PROBLEMS : 123456789 - Jonathan Whitmore

CODE NUMBER: 1

INSTRUCTIONSPHYSICS DEPARTMENT WRITTEN EXAM

PART I

Please take a few minutes to read through all problems before starting the exam.Ask the proctor if you are uncertain about the meaning of any part of any problem.You are to attempt two problems from each section.

The questions are grouped in three sections: classical mechanics, electromagnetism,and math/general. You must attempt two problems from each of these sections,for a total of six problems. Credit will be assigned for six (6) questions only. Eachquestion will be graded on a scale of zero to ten points. Circle the number of eachof the six problems you wish to be graded.

CLASSICAL ELECTRO- MATHEMATICALSECTION : MECHANICS MAGNETISM AND GENERAL

PROBLEMS : 1 2 3 4 5 6 7 8 9

SPECIAL INSTRUCTIONS DURING EXAM

1. You should not have anything close to you other than your pens & pencils,calculator and food items. Please deposit your belongings (books, notes,backpacks, etc.) in a corner of the exam room.

2. Departmental examination paper is provided. Please make sure you:

a. Write the problem number and your ID number on each white papersheet;

b. Write only on one side of the paper;

c. Start each problem on the attached examination sheets;

d. If multiple sheets are used for a problem, please make sure you staplethe sheets together and that your ID number is written on each sheet.

Colored scratch paper is provided and may be discarded when the examinationis over. At the conclusion of the examination period, please staple sheets fromeach problem together. On the top sheet, circle the problem numbers you will besubmitting for grading.

Put everything back into the envelope that will be given to you at the start ofthe exam, and submit it to the proctor. Do not discard any paper.

CODE NUMBER: SCORE: 2

#1 : CLASSICAL MECHANICS

PROBLEM: A point mass m slides frictionlessly, under the influence of gravity, alonga massive ring of radius a and mass M . The ring is affixed by horizontal springsto two fixed vertical surfaces, as depicted in Fig. 1. All motion is within the planeof the figure.

Figure 1: A point mass m slides frictionlessly along a massive ring of radius a andmass M , which is affixed by horizontal springs to two fixed vertical surfaces.

(a) Choose as generalized coordinates the horizontal displacement X of the cen-ter of the ring with respect to equilibrium, and the angle ✓ a radius to the massm makes with respect to the vertical (see Fig. 1). You may assume that atX = 0 the springs are both unstretched. Find the Lagrangian L(X, ✓, X, ✓, t).

(b) Derive the equations of motion.

(c) Find the eigenfrequencies for the small oscillations of this system. You mayfind it convenient to define ⌦2 ⌘ 2k/M , ⌫2 ⌘ g/a, and r ⌘ m/M .



PROBLEM:

A point particle of mass m moves in the vertical (y) direction along an infinitelythin pole with friction coefficient � > 0 (the force of friction is Ff = ��y). Theparticle is also attached to a spring of spring constant K. The particle is attachedto a semi-infinite string (x � 0) of uniform mass density µ and uniform tension ⌧ .A waveform f(ct + x) is incident from the right, and is reflected as g(ct � x). Thesituation is depicted in Fig. 2.

Figure 2: Incident and reflected wave.

(a) What is the speed c of wave propagation in the string in terms of the con-stants ⌧ , µ, K, and �?

(b) What is the equation which relates the reflected Fourier amplitude g(k) tothe incident Fourier amplitude f(k)? Find the reflection amplitude r(k) =g(k)/f(k). Express your answer in terms of the wavevector k and the con-stants K, m, c, �, and ⌧ .

(c) What is fraction ↵(k) of energy dissipated due to friction at incident wavevec-tor k?



PROBLEM:

A particle moves in a potential V (r,�, z).

(a) Derive a differential equation for the action S(r,�, z, t) of the particle, where(r,�, z) are cylindrical coordinates.

(b) What form of V (r,�, z) assures the separability/integrability of the motion?

(c) Calculate S, assuming V = V (r) is independent of (�, z) and comment on themeaning of the constants which enter your solution.


#4 : ELECTROMAGNETISM

PROBLEM:

A non-relativistic particle of charge q and velocity v is traveling through a sta-ble isotropic dielectric medium. Take the particle to be non-relativistic and themedium to have complex dielectric function "(k,!). Derive an expression for thestopping length for the particle by following the following steps:

(a) Calculate the electric field induced in the medium by the particle.

(b) Compute the rate at which the particle loses energy. You may leave the an-swer as an integral.

(c) How is the stopping length ` related to the rate of energy dissipation?

(d) 1. How is the particle energy dissipated?

2. Which frequency makes an especially strong contribution to the stop-ping length?



PROBLEM:

Consider an infinite slab of material in which the region 0 z 12h is filled with

a material with magnetic permeability µ1, and the region 12h < z h is filled

with material with permeability µ2. Other regions are vacuum, with µ = µ0. Theplane z = 0 has current density Kfree = Kx and the plane z = h has current densityKfree = �Kx. Find B, H , and the magnetization M everywhere. Compute r⇥Hand r⇥B and interpret the results.

Figure 3: A portion of the infinite slab.



PROBLEM:

(a) Derive the evolution equation for the magnetic field H in a non-relativisticmoving conductor, with uniform conductivity � and permeability, µ = 1.

(b) Consider a time varying magnetic field H0 e�i!t applied to a conductor at

rest. Estimate the penetration depth of the field.

(c) Calculate the EMF due to unipolar induction between the pole and the equa-tor of a uniformly magnetized conducting sphere of radius a rotating at an-gular frequency ⌦, a constant, around the direction of magnetization. Youmay leave your answer in terms of the B field in the sphere.

Nota bene: Do not assume that the answers to the different parts of this problemare dependent on one another.


#7 : MATH/GENERAL

PROBLEM:

Acoustic waves of a fixed frequency ! in water are conveniently described in termsof a scalar function (the so-called potential) that satisfies the Helmholtz wave equa-tion

(r2 + k2) (r) = 0 , k = !/cw

.

The oscillation amplitudes of pressure �P (r) and fluid velocity v(r) are related to (r) via

�P (r) = i!⇢w

(r) , v(r) = r (r) .

Find the eigenfrequency of radial vibrations of an air bubble in water. Expressthe result in terms of the radius a of the bubble, the sound speeds c

w

, ca

, and theequilibrium mass densities ⇢

w

, ⇢a

of water and air. Assume that the bubble is verysmall (a ⌧ c

a

/! ⌧ cw

/!) so that the pressure Pa

inside of it is nearly uniform at alltimes.

