PHY4604–Introduction 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 can’t do one part of a problem, solve subsequent
parts in terms of unknown answer–define clearly. If you don’t know a formula–ask, 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
q
L̂+ ψ`m = h̄ `(` + 1) − m(m + 1)ψ`m+1
q
L̂− ψ`m = h̄ `(` + 1) − m(m − 1)ψ`m−1

En
"
m
e2
= − 2
2h̄ 4π²0
L̂2 =
=
Hψ =
Hψ =
H =
Z ∞
n2
¶
µ
En =
#2 
1

1
h̄ω n +
2
L̂+ L̂− + L̂2z − h̄L̂z
L̂− L̂+ + L̂2z + h̄L̂z
Eψ
∂ψ
ih̄
∂t
h̄2 2
∇ + V (r)
−
2m
dx xn e−x = n!
0
Z ∞
√
2
dx e−x =
π
−∞
√
Z ∞
π
2 −x2
dx x e
=
2
−∞
³
ψ0 (x) = q
1
− 12
π 1/2 x0
e
µ
ψ1 (x) = q
1
2π 1/2 x0
ψ2 (x) = q
1
8π 1/2 x0
1
x
x0
´2
¶
2x − 12
e
x0
à µ
x
4
x0
¶2
³
x
x0
´2
!
³
− 12
−2 e
x
x0
´2
ψ3 (x) = q
à µ
1
48π 1/2 x0
x
8
x0
¶3
−3/2
R10 = 2a0 e−r/a0
µ
¶
1 r
1 −3/2
R20 = √ a0
1−
e−r/2a0
2 a0
2
1 −3/2 r −r/2a0
R21 = √ a0
e
a0
24
Ã
!
∂
∂
±iφ
L± = h̄e
± + i cot θ
∂θ
∂φ
1
Y00 = √
4π
s
3
Y10 =
cos θ
4π
s
3
sin θe±iφ
8π
"
#
0 1
=
1 0
Y1±1 = ∓
σx
"
σy =
"
σz =
~µ =
h̄
me
me e 4
2h̄2
1eV
a0
=
=
0 −i
i 0
#
#
1 0
0 −1
eg
S
−
2m
1 × 10−34 J − s
9 × 10−31 kg
= 1Rydberg = 13.6eV
= 1.6 × 10−19 J
= h̄2 /(mZe2 )
2
!
−1
x
− 12
e 2
x0
³
x
x0
´2
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
Z
ψ(x, t) =
dp
√
eipx/h̄ φ(p, t)
2πh̄
where φ(p, t) is an “envelope” function describing a distribution of
momenta present in the wave function ψ. If φ is sharply peaked around
one value of p, ψ will look like a plane wave and be spread out over
a large region of space. On the other hand, if φ is roughly constant
in p, ψ will look like a delta function, and be strongly peaked at a
particular x (for given t).
• Ehrenfest theorem
Ehrenfest’s theorem says that the motion of expectation values in
quantum mechanics is classical. More specifically,
dhpi
i
i
dV
dhV i
= − h[H, p]i = − h−ih̄ i = −
,
dt
h̄
h̄
dx
dx
where the right hand side is now just the classical force on a particle,
i.e. this is Newton’s law on the average.
• Pauli principle
Pauli’s principle for 2 particles states that the wave function must
be symmetric or antisymmetric under exchange of particle labels, depending on whether the particles are integer (bosons) or half-integer
(fermion) spin, respectively. Mathematically, if 1 represents all the labels, coordinates or quantum numbers associated with particle 1, and
2 is the same for particle 2, we must have ψ(1, 2) = ±ψ(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 = 0
for 0 ≤ x ≤ a. What is the ground state wavefunction for the 2-particle
system if the particles have parallel spins? Label your answer in terms of
the eigenstates φin (x) of particle i (i = 1, 2), and state what these are.
For the infinite
square well problem as stated, the eigenfunctions are
q
φn (x) = 2/a sin nπx/a. The ground state for 2 particles with parallel
spins will be a spin triplet (S=1),
ψ(1, 2) = (φ0 (x1 )φ1 (x2 ) − φ1 (x1 )φ0 (x2 ))χ↑↑
3
since if both particles were in the single-particle ground state φ0 , such that
the energy was 2E0 , antisymmetry of the overall wavefunction would force
the spin wave function to be antisymmetric, χ↑↓−↓↑ . So the best we can do
with two parallel spins 1/2 is to put one particle in φ0 and one in φ1 , then
antisymmetrize. The lowest energy with parallel spins is E0 + E1 .
