University of California at San Diego – Department of Physics – Prof. John McGreevy TA:
Shauna Kravec
Quantum Mechanics (Physics 212A) Fall 2015
Assignment 5 –
Solutions
Due 12:30pm Wednesday, November 4, 2015
1. Electrons and Stern-Gerlach. Show that it’s impossible to measure the electron
magnetic moment with a Stern-Gerlach apparatus.
Hints: (1) The electron is a charged particle. (2) This problem is intended as an
application of the uncertainty principle.
The issue is going to be that electrons in a magnetic field will feel a Lorentz force
proportional to their velocity v.
Consider a SG device where the magnetic field is in the xz-plane and the electrons
move in the y-direction.
They feel a force due to their magnetic moment: Fspin = µB (∂z Bz )ẑ which causes the
splitting. We wish to measure the amount of splitting to determine µB
Electrons not perfectly in the yz-plane will experience a Lorentz force in the z-direction
Bx ẑ
due to Bx . FL = q vc Bx ẑ with a variance ∆FL = q ∆v
c
For a small ∆x away from the yz-plane we have Bx ≈ ∂x Bx ∆x
(∂z Bz )∆x
But ∇ · B = 0 so ∂x Bx = −∂z Bz implying |∆FL | = q ∆v
c
So we want |Fspin | |∆FL | which implies µB q ∆v
∆x but for the electron this would
c
~
mean ∆v∆x 2 which runs afoul with the uncertainty relation
2. Galilean invariance of the Schrödinger equation.
Show that the free-particle Schrödinger equation maps to itself under the Galilean
boost acting by
~x → ~x0 ≡ ~x + ~v t, t → t0 ≡ t, ψ → ψ 0
where the boosted position-space wavefunction is (prime is not derivative here)
0 0
ψ 0 (x0 , t0 ) ≡ eiϕ(x ,t ) ψ(x, t)
and
ϕ(x, t) ≡ av 2 t + b~v · ~x.
(You must determine a and b.)
2
∇
ψ(x, t) = i∂t ψ(x, t)
The free particle equation is − 2m
Let’s check how the derivatives transform:
1
∂t0 =
∂~
x
∂t0
· ∂~x +
∂t
∂
∂t0 t
= −~v · ∇ + ∂t and ∂x0 =
∂x
∂
∂x0 x
+
∂t
∂
∂x0 t
= ∂x
2
∇
ψ 0 (x, t) = i(−~v · ∇ + ∂t )ψ 0 (x, t)
So the equation transforms to − 2m
But ψ 0 = ei(av
2 t+b~
v ·~
x)
ψ which we can plug in above
∇2
∇
·(ib~v ψ 0 +ei(av
− 2m ψ 0 (x, t) = − 2m
−i~v · (∇ψ 0 ) = bv 2 ψ 0 − ie
i∂t ψ 0 = −av 2 ψ 0 + ie
2 t+b~
v ·~
x)
i(av 2 t+b~v ·~
x)
i(av 2 t+b~v ·~
x)
ib i(av
∇ψ) = − m
e
2 t+b~
v ·~
x)
2 2
~v ·∇ψ+ v2mb ψ 0 +ei(av
2 t+b~
v ·~
x) −∇2
2m
~v · ∇ψ
∂t ψ
Combining these, using the fact ψ satisfies the usual Schrödinger equation, gives the
2 2
2
ib i(av 2 t+b~v ·~
x)
following constraint: v2mb ψ 0 − m
e
~v · ∇ψ = bv 2 ψ 0 − iei(av t+b~v·~x)~v · ∇ψ − av 2 ψ 0
ψ and it’s derivatives are independent so we must demand things cancel independently
b2
2m
− b + a = 0 and − mb + 1 = 0 so b = m and a =
m
2
~
Note if you wanted eiϕ to be of the form ei(ωt+k·~x) that ω = m2 v 2 and k = mv which are
nice classical expressions for the effective energy/momentum of the frame difference
3. Toy model of the hydrogen molecular ion. [from Commins]
A particle of mass m is in a potential
V (x) = W0 (δ(x − a) + δ(x + a))
with W0 < 0.