Hints: (1) Think of a spherically symmetric solution of the Helmholtz equation thatdecays at infinity. (2) Small oscillations of pressure inside the bubble are related tothose of air density by �P

a

= c2a

�⇢a

; from the mass conservation, �⇢a

/⇢a

= �3�a/a.


#8 : MATH/GENERAL

PROBLEM:

The temperature in the Earth’s atmosphere can be assumed to decrease linearlywith height z as T (z) = T (0)�⇤z, where ⇤ is a constant called the temperature lapserate. When the lapse rate exceeds a threshold value, hydrodynamic instabilitiescause air parcels to rise, leading to vertical currents of air. We will be calculatingthis threshold value by treating an air parcel as a separate gas, with its own tem-perature T

p

(z), which interacts adiabatically with the surrounding environment.

(a) Calculate the variation of the atmospheric pressure with height p(z) assum-ing air to be an ideal gas and ⇤z ⌧ T (0).

(b) Calculate the temperature Tp

(z) of the air parcel by assuming that the pres-sure on the parcel is p(z) calculated in (a) and that the relation between T

p

(z)and p(z) can be obtained by treating the parcel as an ideal gas that undergoesan adiabatic reversible transformation. Assume also that T

p

(0) ⇡ T (0).

(c) Instabilities arise when Tp

(z) > T (z). Find the condition for instability as aninequality between ⇤ and � = mg/c

p

, where m is the molecular mass of air, gis gravity and c

p

is the isobaric thermal capacity of air (assume �z ⌧ T (0)).


#9 : MATH/GENERAL

PROBLEM:

Show that if ↵, � real, with 0 < ↵ < 3 and � > 0, then

1Z

0

dxx↵�1 sin(12⇡↵� x)

x2 + �2= 1

2⇡ e��↵�2

Hint: Use contour integration.

CODE NUMBER: 11


PART II


The questions are grouped in two sections: quantum mechanics and statisticalphysics. You must attempt two problems from each of these sections, for a totalof four problems. Credit will be assigned for four (4) questions only. Each ques-tion will be graded on a scale of zero to ten points. Circle the number of each ofthe four problems you wish to be graded.

QUANTUM STATISTICALSECTION : MECHANICS PHYSICS

PROBLEMS : 10 11 12 13 14 15


1. You should not have anything close to you other than your pens, pencils,erasers, calculator and food items. Please deposit your belongings (books,notes, backpacks, etc.) in a corner of the exam room.









#10 : QUANTUM MECHANICS

PROBLEM:

Consider a system of indistinguishable, non-interacting, non-relativistic particlesof mass m in a three-dimensional cube of side a. For each of the following con-figurations, determine the pressure on the walls of the box from its relation to theground state energy.

(a) One particle in the ground state.

(b) Fifteen identical particles with spin 1/2 in their lowest energy configuration.

(c) Fifteen identical particles with spin 1 in their lowest energy configuration.



PROBLEM:

An electron can hop between the four corners of a square with hopping amplitudeb. The Hamiltonian is

H =

0

BB@

0 b 0 bb 0 b 00 b 0 bb 0 b 0

1

CCA , =

0

BB@

1

2

3

4

1

CCA

where i

is the amplitude to find the electron at corner i.

(a) Write down a 4⇥ 4 unitary matrix U corresponding to cyclic permutation ofthe vertices, and verify that [H,U ] = 0.

(b) Find all the energy eigenstates and eigenvectors.

(c) What is the ground state wavefunction and energy if b > 0?

(d) What is the ground state wavefunction and energy if b < 0?

Figure 4: The square, with labeled vertices.



PROBLEM:

Consider four sources {A,B,C,D} each of which produces spin S = 12 particles.

(a) Source A produces particles which are eigenstates of Sz with Sz = ±12~ with

equal probability. Write down the density matrix of the particles. Is this anunpolarized source?

(b) Source B produces particles which are eigenstates of Sx with Sx = ±12~ with

equal probability. Write down the density matrix. Is this an unpolarizedsource?

(c) Source C produces particles which are eigenstates of Sy with Sy = ±12~ with

equal probability. Write down the density matrix. Is this an unpolarizedsource?

(d) Source D is built out of the three sources A, B, and C, and picks particles atrandom with probability pA, pB, and pC (subject to pA + pB + pC = 1). Writedown the density matrix. What values of pA,B,C correspond to an unpolarizedsource?

(e)–(h) Repeat parts (a)–(d) for spin S = 1, where the source A now produces onlythe two states S↵ = ±~, ↵ 2 {x, y, z}, with equal probability, etc. Recall that

Sx

=~p2

0

@0 1 01 0 10 1 0

1

A , Sy

=~p2

0

@0 �i 0i 0 �i0 i 0

1

A , Sz

= ~

0

@1 0 00 0 00 0 �1

1

A .

Nota bene: Parts (a)–(d) for S = 12 are very easy and eight of the ten total points

will be awarded for parts (e)–(h).


#13 : STATISTICAL PHYSICS

PROBLEM:

In 1896 Wilhelm Wien derived a formula

P (!) / !3e�a!/T

for the spectral power of the black-body radiation from the classical statistical me-chanics.

(a) What is the value of the coefficient a for which this classical Wien law yieldsthe same large-! asymptotic behavior as the quantum Planck law? What isthe frequency !0 at which P (!) reaches the maximum? Compare with thePlanck’s prediction ~!0 = 2.82kBT .

(b) The standard explanation for why the sky is blue invokes Rayleigh scatter-ing in the atmosphere, namely, that the spectral power of the scattered lightscales as !4Pin(!) where Pin(!) is the spectral power of the incident light.Derive a formula for the peak frequency of the scattered sun light using theWien law. What is the numerical value of the corresponding wavelength�(!) assuming the Sun’s surface temperature is T = 5700K? How does this� compare with the wavelength of the blue light, 475 nm?

Nota bene: ~c = 1973 eV·A and kB = 8.617⇥ 10�5 eV·K.



PROBLEM:

Recall that in classical statistical mechanics, the microcanonical entropy S(E) enu-merates the logarithm of the number of microstates at energy E, and that dE =TdS where T is temperature. Consider a system where the function S(E) = 1

15 aE5/2+

13 bE

3/2 + cE1/2 with a < 0. Assume 0 E E0 where E0 is the solution of theequation S(E0) = 0.

