2013 fall of vibration mechanics

CODE NUMBER: SCORE: 1

#1 : UNDERGRADUATE MECHANICSPROBLEM: A board of length L and mass M can slide frictionlessly along ahorizontal surface. A small block of mass m initially rests on the board atits right end, as shown in the figure. The coecient of friction between theblock and the board is . Starting from rest, the board is set in motion tothe right with initial speed v0. What is the smallest value of v0 such thatthe block ends up sliding o the left end of the board? Assume the smallblock is suciently narrow relative to L that its width can be neglected.

SOLUTION: The initial speed of the block is v = 0 and the initial speed of theboard is V = v0. The total momentum of the system is conserved, becausethe surface is frictionless. Thus, the total momentum is P = MV +mv =Mv0 at all times. Now while the total momentum P is conserved, the totalenergy E is not, due to the friction between the board and the block. Thekinetic energy of the system is given by E = 12MV

2 + 12mv2 and must be

equal to E0 W , where E0 = 12Mv20 is the initial kinetic energy of thesystem and W = mgd is the work done against friction for the block toslide a distance d to the left relative to the board. The minimum value ofv0 must occur when V = v and d = L. Thus, we have two equations in thetwo unknowns (v0, v):

(M +m)v =Mv012(M +m)v

2 = 12Mv20 mgL .

The solution is

v0 =r2gL

1 +

m

M

.


#2 : UNDERGRADUATE MECHANICSPROBLEM: Two blocks of masses m1 and m2 and three springs with springconstants k1, k2, and k12 are arranged as shown in the figure. All motion ispurely horizontal.(a) Choose as generalized coordinates the displacement of each block fromits equilibrium position, and write the Lagrangian.(b) Find the T and V matrices.(c) Suppose

m1 = 2m , m2 = m , k1 = 4k , k12 = k , k2 = 2k ,

Find the frequencies of small oscillations.(d) Find the normal modes of oscillation. You do not need to normalizethem.

SOLUTION:(a) The Lagrangian is

L = 12m1 x21 + 12m2 x

22 12k1 x21 12k12 (x2 x1)2 12k2 x22

(b) The T and V matrices are

Tij =@2T

@xi @xj=m1 00 m2

, Vij =

@2U

@xi @xj=k1 + k12 k12k12 k12 + k2

(c) We have m1 = 2m, m2 = m, k1 = 4k, k12 = k, and k2 = 2k. Let uswrite !2 !20, where !0

pk/m. Then

!2TV = k2 5 1

1 3

.


The determinant is

det (!2TV) = (22 11+ 14) k2= (2 7) ( 2) k2 .

There are two roots: = 2 and + = 72 , corresponding to the eigenfre-quencies

! =r

2km

, !+ =r

7k2m

(d) The normal modes are determined from (!2aT V) ~(a) = 0. Pluggingin = 2 we have for the normal mode ~()1 1

1 1

()1()2

= 0 ) ~() = C

11

Plugging in = 72 we have for the normal mode ~

(+)

2 11 12

(+)1(+)2

= 0 ) ~(+) = C+

12

The standard normalization (a)i Tij (b)j = ab gives

C =1p3m

, C+ =1p6m

.


#3 : UNDERGRADUATE E & MPROBLEM: A homogeneous magnetic field B is perpendicular to a track ofwidth l, which is inclined at an angle to the horizontal. A frictionlessconducting rod of mass m can move along the two rails of the track asshown in the figure. The resistance of the rod and the rails is negligible.The rod is released from rest. The circuit formed by the rod and the rails isclosed by a coil of inductance L. Find the position of the rod as a functionof time.

