Final So Lns Intro Quantum

7/29/2019 Final So Lns Intro Quantum

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PHY4604Introduction to Quantum Mechanics

Fall 2004

Final Exam SOLUTIONS

December 17, 2004, 7:30 a.m.- 9:30 a.m.

No other materials allowed. If you cant do one part of a problem, solve subsequent

parts in terms of unknown answerdefine clearly. If you dont know a formulaask, I

might be able to help. All parts 10 pts., max=120. Problem 1 required, attempt 2 of

remaining 3 problems; circle which ones you want graded.

Possibly helpful formulae and constants

L+m = h

( + 1) m(m + 1)m+1Lm = h

( + 1) m(m 1)m1

En = m

2h2

e2

40

2 1n2

En = h

n +1

2

L2 = L+L + L2z hLz

= LL+ + L2z + hLz

H = EH = ih

t

H = h2

2m2 + V(r)

0dx xnex = n!

dx ex

2

=

dx x2ex2

=

2

0(x) =1

1/2x0e 1

2

xx0

2

1(x) =1

21/2x0

2x

x0

e 1

2

xx0

2

2(x) =1

81/2x0

4

x

x0

2 2

e 1

2

xx0

2

1


2/9

3(x) =1

481/2x0

8

x

x0

3 12 x

x0

e 1

2

xx0

2

R10 = 2a3/20 e

r/a0

R20 = 12

a3/20

1 12 ra0

er/2a0

R21 =124

a3/20

r

a0er/2a0

L = hei

+ i cot

Y00 =14

Y01 =

3

4cos

Y11 =

3

8sin ei

x =

0 11 0

y =

0 ii 0

z =

1 00 1

= eg

2m S

h = 1 1034J sme = 9 1031kg

mee4

2h2= 1Rydberg = 13.6eV

1eV = 1.6 1019Ja0 = h

2/(mZe2)

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1. Short Answer. Must attempt (only) 4 of 6. Circle answers to be graded.

(a) Identify and discuss:

wave packet

A wave packet is a wave function describing a propagating particle

localized in space. In general, we can write

(x, t) =

dp2h

eipx/h(p, t)

where (p, t) is an envelope function describing a distribution ofmomenta present in the wave function . If is sharply peaked aroundone value of p, will look like a plane wave and be spread out overa large region of space. On the other hand, if is roughly constantin p, will look like a delta function, and be strongly peaked at aparticular x (for given t).

Ehrenfest theorem

Ehrenfests theorem says that the motion of expectation values inquantum mechanics is classical. More specifically,

dpdt

= ih[H, p] = i

hihdV

dx = dV

dx,

where the right hand side is now just the classical force on a particle,i.e. this is Newtons law on the average.

Pauli principlePaulis principle for 2 particles states that the wave function mustbe symmetric or antisymmetric under exchange of particle labels, de-pending on whether the particles are integer (bosons) or half-integer(fermion) spin, respectively. Mathematically, if 1 represents all the la-bels, coordinates or quantum numbers associated with particle 1, and2 is the same for particle 2, we must have (1, 2) = (2, 1), where+ is for bosons and - for fermions (e.g. electrons).

(b) Consider two electrons in a potential well with V = except for V = 0for 0 x a. What is the ground state wavefunction for the 2-particlesystem if the particles have parallel spins? Label your answer in terms ofthe eigenstates in(x) of particle i (i = 1, 2), and state what these are.

For the infinite square well problem as stated, the eigenfunctions are

n(x) =

2/a sin nx/a. The ground state for 2 particles with parallel

spins will be a spin triplet (S=1),

(1, 2) = (0(x1)1(x2)

1(x1)0(x2))

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since if both particles were in the single-particle ground state 0, such thatthe energy was 2E0, antisymmetry of the overall wavefunction would forcethe spin wave function to be antisymmetric, . So the best we can dowith two parallel spins 1/2 is to put one particle in 0 and one in 1, thenantisymmetrize. The lowest energy with parallel spins is E0 + E1.

(c) Given a physical system with Hamiltonian H which has an orthonormal setof eigenfunctions n(x) at time t=0, show that these same eigenfunctionsat a later time t, n(x, t) are still orthonormal.

We need to show that (n(x, t), m(x, t)) = mn for all times t. But this isjust

(n(x, t), m(x, t)) = (eiHt/hn(x), e

iHt/hm(x))

= (eiEnt/hn(x), eiEmt/hm(x))

= ei(EmEn)t/h(n(x), m(x))

= ei(EmEn)t/hmn = mn,

where the last step follows because mn is only nonzero when m = n.

(d) Define the uncertainty in the value of an operator Q as

Q =

Q2 Q2. (1)

Suppose an electron is known to be in an eigenstate of L2 and Lz, i.e.

|

=

|m

. Show explicitly that L2 = 0, Lz=0, but Lx

= 0. Explain.