(a) What is the temperature dependence of entropy S(T ) at low temperatures?Is there any constraint on c for this result to be physical sensible? Does thissystem satisfies the ‘third law of thermodynamics’?

(b) What relation should the variables a, b, c satisfy so that this system is thermo-dynamically stable?

(c) Assuming the condition in (b) is not satisfied, sketch the function S(E). Youshould notice that there exist energies E1, E2 with E1 6= E2 such that

dS

dE

��E1

=dS

dE

��E2

=S(E1)� S(E2)

E1 � E2.

What is the physical significance of the difference |E1 � E2|?



PROBLEM:

Consider a model polymer consisted of N + 1 monomers connected by N freelyjointed links, each of fixed length a. One end of a polymer is fixed, say at theorigin, and a force f is applied to the monomer at the other end.

(a) Show that the Hamiltonian of the system is (up to an additive constant) givenby

H = �NX

i=1

f · ai

,

where ai

= ri

� ri�1 denotes the orientation of the ith link, r

j

being the posi-tion vector of the jth monomer. (You may ignore the kinetic energy contribu-tion since it does not affect the distribution we are interested in.)

(b) Compute the partition function Z(T,f) and obtain the free energy.

(c) Show that the average end-to-end distance R(f) of the polymer is given by

R =D NX

i=1

ai

E,

and obtain R to leading order in f using the result of part (b). Explain thephysical origin of the spring-like elastic behavior of this system.

CODE NUMBER: 1


PART I


The questions are grouped in three sections: classical mechanics, electromagnetism,and math/general. You must attempt two problems from each of these sections,for a total of six problems. Credit will be assigned for six (6) questions only. Eachquestion will be graded on a scale of zero to ten points. Circle the number of eachof the six problems you wish to be graded.

CLASSICAL ELECTRO- MATHEMATICALSECTION : MECHANICS MAGNETISM AND GENERAL

PROBLEMS : 1 2 3 4 5 6 7 8 9


1. You should not have anything close to you other than your pens & pencils,calculator and food items. Please deposit your belongings (books, notes,backpacks, etc.) in a corner of the exam room.










PROBLEM: A point mass m slides frictionlessly, under the influence of gravity, alonga massive ring of radius a and mass M . The ring is affixed by horizontal springsto two fixed vertical surfaces, as depicted in Fig. 1. All motion is within the planeof the figure.

Figure 1: A point mass m slides frictionlessly along a massive ring of radius a andmass M , which is affixed by horizontal springs to two fixed vertical surfaces.

(a) Choose as generalized coordinates the horizontal displacement X of the cen-ter of the ring with respect to equilibrium, and the angle ✓ a radius to the massm makes with respect to the vertical (see Fig. 1). You may assume that atX = 0 the springs are both unstretched. Find the Lagrangian L(X, ✓, ˙X, ˙✓, t).

(b) Derive the equations of motion.

(c) Find the eigenfrequencies for the small oscillations of this system. You mayfind it convenient to define ⌦

2 ⌘ 2k/M , ⌫2 ⌘ g/a, and r ⌘ m/M .

SOLUTION:

(a) The coordinates of the mass point are

x = X + a sin ✓ , y = �a cos ✓ .

The kinetic energy is

T =

1

2

M ˙X2

+

1

2

m�˙X + a cos ✓ ˙✓

�2

+

1

2

ma2 sin2✓ ˙✓2

=

1

2

(M +m)

˙X2

+

1

2

ma2 ˙✓2 +ma cos ✓ ˙X ˙✓ .


The potential energy is

U = kX2 �mga cos ✓ .

Thus, the Lagrangian is

L =

1

2

(M +m)

˙X2

+

1

2

ma2 ˙✓2 +ma cos ✓ ˙X ˙✓ � kX2

+mga cos ✓ .

(b) The equations of motion are

d

dt

✓@L

@q�

◆=

@L

@q�,

for each generalized coordinate q�. For X we have

(M +m)

¨X +ma cos ✓ ¨✓ �ma sin ✓ ˙✓2 = �2kX .

For ✓,ma2 ¨✓ +ma cos ✓ ¨X = �mga sin ✓ .

(c) Linearizing the equations of motion, we have✓(M +m)!2 � 2k ma!2

ma!2 ma2!2 �mga

◆✓ˆX(!)ˆ✓(!)

◆= 0 .

Setting the determinant to zero, we arrive at the quadratic equation

!4

+

�⌦

2

+ (1 + r)⌫2�!2

+ ⌫2⌦2

= 0 ,

and the eigenfrequencies are given by

!2

± =

1

2

�⌦

2

+ (1 + r)⌫2�± 1

2

q�⌦

2 � ⌫2�2

+ 2r(⌦2

+ ⌫2)⌫2 + r2⌫4 .



PROBLEM:

A point particle of mass m moves in the vertical (ˆy) direction along an infinitelythin pole with friction coefficient � > 0 (the force of friction is F

f

= ��y). Theparticle is also attached to a spring of spring constant K. The particle is attachedto a semi-infinite string (x � 0) of uniform mass density µ and uniform tension ⌧ .A waveform f(ct + x) is incident from the right, and is reflected as g(ct � x). Thesituation is depicted in Fig. 2.

Figure 2: Incident and reflected wave.

(a) What is the speed c of wave propagation in the string in terms of the con-stants ⌧ , µ, K, and �?

(b) What is the equation which relates the reflected Fourier amplitude g(k) tothe incident Fourier amplitude ˆf(k)? Find the reflection amplitude r(k) =

g(k)/ ˆf(k). Express your answer in terms of the wavevector k and the con-stants K, m, c, �, and ⌧ .

(c) What is fraction ↵(k) of energy dissipated due to friction at incident wavevec-tor k?

SOLUTION:

(a) From Helmholtz’ equation, c =p⌧/µ.


(b) The wave equation and the boundary conditions are linear, so we may writethe displacement at wavevector k as

yk(x, t) = ˆf(k) eik(ct+x)+ g(k) eik(ct�x) .

