QUALIFYING EXAMINATION, Part 1

1:00 pm – 5:00 pm, Thursday August 31, 2017

Attempt all parts of all four problems.

Please begin your answer to each problem on a separate sheet, write your 3 digit codeand the problem number on each sheet, and then number and staple together the sheetsfor each problem. Each problem is worth 100 points; partial credit will be given.

Calculators and cell phones may NOT be used.

1

Problem 1: Mathematical Methods

(a) (20 points) Evaluate

limx→0

(1

x2− 1

sin2 x

).

(b) (20 points) Find the radius of convergence of the series

∞∑n=12017

(n!)2zn

(2n!).

(c) (30 points) Solve by a power series to find the general solution of the differentialequation

(x2 + 1)d2y

dx2− 2x

dy

dx+ 2y = 0 .

(d) (30 points) Use Cauchy’s theorem to evaluate∫ ∞

0

x sinx

x2 + a2dx

assuming a is real. Justify your choice of the contour integration.

2

Problem 2: Classical Mechanics

A particle of mass m moves along the x-axis in a potential U(x) = −ax2 +λx4, wherea and λ are positive constants.

(a) (25 points) Write down the Hamiltonian H(p, x) describing this system and find ex-pressions for the time derivatives x and p using Hamilton’s equations of motion.

(b) (35 points) Find the equilibrium values of x, and for each one determine if it is stableor unstable. Find the frequency of small oscillations ω around the stable points.

(c) (20 points) Make a qualitative sketch of a phase space orbit [i.e., a trajectory in the(x, p) plane] for a total energy E > 0. How does the plot change for E < 0?

Hint: it might be helpful to first make a qualitative sketch of U(x).

(d) (20 points) State Hamilton’s principle of least action. Consider a trajectory wherethe particle sits stationary at a stable equilibrium point for a total time T . What is theclassical action S along this trajectory?

3

Problem 3: Electromagnetism I

The electric potential for a charge distribution ρ(x′) may be computed from

Φ(x) =

∫ρ(x′)

|x− x′|d3x′ .

At large distances from a localized charge distribution, this may be may be expanded as

Φ(x) =q

r+

p · xr3

+1

2

∑i,j

Qijxixjr5

+ . . . ,

where xi are the cartesian coordinates of x, r = |x|, and Qij is the quadrupole momenttensor of the charge distribution

Qij =

∫ (3x′ix

′j − r′2δij

)ρ(x′) d3x′ .

(a) (20 points) What is the trace of Qij? How many independent components does Qij

have? Explain your answer.

(b) (20 points) What are q and p in the expansion above? Write expressions for q and pthat are analogous to the expression for Qij.

In the following, assume a distribution of four point charges ±e, each at a distance afrom the x and y axes in the z = 0 plane as shown in the figure below.

-e

x

y

z

e

e

-e

a

a

aa

aa

aa

(c) (10 points) What is q for the charge distribution?

4

(d) (10 points) What is p for the charge distribution? Explain your answer.

(e) (20 points) Compute Qij for the charge distribution.

(f) (20 points) What is the electric potential outside the charge distribution due to thecombination of the three terms in the expansion above? Compute the correspondingelectric field.

5

Problem 4: Electromagnetism II

You may solve this problem either in SI or in Gaussian units. Make sure to specify whichat the start.

A plane electromagnetic wave with angular frequency ω is traveling in the +z directionin the vacuum. Assume its electric field is linearly polarized in the x direction, and thatits amplitude is E0.

(a) (10 points) Write expressions for the electric field ~E and the magnetic field ~B as afunction of spatial coordinates and time. Express your answers in terms of E0, ω and c(the speed of light).

For all the remaining parts of this problem, give your answers in time-averaged quantities.

(b) (20 points) Find the power per unit area carried by this wave through a surface normalto the propagation direction.

