + m2µ )
(16.159)
Exercise 10. 1. The speed of the Earth is 30 km/s. What is its rapidity? 2. Show that for small speeds ω ∼
v c
+ O(v 2 /c2 ).
3. Prove the following relationship between the rapidity χ and the Lorentz factor γ: cosh(ω) = γ(v) (16.160) 4. Using the relationship (16.160) to prove that the expressions for Λ(v) and Λ(ω) in Eqs. (6.93) and (6.94) are equivalent. Solution:
CHAPTER 16. SOLUTIONS TO EXERCISES
535
1. The rapidity is defined as: β = tanh−1 (v/c) 3 × 104 m/s −1 = tanh 3 × 108 m/s = tanh−1 10−4 ∼ 10−4
(16.161)
2. Defining = v/c we use d cosh() = sinh() d d sinh() = cosh() d
(16.162)
Taylor expanding about = 0 for small = v/c 1, 1 cosh() = 1 + 2 + O(4 ) 2 1 sinh() = + 3 + O(5 ) 3! sinh() tanh() = ∼ = + O(2 ) cosh() 1 −1 → ω = tanh () ∼
(16.163)
3. By definition:
→
v = tanh(ω) c 1 1 = γ2 = 2 1 − tanh2 (ω) 1 − vc2 (16.164)
but 1 1 = 2 2 sinh 1 − tanh (ω) 1 − cosh2(ω) (ω) cosh2 (ω) cosh2 (ω) − sinh2 (ω) = cosh2 (ω) γ = cosh(ω) =
→
(16.165)
CHAPTER 16. SOLUTIONS TO EXERCISES
536
4. Starting with Eq. (6.93): X
0
= γ(v)
1 − vc
− vc 1
ct x
We know by definition and from previous exercise that: v = tanh(ω) c γ(v) = cosh(ω)
(16.166)
(16.167)
This implies that v γ(v) = tanh(ω) cosh(ω) c sinh(ω) cosh(ω) = cosh(ω) = sinh(ω) Substituting this into Eq. (16.166) yields cosh(ω) − sinh(ω) Λ(v) = − sinh(ω) cosh(ω)
(16.168)
(16.169)
as desired
Exercise 11. vy .
Show that Lx (vx )Ly (vy ) 6= Ly (vy )Lx (vx ) for general vx and
Solution: We can use matrix multiplication use expressions for Lx (vx ) and Ly (vy ) as given in Eq. (6.99) and Eq. (6.100), respectively: γ(vy ) 0 −γ(vy ) vcy 0 γ(vx ) −γ(vx ) vcx 0 0 −γ(vx ) vx γ(vx ) 0 0 0 1 0 0 c Lx (vx )Ly (vy ) = v y γ(vy ) 0 0 0 1 0 −γ(vy ) c 0 0 0 0 1 0 0 0 1 vy γ(vx )γ(vy ) −vx γ(vx ) −γ(vx )vy γ(vy ) c 0 −vx γ(vx )γ(vy ) γ(vx ) vx γ(vx )vy γ(vy ) 0 = (16.170) vy −γ(vy ) c 0 γ(vy ) 0 0 0 0 1
CHAPTER 16. SOLUTIONS TO EXERCISES
537
Multiplying in the opposite order gives: γ(vy ) 0 −γ(vy ) vcy 0 γ(vx ) −γ(vx ) vcx v x 0 1 0 0 γ(vx ) −γ(vx ) c Ly (vy )Lx (vy ) = v y −γ(vy ) 0 γ(v ) 0 0 0 y c 0 0 0 1 0 0 γ(vx )γ(vy ) vx γ(vx )γ(vy ) −vy γ(vy ) 0 −vx γ(vx ) γ(vx ) 0 0 = v v y y −γ(vx )γ(vy ) v γ(v )v γ(v ) γ(v ) 0 x x y y y c c 0 0 0 1
0 0 0 0 1 0 0 1
(16.171) Comparing Eq. (16.170) and Eq. (16.171) we find that Lx Ly 6= Ly Lx .
Exercise 12. Suppose a particle of rest mass m and threemomentum p = (px , 0, 0) is acted on by a lorentz transformation with speed vy in the ydirection.. What is its new fourmomentum vector P0 ? Show that the P · P = P 0 · P 0 = m 2 c2 . Solution:
E c
px P = 0 0 where E =
(16.172)
p p2x + m2 c2 c. Acting on P with a boost in the xdirection gives: E γ(vy ) 0 −γ(vy ) vcy 0 c px 0 1 0 0 0 P = −γ(vy ) vy 0 γ(vy ) 0 0 c 0 0 0 1 0 E γ(vy ) c px = (16.173) −γ(vy ) vy E c c 0
CHAPTER 16. SOLUTIONS TO EXERCISES
538
Therefore 2 E2 2 2 2E − p − γ (v )v y x y 2 c2 c 2 2 2 2 = γ (vy )(1 − v − −px − γ (vy )vy2 p2x 2 1 2 2 E = 1 − v /c − p2x y 1 − vy2 /c2 c2
P0 · P0 = γ 2 (vy )
=
E2 − p2x 2 c
(16.174)
is invariant.
Exercise 13. Eq. (6.78).
Verify Eq. (6.102) for the Lorentz transformation given in
Solution: It is useful to recall first that γ(−u) = (1−(−u)2 /c2 )−1/2 = (1−(u)2 /c2 )−1/2 = γ(u) so that 1 u/c 1 −u/c L(−u)L(u) = γ(−u) γ(u) u/c 1 −u/c 1 1 − u2 /c2 0 = γ 2 (u) 0 1 − u2 /c2 1 0 = (16.175) 0 1 as required.
Exercise 14. Use matrix multiplication to verify that L(v2 )L(v1 ) = L(v3 ) where v3 is given in Eq. (6.103). Solution:
CHAPTER 16. SOLUTIONS TO EXERCISES
539
v1 + v2 1 + v1c2v2
v3 =
(16.176)
so that + vc2 1− 1 + v1c2v2 2 1 + v1c2v2 −
γ(v3 ) =
=
v1 c
1+ = q
1+ 1−
v12 c2
2 v1 + vc2 c v1 v2 2 c2
v1 v2 c2
−
v22 c2
2 !−1/2 !−1/2
+
v12 v22 c2 c2
(16.177)
Now calculate:
L(v1 )L(v2 ) = =
=
= =
1 −v1 /c 1 −v2 /c γ(v1 ) γ(v2 ) −v1 /c 1 −v2 /c 1 v1 v2 v1 +v2 1 1 + c2 − c q q v1 +v2 2 2 − 1 + v1c2v2 v v c 1 − c21 1 − c22 1 + v1c2v2 1 −v3 q −v3 1 v2 v2 v2 v2 1 − c21 − c22 + c21 c22 1 −v3 γ(v3 ) −v3 1 L(v3 ) (16.178)
as required.
Exercise 15. Prove that the Lorentz transformations obey associativity, namely Eq. (6.104). Solution:
CHAPTER 16. SOLUTIONS TO EXERCISES
540
One can either do the matrix multiplication directly, or use the result in the previous exercise so that: L(v2 )L(v3 ) = L(v2+3 ) v2 + v3 where v2+3 = 1 + v2c2v3 and L(v1 )L(v2 ) = L(v1+2 ) v1 + v2 where v1+2 = 1 + v1c2v2
(16.179)
Thus [L(v3 )L(v2 )] L(v1 ) = L(v2+3 )L(v1 ) = L(v(2+3)+1 ) v2 +v3 v1 + 1+ v2 v3 c2 where v(2+3)+1 = v1 v2 +v3 1 + c c 1+ v2 v3 ( c2 ) = L(v3 ) [L(v2 )L(v1 )] = L(v3 )L(v1+2 ) = L(v3+(2+1) ) v2 +v1 v3 + 1+ v2 v1 c2 where v3+(2+1) = v3 v2 +v1 1 + c c 1+ v2 v1 ( c2 )
(16.180)
With a bit of algebra, one can show that v(2+3)+1 = v3+(2+1) v1 + v2 + v3 + v1 vc22 v3 = 1 + v1 v2 +v2c2v3 +v1 v3
(16.181)
Note that this expression is symmetric under the interchange of any of the two speeds.
16.7
General Relativity
No exercises.
CHAPTER 16. SOLUTIONS TO EXERCISES
16.8
541
Introduction to the Quantum
Exercise 1. Verify that λc in Eq. (8.8) has units of length. Solution:
λc =
hc me c2
(16.182)
We know that hc has units Joules ·sec·m/sec = Joules·m and that mc2 has units of energy, or Joules. Thus the ratio hc [λc ] = m e c2 J·m = Joules = m (16.183) has units of length.
Exercise 2. Check the that the units on the left and right hand side of Eq. (16.184) agree. Solution: By definition, the intensity Iclassical is energy per unit time per unit area that emerges from the surface of a black hole, so that it has units Joules/(sec·m3 ). In terms of units Eq. (16.184) is: kB T [Iclassical ] = 2πc 4 λ Joules m/sec · Joules/Kelvin · Joules → = 3 sec · m m4 Joules = (16.184) sec · m3 as required.