(c) Given a physical system with Hamiltonian H which has an orthonormal set
of eigenfunctions ψn (x) at time t=0, show that these same eigenfunctions
at a later time t, ψn (x, t) are still orthonormal.
We need to show that (ψn (x, t), ψm (x, t)) = δmn for all times t. But this is
just
(eiHt/h̄ ψn (x), eiHt/h̄ ψm (x))
(eiEn t/h̄ ψn (x), eiEm t/h̄ ψm (x))
ei(Em −En )t/h̄ (ψn (x), ψm (x))
ei(Em −En )t/h̄ δmn = δmn ,
(ψn (x, t), ψm (x, t)) =
=
=
=
where the last step follows because δmn is only nonzero when m = n.
(d) Define the uncertainty in the value of an operator Q̂ as
q
∆Q =
hQ̂2 i − hQ̂i2 .
(1)
Suppose an electron is known to be in an eigenstate of L̂2 and L̂z , i.e.
|ψi = |`mi. Show explicitly that ∆L2 = 0, ∆Lz =0, but ∆Lx 6= 0. Explain.
Calculate
hL2 i = h`m|L2 |`mi = h̄2 `(` + 1)h`m|`mi = h̄2 `(` + 1)
hLz i = h`m|Lz |`mi = h̄m
h(L2 )2 i = h`m|(L2 )2 |`mi = h̄2 `(` + 1)h`m|L2 |`mi = h̄4 `2 (` + 1)2
³√
´
L+ + L−
1
√
hLx i = h`m|
|`mi = h`m|
..|`m + 1i + ...|`m − 1i = 0
2
2
µ
¶2
L
+
L
L− L+ + L+ L−
+
−
h`m|L2x |`mi = h`m|
|`mi = h`m|
|`mi
2
2
h̄2
=
h`m| (`(` + 1) − m(m + 1)) + (`(` + 1) − m(m − 1)) |`mi
2
= h̄2 (`(` + 1) − m2 ),
so there must be an uncertainty in a measurement of Lx if the system is
prepared in an L2 , Lz eigenstate, since [Lx , Lz ] 6= 0.
(e) Compare the wavelengths of the 2p → 1s transitions in hydrogen (one
proton, no neutrons) and deuterium (one proton, one neutron). Give your
answer in terms of the ratio between the two transition wavelengths λH :
4
λD , which depends only on x ≡ me /mp (neglect the proton-neutron mass
difference, and give only the leading term in powers of x).
The energy which must be carried away by a photon in the transition is
E2 − E1 = hc/λ, so
8h̄3
λH = hc/(E2 − E1 ) =
3µ
µ
4π²0
e2
¶2
.
where µ = mmp /(m + mp ) is equal to m up to an error of order O(10−3 ).
For Deuterium we have µ = 2mmp /(m + 2mp ), so
2(m + mp )
x
λH
=
≈1+
λD
m + 2mp
2
(f) What is the degeneracy of the 2nd excited state (E = (7/2)h̄ω) of the
isotropic 3D simple harmonic oscillator?
The energies of the eigenstates of the 3D SHO are given by E = Ex + Ey +
Ez , where Eα = h̄ω(nα + 1/2), α = x, y, z, where nα are positive integers.
So altogether E = h̄ω(nx + ny + nz + 3/2). To get (7/2)h̄ω, we need two
quanta, which can be distributed in any way among x, y, z. So we have
states labelled |nx ny nz i with possibilities |200i, |020i, |002i, |110i, |101i,
|011i, for a degeneracy of 6.
2. Hydrogen. An electron in a H-atom is in a state described by
1
ψ = √ [2ψ100 + ψ211 + ψ21−1 ]
6
(2)
(a) Calculate the expectation value of L̂z in this state.
µ
hLz i =
¶
1
2ψ100 + ψ211 + ψ21−1 , 2(0h̄)ψ100 + (h̄)ψ211 + (−h̄)ψ21−1 = 0
6
(b) What is the probability a measurement of the energy will yield the value
0.25 Rydberg?
Wave function has amplitude √16 (ψ211 + ψ21−1 ) to be in the n = 2 Bohr
orbit 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.