SOLUTION:The rods equation of motion is

ma = mg sinBlI, (1)

where a is acceleration of the rod and I is the current through the rod. Theinduced voltage in the rod V = Blv, where v is velocity of the rod. Therelationship between the induced voltage and the current is

LdIdt = Blv = Bldxdt . (2)

Since I = 0 and x = 0 at the start of the motion, the above formula givesLI = Blx. Substituting the current I = Blx/L into the equation of motiongives

ma = mg sin B2l2L x. (3)This equation is similar to the equation of motion for a body on a spring.The rod makes harmonic oscillations about the equilibrium position

x0 = mgL sinB2L2 . (4)


The amplitude of the oscillations is A = x0 and the frequency of the oscil-lations is !2 = B

2l2

mL . The position of the rod as a function of time

x(t) = A(1 cos!t) = mgL sinB2L2 (1 cos BlpmL t). (5)


#4 : UNDERGRADUATE E & MPROBLEM:Most planets in our solar system have a magnetic field that extends wellbeyond their atmosphere and contains trapped charged particles. Considera small volume where the field B can be considered uniform.(a) Give an equation for the force F experienced by a particle of charge qmoving with velocity v c in the field B. What is the direction of theforce?(b) The particle in (a) will move in a circle of radius r. Given an expressionfor r for a particle of mass m.(c) Give an equation for the frequency = v/(2r) of the rotation. Whatname do we give this frequency?(d) Discuss how we could detect keV electrons trapped in a 1 Gauss plane-tary field.

SOLUTION:(a) The charged particles experiences the Lorentz force

F = qv B (1)that is perpendicular to both v and B.(b) We equate the centripetal magnetic Lorentz force and the centrifugalforce mv2/r giving r = mv/(qB).(c) The cyclotron frequency is

=qB

2m. (2)

(d) The electrons will emit long wavelength radio waves at the cyclotronfrequency of 2.8 MHz, where we use q = 1.61019C andme = 9.111028g.This radiation can be intense. It was first detected from Jupiter (B=4 G)in 1955.

The cyclotron radiation is polarized. Viewed form the pole of the motionof the electron this radiation will be circularly polarized, and viewed fromthe plane of the circular motion it is linearly polarized. From other directionsit is elliptically polarized.

This radiation can only escape the region of origin if the cyclotron fre-quency exceeds the electron plasma frequency.

We can also detect these electrons if they interact with species in theouter atmosphere to produce auroral emissions over a broad range of wave-lengths from X-rays to infrared, and especially UV.


#5 : UNDERGRADUATE STAT MECHPROBLEM:The heat capacities for gases defined at fixed pressure and fixed volume aredenoted as CP and CV , respectively.

1) Can you explain which one should be larger for ideal gases withoutcalculation?

2) Prove that

CP CV = T@P

@T

V

@V

@T

P

(1)

3) For ideal gases, what is the value of CP CV ? Express the result interms of the number of gas molecules N and the Boltzmann constant kB.

4) Define the constant = CP /CV . Prove that PV is a constant forthe adiabatic process.

SOLUTION:1) Cp is larger than Cv. With increasing temperature from T to T + T ,

during the iso-pressure process, the volume of the gas expands, and thusthe gas does work, while for the iso-thermal process, the work is zero. Forideal gases, the internal energy only depends on temperature, and thus thechange of internal energies are the same for both processes. According toQ = W +U , Q/T is larger in the iso-pressure process, i.e., Cp islarger.

2)

CP = T@S

@T

p

, CV = T@S

@T

V

(2)

From the relation S(T, P ) = S(T, V (T, P )), we have@S

@T

P

=@S

@T

V

+@S

@V

T

@V

@T

P

. (3)

From dF = SdT PdV , we get the Maxwell relation

(@S

@V)T = (

@P

@T)V , (4)

thus CP CV = T@P@T

V

@V@T

P.

3) For the idea gas PV = NkBT ,@P

@T

V

=NkBV

(5)@V

@T

P

=NkBP

, (6)


and thus Cp Cv = T (NkB)2/(PV ) = NkB.4) During the adiabatic process

dS = (CvdT + PdV )/T = 0, PdV + V dP = NkBdT, (7)

thus PdV = CvdT and PdV + NkBdT = CpdT = CvdT = PdV .We have

V dP + PdV = 0 (8)

dPP

= dV

V, (9)

and thus PV is a constant.