Hint: the Poynting vector, which describes the energy flux density (i.e., the energy perunit area per unit time) is given by

~S =

1µ0

( ~E × ~B) (SI units)c4π

( ~E × ~B) (Gaussian units).

(c) (20 points) The wave is incident on a free electron of charge e and mass m whoseaverage position is the origin. Calculate the total average scattered power that is radiatedaway by the electron. Recall the Larmor formula which relates the power P radiated bya point charge q that has an acceleration a

P =

1

4πε023q2a2

c3(SI units)

23q2a2

c3(Gaussian units)

.

(d) (10 points) The total scattering cross section σ is the ratio of the scattered power fromthe electron to the incident power per unit area. Compute σ and express your answer interms of the classical electron radius re defined by

re =

1

4πε0e2

mec2(SI units)

e2

mec2(Gaussian units)

.

Next we work out the angular dependence of the radiate power. Recall that for dipoleradiation, the scattering cross section per unit spatial angle is given by

dσ

dΩ= A sin2 θ ,

6

where θ is the angle between the direction to the observer and the acceleration vector,and dΩ = 2π sin θdθ is the spatial angle (integrated over the the azimuthal direction).

(e) (20 points) Determine the constant A in terms of the classical electron radius re bydoing the appropriate integral and comparing with your result in (d).

(f) (20 points) Assume the observer is in the x − z plane at an angle α to the z axis.What is angular dependence of the scattered radiation? How does this answer change ifthe incident radiation were polarized in the y direction?

7

N2\

€x'dlÈ;ËUri

R

, v,¡=u'ff+u, iH-rH

å3 v'a.=",( W-#)."(#-#)*" |($oau-ä),'v=I*o('#).##.#

Explicit Forms of Vector Operations

Let er, €2, €¡ b€ orthogonal unit vectors associated srith the coordinate directions

specified in the headings on the left, and A1, Az, A¡ be the corresponding

components of A. Then

vú=e, ffi*",ff*^ffiv.¡,:ðrA'+**&ôxt' ôxz- ðxg

:

v x a : e, (H-*)*' (#-#) -* (*-*)v'1,:ffi*ffi*ffi

vú=e' #*",1#** ^håËo .o=+$ tr a,l*ffiS rrt" oA,)*^hq#

vxÀ:er **[*n*oAt-H].., [^h #-] $ t,",r]*", + [#

('ot-#]

u',þ :+ *(t #).r..|6 * (""' #).*- #[No,",t ", ] $ (r S)=i $ t+il

.Ea9(Düìè

AND d FUNCTIONS

Note: A square-root sign is to be understood ovet elter|cocfficicnt, e.g., for -8/15 read -JW'

35. CLEBS CH-G ORDAN COEFFICIENTS, SPHERICAL HARMONIC S,

2x!/2

t- xt/+L +L/21

3/2

+r -r/2o +r/2

3/2 L/2-112 +Ll2I

4: \Æ"* o 2

"l =-l*sir.i,eiö

2x]-

L/3 2/32'/3 -r/3

2+

o -r/2-r +r/2

3

X1

3/2 r/2-L/2 -r/2

+1

F1 +11 1

32+2 +2I

2

+2

w:'/T^G"",'r-å)

",] : -ly- si,'l,eoso eió

"î :i'*

"i,.2 s.2nó

0

+1

2/3 rl37/3 -2/3

+1 0

0+1

/3 2/3/3 -r/3

2

+1

-t -r/2

1

+1

L/2 r/2t/2 -t/2

+2 -1+1 0

0+1

3lZ-3/2

x1

321+1 +1 +1

35. Clebsch-Gordan coefficients 1

Yl^ = ?r) Ytr.

1/1sB/rs2/s

+t- -L00

-1 +1

2ll,2/,:

1

2L0000

r/3 3/s7/6 -3/ro

-L/2 rho

Ll6 1/z r/32/3 o-r/3r/6 -r/2 L/3

3/2xt

+z -Ll2+L +L/2

s/2 3/2-3/2+3/2

+al2 +Ll 1

+1 -100

-1 +1

rls 4/!4ls -Ll!

sl2

321000

-1 0

+3/2 O

+L/2 +1.