At x = 0, Newton’s second law says

my = ��y + ⌧y0 �Ky ,

where y0 = (@y/@x)x=0

is the slope at x = 0. In terms of ˆf(k) and g(k), thisbecomes

�mc2k2

(

ˆf + g) = �i�ck( ˆf + g) + ik⌧( ˆf � g)�K(

ˆf + g) ,

and thus

g(k) =

r(k)z }| {

�(K �mc2k2

) + i(�c� ⌧)k

(K �mc2k2

) + i(�c+ ⌧)kˆf(k) .

(c) The incoming energy per unit wavevector is ⌧k2

�� ˆf(k)��2. The outgoing energy

per unit wavevector is ⌧k2

��g(k)��2. Thus the fraction of energy dissipated at

wavevector k is

↵(k) = 1��r(k)

��2=

4�c⌧k2

(K �mc2k2

)

2

+ (�c+ ⌧)2k2

=

4�c⌧

(Kk�1 �mc2k)2 + (�c+ ⌧)2.



PROBLEM:

A particle moves in a potential V (r,�, z).

(a) Derive a differential equation for the action S(r,�, z, t) of the particle, where(r,�, z) are cylindrical coordinates.

(b) What form of V (r,�, z) assures the separability/integrability of the motion?

(c) Calculate S, assuming V = V (r) is independent of (�, z) and comment on themeaning of the constants which enter your solution.

SOLUTION:

(a) The Lagrangian for the system is

L =

1

2

m�r2 + r2 ˙�2

+ z2�� V (r,�, z) ,

and therefore the Hamiltonian is

H =

p2r2m

+

p2�2mr2

+

p2z2m

+ V (r,�, z) ,

where

pr =@L

@r= mr , p� =

@L

@ ˙�= mr2 ˙� , pz =

@L

@z= mz .

Since the differential of the action S(q, t) is dS = p� dq� �H dt, we have

�@S@t

= H =

1

2m

"✓@S

@r

◆2

+

1

r2

✓@S

@�

◆2

+

✓@S

@z

◆2

#+ V (r,�, z) .

(b) Since @tL = 0, H(q, p, t) = E is a constant of the motion. Clearly if the poten-tial is of the form

V (r,�, z) = R(r) +�(�)

r2+ Z(z) ,


then the HJE is separable and we may write

S = Sr(r) + S�(�) + Sz(z) ,

where

E =

(1

2m

✓@Sr

@r

◆2

+R(r)

)+

1

r2

"1

2m

✓@S�

@�

◆2

+�(�)

#+

"1

2m

✓@Sz

@z

◆2

+Z(z)

#.

The terms in the square brackets must be constants, as is the overall RHS ofthe above equation.

(c) For V = V (r), we have �(�) = Z(z) = 0. We then have p� = @�S� andpz = @zSz are both constants, in which case

E =

1

2m

✓@Sr

@r

◆2

+R(r) +p2�

2mr2+

p2z2m

.

Thus, we may solve

@Sr

@r=

p2m

(E�

Ve↵

(r)z }| {"R(r) +

p2�2mr2

+

p2z2m

# )1/2

and

S =

p2m

rZdr0

hE � V

e↵

(r)i1/2

+ p� �+ pz z + C ,

where C is a constant. Here p� is the angular momentum and pz is the z-component of the linear momentum.



PROBLEM:

A non-relativistic particle of charge q and velocity v is traveling through a sta-ble isotropic dielectric medium. Take the particle to be non-relativistic and themedium to have complex dielectric function "(k,!). Derive an expression for thestopping length for the particle by following the following steps:

(a) Calculate the electric field induced in the medium by the particle.

(b) Compute the rate at which the particle loses energy. You may leave the an-swer as an integral.

(c) How is the stopping length ` related to the rate of energy dissipation?

(d) 1. How is the particle energy dissipated?

2. Which frequency makes an especially strong contribution to the stop-ping length?

SOLUTION:

(a) For a non-relativistic particle, losses are determined by the electrostatic inter-action. We have

⇢ext

= q �(x� vt) , jext

= qv �(x� vt) .

From Maxwell’s equation r · D = r · ("E) = 4⇡⇢ext

and E = �r�, andtaking the Fourier transform, we have Ê(k,!) = �ik ˆ�(k,!) and1

k2 "(k,!) ˆ�(k,!) = 4⇡q �(! � v · k) ) ˆ�(k,!) =4⇡q �(! � v · k)

k2 "(k,!).

(b) For energy loss,

dEdt

=

Zd3xE(x, t) · j(x, t) =

Zd3k

(2⇡)3ˆ|(�k, t) · Êind

(k, t) ,

1In (x, t) space, the dielectric function is a two point function of the form "(x� x0, t� t0) whoseFourier transform with respect to x� x0 and t� t0 is "(k,!).


where ˆ|(k, t) = qv e�iv·k t. One must subtract from Ê(k, t) the self-interactioncontribution, and therefore the induced contribution to Ê(k,!) from the mediumis

Êind

(k,!) = �4⇡iqk

k2

�(! � v · k)

1

"(k,!)� 1

�,

henceÊind

(k, t) = �4⇡iqk

k2

1

"(k,v · k) � 1

�e�iv·k t .

Therefore

dEdt

= �4⇡iq2Z

d3k

(2⇡)3v · kk2

1

"(k,v · k) � 1

�

= �4⇡q2Z

d3k

(2⇡)3v · kk2

"00(k,v · k)��"0(k,v · k)

��2+

��"00(k,v · k)��2

where "(k,!) ⌘ "0(k,!) + i"00(k,!). This is because "0(k,!) is even in fre-quency while "00(k,!) is odd in frequency. Note that the subtraction of theself-interaction term makes no difference in the computation of dE/dt.

(c) The stopping length is given by

1

`= �1

v

1

EdEdt

.

(d) 1. Energy is dissipated via the coupling of the moving charge to the dampedcollective motion of the dielectric medium – hence the dependence on"00(k,!).

2. The collective resonance frequency will have a large impact on `. Inpractice, this is the frequency where "0(k,! = v · k) ' 0.



PROBLEM:

Consider an infinite slab of material in which the region 0 z 1

2

h is filled witha material with magnetic permeability µ

1

, and the region 1

2

h < z h is filledwith material with permeability µ

2

. Other regions are vacuum, with µ = µ0

. Theplane z = 0 has current density Kfree = K ˆx and the plane z = h has current densityKfree = �K ˆx. Find B, H , and the magnetization M everywhere. Compute r⇥Hand r⇥B and interpret the results.

Figure 3: A portion of the infinite slab.