#6 : UNDERGRADUATE STAT MECHPROBLEM:

A zipper has N1 links. Each link is either open with energy " or closedwith energy 0. We require, however, that the zipper can only open from theleft end, and that link ` can only unzip if all links to the left (1, 2, , ` 1)are already open. Each open link has G degenerate states available to it (itcan flop around).

1. Compute the partition function of the zipper at temperature T .Notational request from the grader: please use the name x Ge"/kBT .

2. Find the average number of open links at low temperature, i.e., in thelimit " kBT .

3. Find the average number of open links at high temperature, i.e., inthe limit kBT ".[Note: You can do this part of the problem independently of part 1.]

4. Is there a special temperature at which something interesting happensat large N? What happens there?

[Cultural remark: this is a very simplified model of the unwinding oftwo-stranded DNA molecules see C. Kittel, Amer. J. Physics, 37 917(1969).]

SOLUTION:

1. The state of the system is completely specified by the number of openlinks n, which can go from n = 0, 1, 2, ...N 1. The partition functioncan be summed to give

Z =N1Xn=0

Gnen"/kBT =N1Xn=0

xn =1 xN1 x


where x Ge".2. At any temperature,

hni = x@x lnZ = NxN

1 xN +x

1 x .

kBT =) x 1.In this limit, we can Taylor expand in the Boltzmann factor x. Theleading term is linear in x, and comes from one open link.

hni kBT x+O(x2)

3. At high temperature, every configuration is equally probable, and x!G. The probability of any given state is 1Z where Z is just the numberof states; this is still not so easy to evaluate. For general G, the answeris

hni = x@x lnZ|x=G = NGN

1GN +G

1G .

In the large-G limit hni simplifies to hni G!1! N 1.4. The partition function has a denominator which goes to zero at x = 1,

that is when1 = Ge" , kBT = "lnG.

At any finite N , the actual pole in Z is cancelled by the numerator.As N ! 1, this is not the case and there is a real singularity itbecomes a sharp phase transition. When N = 1, at x = 1 we reach


the radius of convergence about x = 0 of the partition sum. Physically:above this temperature, the entropy of the open bonds wins out overthe energetic cost of opening them, and there is a transition between amostly-zippered state and a mostly-unzippered state. This is strikinglyvisible in the plot of the fraction of open links hniN above.


#7 : UNDERGRADUATE QUANTUMPROBLEM:

A one-dimensional particle of mass m and energy E is incident on the -function potential V (x) = V0(x).

1. Find the reflection and transmission coecients.

2. Find the phase shift of the transmitted wave, and the dierence(E !1) (E ! 0).

3. The scattering amplitudes have a pole at a complex value of momentum~k. Find the location k0 of the pole. What is the physical interpretationof this pole?

SOLUTION:

1.

E =~2k22m

(1)

The wavefunctions are

< = eikx +Reikx > = Teikx (2)

continuous gives

1 +R = T (3)

and

0>(0) 0


2.

T = |T | eitan = c

2k(7)

As E !1, k !1 and(E !1) = 0. (8)

As E ! 0, k ! 0 and

(E ! 0) = 2

(9)

so

(E !1) (E ! 0) = 2

(10)

3. The pole is at

k0 = i c2

(11)

The pole corresponds to a bound state with energy

E = ~2

2m

c2

4= mV

20

2~2 (12)

which is present when c < 0, i.e. V0 < 0.


#8 : UNDERGRADUATE QUANTUMPROBLEM: Consider an electron constrained to move in the xy plane underthe influence of a uniform magnetic field of magnitude B oriented in the +zdirection. The Hamiltonian for this electron is

H =12m

px e

cAx2

+py e

cAy2

where m and e are the mass and charge of the electron, and c is the speedof light.

(a) Find a suitable expression for ~A so that px is a constant of motionfor the above Hamiltonian.