2

-1

+L-L/z

LlS r/2!/5 o

Lls -rl2

slz 3/2+112 +3/2

3/2xL/2

s/27/2

1

-t

./2 r/2

./2 -L/2

m1 ^rlm1 mz I Coefficients

t

312+L/2

2/s 3/s3/s -2/s

2/5 3ls3ls -zls

JJMM

3/10-2/53/ro

-1 -t

2

-2

+312-1+L/2 0

-7/2 +7

0 -rl2-r +L/2

+3/2 -!/2+Ll2 +L/2

0-1-1 0

-2 +L

s/2 3/2 t/2tr/2 +L/2 +L/2

321-1 -1 -1

I

s/2

a;,o: ffirvY"

L/2

LlLo zls3/s LlLs

311-0 -8115

2/s a/2 a/Lo8/rs -L/6-31r0LlLs -L/3 3/s

+

3/2Ll2

3lS 2/52lS -3/s

1

L/4 314rl4 -al4

+r/2 -L/2-r/z +L/2

-L -L/2-2 +L/2

L/2-r/3

Ll6

5/z 3/2-3/2 -3/2

2r00

+r/z -L-t/2 o

-3/2 +L

4/s rlsals -4ls

rl2 u2v2 -a/2

-24

s/2 312 u2-L/2 -L/2 -L/2

32-2 -2

-L/2 -L/2-3/2 +u2

-inø

2/3 L/3

3/LO s/rs3/s -r/Is

LlLo -zls

-2 -t/2

s/2

2!-t -1

3/4 L/41-/4-3/4

?

-t

1

r/6-a/3

L/2

-312-112

-!/2-L-3/2 O

-2

5/2 3/2-312 -3123ls 2/s2/s -3/5

-3/2 -!

s/2-s l2

1

QUALIFYING EXAMINATION, Part 2

1:00 pm – 5:00 pm, Friday September 1, 2017

Attempt all parts of all four problems.

Please begin your answer to each problem on a separate sheet, write your 3 digit codeand the problem number on each sheet, and then number and staple together the sheetsfor each problem. Each problem is worth 100 points; partial credit will be given.

Calculators and cell phones may NOT be used.

1

Problem 1: Quantum Mechanics I

In bound state problems, degeneracy of energy levels is associated with symmetry,sometimes not immediately apparent. For the Coulomb potential, with a rotational SO(3)symmetry, the degeneracy with respect to the angular momentum quantum number l(l = 0, 1, 2, . . . , n − 1, where n is the principal quantum number), is associated with alarger “hidden” SO(4) symmetry.

This problem explores the connection between degeneracy and symmetry for the two-dimensional isotropic harmonic oscillator, whose Hamiltonian is

H =1

2m

(p21 + p22

)+

1

2mω2

(x21 + x22

)= ~ω

(∑j=1,2

a†jaj + 1

),

where aj =√

mω2~

(xj +

ipjmω

)and [aj, a

†k] = δjk. It has a rotational SO(2) symmetry (cor-

responding to rotations around an axis perpendicular to the plane of the oscillator).

(a) (10 points) Study the energy levels of this Hamiltonian by splitting the Hamiltonianaccording to H = H1 +H2, where each part describes a one-dimensional oscillator. Whatis [H1, H2]? Can one find simultaneous eigenstates of the Hj [with eigenvalues ~ω

(nj + 1

2

),

nj ≥ 0 integers]?

(b) (15 points) The total energy eigenvalues of the full system can be labeled by thequantum number n ≡ n1 + n2. What are the possible values of n? What is the energyEn associated with quantum number n? What is the degeneracy of the level associatedwith quantum number n?

