PHY4605–Introduction to Quantum Mechanics II
Spring 2005
Final exam SOLUTIONS
April 22, 2005
1. Short Answer.
(a) Write the expression hψ|ψi = 1 as an explicit integral equation in three
dimensions, assuming
that |ψi represents a wave function ψ(~r). Suppose
P
you have |ψi = n cn |ni where the |ni are a complete set of orthonormal
states. What conditions does the above equation impose on the cn ?
Z
hψ|ψi = d3 r|ψ(r)|2 = 1
X
|cn |2 = 1
⇒
n
(b) How many degenerate levels are there for a hydrogen atom with principal
quantum number n? Are any of these degeneracies lifted by the spin-orbit
interaction? Justify your answer.
degeneracy =
n
X
−1(2` + 1) = n2 ,
`=0
and if we count spin the total degeneracy would be 2n2 . Spin orbit coupling
interaction is proportional to
1
L · S = (J 2 − S 2 − L2 )
2
where J = L + S. So a given n` level is split into different j’s. By the
rules of addition of angular momenta, since s = 1/2, j = ` ± 1/2, j ≥ 0.
So for example in the n = 3 level, the possible j values are +1/2(` = 0),
1/2, 3/2(` = 1), and 3/2, 5/2(` = 2). The degeneracy of each spin-orbit
split level is 2j + 1, so the total degeneracy is
2 + 2 + 4 + 4 + 6 = 18 = 2 · 32
OK
(c) What is the big difference between a particle with spin quantum number
s = 1/2 and one with s = 1?
One behaves as a fermion, one as a boson, respectively. This means in
the presence of a 2nd particle of identical type in each case, the two s =
1/2 particles have to have a wave function which is antisymmetric under
exchange of all particle coordinates, while two s = 1 particles would have
to have a symmetric wave function.
1
(d) Suppose that the operator corresponding to some observable is called Q.
List 2 properties of this operator and/or of its eigenfunctions |ni. The
latter satisfy Q|ni = qn |ni. SupposePfurther that the quantum-mechanical
state of a system is given by |ψi = n cn |ni with several of the cn 6= 0. If
you were to make a single measurement of the observable Q, what would
you get as a result?
Q must be hermitian, which means its eigenvalues must be real. Eigenfunctions corresponding to distinct eigenvalues must be orthogonal. If |ψi
is a mixture of different Q eigenstates, a measurement of Q yields qn with
probability |cn |2 .
(e) Two quantum mechanical particles have orbital angular momentum l = 1
and spin angular momentum s = 0. Suppose that there is some coupling
of the two particles. List the values that the total angular momentum j
of the two-particle system may take on. For each j, state what are the
possible values for the z component.
|` + s| ≤ j ≤ ` + s
so j = `, and Jz has eigenvalues h̄mj , where mj can take on any values
between −` and ` separated by 1.
2. Skewed square well.
Consider an infinite well for which the
bottom is not flat, as sketched here. If
the slope is small, the potential V =
²|x|/a may be considered as a perturbation on the square-well potential
over −a/2 ≤ x ≤ a/2.
(a) Calculate the ground-state energy, correct to first order in perturbation
theory.
q
(0)
Ground state of box of size a: ψ0 = a2 cos πx
.
a
(0)
h̄2 k02
2m
=
h̄2 π 2
.
2ma2
(0) ²|x|
(0)
hψ0 |
|ψ0 i
2²
=
aa
Ground state energy: E0 =
1st order correction
(1)
δE0
=
a
Z
a/2
−a/2
π2 − 4
=²
4π 2
independent of a!
2
dx|x| cos2
πx
a
(b) Calculate the energy of the first excited state, correct to first order in
perturbation theory.
q
(0)
1st excited state of box of size a: ψ0 = a2 sin 2πx
.
a
(0)
Ground state energy: E1 =
h̄2 k12
2m
=
2h̄2 π 2
.
ma2
1st order correction
(1)
δE1
=
(0) ²|x|
(0)
hψ1 |
|ψ1 i
a
2²
=
aa
²
=
4
Z
a/2
−a/2
dx|x| sin2
2πx
a
(c) Calculate the wave function in the ground state, correct to first order in
perturbation theory. (You will probably not be able to get an answer in
closed form, but take the calculation as far as you can–do not evaluate
integrals you encounter here).
ψ0 =
(0)
ψ0
X hm| ²|x| |0i|mi
a
+ δψ0 , δψ0 =
E
− Em
0
m>0
(0)
• Note if m is odd, ψm is odd parity
p ⇒ hm||x||0i = 0.
• so only m’s contribute, |mi = 2/a cos (m + 1)πx/a
Z
4² a/2
πx
(m + 1)πx
²|x|
|0i = 2
dxx cos
cos
⇒ hm|
a
a 0
a
a
µ 2
¶
(m + 2m + 2) − 2(1 + m) cos mπ
2
= −4²
m2 (m + 2)2 π 2
(I didn’t expect you to do these integrals, but on the other hand no
one asked for Gradsteyn-Rhyzhik either.)
• Denominator:
£
¤
2
E0 − Em = h̄2 (k02 − km
)/2m = (h̄2 /2m) (π/a)2 − ((n + 1)π/a)2
• All terms in infinite series for 1st order correction are now specified,
and they can be summed up to some order with error which is easy to
estimate.
(d) At what value of ² does perturbation theory break down? Justify your
answer.
Difference in energy of 1st excited and ground states in zeroth order is
3h̄2 π 2 /(2ma2 ). If the energy shifts in perturbation theory become comparable,
one will get level crossings and must distrust perturbation theory. Comparing,
you find this happens when ² ' h̄2 /(ma2 ).
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3. p-n scattering
The scattering amplitude for p − n scattering is modelled by
f (θ) = χ†f (A + Bσp · σn )χi
(1)
where χf and χi are the final and initial spin states of the n − p system. Possibilities are:
χi , χf = | ↑↑i, | ↑↓i, | ↓↑i, | ↓↓i
(2)
where the first spin refers to the proton and the second to the neutron. Note the
wave functions do not have to be antisymmetrized since p, n distinguishable.
Using
p n
n p
σp · σn = σzp σzn + 2(σ+
σ− + σ+
σ− ),
where