(b) With this choice for ~A, show that the eigenfunctions of H can bewritten in the form

(x, y) = ei~ pxx(y)

where (y) satisfies the Schrodinger equation for a one-dimensional har-monic oscillator whose equilibrium position is y = y0. Find the eectivespring constant k for this oscillator and the shift of the origin y0 in terms ofpx, B,m, e, c.

(c) Find the energy eigenvalues for this system, and indicate degenera-cies.

(d) For the remainder of the problem, suppose we further restrict theparticles to live in a square of side length L. Suppose we demand periodicboundary conditions. What are the possible values of px?

SOLUTION:(a) We can choose a gauge for ~A with ~r ~A = Bz, uniform, so that x doesnot appear:

Ax = By, Ay = 0.(b) The Schrodinger equation is

H (x, y) = E (x, y) (1)

and with the choice of gauge from part (a) we have

H =12m

px +

e

cBy2

+ p2y

.

Plugging in the given ansatz turns px into a number, and (1) becomes:

12m

px +

e

cBy2

+ p2y

e

i~ pxx(y) = Ee

i~ pxx(y)


or p2y2m

+12m

px +

e

cBy2!

(y) = E(y)

or p2y2m

+12m

eB

c

2 y +

c

eBpx2!

(y) = E(y)

which is the Schrodinger equation for a simple harmonic oscillator (SHO) p2y2m

+k

2(y y0)2

!(y) = E(y)

with k = meBmc

2 and y0 = cpxeB .(c) The energy spectrum of the SHO is

En = ~!n+ 12

where

! =r

k

m=

eB

mc= !c,

the cyclotron frequency. Notice that px drops out of the expression for theenergy and so there is a big degeneracy, approximately linear in the systemsize.

(d) Using the boundary condition that the wavefunction should be thesame at the end points ( (x = L) = (x = 0)), we have

px =2`L

, ` 2 Z .


#9 : UNDERGRADUATE GENERAL/MATHPROBLEM:

Imagine a long cylindrical tube of radius R at temperature Twall. Fluid flowsthrough the tube at velocity v. The temperature of the fluid when it entersthe tube is Tfluid. What is the length L that the fluid must travel in thetube so that its temperature reaches Twall?

1. Write down the equation you would use to solve the problem, and theboundary conditions.

2. Write down a back-of-the-envelope estimate of L using dimensionalanalysis. Find L for water if R = 0.5mm, v = 1mm/s, D, the thermaldiusivity is 0.15mm2/s, Tfluid = 21C, and Twall = 37C.

SOLUTION:Basically, all these problems work by dimensional analysis.

(1)@T@t = D

@2T@x2 (1)

(2) From (1), the diusion coecient has units [length]2/[time]. If the fluidwere stationary, it would take a time R2/D to warm up.

Now, its moving at velocity v, so the length you have to travel wouldbe v R2/D. For the given parameters, the characteristic travel distance forthermal equilibration would be order of 1 mm. Since these are estimates forfactors of e, it would be safer to multiply by 10, so the safe estimate wouldbe about 10 mm.


#10 : GENERAL/MATH

PROBLEM:

(a) Recall that (n) R10 ettn dtt (n 1)!. Verify the second equality inI

Z 11

ex2dx = (1/2).

(b) Explicitly evaluate the Gaussian integral I above. Hint: consider I2.(c) The volume element inD-dimensional polar coordinates is dV = rD1drdD1,where dD1 is an angle element. A sphere SD1 of radius r in D di-mensions, given by x21 + . . . x2D = r

2, has surface area D1rD1, whereD1 =

RdD1 is the total solid angle, e.g. 1 = 2 for a circle in D = 2

and 2 = 4 for a sphere in D = 3. Evaluate D1, for general D, in termsof the Gamma function. Hint: consider ID.SOLUTION:

(a) Substitute t = x2, so dt = 2xdx and (1/2) = 2R10 e

x2dx =R11 e

x2dx.