CHAPTER 16. SOLUTIONS TO EXERCISES
542
Exercise 3. Perform the integrals to obtain the last line in Eq. (8.42). Solution: We start from: R∞ dvEP (E) hEi = R0 ∞ dvP (E) 0 mv 2 R∞ 1 2 − 2kB T dv 2 mv e 0 = mv 2 R∞ − 2k T B dv e 0 =
1 kB T 2
(16.185)
In order to use the following integrals Z ∞ 2 dye−by = Z ∞0 2 dyy 2 e−by =
√ π √ 2 b Z ∞ d −by 2 dye db 0 √ r ! d π b = db 2 2 √ π −3/2 b = 4
0
we take b =
2kb T , m
(16.186)
so that: r Z ∞ mv 2 √ m − 2k BT dv e = π2 2kB T 0 (16.187)
and Z
∞
dv
0
1 2 mv e 2
mv 2 − 2k T B
−3/2 √ m π m = 2 4 2kB T
(16.188)
CHAPTER 16. SOLUTIONS TO EXERCISES
543
So that hEi = =
−3/2 √ m m π 2 4 2kB T √ q π m 2 2kB T 1 kB T 2
(16.189)
as required.
Exercise 4. 1. Do a Taylor expansion of the bottom line in Eq. (8.51) assuming khf BT 1 and show that to first order it reproduces the equipartition theorem result for the contribution of a single simple harmonic oscillator to a large system in thermal equilibrium at temperature T . Solution: Start with hf
hEi = e
hf kB T
.
(16.190)
−1
Use the Taylor expansion of an exponentional e = 1 + + 12 2 + O(3 ) where in this case = hf /(kB T )to get: hf 1 + + O(2 ) − 1 hf hf = = hf /(kB T ) = kB T
hEi =
(16.191)
which corresponds with the average energy in Eq. (8.45). 2. Using the Taylor expansion: 1 = 1 + y 2 + y 3 + ... 1−y ∞ X = 1+ ym m=1
(16.192)
CHAPTER 16. SOLUTIONS TO EXERCISES
544
and defining − khfT
y := e
(16.193)
B
we can write ∞ X
e
−m khfT B
1
=
1−e
m=1
−1
− khfT B
− khfT
=
e
B
1−e
(16.194)
− khfT B
Next we use the trick: ∞ X
me
m=1
−mx
∞ d X −mx e =− dx m=1
(16.195)
to write ∞ X m=1
− khfT
hf e
B
∞ X hf −m khfT B = kB T e k T B m=1 ∞ X
d
hf
−m = kB T e kB T hf d kB T m=1
d 1 = kB T −1 − khfT B d khf 1−e BT = kB T
hf kB T
1−e
− khfT
2
B
Combining this with Eq. (16.194) gives Eq. (8.45).
Exercise 5. 1. Verify that Eq. (8.54) follows from Eq. (8.53). Solution:
(16.196)
CHAPTER 16. SOLUTIONS TO EXERCISES
hc = 1241eV · nm = 1241 · 1.6 × 10−19
545
J eV · nm = 2 · 10−16 J · nm eV (16.197)
2 · 10−16 J · nm 1.4 × 10−23 J · K −1 µm nm = 1.4 × 104 = 1.4 × 107 T (K) T (K) m 2πc = 18 × 108 s 2 × 108 ms × 2 · 10−16 J · nm I(λ) = 1.4×104 µm 5 λ exp( λT (K) ) − 1
hc kB T
=
=
4 × 10−8 W atts · m · nm 4 µm λ5 exp( 1.4×10 ) − 1 λT (K)
=
40 Watts 1 14000 5 (λ(µm)) exp( λ(µm)T (K) − 1) m2 µm
(16.198) (16.199)
(16.200)
2. What is I(λ) for λ = 0.6µm at T = 6000K (the temperature of the Sun)? What is I(λ) for λ = 0.6µm at T = 310K (the temperature of the human body)? Solution: For λ = 0.6µm at T = 6000K: I(0.6µm) = =
40 1 Watts 14000 (λ(µm))5 exp( λ(µm)T (K) − 1) m2 µm 40 1 Watts 14000 5 (0.6) exp( (0.6×6000) − 1 m2 µm
1 Watts exp(3.89) − 1 m2 µm Watts = 10.74 2 m µm = 514
(16.201)
CHAPTER 16. SOLUTIONS TO EXERCISES
546
For λ = 0.6µm at T = 310K: I(0.6µm) =
40 1 Watts 14000 (λ(µm))5 exp( λ(µm)T (K) − 1) m2 µm
40 Watts 1 14000 5 (0.6) exp( 0.6×310 ) − 1) m2 µm 1 Watts 40 = 5 (0.6) exp(75) − 1) m2 µm Watts = 514 × 5 × 10−32 2 m µm Watts = 5 × 10−30 2 m µm
=
(16.202)
Exercise 6. Verify that the equation: dI(λ) =0 dλ λmax
(16.203)
yields Eq. (8.49) for λmax . Hint: you may need the fact that the solution to 5(ex − 1) = xex is x ≈ 5. Solution:
I(λ) =
2π hc2 . hc λ5 e kB T λ − 1
dI(λ) 2π hc2 hc 2π k hcT λ = −5 6 hc + e B dλ λ e kB T λ − 1 kB T λ2 λ5 Defining x =
hc , kB T λ
(16.204) hc2 e
hc kB T λ
2 −1
(16.205)
the above derivative vanishes for xmax that satisfies 2π hc2 xmax exmax −5 + xmax −1 =0 (16.206) λ6 exmax − 1 e
CHAPTER 16. SOLUTIONS TO EXERCISES
547
which has xmax ≈ 5 as a solution. This yields hc ≈ 5 kB T λmax →
λmax ≈
hc 5kB T
(16.207)
Solving for λmax and putting in numerical values for the constants gives Eq. (8.49) as required.
Exercise 7. • Verify by explicit calculation that Eq. (8.53) follows from Eq. (8.52). • Verify that for hf /(kB T ) 1 Eq. (8.53) corresponds precisely to Eq. (8.48). Solutions: Starting from: dEEM = 3
π dλ (2L)3 4 λ −12
hf hf kB T
(16.208)
e and using the definition of the intensity:
dEEM (λ) (16.209) ∆t∆A where λ = hc/f , ∆t = L/c is the light transit time of the cavity and ∆A = 6L2 is the surface area of a cubical cavity with sides of length L we get: I(λ)dλ =
hc π 3 dλ I(λ)dλ = 3 hc (2L) 3 λ4 λ e λkB T − 1 2(c/L)6L =
Exercise 8.
2π hc2 dλ . hc λ5 e λkB T − 1
(16.210)
CHAPTER 16. SOLUTIONS TO EXERCISES
548
1. What is the thermal power output (heat due to blackbody radiation) of the Sun, assuming that it is a perfect blackbody at temperature is 6000K and that the radius is 700,000 km? 2. What is the power output of a healthy human, assuming that they are a perfect cylinder of height 3 m and diameter and circumference 0.3 m, and that they are a perfect blackbody at temperature 310K? Is this an over estimate or under estimate? Explain. Solutions:
1. Using Eq. (8.56) and Eq. (8.57) P = σT 4 A Watts (6000K)4 m2 K4 = 74, 000 Watts/m2
= 5.67 × 10−8
(16.211)
The area of the Sun is: 4 2 R π = 13 × (7 × 108 m)2 = 6.4 × 1018 m2
A =
(16.212)
The total thermal power output of the Sun is P = 74, 000 Watts/m2 × 6.4 × 1018 m2 = 5 × 1023 Watts
(16.213)
Pretty impressive! 2. The surface area of a cylinder 3m meters high with radius 0.1 m is: A = = = =
2πR2 + h × 2πR 2 × π(0.1 m)2 + 3 m × 2π × 0.1 m 0.063 m2 + 1.9 m2 2 m2
(16.214)
CHAPTER 16. SOLUTIONS TO EXERCISES
Exercise 9. Verify that for large temperatures the equation: dI(λ) =0 dλ λmax
549
(16.215)
yields Eq. (8.49) for λmax . Solution: From Eq. (16.210) above I(λ) =
hc2 2π λ5 e λkhcB T − 1
(16.216)
so that
dI(λ) 1 hc −5 − = 2πhc2 6 hc hc 2 hc dλ λ e λkB T − 1 (−λ7 kB T )e λkB T e λkB T − 1 hc 2πhc2 hc λkhc T λkB T = −1 + e B (16.217) hc 2 −5λ e kB T 7 λkB T λ e −1 Thus dI/dλ = 0 when: hc hc λkhc T λkB T −1 + e B = 0 −5λ e kB T
(16.218)
For large T , i.e. λkhcB T >> 1, the exponential dominates over 1 in the bracket on the left hand side to leave: λ =
1 hc 5 kB T
(16.219)
Putting in numerical values for h, c and kB yields λ ∼
5 × 10−3 m T (Kelvin)
(16.220)
CHAPTER 16. SOLUTIONS TO EXERCISES
550
Exercise 10. Heisenberg was stopped by a police officer one day while driving his sports car. The policeman asked Heisenberg: “Do you know how fast you were going?”, to which Heisenberg replied: “No, but I know where I am”. “Well let me tell you, says the officer. “You were going 146.5 km/hr”. “Great”, said Heisenberg, “now I am lost”. 1. Assuming that Heisenberg was able to use GPS to locate his position to an accuracy of 10 meters, what was the minimum uncertainty in his position? (Heisenberg’s mass was about 70 kg.) 2. Assuming that the uncertainty in the police officer’s measurement of Heisenberg’s position was in the last digit, i.e. ±.1 km/hr, what was the uncertainty in Heisenberg’s position after the measurement? Solution:
1. Using the uncertainty principle and assuming there is no uncertainty in Heisenberg’s mass:
→
∆p = m∆v h ≥ 4π∆x 6.6 × 10−34 Joule · seconds = 4π10m −34 6.6 × 10 Joule · seconds ∆v ≥ 4π10m · 70 kg −37 = 10 m/s
(16.221)
2. Again, the uncertainty principle implies: h 4π∆p 6.6 × 10−34 Joule · seconds = 4π70kg0.1 m/s = 7.5 × 10−36 m
∆x ≥
(16.222)
CHAPTER 16. SOLUTIONS TO EXERCISES
16.9
551
The Wave Function
Exercise 1. then
Prove that if ψ(x) is defined in terms of φ(k) by Eq. (9.24),
1.