σ+ =

σx + iσy  0 1 
=

2
0 0
(3)

; σ− =

σx − iσy  0 0 
=

2
1 0
(4)
Calculate and tabulate all 16 scattering amplitudes. Make a table of results.
We’ll need the single particle spinor equations
σ+ χ↑ = σ− χ↓ = 0
(5)
σ+ χ ↓= χ↑ , σ− χ↑ = χ↓
(6)
So for 2 spin states∗ |mp mn i ≡ χpmp χnmn
¡
£
¤¢
σp · σn | ↑↑i = σpz σnz + 2 σp+ σn− + σp− σn+ | ↑↑i = | ↑↑i
(7)
σp · σn | ↑↓i = −| ↑↓i + 2| ↓↑i
(8)
σp · σn | ↓↑i = −| ↓↑i + 2| ↑↓i
(9)
σp · σn | ↓↓i = | ↓↓i
(10)
similarly,
∗ written
as χmp0 in text of problem– same thing
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Define V̂ = A + Bσp · σn
V̂ | ↑↑i = A + B| ↑↑i
(11)
V̂ | ↑↓i = (A − B)| ↑↓i + 2B| ↓↑i
(12)
V̂ | ↓↑i = (A − B)| ↓↑i + 2B| ↑↓i
(13)
V̂ | ↓↓i = (A + B)| ↓↓i
(14)
So scattering amplitutde f is a tensor in spin space f(mp mn ;mp mn ) . Allowed values
final
of f :
↑↑
↑↑ A + B
↑↓
↓↑
↓↓
0
0
0
↑↓
0
A−B
2B
0
↓↑
0
2B
A−B
0
↓↓
0
0
0
A+B
initial
For table of cross-sections, take absolute value to the square of all elements.
σ↑↓→↑↓ = |A − B|2
(15)
4. Selection rules (10 pts.) Recall that the selection rule for electric
dipole transitions in hydrogenic atoms, i.e.those with nonzero matrix elements
hφn`m |d|φn0 `0 m0 i, (d = er) are that `0 must be ` ± 1 and that m0 must be m ± 1
(or m).
(a) A hydrogen atom is initially in an n = 4,` = 2 excited state. It decays to
the ground state by a sequence of allowed dipole transitions. How many
steps are required? Write a possible sequence or two.
Since ∆` = 2, there are two steps required, e.g. 42m → 21m0 → 100,
42m → 31m0 → 100.
(b) (b) A hydrogen atom initially in an n = 4, ` = 2 state decays directly
to its ground state. Show that as a dipole transition this is forbidden,
but it is possible as an electric quadrupole transition, with a perturbation
proportional to Qij = e(3ri rj − δij r2 ).
The probability of an electric dipole transition directly to the ground state
from 42m is zero because h42m|r|100i = 0, formally. The “r” comes from
the perturbation to the energy of an atom in the presence of an electric
field in the dipole approximation, V ∝ q² · r, where ² is a unit vector
describing the polarization of the photon. Clearly for a photon with finite
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wavelength coming in at some angle, the electric field can distort the atom
in other ways, and one can expand the distortion in spherical harmonics,
of which the dipole approx. is only the lowest. See, e.g., J.D. Jackson,
Electrodynamics, for a discussion of the multipole expansion. For now, you
don’t need to worry about this; the question is only whether the operator
given can cause transitions from 42m and 100, i.e., can
h42m|Qij |100i 6= 0
for some i, j? The Rr2 term doesn’t help, since the angular part of the
∗
integral will involve Y2m
Y10 dΩ, which is zero just by the orthonormality
of the spherical harmonics. Let’s take the simplest case of i, j = z, z and
consider the first term (with m=0)
Z
Z
2
h420|z |100i ∝ dΩ
Y20∗ cos2 θ Y10 dΩ
Z π
¢
1¡
∝
sin θ dθ
3 cos2 θ − 1 · cos2 θ · 1
2
0
4
= ,
15
so yes, the quadrupole operator can cause transitions between the two
states in question.
.
5. Exchange Symmetry (10 pts.). Suppose you have three particles, one in
state ψa , one in ψb , and one in ψc . Construct a wave function for the three
particles if they are
(a) distinguishable