(a) Find the expression that determines the bound-state energy for even-parity states,
and determine graphically how many even-parity boundstates exist.
We solve the Schrödinger equation in three regions.
Because the potential is symmetric we can focus on eigenstates of parity: P ψ =
ψ(−x) = ±ψ(x)
Let −κ2 = 2mE and we’ll focus on the boundary conditions at x = a
0
0
ψIII (x = a) = ψII (x = a) and ψIII
(x = a) − ψII
(x = a) = 2mW0 ψ(x = a)
2
ψ(x, t)
where ψIII (x) = Ae−κx and ψII (x) = B cosh(κx) for even-parity
Diving these equations by each other gives: −κ(1 + tanh(κa)) = 2mW0
Solving this transcendental equation for κ is a condition on the boundstate energy.
W0 should be negative for an attractive potential.
In the attached Mathematica file I’ve plotted these. There’s exactly one solution
for even parity
(b) What is the wave function for the lowest even-parity bound state? Sketch this
function for large, intermediate, and small a (compared to what?) ?
See the mathematica notebook. a would be small compared to the other length
1
scale in the problem m|W
0|
(c) Repeat the previous parts for odd parity. For what values of W0 is there at least
one such boundstate?
For odd parity one replaces the cosh for sinh in the above analysis. This makes
the transcendental equation:−κ(1 + coth(κa)) = 2mW0
Notice that (1 + coth(κa)) asymtopes to infinity for κ → 0 and there’s only a
boundstate for 2|W0 | > a in which case there is exactly one
(d) Find the even- and odd-parity binding energies for m|W0 |a ~2 . Explain why
these energies move closer together as a → ∞.
In the a → ∞ limit one has effectively two independent delta functions and both
coth and tanh go to 1 so there E = − 21 m(W0 )2 for both cases which is the energy
for the unique boundstate of a single delta function.
(e) Suppose that the two potential centers at a experience some (repulsive) Coulomb
µ
for some constant µ. Including this effect and the energy
interaction, Uc (a) = 2a
of the electron in its groundstate what can you say about the equilibrium value
of a?
µ
1 2
The Schrödinger equation now reads (− 2m
∂x + V (x) + 2a
)ψ = Eψ so the problem
µ
2
is the same if we redefine −κ = 2m(E − 2a )
Equilibrium should be where ∂a E = 0 but κ has a complicated dependency on
a. Instead we can examine this numerically (in units of m = 1) using a special
W (±2a|W0 |e−2a|W0 | )
function expression of κ± =
+|W0 | where W[x] is the product-log
2a
function and κ± refers to the even/odd solution respectively. See the notebook
One notices that for µ |W0 | that there is no stable minimum and that the κ−
solution never has a stable minimum.
4. Numerical quantum mechanics. [optional bonus projects]
(a) Time evolution. Write some code to time-evolve a 1d wavefunction for a particle
in a potential V (x). That is: choose an initial wavefunction ψ(x, 0); choose a
potential V (x), find numerically ψ(x, nδt) for some set of n, and some choice of
δt. Draw pictures.
3
A good strategy is to use the split-step method which approximates the timeevolution operator as
p2
+V (x)
iδt 2m
U(δt) = eiδtH = e
p2
2
= eiδt 2m eiδtV (x) eO(δt) .
Ignoring the O(δt)2 correction, time evolution occurs by action of
p2
p2
p2
p2
U(t = nδt) = U(δt)n ≈ eiδt 2m eiδtV (x) eiδt 2m eiδtV (x) eiδt 2m eiδtV (x) · · · eiδt 2m eiδtV (x) .
Each step is diagonal in either position space or momentum space, so we just need
to fourier transform back and forth after each step.
(b) Eigenvalue problem.
Use the method of shooting to solve numerically for the groundstate wavefunction
of a potential which grows at large |x|.
4