Berry Phase

Georg Manten (georg@manten.biz) 17th May 2017

1 Berry Phase
1.1 Introduction
We consider the general Hamiltonian
H xa ; λi

xa : degrees of freedom of the system / things evolving dynamically

λi : parameters of the Hamiltonian, which are externally adjusted

First, we pick some values for λ, and then after placing the system in some energy
eigenstate |ψi, which we choose for simplicity to be the ground state, we slowly vary
λ. Therefore, the Hamiltonian changes and also the ground state |ψ (λ (t))i.

We make use of the adiabatic theorem, which says, that if we place a system in
a non-degenerate energy eigenstate and vary parameters sufficiently slowly, the sys-
tem will stick to this energy eigenstate and will not be excited to any higher or lower
energystate.

Hereby, it is important how fast you change the parameters, which depends on the
gap between the actual state and the nearest other state.
A very tricky case is level-crossing, where another case becomes degenerate with the
one you are in, after seperating again, it is hard to tell in which linear combination of
states the system is.

If we now vary the parameters slowly and perform a closed path in the space of
parameters with the assumption that we do not pass a point with level-crossing, we
would like to know, which point we are in afterwards.

Due to the adiabatic theorem, we are still in the ground state, and the only uncertainty
is the phase
|ψi −→ eiγ |ψi (1)
This phase could have physical consequence, since we could start with two states and
vary just one of them, and afterwards, eiγ has an effect if we interfere both.
Et
There are two contributions to the phase, the dynamic one e−i } , which is there for
every energy eigenstate, even if we do not change the parameters, and the other one
is the Berry phase.
Seminar The Quantum Hall Effect “, SS 2017, University Heidelberg
”

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1.2 Computing the Berry Phase
The evolution of states is given by the time-dependent Schrödinger equation

∂ |ψi
i} = H (λ (t)) |ψi (2)
∂t
As a first step, we introduce a ground state with some arbitrary but fixed choice of
phase |n (λ)i for every parameter λ. Therefore we can write the ground state as

|ψ (t)i = U (t) |n (λ (t))i (3)

with U (t) a time-dependent phase. If we pick |n (λ (t = 0))i = |ψ (t = 0)i then we

have U (t = 0) = 1.
R
There is always the dynamic contribution e−i dtE0 (t)/} , which we can ignore by set-
ting the ground state energy E0 (t) = 0. For the extra contribution, we plug in our
adiabatic ansatz (3) in (2) and take the overlap with hψ|:
D  E
ψ ψ̇ = U̇ U ? + hn|ṅi = 0


where we have used H (λ) |n (λ)i = 0 to get zero on the right side of the equation.
We get:
∂
U ? U̇ = − hn|ṅi = − hn| |ni λ̇i (4)
∂λi
Here we define the Berry connection
∂
Ai (λ) = −i hn| |ni (5)
∂λi
so we get the differential equation

U̇ = −iAi λ̇i U

which has the solution

 Z 
U (t) = exp −i Ai (λ) λ̇i dt

Therefore, after integrating over a closed curve C, we get:

 I 
iγ i
e = exp −i Ai (λ) dλ (6)
C

This is the Berry phase.

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1.3 The Berry Connection
Since the Berry connection looks like a gauge potential in electromagnetism, we will
investigate in this direction:
In the relativistic electromagnetism, the gauge potential is given by Aµ (x) with
µ = 0, 1, 2, 3 and x are the coordinates from the Minkowski spacetime. The gauge
transformation is given by
Aµ −→ A0µ = Aµ + ∂µ ω (7)
for an arbitrary function ω (x). The field strength tensor is given by
∂Aµ ∂Aν
Fµν = ν
− (8)
∂x ∂xµ
and is invariant under gauge transformations.
Now we compare this to Ai .
In our case, we can think of changing x −→ λi for i = 1, . . . , d which turns Ai into a
one-form.
Since we could pick a different choice of reference state, we get
|n0 (λ)i = eiω(λ) |n (λ)i
For an aribitrary function ω (λ), we get as the new Berry connection the following
expression
∂ ∂ω
Aprime
i = −i hn0 | |n0 i = Ai + (9)
∂λi ∂λi
which looks quite similar to our gauge transformation above.
Therefore, we can calculate the gauge invariant field strength tensor (or curvature
2-form)
∂Ai ∂Aj
Fij (λ) = j
−
∂λ ∂λi
with the help of Ω = dA + A ∧ A and the fact that the Ai ’s commute in this case.
In the presentH context, we want to investigate more to actually calculate the Berry
Phase. Since ∂i ωdλi = 0, the Berry phase is invariant of the transformation, and
with Stokes Theorem we get
 I   Z 
iγ i ij
e = exp −i Ai (λ) dλ = exp −i Fij dS (10)
C S

where S is the two-dimensional surface bound by the path C.