If they are distinguishable we don’t have to worry about spin-statistics
theorem, so we could write
ψ(r1 , r2 , r3 ) = ψa (r1 )ψb (r2 )ψc (r3 )χ1 χ2 χ3 ,
where the χ’s are the spin states of the 3 particles. Note there is no
particular symmetry under interchange of 1 ↔ 2, etc.
(b) fermions
For fermions we have to antisymmetrize:
ψ(r1 , r2 , r3 ) = ψa (r1 )ψb (r2 )ψc (r3 )χ1 χ2 χ3 + ψb (r1 )ψc (r2 )ψa (r3 )χ1 χ2 χ3
+ψc (r1 )ψa (r2 )ψb (r3 )χ1 χ2 χ3 − ψb (r1 )ψa (r2 )ψc (r3 )χ1 χ2 χ3 −
ψa (r1 )ψc (r2 )ψb (r3 )χ1 χ2 χ3 − ψc (r1 )ψb (r2 )ψa (r3 )χ1 χ2 χ3 ,
where I’ve assumed no spin-orbit coupling.
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(c) bosons For bosons we have to symmetrize:
ψ(r1 , r2 , r3 ) = ψa (r1 )ψb (r2 )ψc (r3 )χ1 χ2 χ3 + ψb (r1 )ψc (r2 )ψa (r3 )χ1 χ2 χ3
+ψc (r1 )ψa (r2 )ψb (r3 )χ1 χ2 χ3 + ψb (r1 )ψa (r2 )ψc (r3 )χ1 χ2 χ3 +
ψa (r1 )ψc (r2 )ψb (r3 )χ1 χ2 χ3 + ψc (r1 )ψb (r2 )ψa (r3 )χ1 χ2 χ3 ,
6. Variational principle (5 pts.)Use the variational principle to estimate the
ground state energy of the anharmonic oscillator, H = p̂2 /2m + λx4 , and compare with exact result E0 = 1.06λ1/3 (h̄2 /2m)2/3 .
Let’s try an Gaussian trial wave function, φ = A exp(−λ2 x2 ), with λ as variational parameter. To normalize, we take
r
Z
π −1
2
−2η 2 x2
2
hφ|φi = A
dx e
=A
η =1
2
so A = η 1/2 (2/π)1/4 . The expectation value of the Hamiltonian is then
µ
¶
Z
h̄2 d2
2 2
2 −η 2 x2
4
hφ|H|φi = A e
−
+ λx e−η x dx
2
2m dx
¶
Z µ 2
h̄
2 2
2
4 2
2
4
=A
− (2η x − η ) + λx e−2η x dx
m
·µ 2 ¶ µ
¶
¸
r
r
r
−h̄
π
π
π
3
2
4 1
2
=A
2η ·
−η
+λ
m
2 (2η 2 )3
2η 2
4 (2η 2 )5
r
· 2r
¸
1 h̄
π
π 3λ
= A2
η+
2m 2
2 16η 5
2
3λ
h̄ 2
η +
=
2m
16η 4
q
So minimizing dhHi/dη = 0, we find η = 6 3λm/4h̄2 , and value of hHi at
minimum is
µ ¶4/3
3
hHimin =
λ1/3 (h̄2 /2m)2/3 = 0.68λ1/3 (h̄2 /2m)2/3 ,
4
which is (oops!) smaller than the result I quoted for the exact result, which is
of course impossible. Looks like I got the wrong exact result somewhere.
7. Aharonov-Bohm phase shift (5 pts.) A single electron passes through a
large capacitor of charge Q and capacitance C.Its velocity is v (assume unaltered
by capacitor’s field) and the capacitor has length L. Estimate the change of
phase of the electron’s wave function.
7
From notes, the result of varying potential will be varying phase of electronic
wave function:
Z t
−iS/h̄
ψ(x, t) = ψ0 (x, t)e
, S=
V (t0 )dt0
(16)
0
where ψ0 is wave fctn. in absence of battery. If the electron passes through a
large capacitor with a small gap, we can neglect the fringing fields, and approximate the potential difference between the capacitor plates as usual as V = Q/C,
then the electron experiences a potential which is zero except for a time t = L/v,
when it is Q/C. The phase shift is then S = LQ/(vC).
8. Stern-Gerlach (10 pts.). A Stern-Gerlach apparatus is aligned along the z
direction, and a second one is aligned at 45◦ with respect to this in the z − x
plane. A neutral spin-1/2 particle is prepared with spin k ẑ and then passed
through both in succession.
(a) What is the probability a detector located after the two SG setups finds
the particle to have spin up (k ẑ)?
¡ ¢
The particle passes through the 1st detector with wave function 10 intact.√The second
the eigenfunctions of the operator
√ detector projects onto
◦
Sx / 2 + Sz / 2, i.e. rotated by 45 from z. This operator is represented
in spin space by