Calculate

L2 = m|L2|m = h2( + 1)m|m = h2( + 1)Lz = m|Lz|m = hm

(L2)2 = m|(L2)2|m = h2( + 1)m|L2|m = h42( + 1)2

Lx = m|L+ + L2

|m = 12m|

..|m + 1 + ...|m 1

= 0

m

|L2x

|m

=

m

|L+ + L

22

|m

=

m

|

LL+ + L+L

2 |m

=

h2

2m| (( + 1) m(m + 1)) + (( + 1) m(m 1)) |m

= h2(( + 1) m2),so there must be an uncertainty in a measurement of Lx if the system isprepared in an L2, Lz eigenstate, since [Lx, Lz] = 0.

(e) Compare the wavelengths of the 2p 1s transitions in hydrogen (oneproton, no neutrons) and deuterium (one proton, one neutron). Give youranswer in terms of the ratio between the two transition wavelengths H :

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D, which depends only on x me/mp (neglect the proton-neutron massdifference, and give only the leading term in powers of x).

The energy which must be carried away by a photon in the transition isE2 E1 = hc/, so

H = hc/(E2 E1) = 8h33

40

e2

2 .

where = mmp/(m + mp) is equal to m up to an error of order O(103).For Deuterium we have = 2mmp/(m + 2mp), so

HD

=2(m + mp)

m + 2mp 1 + x

2

(f) What is the degeneracy of the 2nd excited state (E = (7/2)h) of theisotropic 3D simple harmonic oscillator?

The energies of the eigenstates of the 3D SHO are given by E = Ex + Ey +Ez, where E = h(n + 1/2), = x,y,z, where n are positive integers.So altogether E = h(nx + ny + nz + 3/2). To get (7/2)h, we need twoquanta, which can be distributed in any way among x,y,z. So we havestates labelled |nxnynz with possibilities |200, |020, |002, |110, |101,|011, for a degeneracy of 6.

2. Hydrogen. An electron in a H-atom is in a state described by

=1

6[2100 + 211 + 211] (2)

(a) Calculate the expectation value of Lz in this state.

Lz = 16

2100 + 211 + 211, 2(0h)100 + (h)211 + (h)211

= 0

(b) What is the probability a measurement of the energy will yield the value0.25 Rydberg?

Wave function has amplitude 16

(211 + 211) to be in the n = 2 Bohrorbit with energy -0.25 Ryd. Probability is therefore 1/6+1/6 = 1/3.

An additional electron is now added to the H-atom, forming an H ion.

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(c) Write down the Schrodinger equation for the system in terms of the elec-tron coordinates r1 and r2, and show that, if you neglect the Coulombinteraction between the electrons, it separates into two decoupled equa-tions, one for each of the two electrons.

H = h22m

21 e240r1 h22m

22 e240r2 +e2

40|r1 r2| .

If we neglect the last term, the Hamiltonian separates into a sum of twoindependent Hamiltonians, the 1st of which acts only on the coordinatesof the first particle, the second on those of the second. In such a case weknow that (r1, r2) = 1(r1)2(r2) and E = E1 + E2, where each obeysH = E, for = 1, 2.

(d) Continuing to neglect the Coulomb interaction between the 2 electrons,write down a valid 2-electron wavefunction assuming that each of the elec-trons is in a 1s state (be sure to specify the spin state!) Do the same if oneelectron is in a 1s state, one is in a 2s state, and the two spins are parallel.

If both electrons are in 1s states, the orbital part of the 2-electron wave-function (1, 2) is just 1s(r1)1s(r2), which is symmetric under exchange1 2. So we have to multiply by an antisymmetric spin state to satisfyWolfgang:

(1, 2)1s1s = 1s(r1)1s(r2)

If the two spins are parallel (spin exchange symmetric), we must put theelectrons into an antisymmetric orbital linear combination of 1s and 2s inorder to preserve overall antisymmetry:

(1, 2)1s2s = (1s(r1)2s(r2) 1s(r2)2s(r1))

3. Spin. A charge +e, spin- 12

particle with gyromagnetic ratio g is initially in an

eigenstate

|

of Sz corresponding to eigenvalue +h/2.

(a) Evaluate the expectation value of the magnetic moment operator =(ge/2m)S in this state. In which direction does it point?

= (ge/2m)S.Of the three components of S, only the z-component has a nonzero expec-tation values because Sx and Sy can be expressed as S+ S. SinceSz is h/2 in the state given, = (geh/4m)z.

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(b) What is the probability of obtaining a value of h/2 if a measurement of Sxis made on this state?

