University of California at San Diego – Department of Physics – TA: Shauna Kravec
Quantum Mechanics C (Physics 212A) Fall 2015
Worksheet 8 –
Solutions
Announcements
• The 212A web site is:
http://physics.ucsd.edu/∼mcgreevy/f15/ .
Please check it regularly! It contains relevant course information!
Problems
1. Charged Particle on a Ring
Consider the quantum mechanics of a particle of mass m and charge q on a circular
hoop of radius R in the presence of an external magnetic field perpendicular to the
plane of the hoop.
(You can set all these parameters to 1 if you like. I will adopt units where ~ = 1 = c)
Let φ(t) be the angular coordinate of the particle such that φ ≡ φ + 2π
The action is
Z
θ
m 2 2
R φ̇ + ( )φ̇
2
2π
The θ term is a ”topological term” as we’ll see.
S[φ] =
(1)
dt
(a) Show that if there is a azimuthally symmetric magnetic field passing through the
ring, then θ is related to B by
θ = qΦB
where ΦB is the magnetic flux through the ring.
~ =∇×A
~ = Bδ(~x)ẑ =⇒ A
~ = ΦB φ̂
B
2πr
~ · φ̇ = −q Φb φ̇ which determines θ.
Minimal coupling to φ is −qRA
2π
(b) Show that θ is a periodic parameter, i.e. theories with θ and θ → θ + 2π are
equivalent.
See part e
(c) Compute the canonical momentum associated with φ, and determine the Hamiltonian for this system.
θ
θ
Πφ = ∂L
≡ pφ + 2π
= mR2 φ̇ + 2π
∂ φ̇
H = Πφ φ̇ − L =
p2φ
2mR2
=
θ 2
(Πφ − 2π
)
2mR2
=⇒ Ĥ =
1
1
(−i∂φ
2mR2
−
θ 2
)
2π
(d) Show that the θ term doesn’t effect the classical equation of motion for φ.
Euler-Lagrange:
dΠφ
dt
= 0 = mR2 φ̈ and the θ dependence is gone
(e) Find the energies and wavefunctions by solving the Schrödinger equation. What
happens to the eigenstates |n, θi and energies En (θ) as θ → θ + 2π?
2
λn
inφ
An ansatz is pφ ψ(φ) = λn ψ so that En = 2mR
2 which is satisfied by ψ ∝ e
1
θ 2
Yielding En = 2mR
2 (n − 2π ) and ψ(φ) = ψ(φ + 2π)
Note that under θ → θ + 2π that En → En−1 and the total spectrum is left
unchanged. This corresponds to the insertion of unit flux
(f) Suppose I do a gauge transformation A → A + ∇Λ that ”removes” the presence
of the magnetic field. What Λ accomplishes this? Why does the spectrum not
change?
Notice that we have gauge freedom of A → A + ∇Λ and ψ → ψ 0 = e−iqΛ ψ
Nothing stops me from choosing ∇Λ = −A =⇒ Λ = − Φ2πB φ
1
2
The Hamiltonian is then Ĥ = 2mR
2 (−i∂φ ) but notice that the boundary condition
for φ have changed.
θ
ψ 0 (φ + 2π) = ei 2π (φ+2π) ψ(φ + 2π) = eiθ ψ 0 (φ) a twisted boundary condition
So this entire problem is equivalent to a free particle propagating on a ring which
picks up a phase of eiθ after a complete wind. This is the Aharonov-Bohm phase.
If you demanded this phase be 1, completely unobservable, then θ = 2π` for ` ∈ Z
which is what Dirac did for his famous string
(Π −qA)2
Commentary: Actually this is misleading. The Hamiltonian has the form φ2m
and Πφ = p + qA so both A and Π transform under gauge transformation and
the classical Hamiltonian written this way is gauge invariant FOR THIS TRANSFORMATION
However the generic Hamiltonian for a particle in an EM field isn’t gauge invariant!