(b) Using the hint, going to polar coordinates, and substituting t = r2,

I2 =Z Z

dxdye(x2+y2) =

Z Zrdrder

2=

Z 10

dtet = ,

so I = (1/2) = p.(c) Follow the suggestion in the question and go to D-dimensional polarcoordinates, with r2 = x21 + . . . x2D

ID = D/2 =Z

. . .

Zdx1 . . . dxDe

(x21+...x2D) =Z

rD1drZ

dD1er2=

= D1Z 10

er2rD

dr

r= D1

Z 10

ettD/2dt

2t 12D1(D/2),

so D1 = 2D/2/(D/2).


#11 : GRADUATE MECHANICSPROBLEM: A yo-yo of mass M is composed of 2 large disks of radius R andthickness t separated by a distance t with a shaft of radius r. Assume auniform density throughout. Find the tension in the massless string as theyo-yo descends under the influence of gravity.

SOLUTION:Let the density of the yo-yo be , then its moment of inertia and mass arerespectively

I = 2 12tR4 + 12tr4, (1)M = 2 tR2 + tr2, (2)

whenceI = 12M

2R4+r4

2R2+r2

. (3)

The equations of motion of the yo-yo are

Mx = Mg T, (4)I = Tr, (5)

where T is the tension in the string. We also have the constraint x = r.From the above we obtain

T = IMgI+Mr2 =(2R4+r4)Mg

2R4+4R2r2+3r4 . (6)


#12 : GRADUATE MECHANICSPROBLEM: A particle under the action of gravity slides on the inside of asmooth paraboloid of revolution whose axis is vertical. Using the distancefrom the axis, r, and the azimuthal angle as generalized coordinates, find

(a) The Lagrangian of the system.

(b) The generalized momenta and the corresponding Hamiltonian.

(c) The equation of motion for the coordinate r as a function of time.

(d) If ddt = 0, show that the particle can execute small oscillations aboutthe lowest point of the paraboloid, and find the frequency of theseoscillations.

SOLUTION: Suppose the paraboloid of revolution is generated by a parabolawhich in cylindrical coordinates (r,, z) is represented by

z = Ar2, (1)

where A is a positive constant.(a) The Lagrangian of the system is

L = T V = 12m(r2 + r22 + z2)mgz (2)= 12m(1 + 4A

2r2)r2 + 12mr22 Amgr2. (3)

(b) The generalized momenta are

pr = @L@r = m(1 + 4A2r2)r, (4)

p = @L@ = mr2, (5)

and the Hamiltonian is

H = prr + p L (6)= 12m(1 + 4A

2r2)r2 + 12mr22 +Amgr2 (7)

= p2r

2m(1+4A2r2) +p2

2mr2 +Amgr2. (8)

(c) Lagranges equations

ddt

@L@qi

@L@qi = 0 (9)


give

m(1 + 4A2r2)r + 4mA2rr2 mr2 + 2Amgr = 0, (10)mr2 = constant. (11)

Letting the constant be mh and eliminating from Eq. 10, we obtain theequation for r:

(1 + 4A2r2)r3r + 4A2r4r2 + 2Agr4 = h2. (12)

(d) If = 0, Eq. 10 becomes

(1 + 4A2r2)r + 4A2rr2 + 2Agr = 0. (13)

The lowest point of the paraboloid is given by r = 0. For small oscillationsin its vicinity, r, r, r are small quantities. Then to first approximation Eq.13 becomes

r + 2Agr = 0. (14)

Since the coecient of r is positive, the particle executes simple harmonicmotion about r = 0 with angular frequency

! =p2Ag. (15)


#13 :GRADUATE ELECTRODYNAMICS

PROBLEM: A long, straight cylindrical wire, of radius a carries a uniformlydistributed current I. It emits an electron from r = a, with initial, rela-tivistic velocity v0 parallel to its axis. Find the maximum distance rmaxfrom the axis of the wire which the electrons can reach, treating everythingrelativistically.

SOLUTION: Find ~B = b2I/rc and then ~A = bz(2I/c) ln(r/a), so the elec-trons have

L = mc2p

1 v2/c2 + (2I|e|/c2)vz ln(r/a).The energy and pz are conserved:

pz =@L

@vz= mvz + (2I|e|/c2) ln(r/a) = 0mv0.

