Before the start of this lecture, you should be able to:

- Write down the wavefunction after adiabatic evolution.

- Express a magnetic field $\mathbf{B}$ in terms of the vector potential $\mathbf{A}$.

- Flip limits of integration.

!!! summary "Learning goals"

After this lecture you will be able to:

- Write down the Berry phase as a function of an arbitrary parameter.

- Describe how a relative phase can be measured.

- Write down the phase acquired when an electron travels around a flux of magnetic field.

- Relate the Berry phase to the Ahronov-Bohm effect.

## The Berry Phase

In the previous section, we found that the wavefunction acquires two phases during adiabatic evolution. In particular, the geometrical phase is,

$$

\gamma(t) = i \int_0^t \langle{\psi_m| \dot\psi_n} \rangle dt'.

$$

It is known generally as the Berry phase. Perhaps the first question that comes to mind about it is whether it is a phase at all. We can show that it is, but showing that $\gamma(t)$ is in fact *real*.

We know that $\langle{\psi_n | \psi_n} \rangle = 1$. From there we look at

\gamma_n(t) &= i \int_0^t \langle{ \psi_n | \vec{\nabla_R}\psi_n} \rangle \cdot

\dot{R}(t') dt' \\

&= i \int_{\vec{R}(0)}^{\vec{R}(t)} \langle{ \psi_n | \vec{\nabla_R}\psi_n} \rangle

\cdot d\vec{R}.

\end{aligned}

$$

This is a line integral through parameter space which is independent of time.

The Berry phase introduced is independent of how fast the path from $\vec{R}(0)$

to $\vec{R}(t)$ is taken. This is in contrast to $\theta_n(t)$ which has

explicit time dependence, where the more time you take the more phase you

accumulate.

If we consider the case where $\vec{R}(0) = \vec{R}(t)$, then for only one

parameter the Berry phase will always be zero. A non-zero Berry phase arises

only for more than one time-dependent parameter. In the case of more than one

parameter then, the value of the integral depends only on the number of poles

enclosed by the path, discretising the value of $\gamma_n$, and yielding a

result that is path independent.

## Measuring a Phase?

![Splitting mirrors.](figures/2paths.svg)

How can we measure a phase? It may seem difficult since phases often don't contribute to observables that we are interested in measuring. Though it may be impossible to measure an *overall* phase, *relative* phases are

fair game. By splitting a wave packet over two paths of different lengths and using interference, we can get information about this phase.

To see this, consider the two paths depicted in the figure above. The longer path is associated with an accumulated phase of $e^{i \gamma}$ for some $\gamma$. The total wavefunction is then

We see that relative phases do show up non-trivially when going from amplitudes to probabilities, and exploiting these will be our method toward measuring the Berry phase.

## Aharonov-Bohm Effect

![solenoid](figures/solenoid.svg)

First predicted by Ehrenberg and Siday (10 years earlier than Aharonov and Bohm!), this effect is due to the coupling of the electromagnetic potential to an electron's wave function, as we will see.

The setup is depicted in the figure above. The important thing to note here is that the magnetic field $\mathbf{B}$ is non-zero only inside the solenoid. Which means for both electron paths, through *B* or *C*, the magnetic field is constantly zero. The same cannot be said of the vector potential however.

We know that the flux $\phi$ through the solenoid is $\phi=B\cdot \pi r^2$ for $r$ the radius of the solenoid, and that $\mathbf{B} = \nabla \times \mathbf{A}$.

The vector potential $\mathbf{A}$ changes our Hamiltonian to

$$

H = \frac{(\mathbf{p} + e\mathbf{A})^2}{2m} + V(\mathbf{r}).

$$

If you are wondering about this change, the main point is that the momentum operator has changed in the standard way for quantum mechanics to include the vector potential such that the canonical commutation relation $[r_i, p_j] = i\hbar \delta_{ij}$ still holds.

How now do we solve for $\psi$ in $H \psi = E \psi$ for this Hamiltonian? There

is a simple way to solve this using the solution when $\mathbf{A}=0$, which we

denote as $\psi_0$.

We define

$$

\psi (\mathbf{r}) = e^{i g(\mathbf{r})} \psi_0(\mathbf{r}) ,\text{ where }

$g(\mathbf{r})$ is defined as a line integral, which is only well-defined if $\mathbf{B} = \nabla \times \mathbf{A} = 0$. This is precisely the situation we have engineered with the solenoid.

Applying this operator twice then will give $e^{ig} (\mathbf{p}^2 \psi_0)$. And so since $\psi_0$ satisfies $H_0 \psi_0= E_0 \psi_0$, the function $\psi=e^{ig} \psi_0$ satisfies the equivalent equation with $\mathbf{A}$ in it.

Turning back to the beam splitter and the solenoid, we can see that each path will pick up a different phase factor.

$$

\psi_0 e^{ig} \rightarrow \psi_0 e^{\pm i \gamma} , \text{ where }

\gamma = -\frac{e}{\hbar} \int \mathbf{A}(\mathbf{r}) \cdot d \mathbf{r}

$$

If, contrary to the figure, we take two semi-circular paths around the solenoid, we can evaluate what the phase associated with each path is using the circle element on the path $r d\phi$.