5
(c) Write down the Schrödinger equation for the system in terms of the electron coordinates r1 and r2 , and show that, if you neglect the Coulomb
interaction between the electrons, it separates into two decoupled equations, one for each of the two electrons.
h̄2 2
e2
e2
h̄2 2
e2
H = −
∇1 −
−
∇2 −
+
.
2m
4π²0 r1 2m
4π²0 r2 4π²0 |r1 − r2 |
If we neglect the last term, the Hamiltonian separates into a sum of two
independent Hamiltonians, the 1st of which acts only on the coordinates
of the first particle, the second on those of the second. In such a case we
know that ψ(r1 , r2 ) = ψ1 (r1 )ψ2 (r2 ) and E = E1 + E2 , where each obeys
Hα ψα = Eα ψα , for α = 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 electrons is in a 1s state (be sure to specify the spin state!) Do the same if one
electron 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 wavefunction Ψ(1, 2) is just ψ1s (r1 )ψ1s (r2 ), which is symmetric under exchange
1 ↔ 2. So we have to multiply by an antisymmetric spin state to satisfy
Wolfgang:
Ψ(1, 2)1s1s = ψ1s (r1 )ψ1s (r2 )χ↑↓−↓↑
If the two spins are parallel (spin exchange symmetric), we must put the
electrons into an antisymmetric orbital linear combination of 1s and 2s in
order to preserve overall antisymmetry:
Ψ(1, 2)1s2s = (ψ1s (r1 )ψ2s (r2 ) − ψ1s (r2 )ψ2s (r1 ))χ↑↑
3. Spin. A charge +e, spin- 12 particle with gyromagnetic ratio g is initially in an
eigenstate |ψi of Ŝz corresponding to eigenvalue +h̄/2.
ˆ =
(a) Evaluate the expectation value ~µ of the magnetic moment operator ~µ
(ge/2m)S in this state. In which direction does it point?
h~µi = (ge/2m)hSi.
Of the three components of S, only the z-component has a nonzero expectation values because Sx and Sy can be expressed as ∝ S+ ± S− . Since
hSz i is h̄/2 in the state given, h~µi = (geh̄/4m)ẑ.
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(b) What is the probability of obtaining a value of h̄/2 if a measurement of Ŝx
is made on this state?
If we √
expand | ↑iz = (1, 0) in√ terms of the eigenstates of Sx , | ↑ix√=
(1, 1)/ 2 and | ↓ix = (1, −1)/ 2, we see that | ↑iz = (| ↑ix + | ↓ix )/ 2.
The probability
√ of obtaining a value of h̄/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 ŷ-direction.
Show that the time evolution operator for this system may be expressed
in the form
|ψ(t)i = Û (t)|ψ(0)i,
Û (t) = cos θ + iσy sin θ
(3)
where σy is a Pauli matrix, and find the form of θ(t). (Hint: remember
that the exponential of an operator is to be understood as a Taylor series
in that operator, and that σy2 = 1.)
H = −~µ · B = −g(e/2m)B0 Sy = −gµ0 B0 σy /2
U (t) = e−iHt/h̄ = exp[−igµ0 B0 σy /2]
with µ0 = eh̄/(2m). The exponential can be expanded eiX = 1 + iX +
(iX)2 /2! + . . ., and we see that every term will be proportional to an even
or odd power of σy . Even powers give σyn = 1, while odd powers therefore
give σyn = σy . The sum of all the even terms therefore gives cos gµ0 B0 t/2h̄,
and the sum of all the odd terms gives iσy sin gµ0 B0 t/2h̄. So defining
θ = gµ0 B0 t/2h̄, we find the desired result.
(d) Determine the precession period T (define what you mean by period!) and
find the form of the state |ψ(t)i after a time t = T /4. In which direction
does the magnetic moment µ
~ point now?
Let’s define, as in class, the period as the time when the time evolution operator takes a state into minus itself, because it is then again an eigenstate.
