The $\gamma_n(t)$ that we derived above is known 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*.

!!! success "Expected prerequisites"

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

...

...

@@ -47,17 +70,13 @@ enclosed by the path, discretising the value of $\gamma_n$, and yielding a

result that is path independent.

## Measuring a Phase?

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.

![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 Figure\ref{fig:2paths}. The

longer path is associated with an accumulated phase of $e^{i \gamma}$ for some

$\gamma$. The total wavefunction is then

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

$$

\begin{aligned}

...

...

@@ -69,25 +88,17 @@ $$

\end{aligned}

$$

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.

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

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.

## Aharonov-Bohm Effect

The setup is depicted in Figure \ref{fig:ehrenberg_siday}. 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.

![solenoid](figures/solenoid.svg)

![ehrenberg_siday](figures/ehrenberg_siday.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.

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 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

...

...

@@ -95,10 +106,7 @@ $$

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.

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

...

...

@@ -107,14 +115,12 @@ denote as $\psi_0$.

We define

$$

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

\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.

$g(\mathbf{r})$ 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.

It is simple to check that this solution works:

...

...

@@ -129,30 +135,23 @@ $$

\end{aligned}

$$

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 $\psi$

defined in Equation \ref{eq:psi_def} satisfies the equivalent equation with

$\mathbf{A}$ in it.

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 Figure \ref{fig:ehrenberg_siday}, we can

see that each path will pick up a different phase factor.

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 }

\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 $r d\phi$.

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 Equation \ref{eq:interference}, this difference can be measured in experiment

through interference.

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:

...

...

@@ -175,7 +174,7 @@ With the solenoid present we use the same trick as we did above by setting

$\psi$ to

$$

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

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

g(\mathbf{r}) = -\frac{e}{\hbar} \int_{\mathbf{R}}^{\mathbf{r}} \mathbf{A} \cdot d \mathbf{r}

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

\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-independent parameter $\mathbf{R}$ 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 even in a region where there's no $\mathbf{B}$. The magnitude of this phase is proportional to the magnetic flux going through the solenoid. That is,

$$

\gamma = \pm \frac{e \phi}{2\hbar}.

$$

The phase acquired moving around the solenoid can be described as a Berry phase by observing that,

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 Schroedinger equation*:

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>\,

$$

...

...

@@ -11,62 +30,43 @@ 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,

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 Schroedinger equation

is readily soled has

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

is readily solved as,

$$

H \phi_n = E_n \phi_n \hspace{0.5cm} ;

\hspace{0.5cm} \phi_n(t)= \phi_n(0) e^{-i E_n t / \hbar},

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 $\underline{definition}$:

We define $\psi_n(t)$ to be those eigenfunctions that solve the

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.

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

Be careful here: the states $\psi_n(t)$ *do not solve the time-dependent

Schroedinger equation \eqref{eq:TDSE}!* (despite the occurance of $H(t)$ in the equation).

In contrast, the $\psi_n(t)$ that we have just defined solve a stationary

Schr\"{o}dinger 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}$.

!!! 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.

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 $\psi_n(t)$:

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 $\exp (-i E_n t/ \hbar)$,

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 Schr$\"{o}$dinger equation to it. We get

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} =

...

...

@@ -79,33 +79,26 @@ $$

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 Equation \ref{eq:TDSE_superposition}.

What we are left with is

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.

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 Schr\"{o}dinger equation

which we used to define the $\psi_n$ in the first place (Equation

\ref{eq:time_dep_H}).

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 $\textbf{approximation}$ that you must

have been anticipating since reading the title of these notes. We

*neglect* the contributions when $n \neq m$ and just get:

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.

> 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 equation \ref{eq:cm} gives

Solving the equation for the coefficients gives

$$

c_m(t) = c_m(0) e^{i \gamma(t)} ;

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

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 below.)

(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

...

...

@@ -163,38 +143,74 @@ $$

$$

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

### What is slow?

### 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,

In order to obtain the result above, the terms $n\neq m$ have been neglected by considering their contribution small. However, we ask, small compared to what? 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 \bar{\langle \psi_m | \dot H |\psi_n\rangle}

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

$$

* The largest contribution will come from the smallest energy difference.

* The smallest energy difference will contribute the most.

$$

E_m(t) - E-n(t) \approx \bar{E_m - E-n}

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, and an it can be solved as,

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,

The coefficients for $n \neq m$ have been neglected using the argument that they are small. This can be explicitly written by saying that they are much smaller than 1. That is,

* 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.

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