The *difference* between the two paths is then $e \phi/\hbar$. As we saw in the beam splitter, this difference can be measured in experiment through interference.

Some comments:

* $\mathbf{A}$ matters for this derivation, *not* the magnetic field

$\mathbf{B}$!

* $\gamma$ is explicitly gauge invariant (adding some term $\nabla f$ to

$\mathbf{A}$ doesn't change the result)

### Connection to the Berry Phase

If we confine the electron to a box at some point $\mathbf{R}(t)$, and slowly

move the box around the solenoid, we want to find the Berry phase that this

electron acquires as the box moves around the solenoid.

If we had no solenoid, the wavefunction that we would have would be centred

around $\mathbf{R}$, and going around in a circle. This can be written as

$\psi_0(\mathbf{r}-\mathbf{R})$.

With the solenoid present we use the same trick as we did above by setting

But the expectation value of the momentum is 0 for a particle confined in a box!

So we find that

$$

\begin{aligned}

\gamma_n &= - i \oint \frac{ie}{\hbar} \mathbf{A}(\mathbf{R}) \cdot

d\mathbf{R} \\

&= \frac{e}{\hbar} \int \nabla \times \mathbf{A} da \\

&= \frac{e\phi}{\hbar}.

\end{aligned}

$$

## Summary

The Berry phase can be expressed in terms of an arbitrary time-dependent parameter $\mathbf{R}(t)$ as,

$$

\gamma_n(t) = i \int_{\vec{R}(0)}^{\vec{R}(t)} \langle{ \psi_n | \vec{\nabla_R}\psi_n} \rangle

\cdot d\vec{R}.

$$

The Aharonov-Bohm effect arises as an extra phase due to the coupling of the wavefunction with the vector potential when traveling around a solenoid. The acquired phase is proportional to the magnetic flux going through the solenoid. That is,

Before the start of this lecture, you should be able to:

- Write down the Schrodinger equation.

- Distinguish between time-dependent and time-independent Schrodinger equation.

- Solve first order ordinary differential equations.

!!! summary "Learning goals"

After this lecture you will be able to:

- Write down the wavefunction of a system under adiabatic changes.

- Explain the origin of the dynamic and geometrical phases.

- Write down the criteria required for the adiabatic theorem to hold.

## Proof of the theorem

So far we have just *stated* the theorem and shown a couple examples.

Now we will formulate the theorem in a mathematical way and prove it.

Our starting point is the *time-dependent Schrodinger equation*:

$$

i \hbar \frac{\partial}{\partial t} \left|\Psi(t)\right> = H(t) \left|\Psi(t)\right>\,

$$

with a time-dependent Hamiltonian $H(t)$. For notational simplicity, we will

in the following often leave out the explicit bra-ket notation, and simple write $\Psi(t)$ instead

of $\left|\Psi(t)\right>$.

We now want to consider a Hamiltonian $H(t)$ that cheanges slowly in time. But what does *slowly* mean,

slow compared to what? In contrast to the WKB approximation this is less obvious, and

we will find a proper criterion in the course of the proof.

If we have a time-independent Hamiltonian $H$, the time-dependent Schrodinger equation

is readily solved as,

$$

H \phi_n = E_n \phi_n,\quad \phi_n(t)= \phi_n(0) e^{-i E_n t / \hbar},

$$

where we have used $\phi$ to denote the eigenstates of $H$.

To solve the time-dependent problem, we will now start with a definition: We define $\psi_n(t)$ to be those eigenfunctions that solve the

equation

$$

H(t) \psi_n(t) = E_n(t) \psi_n(t)\,

$$

where we now consider time $t$ as a *parameter* of the Hamiltonian that we can fix to some arbitrary value.

!!! warning "Instantaneous eigenstates"

The states $\psi_n(t)$ *do not solve the time-dependent Schrodinger equation!* (despite the occurance of $H(t)$ in the equation). In contrast, the $\psi_n(t)$ that we have just defined solve a stationary Schrodinger equation (we fix $t$ to some value). As a result they must constitute a complete, orthogonal set for any time $t$. That is: $\langle{\psi_n(t) | \psi_m(t)} \rangle = \delta_{nm}$.

If we take two different times $t$ and $t'$, we cannot say anything about the inner product $\langle {\psi_n(t) | \psi_m(t')} \rangle$ in general, since the $\psi_m$ could evolve in any way in principle, such that the overlap between $\psi_n$ and $\psi_m$ is no longer 0.

Because the $\psi_n(t)$ form a complete orthonormal set, we can express the the full quantum state $\Psi(t)$ as a linear combination of the states $\psi_n(t)$:

where $\theta_n(t) = -\frac{1}{\hbar} \int_0^t E_n(t) dt$. What we do here is expand the wave function $\Psi(t)$ in a different basis for every value time $t$ (using the set of $\{\psi_n(t)\}$). This is allowed, for a given value of time $t$ the wave function $\Psi(t)$ can be expanded in any arbitrary basis set.