SOLUTION:

Find H = �K ˆy�⇥(z)�⇥(z� h)

�⌘ Hy(z) ˆy, which satisfies �H?

= 0 and �H ||=

Kfree ⇥ ˆn at the interfaces.

Compute r ⇥ H = �ˆx @@z

Hy(z) = K ˆx��(z) � �(z � h)

�= Jfree, satisfying the

required Maxwell equation.

Find B = µH = B(z) ˆy with

B(z) =

8>>><

>>>:

0 for z < 0

�µ1

K for 0 z 1

2

h

�µ2

K for 1

2

h z h

0 for z > h .


Compute

r⇥B = �ˆx@

@zBy(z) = K ˆx

�µ1

�(z) + (µ2

� µ1

) �(z � 1

2

h)� µ2

�(z � h)�

,

which can be interpreted in terms of µ0

J where J = Jbound + Jfree, and Jbound arethe currents associated with the magnetization.

The magnetization is M = µ�1

0

B �H = �mH ⌘ M(z) ˆy, with M(z) = 0 for z < 0

and z > h, and

M(z) =

(K(µ

0

� µ1

)/µ0

for 0 z 1

2

h

K(µ0

� µ2

)/µ0

for 1

2

h z h .



PROBLEM:

(a) Derive the evolution equation for the magnetic field H in a non-relativisticmoving conductor, with uniform conductivity � and permeability, µ = 1.

(b) Consider a time varying magnetic field H0

e�i!t applied to a conductor atrest. Estimate the penetration depth of the field.

(c) Calculate the EMF due to unipolar induction between the pole and the equa-tor of a uniformly magnetized conducting sphere of radius a rotating at an-gular frequency ⌦, a constant, around the direction of magnetization. Youmay leave your answer in terms of the B field in the sphere.

Nota bene: Do not assume that the answers to the different parts of this problemare dependent on one another.

SOLUTION:

(a) Maxwell’s equations say r ·B = 0, r⇥E = �c�1@tB, and r⇥H = 4⇡c�1J ,with

J = �⇣E +

v

c⇥B

⌘.

With µ = 1, we have H = B. We now have

1

�r⇥ J = r⇥E +

1

cr⇥ (v ⇥B)

= �1

c

@B

@t+

1

cr⇥ (v ⇥B)

=

c

4⇡�r⇥ (r⇥B) = � c

4⇡�r2B .

Thus, we obtain the induction equation,

@B

@t� c2

4⇡�r2B = r⇥ (v ⇥B) .


(b) When v = 0, we have @tB = (c2/4⇡�)r2B, which we may solve by writingB = B

0

eik·x e�i!t. Plugging in,

k2

=

4⇡i�!

c2) k =

✓4⇡�!

c2

◆1/2

e�i⇡/4 ,

where k = k ˆn with ˆn normal to the surface of the conductor. The penetrationdepth is then

` = � 1

Im k=

cp2⇡�!

.

We’ve assumed ! > 0. If ! < 0, replace ! with |!| in the above formula for `.

(c) The wire traverses the path xyz in Fig. 4. We have M , B, and ⌦ are allparallel.

Figure 4:

The EMF E is given by

E =

1

c

Z

xyz

dl · v ⇥B .

But the above integral around the closed curve xyzo must be zero, hence

E =

1

c

Z

xoz

dl · v ⇥B =

1

c

aZ

0

dr ⌦rB =

B⌦a2

2c.


#7 : MATH/GENERAL

PROBLEM:

Acoustic waves of a fixed frequency ! in water are conveniently described in termsof a scalar function (the so-called potential) that satisfies the Helmholtz wave equa-tion

(r2

+ k2

) (r) = 0 , k = !/cw.

The oscillation amplitudes of pressure �P (r) and fluid velocity v(r) are related to (r) via

�P (r) = i!⇢w (r) , v(r) = r (r) .Find the eigenfrequency of radial vibrations of an air bubble in water. Expressthe result in terms of the radius a of the bubble, the sound speeds cw, ca, and theequilibrium mass densities ⇢w, ⇢a of water and air. Assume that the bubble is verysmall (a ⌧ ca/! ⌧ cw/!) so that the pressure Pa inside of it is nearly uniform at alltimes.

Hints: (1) Think of a spherically symmetric solution of the Helmholtz equation thatdecays at infinity. (2) Small oscillations of pressure inside the bubble are related tothose of air density by �Pa = c2a �⇢a; from the mass conservation, �⇢a/⇢a = �3�a/a.

SOLUTION:

Outside of the bubble, r � a, we have an outgoing spherical wave:

=

A

reikr , �P =

iA

r! ⇢w eikr , v = r =

A

r

✓ik � 1

r

◆eikr ˆr .

Neglecting ik compared to 1/r = 1/a at the bubble surface, we get

vr = v · ˆr ' i �P

!⇢war = a .

On the other hand, according to the Hints,

�P (r = a) = �Pa = �⇢a c2

a = �3�a

a⇢a c

2

a .

Finally, we have�i! �a = vr(r = a) .

These equations have a nontrivial solution if

! =

caa

r3⇢a⇢w

.


#8 : MATH/GENERAL

PROBLEM:

The temperature in the Earth’s atmosphere can be assumed to decrease linearlywith height z as T (z) = T (0)�⇤z, where ⇤ is a constant called the temperature lapserate. When the lapse rate exceeds a threshold value, hydrodynamic instabilitiescause air parcels to rise, leading to vertical currents of air. We will be calculatingthis threshold value by treating an air parcel as a separate gas, with its own tem-perature Tp(z), which interacts adiabatically with the surrounding environment.

(a) Calculate the variation of the atmospheric pressure with height p(z) assum-ing air to be an ideal gas and ⇤z ⌧ T (0).

(b) Calculate the temperature Tp(z) of the air parcel by assuming that the pres-sure on the parcel is p(z) calculated in (a) and that the relation between Tp(z)and p(z) can be obtained by treating the parcel as an ideal gas that undergoesan adiabatic reversible transformation. Assume also that Tp(0) ⇡ T (0).

(c) Instabilities arise when Tp(z) > T (z). Find the condition for instability as aninequality between ⇤ and � = mg/cp, where m is the molecular mass of air, gis gravity and cp is the isobaric thermal capacity of air (assume �z ⌧ T (0)).