H = mc2 = 0mc2

with = 1/p

1 v2/c2 and 0 1/p

1 v20/c2. So = 0 and rmax iswhere r = 0, which means that vz = v0 (half-period of cyclotron rotation),which gives

rmax = a exp(0mv0c2/I|e|).


#14 :GRADUATE ELECTRODYNAMICS

PROBLEM: Consider a hollow spherical shell of radius a and surface chargedensity .

(a) Derive the magnetic field, ~B both inside and outside the shell, if it isspinning at frequency ! around an axis through its center. One way to solvethis is to write ~B = rmag, note that mag solves Laplaces equation, andthat the rotational symmetry is broken only by the vector ~!, so only the` = 1 harmonic contributes. If you solve the question this way, be sure tonote and use all the matching conditions above and below r = a.

(b) How much work is required to get the shell spinning at frequency !,starting from the shell at rest? Show that your answer fits with assigningan additional moment of inertia associated with electrodynamics, Itotal =Imech + IE&M , with Imech the usual moment of inertia of a hollow sphere.Dont bother to work out Imech, its 2mr2/3 and not relevant for computingthe quantity of interest, IE&M .

SOLUTION:

(a) Following the hints in the question, we have

outmag = C cos /r2, inmag = Dr cos = Dz.

Gauss law implies that ~Bbr must be continuous at the surface, and thecurl ~B Maxwell equation implies that br ( ~Bout ~Bin) = 4 ~K/c, with~K = ~v = ~! abr. It follows that ~Bout is that of a magnetic dipole, and~Bin is a constant:

~Bin =2~ma3

, ~Bout =3(br ~m)br ~m

r3,

with ~m = 43~!a4

c .

(b) The work required is the additional energy associated with the magneticfield of the spinning sphere,

WE&M = Ufield =Zd3x ~B2/8

Plugging in the above ~B and doing the volume integral gives

WE&M =12a3 ~m2 = 12(4/3)

2~!22a5/c2 12IE&M~!2,


withIE&M = (4/3)22a5/c2.

(Easy to check that the units are correct.)


#15 : GRADUATE STATISTICAL MECHANICSPROBLEM: The energy flux emitted from the surface of a perfect blackbodyis JE = T

4, where = 5.67 108W/m2K4 is Stefans constant and T isthe temperature.

(a) Find a corresponding expression for the entropy flux, JS .

(b) Idealizing the earth as a perfect blackbody, the average absorbed solarenergy flux is E = 342W/m

2. Derive an expression for the entropyflux of the radiation emitted by the earth, and provide an estimate ofits value in MKS units.

(c) Derive an expression for the ratio of the entropy flux of radiated ter-restrial photons to the entropy flux of solar photons incident on theearth? Express your answer in terms of the surface temperatures ofthe earth and the sun.

SOLUTION:

(a) The energy flux is

JE =cE

4V

/2Z0

d sin cos 2Z0

d =cE

4V,


where E is the total energy. Mutatis mutandis, we have that JS =cS/4V . We relate E and S using thermodynamics: dE = T dSp dV .From JE = T

4 we then have

JS = 43T3 .

(b) The total terrestrial entropy flux is

S =4E3Te

4 342W/m2

3 278K = 1.64W/m2K ,

where we have approximated the mean surface temperature of theearth as Te 278K, which is what one finds assuming the earth isa perfect blackbody. (The actual mean surface temperature is about287K.)

(c) First, recall the argument for the steady state temperature of the earth.The energy of the absorbed solar radiation is (T 4)(4R2)(R2e/4a2e),which is obtained by multiplying the total energy output of the sunby the ratio of the earths cross section R2e to the area of a spherewhose radius is ae = 1AU. Setting this to the energy output of theearth, (T 4e )(4R2e) yields the expression Te = T

pR/2ae. When

comparing the entropy flux of the solar radiation to that emitted bythe earth, we use the same expressions, except we replace T 4 by 43T

3

throughout. Therefore the ratio of entropy fluxes is

=T 3e

T 3(R2/4a2e)=

TTe 21 ,

taking T 5780K.