1.4 A Spin in a Magnetic Field

~ The Hamiltonian
As an example, we consider a spin in a magnetic field B.
~σ + B
H = −B~

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~σ : Pauli matrix vector
~
B = kBk.

We get two eigenvalues 0 and 2B and therefore two eigenstates, the ground state
|↓i and the excited state |↑i with

H |↓i = 0 and H |↑i = 2B |↑i

~ and given B
If we now take λi ≡ B ~ in polar coordinates
 
B sin (θ) cos (φ)
~ =  B sin (θ) sin (φ) 
B
B cos (θ)

with θ ∈ [0, π] and φ ∈ [0, 2π) we get

e−iφ sin (θ)

 
cos (θ) − 1
H = −B
eiφ sin (θ) − cos (θ) − 1

and the normalised eigenstates

 −iφ   −iφ 
e sin (θ/2) e cos (θ/2)
|↓i = and |↑i =
− cos (−θ/2) sin (θ/2)

We can now compute the Berry phase (and do not take θ = π so that φ is well-defined)
and start with
 
∂ ∂ 2 θ
Aθ = −i h↓| |↓i = 0 and Aφ = −i h↓| |↓i = − sin
∂θ ∂φ 2

For this result, we get

∂Aφ ∂Aθ
Fθφ = − = − sin (θ)
∂θ ∂φ
as the only entry of the antisymmetric tensor. Transforming this into cartesian coor-
dinates gives us
k
~ = −ijk B
 
Fij B
~ 3
2kBk

which looks like a magnetic monopole at B ~ = 0, but in the space of magnetic fields.
In the origin, the field strength tensor gets singular and the two energy levels coincide.
The magnetic pole has charge g = − 21 , since we get
Z
Fij dS ij = 4πg = −2π (11)
S2

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for any two-sphere S 2 , since Fθφ does not depend on the radius. With this we can
compute the Berry phase as long as the path C does not cross the origin. Suppose the
area S surrounded by C has the solid angle Ω.
Then we get:
 Z   
iΩ
eiγ = exp −i Fij dS ij = exp (12)
S 2

One could also calculate the integral over S 0 = S 2 − S with the angle Ω0 = 4π − Ω and
get
 Z   
0 −i (4π − Ω)
eiγ = exp −i Fij dS ij = exp = eiγ (13)
S0 2

This requires that 2g ∈ Z.

1.5 Particles Moving Around a Flux Tube

As assumed in electromagnetism, the gauge potential Aµ is an unphysical quantity.
In this section we will discuss a set-up where the gauge potential appears in the
Hamiltonian, invariant under gauge tranformations.
Consider the situation shown in the picture:

Figure 1: A particle moving around a solenoid

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The magnetic flux, with A the area of the solenoid, is Φ = BA. While the magnetic
field is zero, the vector potential is not and this follows with Stokes Theorem:

I Z
~ · d~r =
A ~ · dS
B ~=Φ

Solving this, we get

Φ
Aφ =
2πr
Consider now a quantum particle, which lies on a ring of radius r outside the cylinder
with angle φ ∈ [0, 2π) and has the Hamiltonian

 2
1 2 1 ∂ eΦ
(pφ + eAφ ) = −i} +
2m 2mr2 ∂φ 2π
since its only degree of freedom is the angle φ ∈ [0, 2π).
The energy eigenstates are simply
1
φ= √ einφ n∈Z
2πr
with n ∈ Z because it has to be periodic. The energy eigenvalues are
}2
 
Φ
En = n + n∈Z
2mr2 Φ0
with Φ0 = 2π}
e . Note, that if Φ is a multiple of Φ0 , then the spectrum is unaffected by
the solenoid and if not, the spectrum gets shifted.
The spectrum looks like this:

Figure 2: Energy spectrum of a particle moving around a solenoid

Now, the following situation can occur: We start with a turned off solenoid in the
particle’s ground state and then slowly increase the flux. By the adiabatic theorem,
the particle stays in the ground state, but after some time, we reach Φ = Φ0 and is
now in the state, we just called the n = 1-state.
This effect is called spectral flow.