Z
∞
hψψi =
dkφ∗ (k)φ(k)
(16.223)
−∞
2. ∞
Z ∞ ∂ψ(x) dxψ (x)(−i~) = dkφ∗ (k)~kφ(k) ∂x −∞ −∞ = hψ~kψi (16.224)
Z
∗
You will need to use Eq. (15.36) in Sec. 15.3.3. Solution:
1. We have Z
∞
hψψi = Z−∞ ∞
= = = = =
dxψ ∗ (x)ψ(x)
Z ∞ 1 dxψ (x) √ dkφ(k)eikx 2π Z−∞ Z ∞ −∞ Z ∞ ∞ 1 1 ˜ ∗ −i kx ˜ (k)e ˜ √ dkφ dkφ(k)eikx dx √ 2π −∞ 2π −∞ Z−∞ Z ∞ Z ∞ ∞ 1 ˜ ˜ ∗ (k) ˜ dkφ dkφ() dxei(−k)x 2π −∞ Z−∞ Z−∞ ∞ ∞ ˜ ∗ (k) ˜ ˜ dkφ dkφ(k)δ(k − k) −∞ −∞ Z ∞ dkφ∗ (k)φ(k) (16.225) −∞
∗
CHAPTER 16. SOLUTIONS TO EXERCISES
552
2. Z
∞
∂ dxψ (x)(−i~) ψ(x) = ∂x −∞ ∗
= = = = = =
∞
Z ∞ ∂ 1 dxψ (x)(−i~) √ dkφ(p)eikx ∂x 2π Z−∞ Z ∞ −∞ ∞ 1 ∂ dxψ ∗ (x)(−i~) √ dkφ(k) eikx ∂x 2π Z−∞ Z ∞ −∞ ∞ 1 dkφ(k)(−i~)(ik)eipx dxψ ∗ (x) √ 2π −∞ Z ∞ −∞ Z ∞ Z ∞ 1 1 ˜ ∗ ˜ −ikx ˜ √ dx √ dkφ (k)e dk(~k)φ(k)eikx ~ 2π 2π −∞ −∞ Z ∞−∞ Z ∞ Z ∞ 1 ˜ ˜ ∗ (k) ˜ dxei(k−k)x dkφ dkkφ(k) 2π −∞ Z−∞ Z−∞ ∞ ∞ ˜ ∗ (k) ˜ ˜ dkφ dkkφ(k)δ(k − k) −∞ Z−∞ ∞ dkφ∗ (k)~kφ(k) Z
∗
−∞
= hφφi
(16.226)
√ Exercise 2. Verify Eq. (9.51) explicitly for a Gaussian of width b 2 and mean position x0 = 0. You can use the generalization of the result in Exercise 1, namely Z ∞ n n hψ(x)ˆ p ψ(x)i = ~ φ∗ (k)k n φ(k) (16.227) −∞
and the Fourier transform of the Gaussian given in Eq. (15.54). Solution: ∆(p2 ) =
p
hp4 i − (hp2 i)2
(16.228)
From Eq. (15.54), s φ(k) =
b 2 2 √ e−k b /2 π
(16.229)
CHAPTER 16. SOLUTIONS TO EXERCISES
553
so that hp2 i = ~2
Z
s
∞
dk −∞
= = = = =
b 2 2 √ e−k b /2 k 2 π
s
b 2 2 √ e−k b /2 π
Z ∞ b 2 2 dkk 2 e−k b ~ √ π −∞ Z ∞ ∂ 2 2 2 b −~ √ dke−k b 2 π ∂b −∞ pπ ∂ b b2 −~2 √ 2 π ∂b 1 1 2 ~b 2 (b2 )3/2 1 ~2 2 b2 2
(16.230)
and hp4 i = ~4
Z
s
∞
dk −∞
b = ~ √ π
= = =
∞
s
b 2 2 √ e−k b /2 π
2 2
dkk 4 e−k b −∞ Z ∞ ∂ 4 2 2 4 b dke−k b (−) ~ √ 2 π ∂b −∞ q ∂ 2 (b2π)3/2 b −~4 √ π ∂(b2 )2 1 1 2 ~b 2 (b2 )2 1 ~4 4 b4 4
=
Z
b 2 2 √ e−k b /2 k 4 π
(16.231)
Therefore: p hp4 i − (hp2 i)2 r 1 1 = ~2 − 4b4 4b4 = 0
∆(p2 ) = ~2
(16.232)
CHAPTER 16. SOLUTIONS TO EXERCISES
554
This is consistent with Eq. (9.51) because we know that for a Gaussian that has x → −x symmetry, hpi = 0. This was also verified in Eq. (9.36)
Exercise 3. In March, 2013, it was announced that the data obtained previously by two independent measurements using the Large Hadron Collider at CERN, Switzerland, had detected the Higgs boson. This illusive particle was the last missing piece to the standard model that unified the electromagnetic and weak interactions, and was the first fundamental scalar particle to be detected in nature. The Higgs is an unstable particle and its mass was measured to be 125 ± 0.2 GeV. Based on the uncertainty in energy, use the energytime uncertainty principle to deduce the minimum time it would take for the Higgs to undergo a major change in state, i.e. decay to other elementary particles. Solution: Starting with the uncertainty in energy: ∆E = 0.2GeV
(16.233)
We use the energytime uncertainty principle to get: ~ 2∆E 1.05 × 10−34 J · sec = 0.2GeV
∆t ≥
(16.234)
Using the conversion: 1GeV = 109 eV = 1.6 × 10−10 J
(16.235)
1.05 × 10−34 J · sec 0.2 × 1.6 × 10−10 J = 3.2 × 10−24 sec
(16.236)
we get: ∆t ≥
The observed lifetime of the Higgs is about 10−22 so it obeys the bound from the energy time uncertainty relation.
CHAPTER 16. SOLUTIONS TO EXERCISES
16.10
555
The Schr¨ odinger Equation
Exercise 1. On a single graph sketch the wave in Eq. (10.1) over one wavelengths at t = 0, ωt = π2 and ωt = π. You can take A = 1. Solution:
Figure 16.8: Graphs of Eq. (10.1), A=1, ωt = 0 in blue, ωt = π/2 in red ωt = π in green.
Exercise 2. Show that the normalization constant An for the wave function Eq. (10.22) is: 2 An 2 = (16.237) L Solution:
CHAPTER 16. SOLUTIONS TO EXERCISES
2
Z
hψEn ψEn i = An 
L
dx sin 0
2
556
2πnx L
(16.238)
, dy = 2πndx and integration endpoint change variables in integral to y = 2πnx L L changes from x = L to y = 2πn so that Z 2πn 2 L hψEn ψEn i = An  dy sin2 (y) 2πn 0 L (nπ) = An 2 2πn L = An 2 2 = 1 for normalization (16.239) This yields Eq. (16.237).
Exercise 3. Verify that Eq. (10.31) satisfies the Schr¨odinger equation, (10.30) if and only if the constant C is related to the parameters of the harmonic oscillator by: mΩ C= (16.240) 2~ and that the corresponding energy E of the state is 1 E0 = ~Ω 2 Solution:
(16.241)
CHAPTER 16. SOLUTIONS TO EXERCISES
557
2
Insert ψ0 (x) = Ae−Cx into Eq. (10.30) to get: 2 −Cx2 ~2 d (Ae 1 2 2 − + Kx2 Ae−Cx = EAe−Cx 2 2m dx 2 −Cx2 ~2 d (−2Cxe + 1 Kx2 Ae−Cx2 = EAe−Cx2 − A 2m dx 2 −
1 ~2 2 2 2 2 A (−2Ce−Cx + 4x2 C 2 e−Cx + Kx2 Ae−Cx = EAe−Cx 2m 2 (16.242)
In order for the left hand side to equal the right hand side we need to match 2 2 the right hand side the coefficients of e−Cx and x2 e−Cx must be the same. The two conditions are: ~2 = E 2m ~2 K −4C 2 + = =0 2m 2 2C
(16.243)
This implies that r
C = = = E = =
mK 2 √ 4~ mk 2~ Ω m 2~ ~2 Ω 2 m 2m 2~ 1 ~Ω 2
(16.244)
(16.245)
Exercise 4. Consider a simple harmonic oscillator of mass 1 kg, with initial amplitude 1 meter and spring constant 2 Newtons per meter. If its energy is
CHAPTER 16. SOLUTIONS TO EXERCISES
558
quantized according to (10.36) above, what is the value of the integer n? Solution: The energy of an oscillator with amplitude 1 meter and spring constant 2 N/m is: 1 k A2 2 2 × 12 = kg m2 / s2 2 = 1Joule
E =
(16.246)
The angular frequency of the SHO is: r
k s m 2N = m · 1 kg √ = 2 sec−1
ω =
(16.247)
The energy spectrum of the quantum SHO is: 1 En = (n + )~ ω 2
(16.248)
In the present case En = 1 Joule, so n+
1 1J = √ 2 ~ 2 s−1 =
1J J · s · s−1
10−34
6.6 × ∼ 1.5 × 1033
(16.249)
It is safe to neglect the 1/2, and the 1.5, given how large n is. It must be an integer whose value is of the order of 1033 . Very large!