We could have absorbed the exponential factor involving $\theta$ into the $c_n(t)$, but as we will see it is convenient to include it explicitly. This term may look strange, but in fact it is a straightforward generalisation of the phase factor that an eigenstate picks up, which would be $\exp (-i E_n t/ \hbar)$,

when using a constant Hamiltonian.

Now we have an expression for $\Psi(t)$ and we want to see what that gives us when we apply the time-dependent Schrodinger equation to it. We get

$$

i \hbar \sum_n \left[ \dot{c}_n \psi_n + c_n \dot{\psi}_n +

c_n \psi_n i \dot{\theta}_n \right] e^{i \theta_n} =

\sum_n c_n(t) H(t) \psi_n(t) e^{i \theta_n(t)}.

$$

Now we remember that we defined $\psi_n$ to satisfy $H \psi_n = E_n \psi_n$, and

we recognize that

$$

i \dot{\theta}_n = -\frac{i}{\hbar} E_n(t)

$$

so the $\theta$ derivative term on the left-hand side cancels with the right-hand side of the previous equation. What we are left with is

where we used the orthogonality of $\psi_n$ and $\psi_m$ to eliminate the sum on the left-hand side.

In order to make progress evaluating what $\langle{\psi_m| \dot\psi_n} \rangle$ might be, we take the derivative of the time-independent Schrodinger equation which we used to define the $\psi_n$ in the first place. That is,

$$

\dot{H}\psi_n + H \dot{\psi}_n = \dot{E}_n \psi_n + E_n \dot{\psi}_n

When we have slow, gradual change being applied to the system, $\dot{H}$ is small. Now we are finally ready to make the **approximation** that you must have been anticipating since reading the title of these notes. We *neglect* the contributions when $n \neq m$ and just get:

What is considered *slow* is determined by $E_n - E_m$. If the energy difference is *large*, then a gradual change could mean something that is much faster than we might expect. We also see that in cases of degeneracy (when $E_n = E_m$ despite $n \neq m$), there isn't *any* definition of slow that is slow enough. The adiabatic approximation breaks down here. Similarly if we change the system in such a way that two energy levels that were separated come together or switch places, the approximation will again break down.

Solving the equation for the coefficients gives

$$

c_m(t) = c_m(0) e^{i \gamma(t)}, \quad \text{ where } \gamma(t) = i \int_0^t \langle{\psi_m| \dot\psi_n} \rangle dt'

$$

(You may notice that the factor $i$ appears twice in the above expression, both in the expression $e^{i \gamma(t)}$ and in the definition of $\gamma(t)$. This is not a mistake, as we will see later.)

So if we start in the $n^{\text{th}}$ eigenstate of the initial Hamiltonian,

that means that $\Psi = \psi_n$. So $c_n(0) = 1$ and all $c_m(0) = 0$ for $n

\neq m$. Then the full wavefunction after the gradual change has taken place is

The $n^{\text{th}}$ eigenstate stays in the $n^{\text{th}}$ eigenstate.

### Adiabatic criteria: how slow is *slow*?

In order to obtain the result above, the terms $n\neq m$ have been neglected by considering their contribution small. In principle, it should be smaller than the time associated with the energy difference of the first excitation. That is,

This result can be derived in the style of perturbation theory. Let us evaluate the first order solution that satiesfies $c_m(t) = 1$ and $c_{n\neq m}(t)=0$. That is,

There are several time dependent quantities in the previous expression. However, we approximate some of them by using the following bounds:

* The largest contribution from the rate of change in the Hamiltonian will come from the largest eigenvalue.

$$

\langle \psi_m (t) | \dot H |\psi_n(t) \rangle \approx \overline{\langle \psi_m | \dot H |\psi_n\rangle}

$$

* The smallest energy difference will contribute the most.

$$

E_m(t) - E_n(t) \approx \overline{E_m - E_n}

$$

After such approximations, these quantities are replaced in the formula for the coefficients. Since they are not time-dependent anymore, it can be solved as,

We have found an expression for the coefficients where $n \neq m$. The addibatic approximation assumes that these coefficients are small. This criteria can be explicitly written as,

* The system remains in the instantaneous eigenstate.

* It acquires two phases shifts depending on the evolution. The dynamical phase, $\theta_n(t)$, and the geometrical phase $\gamma_n(t)$. They are given as,

$$

\theta_n(t) = i \int_0^t E_n(t') \rangle dt', \quad \gamma_n(t) = i \int_0^t \langle{\psi_n| \dot\psi_n} \rangle dt'.

$$

In order for this description to hold, we require that the system evolves slowly enough. That is, the following condition must be satisfied at all times:

$$

\frac{\hbar \overline{\langle \psi_m | \dot H |\psi_n\rangle} }{\overline{E_m - E_n}^2} << 1, \quad n \neq m.

$$

Here, the overline indicates the largest matrix element, and the smallest energy difference.