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1.6 The Aharonov-Bohm Effect
Now we consider a similar situation, where the magnetic field is localised to a region
like the solenoid, but this time, our particle is inside a small box. We just include a
potential V (~x) into the Hamiltonian in order to trap the particle with an infinitely
high potential outside the box.
Small in this case means, that the gauge potential is constant inside the box. If we
start by placing the center of the box at ~x = X,~ the Hamiltonian is given by

1  2  
H= −i}∇~ + eA ~ (X) + V ~x − X ~
2m
 
After starting with the center of the box at ~x = X ~ 0 and gauge so that A ~ X
~ 0 = 0 we
 
get the particle in the box with the ground state ψ ~x − X ~ 0 localised around ~x = X
~ 0.
 
As we now slowly move the box, we have the potential A ~ ~x − X~ in the Hamiltonian
and the solution for the SGE
Z ~x=X~ !
 
~ = exp − ie 
~ (~x) · d~x ψ ~x − X ~0

ψ ~x − X A
} ~x=X~ 0

If we now make a loop C with the box, we get:

 I 

~

iγ

~

iγ ie ~
ψ ~x − X0 → e ψ ~x − X0 ; e = exp − A (~x) · d~x (14)
} C
where in this case, our Berry connection is a real electromagnetic potential given by
  e  
A~ X~ = A ~ ~x = X ~
}
The electron has charge q = −e. In the general case, where a particle with charge q
travels around an area with flux Φ, we get the Aharonov-Bohm phase
eiqΦ/}
There is an experiment showing the Aharonov-Bohm effect:

It is a variant of the double split experiment, with the addition, that between the
slits, there is a solenoid hidden on the other side of the wall with flux Φ. The wave-
~
function never touches the magnetic field B,but induces a phase difference eiγ between
the two possible ways a particle could take. The phase difference can then be seen in
the interference pattern on the screen. It stay the same, if Φ is a multiple of Φ0 , and
changes in the other cases.

1.7 Non-Abelian Berry Connection

We now talk about the case where the ground state is N-times degenerate and remains
like this for all values of λ. As before, we place the system in one of the N states

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and vary the parameters along a closed path, but now, since the adiabatic theorem
only tells us that the system stays in a particular energy eigenstate, the system is
somewhere in the N-dimensional subspace and we have to compute a unitary matrix
U ⊂ U (N ) instead of just a phase.

We therefore assume E = 0 as before to get rid of the dynamic phase as before

and the time-dependent SGE is

∂ |ψi
i = H (λ (t)) |ψi = 0 (15)
∂t
But this time, we introduce an N-dimensional basis of the ground states

|na (λ)i a = 1, . . . , N

and get

|ψa (t)i = Uab |nb (λ (t))i

As before, after plugging this into the SGE we get

 E
 ˙
ψa = U˙ab |nb i + Uab |n˙b i = 0

and from there

† ˙ ∂
Uac Uab = − hnb |n˙c i = − hnb | |nc i λ̇i
∂λi
This leads us to the non-Abelian Berry connection
∂
(Ai )ab = −i hnb | |na i
∂λi
which can be later be seen as an element of the Lie algebra u (N ).
Since we could have picked another choice of basis vectors at each point, we would
have

|n0a (λ)i = Ωab |nb (λ)i

with Ω (λ) ⊂ U (N ) a unitary rotation. As a berry connection we get

∂Ω †
A0i = ΩAi Ω† + i Ω (16)
∂λi
We can now construct the field strength
∂Ai ∂Aj
Fij = j
− − i[Ai , Aj ]
∂λ ∂λi

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which is not a gauge-invariant quantity, since it transforms like this:

0
Fij = ΩFij Ω†

One could take the trace of this to get one, but we will anyways compute the Berry
phase.
Since the Ai (λ) does not have to commute for different values of λ, we have to solution
 I 
U = P exp −i Ai dλi

where P means ”path ordered”: here, if we expand the matrix exponential, we order
the products so that Ai (λ) are later in the path are placed to the right. This matrix
is called the Berry holonomy.

References
[1] David Tong (ed.): The Quantum Hall Effect. TIFR Infosys Lectures, Cambridge,
2016.