Exercise 5. An electron is in an electron trap of length L = 0.6 microns.
CHAPTER 16. SOLUTIONS TO EXERCISES
559
1. What is the energy of the electron in its ground state? 2. What are the energy and frequency of the photon it would have to absorb in order to go from the ground state to the n = 4 state? 3. What frequency photon would it subsequently emit if it dropped into the n = 2 state? Solution:
1. Consider the electron trap to be an infinite square well of length L = 0.6 microns. The energy levels are: En =
~2 π 2 2 n 2mL2
(16.250)
For an electron, mc2 = 511 keV. We can also use the formula 2π~ = hc = 1241 eV · nm to simplify the expression for the ground state n = 1: ~2 c2 π 2 n2 2mc2 L2 (1241 eV)2 · nm = 5.11 × 105 eV · 600nm = 5 × 10−3 eV
E1 =
(16.251)
2. The excited states are related to the ground state by: En = n2 E1
(16.252)
So the energy of the n = 4 state is: E4 = 42 × E1 = 8 × 10−2 eV
(16.253)
The frequency of the n = 4 state is: f4 =
E4 h
.08 eV 4.1 × 10−15 eV · s = 2 × 1013 Hz
=
(16.254)
CHAPTER 16. SOLUTIONS TO EXERCISES
560
3. The energy emitted or absorbed is the difference in energy between the energies of the final and initial state. If the final energy is lower, as in this case energy is emitted. ∆E := Ef − Ei = E0 (n2f − n2i ) = (22 − 42 )E0 = −6 × 10−2 eV
(16.255)
The negative sign means the energy is emitted. The emitted photon therefore has energy Eγ = ∆E so the frequency is: fγ =
E4 h .06 eV 4.1 × 10−15 eV · s = 1.5 × 1013 Hz =
(16.256)
Exercise 6. Verify explicitly that Eq. (10.47) solves Eq. (10.45). Solution: Substituting ψnout (x) = Ceκx + De−κx
(16.257)
into the left hand side of Eq. (10.45) (dropping the subscript n) yields d2 (Ceκx + De−κx ) d (κCeκx + (−κ)De−κx ) = dx2 dx d (κ2 Ceκx + (−κ)2 De−κx ) = dx 2 κx = κ Ce + De−κx as required
(16.258)
CHAPTER 16. SOLUTIONS TO EXERCISES
561
Exercise 7. In one form of nuclear fission, Uranium nuclei decay into Thorium nuclei by emitting an alpha particle, as follows: 226 88 U
4 → 222 86 T h + 2 He
(16.259)
The kinetic energy of alpha particle in this decays between about 4 and 5 MeV. Does the alpha particle emerge with relativistic velocity? Justify your answer. Solution: In order to determine whether the velocity of the alpha particle is relativistic you must can either calculate its speed or compare its kinetic energy to its rest mass. This case the latter is easier. The rest mass of the alpha particle, which consists of two protons and two neutrons is approximately 4 × 1Gev, where we are ignoring the difference in rest mass between the two. So for kinetic energy say 5M eV , we have: 1 mα γ(v)v 2 K = mα c2 2 m α c2 v2 = γ(v) 2 c 5 1 = 4000
(16.260)
This implies directly that v/c 1
Exercise 8. Verify that the correspondence principle holds for the Simple Harmonic Oscillator in Sec. 10.4.3 Solution: We wish to show that the energy spectrum become approximately continuous
CHAPTER 16. SOLUTIONS TO EXERCISES
562
for the simple harmonic oscillator. To this end we check: ∆En = En+1 − En = ~Ω∆n ∆En ~Ω∆n = En ~Ω(n + 1/2) ∆n = n + 1/2 ∆n → 0 as →0 n
(16.261)
Exercise 9. Use the Bohr Correspondence Principle to derive the semiclassical energy spectrum for the simple harmonic oscillator. Solution: For the SHO, f = 2π
p K/m is constant, so that dEn = f hdn
(16.262)
is trivially integrated to yield: En = f hn + constant
(16.263)
which is the correct spectrum. In this case the constant is not zero (it is 1/2), but this is irrelevant for large n and cannot be derived using the Bohr Correspondence Principle.
Exercise 10. Show that if one tries to solve the TDSE with a time dependent wave function of the form: Ψ(x, t) = ψ(x)f (t) then the TDSE (10.78) implies:
(16.264)
CHAPTER 16. SOLUTIONS TO EXERCISES
563
1. The time dependent part of the wave function is of the form: Et
f (t) = e−i ~
(16.265)
and 2. the spatial part of the wave function ψ(x) obeys the time independent Schr¨odinger equation: ˆ Hψ(x) = Eψ(x)
(16.266)
where E is an arbitrary constant, independent of x and t. This exercise essentially uses separation of variables to derive from the equations for stationary states from the TDSE assuming only that the stationary state wave function is in the form of a standing wave. Solution: We insert Ψ(x, t) from Eq. (16.264) into Eq. (10.78) to obtain: i~
∂ψ(x)f (t) ~2 ∂ 2 ψ(x)f (t) = − + V (x)ψ(x)f (t) ∂t 2m ∂x2 (16.267)
We can rewrite Eq. (16.267) as ~2 ∂ 2 ψ(x) ∂f (t) = f (t) − + V (x)ψ(x) i~ψ(x) ∂t 2m ∂x2 Dividing both sides by f (t)ψ(x) we get: 1 ∂f (t) 1 ~2 ∂ 2 ψ(x) i~ = − + V (x)ψ(x) f (t) ∂t ψ(x) 2m ∂x2
(16.268)
(16.269)
The left hand side is a function of t only, while the right hand side is a function of x only, so the only way they can be equal is if both sides are equal to the same constant, A that is independent of both x and t. We get two equations: 1 ∂f (t) = A f (t) ∂t 1 ~2 ∂ 2 ψ(x) − + V (x)ψ(x) = A ψ(x) 2m ∂x2 i~
(16.270)
CHAPTER 16. SOLUTIONS TO EXERCISES
564
or i~ −
∂f (t) = Af (t) ∂t
~2 ∂ 2 ψ(x) + V (x)ψ(x) 2m ∂x2
= Aψ(x)
(16.271)
The first of these has the solution: f (t) = De−
iAt ~
(16.272)
while the second is the time independent Schr¨odinger equation with E = A.
Exercise 11. Consider the coherent state given in Eq. (10.83) 1. Verify that it satisfies the TDSE with V (x) = 21 Kx2 for appropriate choice of the function θ(t). 2. Show that the expectation value of position and momentum are given by xα (t) and pα (t) respectively. Solution:
1. We first note that xα (t) and pα (t) are functions of t that satisfy the SHO equations of motion for specific initial conditions. They are independent of the spatial coordinate x. Recall also that K = mω 2 . Start with the TDSE: i~
~2 ∂ 2 Ψ(x, t) ∂Ψ(x, t) = − + V (x)Ψ(x, t) ∂t 2m ∂x2
(16.273)
Substitute the coherent state Ψ(x, t) into Eq. (16.273), where Ψα (x, t) = where
mω 1/4
i
mω
2 +iθ(t)
e ~ pα (t)x− 2~ (x−xα (t))
π~ xα (t) := x0 cos(ωt) dxα (t) pα (t) := m = −x0 mω sin(ωt) dt
(16.274)
CHAPTER 16. SOLUTIONS TO EXERCISES
565
where iθ(t) is a time dependent phase to be determined. and use V (x) = 12 Kx2 to get: h i i h i i pα (t)x− mω (x−xα (t))2 +iθ(t) pα (t)x− mω (x−xα (t))2 2 ~ 2~ ~ 2~ ∂ e ~2 ∂ e i~ + ∂t 2m ∂x2 h i i mω 1 2 Kx2 e ~ pα (t)x− 2~ (x−xα (t)) = 0 − 2 (16.275) 1/4 . where we have divided by the overall normalization factor mω π~ Solving verifying Eq. (16.275) is a bit tedious so we consider each of the three terms separately. The first term is: h i i mω 2 ∂ e ~ pα (t)x− 2~ (x−xα (t)) +iθ(t) i~ ∂t i mω i 2mω 2 ˙ = i~ p˙α (t)x + (x − xα (t))x˙ α (t) + iθ(t) e ~ pα (t)x− 2~ (x−xα (t)) ~ 2~ h i i mω 2 ˙ = −p˙α (t)x + imω(x − xα (t))x˙ α (t) − ~θ(t) e ~ pα (t)x− 2~ (x−xα (t)) h i i mω 2 ˙ = +mω 2 xα (t)x + iω(x − xα (t))pα (t) − ~θ(t) e ~ pα (t)x− 2~ (x−xα (t)) (16.276) Where we have used p˙α (t) = −mω 2 xα (t) and mx˙ α (t) = pα (t) to get the first term in the last line of the above. The second term is: h i i p (t)x− mω (x−xα (t))2 2 ~ α 2~ e 2 ∂ ~ + 2m ∂x2 2 i mω ~ ∂ i 2mω 2 p (t)x− (x−x (t)) α 2~ pα (t) − (x − xα (t)) e ~ α = + 2m ∂x ~ 2~ " 2 # i mω ~2 mω i mω 2 = + − + pα (t) − (x − xα (t)) e ~ pα (t)x− 2~ (x−xα (t)) 2m ~ ~ ~ i mω ~ω p2α mω 2 2 2 = − − iωpα (t)(x − xα (t)) + (x − xα (t)) e ~ pα (t)x− 2~ (x−xα (t)) 2 2m 2 (16.277)
CHAPTER 16. SOLUTIONS TO EXERCISES
566
Adding the first and second terms together, dropping the overall exponential factor yields: h i ˙ +mω 2 xα (t)x + iω(x − xα (t))pα (t) − ~θ(t) ~ω p2α mω 2 2 + − − iωpα (t)(x − xα (t)) + (x − xα (t)) 2 2m 2 2 2 mω 2 2 ˙ + ~ω − pα (t) + mω x2 (t) = x − ~θ(t) (16.278) 2 2 2m 2 α If we choose θ(t) to satisfy: ~ω p2α (t) mω 2 2 − + x (t) = 0 2Z 2m 2 α 1 ~ω p2α (t) mω 2 2 θ(t) = dt − + x (t) + constant ~ 2 2m 2 α (16.279)
˙ −~θ(t) + →
and use mω 2 = K, then Eq. (16.275) is satisfied and Eq. (16.274) obeys the time dependent Sch¨rodinger equation. We now consider the expectation value of momentum: Z ∞ mω 1/4 −i pα (t)x− mω (x−xα (t))2 −iθ(t) 2~ dx hψxψi = e~ × π~ ∞ mω 1/4 i pα (t)x− mω (x−xα (t))2 +iθ(t) 2~ e~ x π~ mω 1/2 Z ∞ mω 2 = dx xe− ~ (x−xα (t)) (16.280) π~ ∞ The last expression is the expectation value of x for a Gaussian, pas given in Eq. (15.28) of Sec. Eq. (15.3), with x0 = xα (t) and b = ~/(mω) so: hψxψi = xα (t) as required.