#16 : GRADUATE STATISTICAL MECHANICSPROBLEM: The Landau expansion of the free energy density of a particularsystem is given by

f(T,m) = (T T0)m2 T0m3 + T0m4 ,where m is the order parameter and T0 is some temperature scale, and wherewe have set kB 1.

(a) Find the critical temperature Tc. Is the transition first or second order?

(b) Sketch the equilibrium magnetization m(T ).

(c) Find m(Tc ), i.e. the value of m just below the critical temperature.

SOLUTION:

Figure 1: Sketch for part (b).

(a) The free energy is of the general form f(m) = 12am2 13ym3 + 14bm4,

with T dependence implicit in the coecients:

a = 2(T T0) , y = 3T0 , b = 4T0 .Setting f 0(m) = 0 we have (a ym + bm2)m = 0, yielding threesolutions, one of which lies at m = 0. The other two solutions areroots of the quadratic factor:

m =y

2br y

2b

2 ab.


These solutions are both real and positive if a < y2/4b. Since thefree energy increases without bound for m ! 1, we must havethat m is a local maximum and m+ a minimum. To see if it is theglobal minimum, we must compare f(m+) with f(0). To do this, setf(m) = f(0), which yields 2a 43ym + bm2 = 0, and subtract fromthe equation a ym+ bm2 = 0 to find m = 3a/y. This is the value ofm for which the two minima are degenerate. Equating this with theexpression for m+, we obtain a = 2y2/9b, which is our equation forTc. Solving this equation for T , we obtain Tc = 54 T0.

(b) A sketch is provided above. Note that m(T ) drops discontinuously tozero at the transition.

(c) Evaluating m = 3a/y at T = Tc, we have m(Tc ) = 12 .


#17 : GRADUATE QUANTUMPROBLEM:

1. An electron gun produces electrons randomly polarized with spins upor down along one of the three possible radomly selected orthogonalaxes 1, 2, 3 (i.e. x, y, z), with probabilities pi," and pi,#, i = 1, 2, 3. Tosimplify the final results, it is better to rewrite these in terms of di andi defined by

pi," =1

2di +

1

2i pi,# =

1

2di 1

2i i = 1, 2, 3 (1)

Probabilites must be non-negative, so di 0 and |i| di.(a) Write down the resultant electron spin density matrix in the

basis |"i, |#i with respect to the z axis.(b) Any 2 2 matrix can be written as

= a1+ b (2)in terms of the unit matrix and 3 Pauli matrices. Determine aand b for from part (a).

(c) A second electron gun produces electrons with spins up or downalong a single axis in the direction n with probabilities (1)/2.Find n and so that the electron ensemble produced by thesecond gun is the same as that produced by the first gun.

SOLUTION:

(a) The density matrix is

p1,"1

2

1 11 1

+ p1,#

1

2

1 11 1

+ p2,"

1

2

1 ii 1

+ p2,#

1

2

1 ii 1

+ p3,"

1 00 0

+ p3,#

0 00 1

=

1

2

(d1 + d2 + d3) + 3 1 i2

1 + i2 (d1 + d2 + d3) 3

=1

2

1 + 3 1 i21 + i2 1 3

(3)


(b) By inspection,

a =1

2b =

1

2 (4)

(c) This is the same as the density matrix produced by the secondelectron gun if n is parallel to , and D = ||.A simple way to see this is to a go to a rotated coordinate systemwith z0 axis along .


#18 : GRADUATE QUANTUMPROBLEM: The Hamiltonian for a quantum mechanical rigid body is

H = 12

L21I1

+L22I2

+L23I3

, (1)

where I1, I2, I3 are the principal moments of inertia, and (L1,L2,L3) (Lx,Ly,Lz) are the angular momentum operators, satisfying the commuta-tion relations

[Li,Lj ] = i~ijkLk . (2)

You may assume the angular momentum takes on only integer values. Hamil-tonians of the form (1) describe the rotational spectrum of molecules.