If we expand | z = (1, 0) in terms of the eigenstates of Sx, | x =(1, 1)/

2 and | x = (1, 1)/

2, we see that | z = (| x + | x)/

2.

The probability of obtaining a value of h/2 in a measurement of Sx is

therefore (1/2)2 = 1/2.(c) At t = 0 a homogeneous magnetic field B0 is applied in the y-direction.

Show that the time evolution operator for this system may be expressedin the form

|(t) = U(t)|(0), U(t) = cos + iy sin (3)where y is a Pauli matrix, and find the form of (t). (Hint: rememberthat the exponential of an operator is to be understood as a Taylor seriesin that operator, and that 2y = 1.)

H = B = g(e/2m)B0Sy = g0B0y/2U(t) = eiHt/h = exp[ig0B0y/2]

with 0 = eh/(2m). The exponential can be expanded eiX = 1 + iX +

(iX)2/2! + . . ., and we see that every term will be proportional to an evenor odd power of y. Even powers give

ny = 1, while odd powers therefore

give ny = y. The sum of all the even terms therefore gives cos g0B0t/2h,and the sum of all the odd terms gives iy sin g0B0t/2h. So defining = g0B0t/2h, we find the desired result.

(d) Determine the precession period T (define what you mean by period!) andfind the form of the state |(t) after a time t = T/4. In which directiondoes the magnetic moment point now?

Lets define, as in class, the period as the time when the time evolution op-erator takes a state into minus itself, because it is then again an eigenstate.Here we start with an eigenstate of Sz with eigenvalue h/2, and waituntil it evolves into , which is again an eigenstate of Sz. This occurswhen U =

1, i.e. when = , or T = 2h/(g0B0). When t = T/4, we

have U = cos /4 + i sin /4y. So

|(T/4) = (cos /4 + sin(/4)y)|(0)=

12

1 00 1

+ i

0 ii 0

10

=1

2

1

1

=1

2(| z | z)

which is an eigenstate ofSx with eigenvalue h/2. So (T /4)|S|(T /4) x.

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4. Simple harmonic oscillator.

(a) Write down the Schrodinger equation for a 1D simple harmonic oscillator,i.e. a mass m particle oscillating with classical angular frequency . Theground state 0 is given on the first page of the exam; it has energy eigen-value h/2. Show explicitly that 0 is an eigenstate and find the natural

length scale x0 in terms of m, , and h (show your work!).

H0 = ( h2

2m

2

x2+

1

2m2x2)

11/2x0

e 1

2

xx0

2=

h

20

Calculate:

xe 1

2

xx0

2= x

x20e 1

2

xx0

2

2

x2e 1

2

xx0

2=

x2

x40 1

x20

e 1

2

xx0

2,

so we get a solution if

h2

2m

x2

x40 1

x20

+

1

2m2x2 =

h

2

x20 =h

m

(b) Calculate the expectation value of the kinetic energy p2/2m in the groundstate 0 (n=0).

p2 = h22m

2

x2, so

p2/2m = h2

2m0|

2

x2|0

= h2

2m

1

1/2x0

dxx2

x40 1

x20

e

xx02

= h2

2m

1

1/2x0

1

x40(x30

/2) 1

x20(x0

)

=h2

2m

1

2x20=

h

4.

Note this is consistent with classical virial theorem, where the energy isshared equally between kinetic and potential energy !

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(c) Suppose the harmonic oscillator is in its ground state at time t = 0 whenthe particle absorbs a particle of mass 3m (neglect any momentum trans-fer), thus instantaneously changing its mass to 4m. What is the probabilityit remains in its ground state?

Just like the Helium problem we did in class, the only thing that changes

in the eigenfunctions is the x0, which is proportional to the inverse squareroot of the mass of the oscillator, xafter0 = x

before0 /2. Probability it remains

in ground state is then |before0 |after0 |2. The inner product is

before0 |after0 =1

xbefore0 xafter0

dx e

x22

1

(xbefore0

)2+ 1

(xafter0

)2

=

2(xbefore0 )

2

2

5xbefore0 =

25

,

so probability is 4/5 to stay in ground state.

(d) Consider the two eigenstates of the 3D simple harmonic oscillator,

a = 0(x)0(y)0(z) ; b =1

2[1(x)0(y)0(z) + i0(x)1(y)0(z)]

(4)Give the energy corresponding to each eigenvector, and specify the totalorbital angular momentum quantum number and its z-component m foreach.

Use again theorem which says if H is sum of independent Hs, E = Ex +Ey + Ez. So Ea = 3h/2 (ground state) and Eb = 5h/2. a(x,y,z) exp r2, spherically symmetric so = 0, m = 0. b is (x + iy)exp r2,but this is Y11, so = 1, m = 1.

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Final So Lns Intro Quantum