(16.281)
CHAPTER 16. SOLUTIONS TO EXERCISES
567
2. Similarly, we have: ∗ Z ∞ mω 1/4 −i pα (t)x− mω (x−xα (t))2 −iθ(t) 2~ dx hψˆ pψi = e~ × π~ ∞ mω 1/4 i pα (t)x− mω (x−xα (t))2 +iθ(t) ∂ ~ 2~ e −i~ ∂x π~ ∗ Z ∞ mω 1/4 −i pα (t)x− mω (x−xα (t))2 −iθ(t) 2~ dx = e~ × π~ ∞ mω mω 1/4 i pα (t)x− mω (x−xα (t))2 +iθ(t) i 2~ (x − xα (t)) e~ −i~ pα (t) − 2 ~ 2~ π~ mω = hψpα (t) − 2 (x − xα (t))ψi 2~ mω = pα (t)hψψi − 2 hψ(x − xα (t))ψi (16.282) 2~ The second term in the last line vanishes because as we have proved above hψxψi = xα (t)i, and the first term is just pα (t) as required because the state is normalized.
Exercise 12. Show that for all ψ(x): [ˆ x, pˆ] ψ(x) = i~ψ(x)
(16.283)
Solution: −i~∂ψ(x) −i~∂xψ(x) − ∂x ∂x ∂ψ(x) −i~∂x partialψ(x) = −i~x − ψ(x) − i~ x ∂x ∂x ∂x
[ˆ x, pˆ] ψ(x) = x
= i~ψ(x)
(16.284)
CHAPTER 16. SOLUTIONS TO EXERCISES
568
Exercise 13. Suppose x(t) = A cos(ωt + φ) solves the simple harmonic oscillator equation of motion: d2 x(t) = −ωx2 (t) dt2
(16.285)
Show that if you do a shift x(t) → x˜(t) = x(t) + x0 then x˜(t) no longer solves Eq. (16.285) Solution: We shift the solution to the SHO as follows x(t) → x(t) + x0 = A cos(ωt + φ) + x0
(16.286)
and put this into the SHO equation: d2 x(t) d2 x(t) + x0 2 + ω(x(t) + x ) (t) = + ωx2 (t) + 2x(t)x0 + x20 0 dt2 dt2 = 0 + 2x(t)x0 + x20 6= 0 unless x0 = 0 (16.287)
Exercise 14. (Supplementary) 1. Repeat Example 4 in Sec. 10.8 to show that P = ψ ∗ ψ does not result in conservation of probability for the KleinGordon equation Eq. (10.120). Hint: It is easier to show that the second derivative of the integrated probability density is nonzero. You may assume boundary conditions for which boundary terms (i.e. those evaluated at spatial infinity) vanish. 2. Use the same techniques to prove that PR in Eq. (16.292) does yield time independent “probabilities” for equation Eq. (10.120). Solution:
CHAPTER 16. SOLUTIONS TO EXERCISES 1. We first evaluate: Z ∞ ∗ Z ∞ d ∂ψ ∗ ∗ ∂ψ dxψ ψ = dx ψ+ψ dt ∂t ∂t −∞ −∞
569
(16.288)
and d2 dt2
Z
∞ ∗
dxψ ψ
=
−∞
= =
=
∗ Z d2 ∞ ∂ψ ∗ ∂ψ dx ψ+ψ dt2 −∞ ∂t ∂t 2 ∗ ∗ Z ∞ Z ∞ 2 ∂ ψ ∂ψ ∂ψ ∗∂ ψ dx dx +2 ψ+ψ ∂t2 ∂t2 ∂t ∂t −∞ −∞ 2 ∗ Z ∞ Z ∞ 2 2 ∂ ψ ∗∂ ψ ∗∂ ψ ψ+ψ −2 dx ψ dx ∂t2 ∂t2 ∂t2 −∞ −∞ Z ∞ d ∂ψ +2 dx ψ ∗ dt −∞ ∂t 2 ∗ Z ∞ Z 2 ∂ ψ d ∞ ∗∂ ψ ∗ ∂ψ dx ψ−ψ dx ψ +2 ∂t2 ∂t2 dt −∞ ∂t −∞ (16.289)
where to get the last line differentiated by parts. Using the KleinGordon equation: 2 2 ∂ (16.290) −~ 2 ψ = −~2 c2 ∇2 + (mc2 )2 ψ , ∂t to replace second time derivatives this gives: ∗ Z ∞ Z ∞ (mc2 )2 d2 (mc2 )2 ∗ 2 2 ∗ 2 2 dx −c ∇ ψ + dxψ ψ = ψ ψ − ψ −c ∇ + ψ dt2 ~2 ~2 −∞ −∞ Z d ∞ ∗ ∂ψ dx ψ +2 dt −∞ ∂t Z ∞ Z d ∞ 2 2 ∗ ∗ 2 ∗ ∂ψ = −c dx ∇ ψ − ψ ∇ ψ + 2 dx ψ (16.291) dt −∞ ∂t −∞ The first term in the last line vanishes after integration by parts for the same reason as in Example 4 but the last term is in general nonzero.
CHAPTER 16. SOLUTIONS TO EXERCISES
570
2. Using the relativistic probability density Eq. (16.292): Z Z ∞ d ∞ −i~ ∂ ∂ ∂ ? ? ∂ dxPR = dx ψ ψ − ψ ψ , dt −∞ 2mc2 −∞ ∂t ∂t ∂t ∂t Z ∞ 2 −i~ ∂2 ? ? ∂ = dx ψ 2 ψ − ψ 2 ψ , 2mc2 −∞ ∂t ∂t (16.292) where the terms containing first derivatives of the wave functions have cancelled out. Now one can again replace the second time derivatives of the wave functions by the right hand side of the KleinGordon equation, differentiate twice by parts and, assuming the usual vanishing boundary conditions at infinity, what is left cancels to give zero.
16.11
The Hydrogen Atom
Exercise 1. Verify that Eq. (11.28), in conjunction with the Bohr correspondence principle, gives the correct energy spectrum for the hydrogen atom. Solution: Substituting Eq. (11.28) in Eq.(11.26), we get 3
dEn h (−2En ) 2 √ = fclass = dn 2πke2 m Z →
3
En−3/2
dEn
~ (−2) 2 n √ = ke2 m 3
−2 En−1/2 En
~ (−2) 2 n √ = ke2 m m (ke2 )2 1 = − 2~2 n2
(16.293)
CHAPTER 16. SOLUTIONS TO EXERCISES
571
Exercise 2. Show that plugging the separated function Eq. (11.40) into Eq. (11.39) yields the following: ~2 1 ∂ 2 ∂R(r) r − ke2 r − r2 E − 2me R(r) ∂r ∂r ~2 1 1 ∂ 2 χ(θ, φ) 1 ∂ ∂χ(θ, φ) 1 = sin(θ) + 2me sin2 (θ) χ(θ, φ) ∂φ2 ∂θ sin2 (θ) χ(θ, φ) ∂θ (16.294) Solution: Substituting ψ(r, θ, φ) = R(r)χ(θ, φ) in ~2 ∂ 2 ∂ψ(r, θ, φ) r − ke2 r ψ(r, θ, φ) − r2 Eψ(r, θ, φ) − 2me ∂r ∂r ~2 1 ∂ 2 ψ(r, θ, φ) 1 ∂ ∂ψ(r, θ, φ) = + sin(θ) 2me sin2 (θ) ∂φ2 ∂θ sin2 (θ) ∂θ
, we get
.