Here we start with an eigenstate ψ of Sz with eigenvalue h̄/2, and wait
until it evolves into −ψ, which is again an eigenstate of Sz . This occurs
when U = −1, i.e. when θ = π, or T = 2πh̄/(gµ0 B0 ). When t = T /4, we
have U = cos π/4 + i sin π/4σy . So
|ψ(T /4)i = (cos π/4 + sin(π/4)σy )|ψ(0)i
Ã"
#
"
#! "
#
1
1 0
0 −i
1
= √
+i
0 1
i 0
0
2
"
#
1
1
1
= √ (| ↑iz − | ↓iz )
= √
2 −1
2
which is an eigenstate of Sx with eigenvalue −h̄/2. So hψ(T /4)|S|ψ(T /4)i ∝
−̂x.
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4. Simple harmonic oscillator.
(a) Write down the Schrödinger equation for a 1D simple harmonic oscillator,
i.e. a mass m particle oscillating with classical angular frequency ω. The
ground state ψ0 is given on the first page of the exam; it has energy eigenvalue h̄ω/2. Show explicitly that ψ0 is an eigenstate and find the natural
length scale x0 in terms of m, ω, and h̄ (show your work!).
Hψ0
− 12
h̄2 ∂ 2
1
1
2 2
q
e
= (−
+
mω
x
)
2m ∂x2 2
π 1/2 x0
³
x
x0
´2
=
h̄ω
ψ0
2
Calculate:
∂ − 12
e
∂x
∂ 2 − 12
e
∂x2
³
x
x0
³
x
x0
´2
´2
x −1
= − 2e 2
x0
Ã
=
³
x
x0
´2
!
−1
x2
1
− 2 e 2
4
x0 x0
³
x
x0
´2
,
so we get a solution if
h̄2
−
2m
Ã
x2
1
− 2
4
x0 x0
!
h̄ω
1
+ mω 2 x2 =
2
2
h̄
⇒ x20 =
mω
(b) Calculate the expectation value of the kinetic energy p̂2 /2m in the ground
state ψ0 (n=0).
2
2
h̄ ∂
p2 = − 2m
, so
∂x2
h̄2
∂2
hψ0 | 2 |ψ0 i
2m
∂x
Ã
! ³ ´2
2
− xx
x2
1 Z
h̄
1
0
dx
= −
−
e
2m π 1/2 x0
x40 x20
Ã
!
√
h̄2
1
1
1 3√
= −
(x π/2) − 2 (x0 π)
2m π 1/2 x0 x40 0
x0
hp2 /2mi = −
h̄2 1
h̄ω
.
=
=
2
2m 2x0
4
Note this is consistent with classical virial theorem, where the energy is
shared equally between kinetic and potential energy !
8
(c) Suppose the harmonic oscillator is in its ground state at time t = 0 when
the particle absorbs a particle of mass 3m (neglect any momentum transfer), thus instantaneously changing its mass to 4m. What is the probability
it 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 square
ter
ore
root of the mass of the oscillator, xaf
= xbef
/2. Probability it remains
0
0
in ground state is then |hψ0bef ore |ψ0af ter i|2 . The inner product is
µ
1
hψ0bef ore |ψ0af ter i = q
bef ore af ter
πx0
x0
=
v
u
u
t
2
ore 2
π(xbef
)
0
s
Z
dx e
2
− x2
¶
1
1
bef ore 2 + af ter 2
(x
)
(x
)
0
0
2π bef ore
2
x0
=√ ,
5
5
so probability is 4/5 to stay in ground state.
(d) Consider the two eigenstates of the 3D simple harmonic oscillator,
1
; ψb = √ [ψ1 (x)ψ0 (y)ψ0 (z) + iψ0 (x)ψ1 (y)ψ0 (z)]
2
(4)
Give the energy corresponding to each eigenvector, and specify the total
orbital angular momentum quantum number ` and its z-component m for
each.
ψa = ψ0 (x)ψ0 (y)ψ0 (z)
Use again theorem which says if H is sum of independent H’s, E = Ex +
Ey + Ez . So Ea = 3h̄ω/2 (ground state) and Eb = 5h̄ω/2. ψa (x, y, z) ∝
exp −r2 , spherically symmetric so ` = 0, m = 0. ψb is ∝ (x + iy) exp −r2 ,
but this is ∝ Y11 , so ` = 1, m = 1.
9