(c) (15 points) To explain the degeneracy in (b) via symmetry, define the four operatorsAjk = a†jak, where A11 + A22 = H/(~ω)− 1. Three other Hermitian linear combinationscan be formed from the Ajk. One of them is A11 − A22. What are the other two? Do allthree commute with H? If so, they can be taken to be the generators of the symmetryyou are looking for.

(d) (10 points) One of the Hermitian combinations in (c) can be identified with the an-gular momentum ~L ≡ x1p2 − x2p1. Find which is the particular combination that givesL. The three generators in (c) do not mutually commute. Thus only one of them can besimultaneously diagonalized with H. Take it to be L.

(e) (20 points) The commutation relations among the three Hermitian generators youfound in (c) are that of a three-dimensional angular momentum. That is, [Jx, Jy] = iJz,etc., where Jz = L/2, and Jx and Jy are proportional to the other two combinationsyou found in (c). (Note that only Jz is a physical angular momentum here.) Show thatthe operators Jx and Jy, defined with appropriate proportionality coefficients, satisfy

2

[Jx, Jy] = iJz.

Hint: it is convenient to first calculate the commutators [Ajk, Alm].

(f) (15 points) The simultaneous eigenvalues of J2 are then j(j+1), where it can be shownthat j = n/2 (you need not show this). From the properties of the angular momentumalgebra, what is the degeneracy of each energy level in terms of j? Compare to your resultin part (b). What are the possible eigenvalues of L in each degenerate level with quantumnumber n?

(g) (15 points) What is the larger symmetry group responsible for the degeneracy?

3

Problem 2: Quantum Mechanics II

Consider a three-level system described by the Hamiltonian

H

~= Ω1 |e〉 〈g1|+ Ω∗1 |g1〉 〈e|+ Ω2 |e〉 〈g2|+ Ω∗2 |g2〉 〈e| ,

where Ω1 and Ω2 are complex coupling parameters, and the three levels are described bythe normalized states |g1〉, |g2〉 and |e〉.

(a) (25 points) For real Ω1 = Ω2 = Ω > 0, find all the eigenenergies and eigenstates.

Hint: describe the Hamiltonian by a 3× 3 matrix.

(b) (25 points) For real Ω1 = Ω2 = Ω > 0, assume an initial state |ψ (0)〉 = |g1〉 and calcu-late the state |ψ (t)〉 at time t. Find the minimal time T > 0 when |ψ (T )〉 is given by |g2〉.

(c) (20 points) For general complex coupling parameters Ω1 and Ω2, show that there isalways an eigenstate with zero eigenenergy. Write an explicit expression for the normal-ized eigenstate of zero energy.

(d) (30 points) We may generalize the result in part (c) to an n-level system (n ≥ 3)described by the Hamiltonian

H

~=

n−1∑j=1

[Ωj |e〉 〈gj|+ Ω∗j |gj〉 〈e|

],

where the n levels are described by the normalized states |gj〉 (j = 1, . . . , n − 1) and |e〉.Show that there are (n− 2) eigenstates of H with zero eigenenergy.

Hint: consider a general state |φ〉 that is a linear combination of the n states describingthe n levels, i.e., φ =

∑n−1i=1 ci|gi〉 + ce|e〉, and find the conditions that the coefficients ci

(i = 1, . . . , n− 1) and ce must satisfy for |φ〉 to be an eigenstate of zero energy.

4

Problem 3: Statistical Mechanics I

Consider a linear chain of N beads as shown in the figure. Each bead is connectedto its two neighbors via springs of unstretched length a and spring constant κ, exceptfor bead 1 and bead N , which are connected, also via springs of spring constant κ andunstretched length a to two fixed points, separated by a distance (N + 1)a. The chain isin thermal equilibrium with a heat bath at temperature T .

x1 x2 xn xNx0=0 xN+1=0

The potential energy of the chain is

V (x1, x2, . . . , xN) =N∑n=0

1

2κ(xn+1 − xn)2 ,

where xn is the displacement of bead n from its position in the completely unstretchedconfiguration of the chain. Note that x0 = xN+1 = 0. In the following, we denote by xthe N -dimensional vector of displacements xn(n = 1, . . . , N).