~2 χ(θ, φ) d 2 dR(r) r − ke2 r R(r) χ(θ, φ) − r2 E R(r) χ(θ, φ) − 2me dr dr ~2 R(r) 1 ∂ 1 ∂ 2 χ(θ, φ) ∂χ(θ, φ) = + sin(θ) 2me ∂θ sin2 (θ) ∂φ2 sin2 (θ) ∂θ
Dividing by R(r)χ(θ, φ) yields ~2 1 d 2 dR(r) − r − ke2 r − r2 E 2me R(r) dr dr ~2 1 1 ∂ 2 χ(θ, φ) 1 ∂ ∂χ(θ, φ) = + χ(θ, φ) sin(θ) 2me sin2 (θ) χ(θ, φ) ∂φ2 ∂θ ∂θ sin2 (θ) .
CHAPTER 16. SOLUTIONS TO EXERCISES
572
Exercise 3. Verify that the most general solution to Eq. (11.52) is: gm (φ) = Dm eimφ + Bm e−imφ ,
(16.295)
where we have labelled the solution by the constant m for future consideration. Solution: The most general solution to a second order ordinary differential equation has two free parameters. If the equation is complex then there are two free complex parameters. Since Dm and BM are arbitrary complex parameters we just need to verify that the following solves reqeq:PhiEquation: gm (φ) = Dm eimφ + Bm e−imφ 00 → gm (φ) = (im)2 Dm eimφ + (−im)2 Bm e−imφ = −m2 Dm eimφ + Bm e−imφ = −m2 gm (φ) d2 g(φ) = m2 g(φ) →− 2 dφ
(16.296)
Exercise 4. me 1. Verify that the 2~ 2 value is 13.6 eV
e2 4π0
has units of energy, and that its numerical
Solution: Coulomb’s law for the magnitude of the force between two electrons a distance r apart is: e2 Fe = 1/4π0 2 (16.297) r that e2 /4π0 has the dimensions of Force × length2 = mass × (velocity)2 × 2 2 me e length. Therefore 2~2 4π has the dimensions of (mass)2 (velocity)4 × 0
CHAPTER 16. SOLUTIONS TO EXERCISES
573
(length)2 × mass/(mass × velocity × length)2 = mass × (velocity)2 which is energy! Write E1  =
e2 4π0
me 2~2
2
e2 4π0 ~ c
m e c2 2
=
2
=
me c2 2
× α2 , where α ' 1/137
is the (dimensionless) finestructure constant and me c2 ' 0.5 MeV is the electron rest energy. The fine structure constant plays a very important role in the quantum version of electrodynamics, called quantum electrodynamics, or QED, for short. Plugging in these numbers, we get E1  = 13.3 eV. The small difference with 13.6 eV is due to the use of approximate values.
Exercise 5. Calculate hri for the following states: 1. n = 1, l = 0 2. n = 2, l = 0 3. n = 2, l = 1 4. n = 3, l = 2 Note: you should not have to do any angular integrals. Why? Solution:
Z
π
hrilm =
Z
2π
dφ
dθ sin(θ) 0 Z
0 Z
π
dθ sin(θ)
= 0

Z
∞ ? dr r2 ψnl r ψnl
0 2π ? dφ Ylm (θ, φ)Ylm (θ, φ)
Z
0
∞
dr r3 Rnl 2 .
0
{z
=1
}
We do not need to do R ∞the angular integrals because the Ylm are normalized. Next we note that 0 xn e−x = n! for integer n. We change integration variable to x = 2r/n r0 where r0 is the Bohr radius given in Eq. (11.23) to
CHAPTER 16. SOLUTIONS TO EXERCISES get: Z
∞
dr r3 R10 2
hri10 =
0 Z 4 ∞ dr r3 e−2r/r0 = 3 r0 0 Z r0 ∞ = dx x3 e−x 4 0 3r0 = 2
Z
∞
dr r3 R20 2 0 Z ∞ 1 r r2 3 dr r 1 − + 2 e−r/r0 = 3 2r0 0 r0 4r0 Z ∞ x5 −x r0 3 4 dx x − x + e = 2 0 4 = 6r0
hri20 =
Z
∞
dr r3 R21 2 0 Z ∞ 1 dr r5 e−r/r0 = 24 r05 0 Z r0 ∞ dx x5 e−x = 24 0 = 5r0
hri21 =
Z
∞
dr r3 R32 2 0 Z ∞ 8 = dr r7 e−2r/3r0 7 3645 r0 0 Z ∞ 8 × 38 r0 dx x7 e−x = 3645 × 28 0 21r0 = 2
hri32 =
574
CHAPTER 16. SOLUTIONS TO EXERCISES
575
The result for a hpair (n, l) of principle i and orbital quantum numbers is given 2 `(`+1) , where Z is the number of protons in a by hrin,l = n Zr0 1 + 12 1 − n2 Hydrogenlike atom (Z = 1 for the Hydrogen atom).
Exercise 6. Verify that ~ has the same units as angular momentum. Solution: The units of ~ are Joules·seconds = kg·m2 /s. Angular momentum is r × p so it has units of m× kg·m2 /s, so they are the same.
Exercise 7. How many electrons does aluminum (Z=13) have in its outermost subshell shell? Is there room for more? If so, how many? Solution: Aluminum has 13 protons and 13 electrons. The electrons are distributed as follows: 1s2 2s2 2p6 3s2 3p1 The outer most shell n = 3 has room for (3 × 2 + 1) × 2 = 14 electrons but aluminum only has 3 in this shell. The 3p subshell has 1 electron but can accommodate (2 × 1 + 1) × 2 electrons, so there is room for 5 more.
16.12
Nuclear Physics
Exercise 1. 1. Osmium, Z = 76 is the densest stable element known. It has many several stable isotopes, the most common being 192 76 Os. The density of 3 Os is about 23, 000 kg/m . What fraction of a given volume V of Os is
CHAPTER 16. SOLUTIONS TO EXERCISES
576
taken up by the nuclei? Is this consistent with what you would expect from the Rutherford model of the atom? 2. Compare your answer above to the volume of the solar system taken up by the Sun. Consider the radius of the solar system to be the mean radius of Neptune’s orbit, which is about 4.5 billion kilometers. The sun’s radius is about 700,000 km. Solution:
1. The mass of one cubic meter of Os is 23,000 kg. The mass of a nucleon is about 1.7 × 10−27 kg. Therefore one cubic meter of Os contains: nucleon number =
23, 000 = 1.3 × 1031 nucleons (16.298) 1.7 × 10−27
Each nucleon has a radius of about 10−15 m and so the total volume of Os nuclei is : 4π −45 3 VOs = 1.3 × 1031 × 10 m = 5.5 × 10−14 m3 3
(16.299)
The ratio of the nuclear volume to total volume is: VOs = 5.5 × 10−14 1 m3
(16.300)
This is consistent with the Rutherford model of the atom in which the atom is mostly empty space with tiny electrons orbiting nuclei much like planets orbit the solar system. 2. The volume of the sun compared to the volume of the solar system is roughly: 3 7 × 105 km ∼ 4 × 10−12 (16.301) 4.5 × 109 km which is not too different!
CHAPTER 16. SOLUTIONS TO EXERCISES
577
Exercise 2. Are charge and lepton number conserved by the decay in Eq. (12.7)? Justify your answer. Solution:
1. Charge is conserved because the total charge in the LHS of the equation is +29 e, while that on the RHS is +28 e (Ni nucleus) + e (the positron)= +29 e, 2. Lepton number is conserved because on the RHS of the equation the positron has a lepton number of −1 while that of the neutrino is +1. Therefore they add up to zero and no net lepton number is created in the reaction.
Exercise 3. Is the decay in Eq. (12.9) relativistic? Justify your answer. Solution: The rest energy of an alpha particle is E = M c2 = 4 × 1.7 ×{z10−27} ×(3 × proton mass 8 2 −11 10 ) J = 61.2 × 10 J. Its kinetic energy in this decay is K = 4.5 MeV = 4.5 × 10−13 J. That is, K/E = 7 × 10−4 1. Therefore it is not relativistic.
Exercise 4. Derive Eq. (12.17) from Eq. (12.16). Solution:
ln(2) λ ln(2) → λ = . T1/2 T1/2 =
(16.302)
CHAPTER 16. SOLUTIONS TO EXERCISES
578
Substituting in the expression for N (t) ln(2)
N (t) = N0 e
−T
1/2
t
−t/T1/2
= N0 eln(2) = N0 2−t/T1/2 .
(16.303)
Exercise 5. In 2019, paleoanthropologists showed that a skull found in the 1970’s in southern Greece was in fact the oldest human fossil found outside of Asia. The skull was proven by radioactive dating to by 210,000 years old. If the skull was composed primarily of carbon and weighed 2 kg, how many 14 6 C atoms did it contain? Solution: The mass of a C atom is 2 × 10−23 kg. Therefore the number of C atoms in 2 kg is N = 1023 . The ratio of the number of 12 C atoms to 14 C atoms in a living organism is about 1.35 × 10−12 to 1. In other words, the sample had about 1023 × 1.35 × 10−12 = 1.35 × 1011 atoms at the time of its demise (note that we neglect the contribution of 14 C atoms to the mass of the sample). Since the halflife of 14 C is 5, 700 years, the number of 14 C left is 1.35 × 1011 × 2−210,000/57,000 = 1010 atoms.