SOLUTION:

(a) We have dp = �⇢g dz where ⇢ is density of air. By the ideal gas equation,p(z) = ⇢RT (z)/m. Dividing these two equations,

dp

p= �mg

R

dz

T (0)� ⇤z

which gives

p(z) = p(0)

✓1� ⇤z

T (0)

◆mg/R⇤

⇡ p(0) exp

✓� mgz

RT (0)

◆


(b) The parcel acquires the pressure of the surroundings, which changes withheight. By the adiabatic equation,

dTp

Tp

=

� � 1

�

dp

p

where � = cp/cv. Using cp � cv = R and equation (1), we have

dTp

Tp

= �mg

cp

dz

T (0)� ⇤z

which gives

Tp(z) = T (0)

✓1� ⇤z

T (0)

◆mg/cp

⇤

⇡ T (0)� �z

where � = mg/cp.

(c) The air parcel rises if its temperature is higher than the surrounding temper-ature, i.e. Tp(z) > T (z), which happens when ⇤ > �.


#9 : MATH/GENERAL

PROBLEM:

Show that if ↵, � real, with 0 < ↵ < 3 and � > 0, then

1Z

0

dxx↵�1

sin(

1

2

⇡↵� x)

x2

+ �2

=

1

2

⇡ e��↵�2

Hint: Use contour integration.

SOLUTION:

We consider the contour integralI

dzz↵�1e�z

�2 � z2

over a counterclockwise semicircle contour encompassing the Re(z) � 0 region.We use the determination of the power with a branch cut from 0 to �1, i.e. theargument takes values ⇡ (�⇡) on the positive (negative) imaginary axis, respec-tively. The integral over the large semi circle of radius R goes to zero as R ! 1because of the exponential. On the contour over the imaginary axis from i1 to 0,the integral is (set z = x ei⇡/2)

I1

⌘ �ei⇡↵/21Z

0

dxx↵�1e�ix

x2

+ �2

The integral from 0 to �i1 is (set z = x e�i⇡/2), we get

I2

⌘ e�i⇡↵/2

1Z

0

dxx↵�1eix

x2

+ �2

The integrand has a pole at �, with the residue �1

2

�↵�2e�� . The final result followsfrom the integral theorem.

CODE NUMBER: 18


PART II


The questions are grouped in two sections: quantum mechanics and statisticalphysics. You must attempt two problems from each of these sections, for a totalof four problems. Credit will be assigned for four (4) questions only. Each ques-tion will be graded on a scale of zero to ten points. Circle the number of each ofthe four problems you wish to be graded.

QUANTUM STATISTICALSECTION : MECHANICS PHYSICS

PROBLEMS : 10 11 12 13 14 15


1. You should not have anything close to you other than your pens, pencils,erasers, calculator and food items. Please deposit your belongings (books,notes, backpacks, etc.) in a corner of the exam room.










PROBLEM:

Consider a system of indistinguishable, non-interacting, non-relativistic particlesof mass m in a three-dimensional cube of side a. For each of the following con-figurations, determine the pressure on the walls of the box from its relation to theground state energy.

(a) One particle in the ground state.

(b) Fifteen identical particles with spin 1/2 in their lowest energy configuration.

(c) Fifteen identical particles with spin 1 in their lowest energy configuration.

SOLUTION:

The Hamiltonian is H = p2/2m and the wavefunctions must vanish at the bound-aries of the cube, say when any of the coordinates is 0 or a. A good basis of energyeigenfunctions is then

nx

,ny

,nz

(x, y, z) = sin

⇣⇡nx

ax⌘sin

⇣⇡ny

ay⌘sin

⇣⇡nz

az⌘

with each ni a positive integer. I mention the wavefunctions to emphasize that eachof the integers ni must be positive. The energies are E

n

= (n2

x+n2

y+n2

z) ⇡2~2/2ma2 .

(a) The ground state has E = E111

⌘ E0

, where E0

⌘ 3⇡2~2/2ma2. The pressureis given by

P = �@E@V

= � @a

@V|{z}=

1

3a2

@E

@a=

⇡2~2ma5

⌘ P0

with P0

⌘ ⇡2~2/ma5 = 2E0

/3a3.

(b) Spin-12

particles are fermions. The ground state of N non-interacting fermionsis obtained by filling the lowest N levels. In units of E

0

, the lowest energylevels are given in the table below. So the fifteenth level has n = (3, 1, 1)(plus permutations), and the energy of the ground state is

E15⇥(S=1

2

)

= (2 · 3 + 2 · 3 · 6 + 2 · 3 · 9 + 1 · 11)⇥ ⇡2~22ma2

=

107

3

E0


(nx, ny, nz) E/E0

degeneracy number of spin-12

particles which fit(1, 1, 1) 3 2S + 1 2

(2, 1, 1) , (1, 2, 1) , (1, 1, 2) 6 3(2S + 1) 8

(2, 2, 1) , (2, 1, 2) , (1, 2, 2) 9 3(2S + 1) 14

(3, 1, 1) , (1, 3, 1) , (1, 1, 3) 11 2(2S + 1) 20

(2, 2, 2) 12 2S + 1 22

......

......

and the pressure isP =

107

3

P0

.

(c) Spin-1 particles are bosons and in the ground state non-interacting bosons alloccupy the lowest single-particle level. The spin degeneracy doesn’t matterhere. So

E15⇥(S=1)

= 15E0

and the pressure is P = 15P0

.



PROBLEM:

An electron can hop between the four corners of a square with hopping amplitudeb. The Hamiltonian is

H =

0

BB@

0 b 0 bb 0 b 0

0 b 0 bb 0 b 0

1

CCA , =

0

BB@

1

2

3

4

1

CCA

where i is the amplitude to find the electron at corner i.

(a) Write down a 4⇥ 4 unitary matrix U corresponding to cyclic permutation ofthe vertices, and verify that [H,U ] = 0.

(b) Find all the energy eigenstates and eigenvectors.

(c) What is the ground state wavefunction and energy if b > 0?

(d) What is the ground state wavefunction and energy if b < 0?

Figure 5: The square, with labeled vertices.

