Vector Potential for a Particle on the ring and

Unitary transform of the Hamiltonian

25th July, 2018

Supervisor: Siddhartha Lal

Department of Physics, IISER-Kolkata

1
Contents
Abstract 2

1 Classical Particle in a Ring 3

2 Quantum Particle in a Ring 3

3 Unitary Transform of the Hamiltonian 5

4 Conclusion 6

Bibliography 7

1
Abstract
Consider a charged particle on a ring. A thin infinitely long solenoid passes
through it’s centre. The magnetic field due to the infinite solenoid is confined
inside the solenoid(Figure 1). Hence, the field about the ring is zero. According
to Maxwell’s equations,

~ =0
∇B

This implies that there exists a vector potential for the magnetic field such
that,

~ =∇×A
⇒B ~

This equation tells us that A~ can be non-zero where the field is zero. In classical
electrodynamics, A is not directly measurable. For a classical particle on a ring,
field on the ring is zero. This means that the effect of the field on the motion of
a charged particle is zero. A~ does not affect the equations of motion as it can be
canonically removed. However, for quantum particle on a ring, even though the
~ is non zero and it affects the quantum spectrum. The sensitivity
field is zero,A
of the quantum model to A ~ will be established.

Figure 1: Infinite long solenoid

2
1 Classical Particle in a Ring
The Lorentz force on the charged particle in a magnetic field is given by,
~
FL = q(~v × B)

Since the B~ = 0, the force on the particle due to field is zero. Hence, it does
not affect the equation of motion of the particle.
The Lagrangian for the classical particle is given by
M ˙2
L= φ + Aφ̇
2
where M is the moment of Inertia of the particle and φ is the angular posi-
tion.The Hamiltonian is given by,
1 2
H = pφ̇ − L = (p − A)
2M
where p is the conjugated momentum of φ. The corresponding Hamiltonian
equations of motion are,

1
φ̇ = (p − A) (1)
M
ṗ = 0 (2)

Equation (1) tells us that, if a particle is given a initial velocity, it moves with
constant angular velocity. From eqution (2), rate of change in conjugate momen-
tum is zero. The conjugate momentum here, is actually the angular momentum.
Rate of change of angular momentum is torque and it is zero. We can eliminate
A from the Hamiltonian by a canonical transformation, p 7−→ p + A , without
changing the equations of motion. Thus the vector potential of magnetic field
does not affect the motion of the particle in the ring.

2 Quantum Particle in a Ring

b = 1 (−i ∂ − A)2
H
2M ∂φ
We take } = 1. Now, we solve the time independent Schrödinger equation
Hψ = Eψ imposing the periodic boundary conditions

ψ(φ + 2π) = ψ(φ)

3
we get,

ψm = eimφ ,
1
Em = (m − A)2 , mZ.
2M
With changing values of A we can observe Spectral flow. The figure below
represents the energy spectrum on the classical energy function i.e., Em vs m.
As the values of A increases or decreases, the energy spectrum moves right or
left along the parabola. The eigenstates at infinity are viewed as if they jump
from one arm of the parabola to the other, thus creating the spectral flow.

Figure 2: Spectral Flow (E vs m)

Φ
We can think of A as, A = Φo , where Φ is the magnetic flux and Φo = 2π }c
e
is the flux quantum. Let
t2 φ2
4φ
Z Z
Sp = Aφ̇dt = Adφ = 2πA
t1 φ1 2π

4
 R
. S is the potential term of action, which is given by S = dtL , where L is the
Lagrangian. It is path independent and changes by 2πA every time the particle
goes one full round, counter-clockwise. We can see that the Hamiltonian has A
in it. We can gauge transform the Hamiltonian to eliminate A from it.

3 Unitary Transform of the Hamiltonian

⇒ U = e−iAφ

is an unitary operator. Now, we perform the unitary transformation to eliminate

H 7−→ U HU + = H 0
ψ 7−→ U ψ = ψe−iAφ = ψ 0
1 ∂ 1 ∂
H 0 = e−iAφ [ (−i − A)2 ]eiaφ = (−i )2
2M ∂φ 2M ∂φ

Though we eliminated A from the Hamiltonian, it enters the model through

ψ 0 (φ + 2π) = e−i2πA ψ 0 (φ)

. Imposing these boundary conditions, we get

0
ψm = ei(m−A)φ ,
1
Em = (m − A)2 , mZ
2M
The eigenstates are change due the transformation performed on Hilbert
space. However, the eigenvalues, being observables, are same before and after
transformation.
The transformation can be visualised with the mobious strip.If A is an inte-
ger, the boundary conditions become ψ 0 (φ + 2π) = e−in2π ψ 0 (φ), n is an in-
teger. The case is trivial. This, on the mobious strip is when it has two
phases or n full twists. When A is half integer, the boundary conditions
are ψ 0 (φ + 2π) = e−inπ ψ 0 (φ) i.e, ψ 0 (φ + 2π) = ψ 0 (φ) when n is even and
ψ 0 (φ + 2π) = −ψ 0 (φ) when n is odd. This is observed as doubly degenerate
ground state eigenvalues. The mobious strip representation for this case is n
times half- twisted strip.There is only one phase for a half twisted strip. Thus
every value of m, for A= n2 , there are two degenerate states.

5
 Figure 3: Mobious Strip

For all the other continuous values that A can take, we obtain twisted bound-
ary conditions. Here, the strip is cut and reattached at an arbitrary angle. This
represents the phase change that occurs as the particle moves around the ring.
This phase is explained by the Aharanov-Bohm effect. There are electromag-
netic effects in the region where the field is zero, even though A is not directly
measurable.

4 Conclusion
Though, the vector potential of magnetic field, A~ is not classically measurable,
it’s mysterious effects are seen in quantum mechanics. It is not possible to re-
move the effect of A-term in quantum mechanics. It can be formally removed
from the Hamiltonian through the unitary transformation. As this transforma-
tion is done on Hilbert space, the states are also transformed. This produces
the twisted boundary conditions which reintroduces A into the picture. Even
though the eigenstates are transformed, the eigenvalues are not, because they
are observables.
These twisted boundary conditions represent the phase change that is ob-
served, every time the particle goes one full round. This is the most important
phenomena in particle in a ring.Spectral flow is also observed due to changing
values of A. As the magnitude of A increases, the quantum energy spectrum
moves to the left and vice versa.

6
Bibliography
1.Abanov,Alexander G,(2017).Topology, geometry and quantum interference in
condensed matter physics,Stony Brook University, Stony Brook, NY

2.von,Projektarbeit & Danilevich,Anatoly,2007,1D Quantum Rings ,Lehrstuhl

3. Griffith, David S(Second Edition),1995,Introduction to Quantum Mechan-