16.13
Mysteries of the Quantum World
Exercise 1. Verify that the choice in Eq. (13.29) satisfies aa∗ + bb∗ = 1. Solution: From Eq. (13.29) a = eiχ cos(θ) b = eiχ eiφ sin(θ)
(16.304) (16.305)
CHAPTER 16. SOLUTIONS TO EXERCISES
579
We need to check: a a? + b b? = a2 + b2 = (eiχ cos θ) (e−iχ cos θ) + (eiχ eiφ sin θ) (e−iχ e−iφ sin θ) = cos2 θ + sin2 θ = 1 (16.306) as required.
Exercise 2. 1. Repeat Example 1 for electrons in states with: (a) Spin down along the xaxis. Solution: Spin down along the negative xaxis: θ = π/2, φ = π. Therefore: −ix = cos(π/4)+iz + eiπ sin(π/4)−iz 1 1 = √ +iz − √ −iz 2 2 Therefore P =
z h+−ix
2 1 1 = √ z h++iz + √ z h+−iz 2  {z } 2  {z } =1
=0
1 = 2
(16.307)
(b) Spin up along the yaxis, and spin down along the yaxis. Solution: Spin up along the yaxis: θ = π/2, φ = π/2 Therefore: +iy = cos(π/4)+iz + eiπ/2 sin(π/4)−iz 1 i = √ +iz + √ −iz 2 2
CHAPTER 16. SOLUTIONS TO EXERCISES
580
Therefore P =
z h+−ix
2 1 i = √ z h++iz + √ z h+−iz 2 2 1 = (16.308) 2 Spin down along the yaxis: θ = π/2, φ = 3π/2 Therefore: +iy = cos(π/4)+iz + ei3π/2 sin(π/4)−iz 1 i = √ +iz − √ −iz 2 2 Therefore P =
z h+−ix
2 1 i = √ z h++iz − √ z h+−iz 2 2 1 = 2
(16.309)
(c) Spin up along an axis in the y − z plane rotated 120◦ clockwise with respect to the zaxis and spin down along the same axis. Solution: In this case, θ = 2π/3 and assume φ = 0 and χ = 0. Then the spin state along that direction is given by √ 1 3 −iz , ψ(2π/3, 0, 0)i = − +iz + 2 2 (16.310) and the required probability of measuring a spin up along the zaxis is P = h+ψ(2π/3, 0, 0)i2 2 √ 1 3 = − z h++iz + z h+−iz 2 2 =
1 . 4
(16.311)
CHAPTER 16. SOLUTIONS TO EXERCISES
581
Similarly, the probability of measuring a spin down along the zaxis is P = h−ψ(2π/3, 0, 0)i2 2 √ 1 3 = − z h−+iz + z h−−iz 2 2 =
3 . 4
(16.312)
2. Suppose an electron is known to have spin up along the zaxis. What is the probability of measuring spin up along an axis rotated clockwise away from the zaxis by an angle of 120◦ ? What is the probability of measuring spin down along the same axis? Spin up along an axis in the y − z plane, rotated 120◦ clockwise with respect to the zaxis: θ = π/2, φ = 2π/3. Therefore: +iy = cos(π/4)+iz + ei2π/3 sin(π/4)−iz √ 1 1 = √ +iz + (1 + 3 i)−iz 2 2 Therefore P =
z h+−ix
2 √ 1 1 = √ z h++iz + (1 + 3 i)z h+−iz 2 2 1 = 2
(16.313)
Exercise 3. Prove that the norm of a spin vector is preserved under an arbitrary unitary transformation. Solution: Consider again the most general spin state: ψ(θ, φ, χ)i = ψ+ +iz + ψ− −iz (16.314)
CHAPTER 16. SOLUTIONS TO EXERCISES
582
A unitary transformation U=
a b c d
(16.315)
is one for which U † U = 1 or †
a∗ b∗
1 0
U U := =
c∗ d∗ 0 1
a b c d
(16.316)
Under a unitary transformation: 0 ψ+ ψ i = 0 ψ− = U ψ(θ, φ, χ)i a b ψ+ = ψ− c d
0
(16.317)
The norm of hψ 0 ψ 0 i = =
0 ∗ 0 0 ∗ 0 (ψ+ ) ψ+ + (ψ− ) ψ− 0 ψ+ 0 ∗ 0 ∗ (ψ+ ) , (ψ− ) 0 ψ−
(16.318)
One can write 0 ∗ 0 ∗ (ψ+ ) , (ψ− )
† 0 ψ+ = 0 ψ− † a b ψ+ = ψ− c d
= ((ψ+ )∗ (ψ− )∗ ) U †
(16.319)
where we have used the relationship in linear algebra that for any two matrices, A and B: (AB)† = B † A† (16.320) Finally we can use this to get: 0
0
hψ ψ i =
∗ ψ+
∗ ψ−
†
U U
ψ+ ψ−
= hψ(θ, φ, χ)U † U ψ(θ, φ, χ)i = hψ(θ, φ, χ)ψ(θ, φ, χ)i = 1 .
(16.321)
CHAPTER 16. SOLUTIONS TO EXERCISES
583
where we have used the unitarity of U to get to the last line.
Exercise 4. 1. Verify that Sˆx , Sˆy and Sˆz are hermitian. Solution: Consider ~ Sˆy = 2
0 −i i 0
.
(16.322)
Then, ~ Sˆy† = 2
0 (−i)?
i? 0
~ = 2
0 −i i 0
= Sˆy .
(16.323)
Similar results hold for Sˆx and Sˆz . 2. Show that the spin operators defined in Eqs. Eq. (13.46)Eq. (13.48) satisfy the following: [Sˆx , Sˆy ]ψi := Sˆx Sˆy − Sˆy Sˆx ψi = i~Sˆz ψi
(16.324)
when acting on any spin vector ψi. Solution: We have Sˆx Sˆy Sˆz
~ = 2 ~ = 2 ~ = 2
0 1
1 0
(16.325)
0 −i i 0
1 0 0 −1
(16.326) (16.327) (16.328)
CHAPTER 16. SOLUTIONS TO EXERCISES
584
We can use matrix multiplication to calculate: ~ ~ 0 1 0 −i Sˆx Sˆy = 2 1 0 2 i 0 2 ~ i 0 = 0 −i 2 i~ ˆ = Sz 2
(16.329)
and ~ 0 −i ~ 0 ˆ ˆ Sy Sx = 2 i 0 2 1 2 ~ −i 0 = 0 i 2 i~ = − Sˆz 2
1 0
(16.330)
So that i~ i~ − − ))) Sˆz ψi 2 2 = i~Sˆz
[Sˆx , Sˆy ]ψi = i
(16.331)
Exercise 5. In the case of two electrons, there is only one single extra real parameter for the nonproduct states as compared to the product states. How many extra parameters are there for 3 electron states? How many extra parameters are there for n electron states? Solution: A threeelectron state can be written as a direct product of three one electron states. Each of the three one electron states has 2 complex parameters. One of these two complex parameters can be pulled out as an overall factor in each state. Thus, although there are a total of six complex parameters or twelve real parameters in the product of three states, the three overall factors in each
CHAPTER 16. SOLUTIONS TO EXERCISES
585
state combine to a single overall factor whose magnitude is the normalization constant and phase is arbitrary. Thus one is left with 6 − 3 = 3 complex parameters, or 6 real parameters. On the other hand the nonproduct three electron state is an arbitrary, normalized linear combination of all possible product states involving the individual electron basis states. Each electron has 2 basis states, so there are 2 × 2 × 2 different ways to make a product, which is 8 different basis vectors from which we can construct nonproduct states. There are therefore 8 complex parameters or sixteen real parameters  one overall phase  one normalization constant, for a net 14 real parameters. The space of possible nonproduct states is larger that the space of products states for three electrons. The generalization to n electrons is straightforward. Each one particle state can be written as ψ(i)i = ai +, iiz + bi −, iiz , i = 1, . . . , n
(16.332)
Each such single particle state has initial two complex parameters as before, with one complex parameter going to normalization (real part) and overall phase, leaving one complex or two real parameters. Then the most general product state is: Ψ(1, 2, . . . , n) =
n Y
⊗ψ(i)
(16.333)
⊗ (ai +, iiz + bi −, iiz ) .
(16.334)
i=1
=
n Y i=1
The product states have n complex parameters, one for each state in the product, or 2n real parameters. This formula matches the numbers derived above for two particle and three particle product states. The number of different basis vectors for nonproduct states is 2n . The most general nonproduct state is a linear combination of basis vectors obtained by choosing one of 2 possible possible one particle basis vectors for the first particle, times 2 possible one particle basis vectors for the second, etc. The total number of different basis vectors is therefore 2 × 2 × 2.....n times = 2n . From this general state one can can take off 1 complex parameter for normalization and phase. The nonproduct states therefore have 2n − 1 complex parameters or 2n+1 − 2 = 2 × (2n − 1) real parameters. This formula
CHAPTER 16. SOLUTIONS TO EXERCISES
586
also works for the two and three particle nonproduct states obtained above. Note that the size of the parameter space of nonproduct states grows exponentially with the number of electrons! The size of the parameter space for product states only grows linearly. For example, for just 10 electrons, there are 2 × (210 − 1) = 2 × (1024 − 1) = 2046 parameters for nonproduct states and only 20 for product states.