SOLUTION:

(a) The matrix which permutes the corner indices is

U =

0

BB@

0 1 0 0

0 0 1 0

0 0 0 1

1 0 0 0

1

CCA , U =

0

BB@

0 1 0 0

0 0 1 0

0 0 0 1

1 0 0 0

1

CCA

0

BB@

1

2

3

4

1

CCA =

0

BB@

2

3

4

1

1

CCA ⌘ e ,


and that

U †U =

0

BB@

0 0 0 1

1 0 0 0

0 1 0 0

0 0 1 0

1

CCA

0

BB@

0 1 0 0

0 0 1 0

0 0 0 1

1 0 0 0

1

CCA =

0

BB@

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

1

CCA ,

so U is unitary. Furthermore,⇥U,H

⇤= 0:

Uz }| {0

BB@

0 1 0 0

0 0 1 0

0 0 0 1

1 0 0 0

1

CCA

Hz }| {0

BB@

0 b 0 bb 0 b 0

0 b 0 bb 0 b 0

1

CCA=

0

BB@

b 0 b 0

0 b 0 bb 0 b 0

0 b 0 b

1

CCA =

Hz }| {0

BB@

0 b 0 bb 0 b 0

0 b 0 bb 0 b 0

1

CCA

Uz }| {0

BB@

0 1 0 0

0 0 1 0

0 0 0 1

1 0 0 0

1

CCA .

(b) Since⇥U,H

⇤= 0, U and H share their eigenvectors. And since U is unitary,

its eigenvalues are unimodular complex numbers of the form � = ei↵. Thus,

e = U = ei↵ ,

which says

2

= ei↵ 1

, 3

= ei↵ 2

, 4

= ei↵ 3

, 1

= ei↵ 4

,

from which we conclude e4i↵ = 1, hence the allowed values of ↵ are ↵ =

1

2

n⇡with n 2 {0, 1, 2, 3}. The normalized eigenvectors of U are then given by

=

1

2

0

BB@

1

ei↵

e2i↵

e3i↵

1

CCA .

These must also be eigenvectors of H :0

BB@

0 b 0 bb 0 b 0

0 b 0 bb 0 b 0

1

CCA

0

BB@

1

ei↵

e2i↵

e3i↵

1

CCA = b

0

BB@

ei↵ + e3i↵

1 + e2i↵

ei↵ + e3i↵

1 + e2i↵

1

CCA = b�ei↵ + e�i↵

�

0

BB@

1

ei↵

e2i↵

e3i↵

1

CCA

since e3i↵ = e�i↵, and therefore En = 2b cos�1

2

n⇡�. The allowed value of ↵ are

↵ =

1

2

n⇡ with n 2 {0, 1, 2, 3}. We then have

n=0

=

1

2

0

BB@

1

1

1

1

1

CCA , n=1

=

1

2

0

BB@

1

i�1

�i

1

CCA , n=2

=

1

2

0

BB@

1

�1

1

�1

1

CCA , n=3

=

1

2

0

BB@

1

�i�1

i

1

CCA

and

En=0

= 2b , En=1

= 0 , En=2

= �2b , En=3

= 0 .


(c) For b > 0, the ground state is n=2

, with E = �2b.

(d) For b < 0, the ground state is n=0

, with E = +2b = �2|b|.



PROBLEM:

Consider four sources {A,B,C,D} each of which produces spin S =

1

2

particles.

(a) Source A produces particles which are eigenstates of Sz with Sz= ±1

2

~ withequal probability. Write down the density matrix of the particles. Is this anunpolarized source?

(b) Source B produces particles which are eigenstates of Sx with Sx= ±1

2

~ withequal probability. Write down the density matrix. Is this an unpolarizedsource?

(c) Source C produces particles which are eigenstates of Sy with Sy= ±1

2

~ withequal probability. Write down the density matrix. Is this an unpolarizedsource?

(d) Source D is built out of the three sources A, B, and C, and picks particles atrandom with probability p

A

, pB

, and pC

(subject to pA

+ pB

+ pC

= 1). Writedown the density matrix. What values of p

A,B,C correspond to an unpolarizedsource?

(e)–(h) Repeat parts (a)–(d) for spin S = 1, where the source A now produces onlythe two states S↵

= ±~, ↵ 2 {x, y, z}, with equal probability, etc. Recall that

Sx =

~p2

0

@0 1 0

1 0 1

0 1 0

1

A , Sy =~p2

0

@0 �i 0

i 0 �i0 i 0

1

A , Sz = ~

0

@1 0 0

0 0 0

0 0 �1

1

A .

Nota bene: Parts (a)–(d) for S =

1

2

are very easy and eight of the ten total pointswill be awarded for parts (e)–(h).

SOLUTION:

(a)–(d) The density matrix in all cases is

% =

✓1

2

0

0

1

2

◆.

All sources are unpolarized, and any values of pA,B,C which sum to unity give

an unpolarized source.


(e)–(g) Let us denote the eigenvectors of Sa by µa , where a 2 {x, y.z} and µ 2

{�1, 0, 1}. The corresponding eigenvalues are µ~. For each of the sources A,B, and C, we have

%a =1

2

�� +1

a

↵⌦ +1

a

��+

1

2

�� 1

a

↵⌦ �1

a

�� .

All we need are the eigenvectors. Sz is diagonal and it is a trivial matter toelicit its eigenvectors. For Sx,y, we have

+1

x =

1

2

0

@1p2

1

1

A , �1

x =

1

2

0

@1

�p2

1

1

A , +1

y =

1

2

0

@1p2 i

�1

1

A , �1

y =

1

2

0

@1

�p2 i

�1

1

A .

Constructing the density matrices, we have

%A

=

1

2

0

@1 0 0

0 0 0

0 0 1

1

A , %B

=

1

4

0

@1 0 1

0 2 0

1 0 1

1

A , %C

=

1

4

0

@1 0 �1

0 2 0

�1 0 1

1

A .

Note that Tr% = 1 in all cases. An unpolarized source has % =

1

3

I, so neitherA, nor B, nor C is unpolarized.

(h) The density matrix for source D is

%D

= pA

%A

+pB

%B

+pC

%C

=

0

@1

2

pA

+

1

4

(pB

+ pC

) 0

1

4

(pB

� pC

)

0

1

2

(pB

+ pC

) 0

1

4

(pB

� pC

) 0

1

2

pA

+

1

4

(pB

+ pC

)

1

A .

Note that Tr%D

= pA

+ pB

+ pC

= 1. Source D is unpolarized when %D =

1

3

I,which requires

pA

= pB

= pC

=

1

3

.



PROBLEM:

In 1896 Wilhelm Wien derived a formula

P (!) / !3e�a!/T

for the spectral power of the black-body radiation from the classical statistical me-chanics.