(a) (Very easy) First, consider the case I1 = I2 = I3 = I (a sphericaltop, such as methane). Write down a formula for the energy levels in termsof an appropriate quantum number.

(b) (Easy) Next, consider the case I1 = I2 = I? 6= I3 (a symmet-ric top, such as ammonia). Show that L3 is a constant of motion. Writedown another constant of motion which commutes with L3, and express theHamiltonian as a function of the two constants of motion. Then write downa formula for the energy levels as a function of the two good quantum num-bers. Indicate the allowed ranges of these quantum numbers. Also indicateany degeneracies.

(c) (A little harder) Now consider the case of a slightly asymmetric top,i.e. one for which

I1 = I? , I2 = I? + (3)where is small. What are the good quantum numbers in this case? Findthe shifts in the energy levels relative to those of the symmetric top, to firstorder in the small quantity . List the energy shifts for all values of the twoquantum numbers in part (b).Hints: (1) Express the perturbing Hamiltonian HHsymmetric in terms ofraising and lowering operators. (2) A useful formula:

L+|`,mi = ~p(`m)(`+m 1)|`,m+ 1i . (4)

(3) You will need to use degenerate perturbation theory.

SOLUTION:


(a) For the spherical case,

H =12IL21 + L

22 + L

23

=

~L2

2I(5)

and soE` =

~2`(`+ 1)2I

. (6)

(b) For the symmetric top,

H =12

L21 + L22

I?+L23I3

=

12

~L2 L23

I?+L23I3

!. (7)

Since [L3, ~L2] = 0, both L3 and ~L2 are constants of the motion ([L3,H] =0 = [~L2,H]). For a given `, the eigenvalues of L3 take values m = `,`+1, ...0, ...` 1, `, and the energies are:

E`,m =~22

`(`+ 1)m2

I?+

m2

I3

,

` = 0, 1, ...1m = `, ...0, ...` .

(c) The perturbing Hamiltonian is

H HHsymmetric = 12

1I? L

21 +

1I? +

L22 1I?

L21 + L

22

. (8)

Using 1I? =1I? I2? + ..., we have

H =

2I2?

L21 L22

. (9)

Now lets write this in terms of raising and lowering operators:

L L1 iL2,L1 = 12(L+ + L),L2 =12i(L+ L)

which satisfy (according to (2))

[L3,L] = L, [L+,L] = 2L3.In terms of these,

H =

4I2?

L2+ + L

2. (10)

This operator changesm by 2 (i.e. |m| = 2 ). This means that its firstorder correction to the energies is zero for any state that isnt degenerate


with some other state related to it by |m| = 2. The only degeneracy in thespectrum comes from E`,m = E`,m. So the condition for nonzero energyshift is only met by m = 1 states (and any `). So:

E`,m = 0, m 6= 1.For m = 1, we have to use degenerate perturbation theory, that is, we haveto diagonalize the matrix obtained by acting with H in the degeneratesubspace. For each `, this matrix h is 2 2, indexed by m = 1, and itsnonzero matrix elements are:

4I2?h`, 1|(L2+ + L2)|`,1i =

4I2?h`,1|(L2+ + L2)|`,+1i .

So within the subspace, the matrix to diagonalize is

h =0 0

whose eigenvalues are . Using (4),

L2+|`,1i = ~2`(`+ 1)|`,+1i,

and = ~2

4I2?`(`+ 1).

The energy shifts are therefore

E`,m = ~24I2? `(`+ 1), for m = 1E`,m = 0 else. (11)


#19 : GRADUATE GENERAL/MATH

PROBLEM:

Solve the Laplace equation r2f = 0 for function f(r, ) inside a unit circlethat obeys the boundary conditions

f(1, ) = 1 , 2

2013 fall of vibration mechanics

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