Exercise 6. Consider the following twoelectron state: 1 −i π6 Ψ(1, 2) = A 300i + 01i + 2e 11i 2
(16.335)
1. Find the normalization constant A 2. Is it a product state? Justify your answer. Solution:
(a) Since the basis vectors on the RHS are mutually orthogonal, hΨ(1, 2)Ψ(1, 2)i = A2 32 + 2−2 + 22 . (16.336) √ Setting this equal to 1 gives A = 1/ 13.25 = 0.275. (b) It is a nonproduct state, as it cannot be written as (a1 0i + b1 1i)(a2 0i + b2 1i). The easiest way to see it is to recognize that for the state in Eq. (16.335), the product b1 a2 must be zero, since the state does not have a 10i term. This means either b1 = 0, which precludes the 11i terms or a2 = 0, which precludes the 00i term. In either case the state cannot be written as a product state.
Exercise 7.
CHAPTER 16. SOLUTIONS TO EXERCISES
587
1. Complete the algebra to show that Eq. (13.65) follows from the last line of Eq. (13.68). Solution: The last line of Eq. (13.68) is 1 1 1 1 1 √ +; 1iz + √ −; 1iz √ +; 2iz − √ −; 2iz  Ψ(1, 2) = √ 2 2 2 2 2 1 1 1 1 √ +; 1iz − √ −; 1iz √ +; 2iz + √ −; 2iz − . 2 2 2 2 (16.337) Multiplying through gives: Ψ(1, 2) =
1 √ (+; 1iz +; 2iz − +; 1iz −; 2iz 2 2 +−; 1iz +; 2iz − −; 1iz −; 2iz −+; 1iz +; 2iz  − +; 1iz 1−; 2iz + −; 1iz +; 2iz + −; 1iz −; 2iz ) . (16.338)
The first term cancels with the fifth term and the fourth term cancels with the last term. The remaining terms produce Eq. (13.65) as required. 2. Show that Ψ(1, 2) takes the same form up to an overall phase when expressed in terms of basis vectors +in and −in that point up and down along an arbitrary axis given by: +in = eiχ cos(θ/2)+iz + eiφ sin(θ/2)−iz (16.339) −in = eiχ sin(θ/2)+iz − eiφ cos(θ/2)−iz (16.340) Solution: We can invert Eq. (16.339) and Eq. (16.340) by first writing them in matrix form: +in cos(θ/2) eiφ sin(θ/2) +iz iχ = e −in sin(θ/2) −eiφ cos(θ/2) −iz +iz = M (16.341) −iz
CHAPTER 16. SOLUTIONS TO EXERCISES
588
Now invert the matrix M to give: +in +iz −1 = M −in −iz cos(θ/2) sin(θ/2) +in −iχ = e e−iφ sin(θ/2) −e−iφ cos(θ/2) −in (16.342) so that +iz = e−iχ (cos(θ/2)+in + sin(θ/2)−in ) (16.343) −iχ −iφ −iφ −iz = e e sin(θ/2)+in − e cos(θ/2)−in (16.344) Substitute the above into the expression for Ψ(1, 2): 1 Ψ(1, 2) = √ (+; 1iz −; 2iz − −; 1iz +; 2iz ) 2 −2iχ e = √ (cos(θ/2)+; 1in + sin(θ/2)−; 1in ) 2 × e−iφ sin(θ/2)+; 2in − e−iφ cos(θ/2)−; 2in e−2χ − √ e−iφ sin(θ/2)+; 1in − e−iφ cos(θ/2)−; 1in 2 × (cos(θ/2)+; 2in + sin(θ/2)−; 2in ) e−2iχ−iφ cos2 (θ/2) + sin2 (θ/2) (+; 1in −; 2in − −; 1in +; 2in ) = − √ 2 1 (16.345) = −e−2iχ−iφ √ (+; 1in −; 2in − −; 1in +; 2in ) 2 as required. 3. Show that the expressions for ±in for a vector n in the x − z plane that is at an angle θ = ±120◦ clockwise relative to the zaxis are up to an overall phase: 1 iπ/3 1 √ −iz +i±120◦ = +iz ± e (16.346) 2 2 1 1 −i±120◦ = ∓eiπ/3 √ +iz + −iz (16.347) 2 2
CHAPTER 16. SOLUTIONS TO EXERCISES
589
Solution: For an axis n in the y − z plane with φ = +120◦ = 2π/3 radians, θ = 90◦ = π/2 radians. For an axis n in the y − z plane and φ = −120◦ we must recall that the range of φ must be 0 → π radians, while θ goes from 0 → 2π, so we take φ = 2π/3 radians, θ = 3π/2 radians. Thus: +i+120◦ = eiχ cos(π/4)+iz + ei2π/6 sin(π/4)−iz 1 iχ iπ/3 1 √ +iz + e ( √ −iz = e (16.348) 2 2 and +i−120◦ = eiχ cos(3π/4)+iz + ei2π/6 sin(3π/4)−iz 1 iπ/3 1 iχ √ +iz − e ( √ −iz (16.349) = e 2 2 Similarly, −i+120◦ = eiχ sin(π/4)−iz + ei2π/6 cos(π/4)−iz 1 iχ iπ/3 1 = e − √ +iz + e ( √ −iz (16.350) 2 2 and +i−120◦ = eiχ sin(3π/4)+iz − ei2π/6 cos(3π/4)−iz 1 iχ iπ/3 1 = e − √ +iz − e ( √ −iz 2 2 1 iχ iπ/3 1 (16.351) = e − √ +iz − e ( √ −iz 2 2 Both expressions are correct up to an overall phase.
Exercise 8. Prove that S 2 Sx − Sx S 2 = 0
(16.352)
CHAPTER 16. SOLUTIONS TO EXERCISES
590
The analogous relations for the y and z components of spin follow by symmetry. Solution:
[SX , S 2 ] = [Sx , Sx2 + Sy2 + Sz2 ] = = = =
0 + [Sx , Sy2 ] + [Sx , Sz2 ] [Sx , Sy ] Sy + Sy [Sx , Sy ] + [Sx , Sz ] Sz + Sz [Sx , Sz ] i~ (Sz Sy + Sy Sz + (−Sy )Sz + Sz (−Sy )) 0
Exercise 9. Using the individual single particle identities Eq. (13.101)Eq. (13.104) prove the following: 1. Sx(1) Sx(2) X(1, 2)
4 ~ 1 = 2
(16.353)
2. X(1, 2), Y (1, 2) and Z(1, 2) are mutually commuting. 3.
6 ~ X(1, 2)Y (1, 2)Z(1, 2) = − 1 2
(16.354)
Solution:
1. Sx(1) Sx(2) X(1, 2) = Sx(1) Sx(2) Sx(1) Sx(2) = Sx(1) Sx(1) Sx(2) Sx(2) 2 2 ~ ~ = 1 1 2 2 4 ~ = 1 2
(16.355)
CHAPTER 16. SOLUTIONS TO EXERCISES
591
where we have used the fact that spin operators for different particles commute to get the second line, and Eq. (13.103) to get the third line. 2. X(1, 2)Y (1, 2) = Sx(1) Sx(2) Sy(1) Sy(2) = Sx(1) Sy(1) Sy(2) Sy(2) ~ ~ = i Sz(1) i Sz(2) 2 2 2 ~ = − Z(1, 2) 2
(16.356)
where the third line was obtained from Eq. (13.102). On the other hand Y (1, 2)X(1, 2) = Sy(1) Sy(2) Sx(1) Sx(2) = Sy(1) Sx(1) Sy(2) Sx(2) ~ ~ = −i Sz(1) (−i) Sz(2) 2 2 2 ~ Z(1, 2) = − 2
(16.357)
which equals the previous expression, so that: [X(1, 2), Y (1, 2)] = X(1, 2)Y (1, 2) − Y (1, 2)X(1, 2) = 0
(16.358)
3. 2 ~ X(1, 2)Y (1, 2)Z(1, 2) = − Z(1, 2)Z(1, 2) 2
(16.359)
where we have used the result from the previous answer. Z(1, 2)Z(1, 2) = Sz(1) Sz(2) Sz(1) Sz(2) = Sz(1) Sz(1) Sz(2) Sz(2) 2 2 ~ ~ = 1 2 2
(16.360)
Putting this into the previous equation gives the desired result.
CHAPTER 16. SOLUTIONS TO EXERCISES
16.14
592
Conclusions
No Exercises
16.15
Appendix: Mathematical Background
Exercise 1. Derive Eq.(15.16) from Eq.(15.15). Solution: Exercise 1
h(x − hxi)2 i = hx2 − 2xhxi + hxi2 i = hx2 i − 2hxihxi + hxi2 = hx2 i − hxi2
(16.361)
Exercise 2. Show using Euler’s formula that the Fourier series above can be written in complex form: f (x) =
∞ X
cn eikn x
(16.362)
n=−∞
where
(16.363)
cn
(16.364)
c−n = c∗n
Solution:
2πn L an − ibn = 2 an + ibn = 2
kn =
(16.365)
CHAPTER 16. SOLUTIONS TO EXERCISES
593
Let us start with expression (15.38) y(x) =
∞ X 1 a0 + (an cos(kn x) + bn . sin(kn x)) 2 n=0
(16.366)
Using the Euler formula eiθ = cos θ + i sin θ, one gets y(x) =
∞ 1 1 X a0 + (an + bn )eikn x + i (an − bn )e−ikn x 2 2 n=0
∞ 1 1 X = a0 + (an + ibn )eikn x + (an − ibn )eikn x 2 2 n=−∞
=
∞ X
cn e−ikn x ,
n=−∞
where c0 ≡ a0 /2, cn = an + ibn , c−n = c?n = an − ibn .
(16.367)