In solving this problem, assume classical statistical mechanics.

(a) (15 points) Write an expression for the classical probability density P (x) (up to anoverall normalization) for a general potential V (x). Justify your answer.

(b) (20 points) We can rewrite the potential energy above as

V (x) =1

2xTVx,

where V is the potential energy matrix (of constants), and xT is the transpose of thecolumn vector x. The matrix V can be diagonalized by an orthogonal transformation viathe transformation matrix R whose elements are

Rmn =

√2

N + 1sin

(πmn

N + 1

).

5

Then the corresponding vector of normal coordinates y is given by y = Rx, i.e., ym =∑Nn=1Rmnxn(m = 1, . . . , N). The eigenvalues of V are given by

λm = 2κ

(1− cos

mπ

N + 1

), (m = 1, 2, 3, ..., N) .

Find the normalized probability density P (ym) of a normal coordinate ym.

(c) (20 points) Determine the mean square amplitude of each normal coordinate, i.e.,determine 〈y2m〉 for each m.

(d) (10 points) What is the correlation between different normal mode coordinates, i.e.,what is 〈ymyn〉, when m 6= n?

(e) (35 points) Determine the mean square displacement of bead n, i.e., 〈x2n〉. Sketch yoursolution as a function of n.

The following relation may be useful (for 1 ≤ n ≤ N)

N∑m=1

sin2

(πmn

N + 1

)1

1− cos(mπN+1

) = (N + 1)n

(1− n

N + 1

).

6

Problem 4: Statistical Mechanics II

Consider a gas of N identical fermionic atoms of spin 1/2 and mass m in a container ofvolume V . The walls of the container have N0 adsorption sites, each of which can adsorbone atom. Assume that the energy of an atom adsorbed to the wall is −∆, with ∆ > 0,and the energy of the atoms inside the container is the ordinary non-relativistic kineticenergy. The system is at equilibrium at temperature T and chemical potential µ.

You might find the following function of the fugacity z ≡ eβµ (where β = 1kT

) useful

f3/2(z) ≡ 4√π

∫ ∞0

dxx2

z−1ex2 + 1.

This function has the property

f3/2(z) ≈ z for z 1 .

(a) (15 points) What is the average number of atoms, 〈Na〉, that are adsorbed to the wallsas a function of the fugacity z ≡ eβµ and temperature T?

(b) (25 points) What is the average number of atoms, 〈Nf〉, that are moving freely in thevolume V as a function of the fugacity z and temperature T? The answer can be givenin terms of an integral that does not need to be evaluated explicitly.

(c) (25 points) Suppose that N > N0. What are the average numbers of the adsorbedand free atoms at T = 0? Calculate the Fermi energy of this system.

(d) (25 points) Again, suppose that N > N0. What are 〈Na〉 and 〈Nf〉 as T →∞? Notethat for large temperatures the fugacity is small, i.e., z 1.

(e) (10 points) This time, assume that N = N0. What is the Fermi energy of the system?

7

N2\

€x'dlÈ;ËUri

R

, v,¡=u'ff+u, iH-rH

å3 v'a.=",( W-#)."(#-#)*" |($oau-ä),'v=I*o('#).##.#

Explicit Forms of Vector Operations

Let er, €2, €¡ b€ orthogonal unit vectors associated srith the coordinate directions

specified in the headings on the left, and A1, Az, A¡ be the corresponding

components of A. Then

vú=e, ffi*",ff*^ffiv.¡,:ðrA'+**&ôxt' ôxz- ðxg

:

v x a : e, (H-*)*' (#-#) -* (*-*)v'1,:ffi*ffi*ffi

vú=e' #*",1#** ^håËo .o=+$ tr a,l*ffiS rrt" oA,)*^hq#

vxÀ:er **[*n*oAt-H].., [^h #-] $ t,",r]*", + [#

('ot-#]

u',þ :+ *(t #).r..|6 * (""' #).*- #[No,",t ", ] $ (r S)=i $ t+il

.Ea9(Düìè

AND d FUNCTIONS

Note: A square-root sign is to be understood ovet elter|cocfficicnt, e.g., for -8/15 read -JW'

35. CLEBS CH-G ORDAN COEFFICIENTS, SPHERICAL HARMONIC S,

2x!/2

t- xt/+L +L/21

3/2

+r -r/2o +r/2

3/2 L/2-112 +Ll2I

4: \Æ"* o 2

"l =-l*sir.i,eiö

2x]-

L/3 2/32'/3 -r/3

2+

o -r/2-r +r/2

3

X1

3/2 r/2-L/2 -r/2

+1

F1 +11 1

32+2 +2I

2

+2

w:'/T^G"",'r-å)

",] : -ly- si,'l,eoso eió

"î :i'*

"i,.2 s.2nó

0

+1

2/3 rl37/3 -2/3

+1 0

0+1

/3 2/3/3 -r/3

2

+1

-t -r/2

1

+1

L/2 r/2t/2 -t/2

+2 -1+1 0

0+1

3lZ-3/2

x1

321+1 +1 +1

35. Clebsch-Gordan coefficients 1

Yl^ = ?r) Ytr.

1/1sB/rs2/s

+t- -L00

-1 +1

2ll,2/,:

1

2L0000

r/3 3/s7/6 -3/ro

-L/2 rho

Ll6 1/z r/32/3 o-r/3r/6 -r/2 L/3

3/2xt

+z -Ll2+L +L/2

s/2 3/2-3/2+3/2

+al2 +Ll 1

+1 -100

-1 +1

rls 4/!4ls -Ll!

sl2

321000

-1 0

+3/2 O

+L/2 +1.

2

-1

+L-L/z

LlS r/2!/5 o

Lls -rl2

slz 3/2+112 +3/2

3/2xL/2

s/27/2

1

-t

./2 r/2

./2 -L/2

m1 ^rlm1 mz I Coefficients

t

312+L/2

2/s 3/s3/s -2/s

2/5 3ls3ls -zls

JJMM

3/10-2/53/ro

-1 -t

2

-2

+312-1+L/2 0

-7/2 +7

0 -rl2-r +L/2

+3/2 -!/2+Ll2 +L/2

0-1-1 0

-2 +L

s/2 3/2 t/2tr/2 +L/2 +L/2

321-1 -1 -1

I

s/2

a;,o: ffirvY"

L/2

LlLo zls3/s LlLs

311-0 -8115

2/s a/2 a/Lo8/rs -L/6-31r0LlLs -L/3 3/s

+

3/2Ll2

3lS 2/52lS -3/s

1

L/4 314rl4 -al4

+r/2 -L/2-r/z +L/2

-L -L/2-2 +L/2

L/2-r/3

Ll6

5/z 3/2-3/2 -3/2

2r00

+r/z -L-t/2 o

-3/2 +L

4/s rlsals -4ls

rl2 u2v2 -a/2

-24

s/2 312 u2-L/2 -L/2 -L/2

32-2 -2

-L/2 -L/2-3/2 +u2

-inø

2/3 L/3

3/LO s/rs3/s -r/Is

LlLo -zls

-2 -t/2

s/2

2!-t -1

3/4 L/41-/4-3/4

?

-t

1

r/6-a/3

L/2

-312-112

-!/2-L-3/2 O

-2

5/2 3/2-312 -3123ls 2/s2/s -3/5

-3/2 -!

s/2-s l2

1