1
h̄ 1  1 1 
S 0 = √ (Sx + Sz ) = √ 

2 2 1 −1
2
|ψ↑0 i
= √
1
which has normalized eigenvectors
√
2(2+ 2)
³ √ ´
¡
¢
1− 2
√ 1√
. The state 10 can be decomposed
1
³
√ ´
1+ 2
1
and |ψ↓0 i =
2(2− 2)
Ã
√ !
√ !
µ ¶ Ãp
1
2+ 2
1− 2
|ψ↓0 i
|ψ↑0 i + √
=
0
2
4 − 22
so
detector with probability (2 +
√ the particle will pass through the second
0
2)/4 =85%. The state will then be |ψ↑ i, and we want to measure with
a third detector along z. From the
of this state,
µ definition
¶2 we see that
√ q
√
¡1¢
the probability of measuring 0 is 1 + 2/ 2(2 + 2) =85%. Total
probability is therefore 73%.
(b) What is the probability a detector measures the spin to be along the ŷdirection?
8
¡ ¢ √
The up eigenvector of Sy in the original coordinate system is −i
/ 2.
1
Therefore if we measure after it makes it through the 2nd detector in
state |ψ↑0 i we take the inner product of the two states squared to find
√
√
(4 + 2 2)/(4(2 + 2)) = 50%.
The initial spin state is now changed to definitely k x̂.
(c) Redo question a)
By symmetry must be the same.
9