There’s a term qA0 which is absent in this problem which makes H → H 0 =
H + q∂0 Λ. Similarly the Lagrangian changes by a total derivative, making the
action invariant up to boundary terms.
So OK, does this invalidate the discussion above about twisted boundary conditions? I don’t think so. From the point of view of just an operator on the
Hilbert space ∂φ2 needn’t necessarily transform under gauge transformation. The
Schroedinger equation you write is invariant either way.
θ
= 2m+1
for m ∈ Z the spectrum is two-fold degenerate
(g) Show that for 2π
2
|mi and |m + 1i are degenerate as (m − 2m+1
)2 = 14 = ((m + 1) − 2m+1
)2
2
2
~ = (Ex cos φ, Ey sin φ, 0)
Now suppose I turn on an electric field in the xy-plane. E
(h) Write the Hamiltonian for the system in the presence of this field
There’s a new term H = H0 + H1 where H1 = −qV = −qR(Ex cos φ + Ey sin φ)
= − eR
[(Ex − iEy )eiφ + (Ex + iEy )e−iφ ]
2
2
θ
(i) Consider 2π
= 21 + δ for δ 1 so that |0i and |1i are nearly degenerate. Write
an effective Hamiltonian for this 2-state system, including the electric field.
I’m going to write everything in the |0i and |1i basis. The H0 piece is easy as it’s
diagonal in this basis.
1
( 2 + δ)2
0
1
H0 = 2mR2
0
( 21 − δ)2
Examining H1 the term with eiφ would transform |0i → |1i as recall the wavefunctions are einφ . Similarly the term e−iφ takes |1i → |0i
0
(Ex + iEy )
qR
H1 = − 2
(Ex − iEy )
0
(j) Calculate the ’precession’ frequency of the particle in this field keeping with the
2-state system idea.
~ ·S
~ so that ω0 = |H|
~
We wish to cast Hef f = H
~ = −qREx , qREy , δ 2
Hef f = δ 2 σ z − qREx σ x + qREy σ y =⇒ H
2mR
2
2
mR
(k) Now suppose Ey = Ex = E cos ωt varies in time. Suppose at t = 0 the particle
is in state |0i. What’s the probability of the particle to be in |1i at a later time
T ? Show this probability is periodic and find the period T0 . What’s a physical
constraint on ω for this picture to be valid?
ω
It’ll be useful to transform to a rotated frame: |ψω (t)i ≡ e−i 2 Ẑt |ψ(t)i
ω
Then i∂t |ψω i = ω2 Ẑ|ψω i + e−i 2 Ẑt H(t)|ψ(t)i = (Hω (t) + ω2 Ẑ)|ψω i
ω
ω
Where Hω ≡ e−i 2 Ẑt H(t)ei 2 Ẑt
Clearly this commutes the the σ z piece. A direct calculation for the rest of H
qRE x
δ
z
gives: Hω = 2mR
σ The time dependence is cancelled!
2σ −
2
ω
So we can easily solve |ψω (t)i = Uω (t)|ψ(0)i for Uω = e−i(Hω + 2 Ẑ)t
ω
The probability is P = |h1|ψ(t)i|2 where h1|ψ(t)i = h1|ei 2 Ẑt Uω (t)|0i
ω
δ
ω
qRE
= h1|e−i 2 t e−i[( 2mR2 + 2 )Ẑ− 2 X̂]t |0i
qRE 2
δ
ω 2
Define Ω2 ≡ ( 2mR
≡ α2 + β 2 such that Uω = e−i(tΩ)n̂·σ for
2 + 2 ) + (− 2 )
n̂ = Ω1 (α, β)
Then Uω |0i = [cos(tΩ)1 − i sin(tΩ)( Ωα Ẑ + Ωβ X̂)]|0i
2
By orthogonality P = Ωβ 2 sin2 (tΩ) with period T0 = Ωπ
Recall that the accuracy of our effective hamiltonian depends on the gap between
δ
|1i and |2i being large so ω mR
2
3