(a) What is the value of the coefficient a for which this classical Wien law yieldsthe same large-! asymptotic behavior as the quantum Planck law? What isthe frequency !

0

at which P (!) reaches the maximum? Compare with thePlanck’s prediction ~!

0

= 2.82kBT .

(b) The standard explanation for why the sky is blue invokes Rayleigh scatter-ing in the atmosphere, namely, that the spectral power of the scattered lightscales as !4P

in

(!) where Pin

(!) is the spectral power of the incident light.Derive a formula for the peak frequency of the scattered sun light using theWien law. What is the numerical value of the corresponding wavelength�(!) assuming the Sun’s surface temperature is T = 5700K? How does this� compare with the wavelength of the blue light, 475 nm?

Nota bene: ~c = 1973 eV·A and kB = 8.617⇥ 10

�5 eV·K.

SOLUTION:

(a) The Plank law isP (!) / !3

exp(~!/kBT )�1

,

which agrees with the Wien law if a = ~/kB . To find !0

, set P 0(!

0

) = 0. UsingWien’s form for P (!), we find

!0

=

3T

a=

3kBT

~ ,

which is only 6% higher than Planck’s correct result.


(b) The scattered spectrum !7e�a!/T has its maximum at ! = 7T/a = 7kBT/~.The corresponding wavelength is

� =

2⇡c

!=

2⇡

7

~ckBT

= 360 nm .

This is ultraviolet (UVA band) rather than blue light. It is visible to insects,birds, and, apparently, reindeer.



PROBLEM:

Recall that in classical statistical mechanics, the microcanonical entropy S(E) enu-merates the logarithm of the number of microstates at energy E, and that dE =

TdS where T is temperature. Consider a system where the function S(E) =

1

15

aE5/2+

1

3

bE3/2+ cE1/2 with a < 0. Assume 0 E E

0

where E0

is the solution of theequation S(E

0

) = 0.

(a) What is the temperature dependence of entropy S(T ) at low temperatures?Is there any constraint on c for this result to be physical sensible? Does thissystem satisfies the ‘third law of thermodynamics’?

(b) What relation should the variables a, b, c satisfy so that this system is thermo-dynamically stable?

(c) Assuming the condition in (b) is not satisfied, sketch the function S(E). Youshould notice that there exist energies E

1

, E2

with E1

6= E2

such that

dS

dE

��E1

=

dS

dE

��E2

=

S(E1

)� S(E2

)

E1

� E2

.

What is the physical significance of the difference |E1

� E2

|?

SOLUTION:

(a) We havedS

dE=

1

6

aE3/2+

1

2

bE1/2+

1

2

cE�1/2 ,

thus at low temperatures, 1

T=

dSdE

gives E ⇡ 1

4

c2T 2, whence S(T ) ⇡ 1

2

c2T .Since entropy must be positive, c > 0. Indeed, the system does satisfy theThird Law since S(T ) ! 0 as T ! 0.

(b) Thermodynamic stability requires that the specific heat be positive, i.e. S 00(E) <

0. For small E this is already guaranteed sincepE is a concave function.

Similarly, at large values of E, a < 0 again ensures that S(E) is concave. Forintermediate values, some analysis is required. We have

S 00(E) =

aE2

+ bE � c

4E3/2.


Thus, given a < 0 and c > 0, in order for the above expression to be alwayspositive, it is necessary that the roots of S 00

(E) be nonreal. This is tantamountto the condition that the discriminant of the quadratic form in the numeratorbe negative, i.e. we must require b2 + 4ac = b2 � 4|a|c < 0.

(c) When the condition b2 < 4|a|c is not satisfied, at low and high E, S(E) isconcave, while it is convex at intermediate E. Any such function has twodistinct points where the slope (= T�1) is identical and the tangent vectorpasses through the two points. Thus there exist energies E

1

, E2

with E1

6= E2

such thatdS

dE

��E1

=

dS

dE

��E2

=

S(E1

)� S(E2

)

E1

� E2

.

The violation of stability, and correspondingly, the negative specific heat im-plies that the system phase separates in the intermediate regime E

1

< E < E2

and if one changes the temperature continuously while restricting oneself tonon-negative specific heat (Maxwell’s construction), the entropy jumps whilethe temperature changes continuously. This corresponds to a first order tran-sition (e.g. boiling of water). The difference |E

1

�E2

| = T |S1

�S2

| correspondsto the latent heat for this transition.



PROBLEM:

Consider a model polymer consisted of N + 1 monomers connected by N freelyjointed links, each of fixed length a. One end of a polymer is fixed, say at theorigin, and a force f is applied to the monomer at the other end.

(a) Show that the Hamiltonian of the system is (up to an additive constant) givenby

H = �NX

i=1

f · ai ,

where ai = ri � ri�1

denotes the orientation of the ith link, rj being the posi-tion vector of the jth monomer. (You may ignore the kinetic energy contribu-tion since it does not affect the distribution we are interested in.)

(b) Compute the partition function Z(T,f) and obtain the free energy.

(c) Show that the average end-to-end distance R(f) of the polymer is given by

R =

D NX

i=1

ai

E,

and obtain R to leading order in f using the result of part (b). Explain thephysical origin of the spring-like elastic behavior of this system.

SOLUTION:

(a) The Hamiltonian H can be written as

H = �f · rN+1

= �f ·N+1X

i=1

�ri � ri�1

�= �f ·

NX

i=1

ai .

(b) We have Z =

QNi=1

Zi where

Zi =

Zd3ai �

�|ai|� a

�e��H

i

=

Zd3ai �

�|ai|� a

�e�f ·ai

=

1Z

0

dr r2⇡Z

0

d✓ sin ✓

2⇡Z

0

d� ��|ai|� a

�e�fa cos ✓ =

4⇡a

�fsinh(�fa) .


The free energy of each link is then

Fi = ��1

lnZi = ��1

ln

✓4⇡a

�fsinh(�fa)

◆.

(c) Assume without loss of generality that f is in the ˆz direction. Then

hazi i = � 1

fhHii =

1

f

@ lnZi

@�

= � 1

�f+ a coth(a�f) = 1

3

a2�f � 1

45

a4�3f 3

+ . . .

and thereforehRzi ⇡ Na2f

3kBT.

Note this form is that of Hooke’s law with zero equilibrium extension andspring constant k = 3kBT/Na2.

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