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# The Berry Phase
!!! 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
$$
\frac{d}{dt} \left( \langle{\psi_n}\rangle \right) =
\langle{\dot \psi_n | \psi_n} \rangle + \langle{ \psi_n | \dot\psi_n} \rangle =
\left(\langle{ \psi_n | \dot\psi_n} \rangle\right)^* + \langle{ \psi_n | \dot\psi_n} \rangle =
2 \Re \left( \langle{ \psi_n | \dot\psi_n} \rangle \right) = 0
$$
So the real part of $\langle{ \psi_n | \dot\psi_n} \rangle$ is 0, meaning that the Berry
phase is real. Typically, $H(t)$ gets its time dependence through some parameter
which depends on $t$, for example on the width of the infinite square well. In
that case we express $H$ as $H(\vec{R}(t))$ where $\vec{R}$ is just the set of
parameters that depend on $t$. Then
$$
\dot{\psi}_n = (\vec{\nabla}_{R} \psi_n) \cdot \dot{R}
$$
and so
$$
\begin{aligned}
\gamma_n(t) &= i \int_0^t \langle{ \psi_n | \vec{\nabla_R}\psi_n} \rangle \cdot
\dot{R}(t') dt' \\
&= i \int_{\vec{R}(0)}^{\vec{R}(t)} \langle{ \psi_n | \vec{\nabla_R}\psi_n} \rangle
\cdot d\vec{R}.
\end{aligned}
$$
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
$$
\begin{aligned}
\psi_{tot} &= \frac{1}{2} \psi_0 + \frac{1}{2} \psi_0 e^{i \gamma} \\
|\psi_{tot}|^2 &= \frac{1}{4} |\psi_0|^2 \left( 1 + e^{i \gamma} \right)
\left( 1 + e^{-i \gamma} \right) \\
&= \frac{1}{2} |\psi_0|^2 (1 + \cos(\gamma)) \\
&= |\psi_0|^2 \cos(\gamma/2)
\end{aligned}
$$
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.
## 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
$$
\psi (\mathbf{r}) = e^{i g(\mathbf{r})} \psi_0(\mathbf{r}) ,\text{ where }
g(\mathbf{r}) = -\frac{e}{\hbar} \int_{\mathbf{r}_0}^{\mathbf{r}}
\mathbf{A}(\mathbf{r}) \cdot d \mathbf{r}
$$
$g(\mathbf{r})$ 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:
$$
\begin{aligned}
(\mathbf{p} + e\mathbf{A})\psi &= (-i\hbar\nabla + e\mathbf{A})\psi \\
&= -i\hbar e^{i g} i (\nabla g) \psi_0 - i\hbar e^{ig} \nabla \psi_0
+ e \mathbf{A} e^{ig}\psi_0 \\
&= -i\hbar e^{i g} i (-\frac{eA}{\hbar}) \psi_0 - i\hbar e^{ig} \nabla \psi_0
+ e \mathbf{A} e^{ig}\psi_0 \\
&= e^{ig}(\mathbf{p}\psi_0)
\end{aligned}
$$
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 $\psi=e^{ig} \psi_0$ 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 $r d\phi$.
$$
\gamma = -\frac{e}{\hbar} \int \frac{\phi}{2 \pi r} \hat{\phi} \cdot r d\phi \hat{\phi} =
-\frac{e \phi}{ 2 \pi \hbar} \int d\phi = \pm \frac{e \phi}{2\hbar}.
$$
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
$\psi$ to
$$
\psi (\mathbf{r}) = e^{i g(\mathbf{r})} \psi_0(\mathbf{r} - \mathbf{R}), \text{ where }
g(\mathbf{r}) = -\frac{e}{\hbar} \int_{\mathbf{R}}^{\mathbf{r}} \mathbf{A} \cdot d \mathbf{r}
$$
Then,
$$
\nabla_{\mathbf{R}} \psi(\mathbf{r}) = -i \frac{e}{\hbar} \mathbf{A}(\mathbf{R})
e^{i g(\mathbf{r})} \psi_0(\mathbf{r} - \mathbf{R})
- e^{i g(\mathbf{r})} \nabla_{\mathbf{r}} \psi_0(\mathbf{r} - \mathbf{R})
$$
where we used the fact that $\nabla_{\mathbf{R}} \psi_0(\mathbf{r} -
\mathbf{R}) = -\nabla_{\mathbf{r}} \psi_0(\mathbf{r} - \mathbf{R})$.
We can now use this to evaluate inner product inside the integral of the Berry
phase:
$$
\begin{aligned}
\langle \psi| \nabla_{\mathbf{R}} \psi \rangle &= \int e^{-ig} \psi_0(\mathbf{r} -
\mathbf{R}) \left[ -\frac{e}{\hbar}\mathbf{A}(\mathbf{R})
e^{ig}\psi_0(\mathbf{r} - \mathbf{R}) - e^{ig} \nabla_{\mathbf{r}}
\psi_0(\mathbf{r} - \mathbf{R})\right] d^3r \\
&= -\frac{ie}{\hbar} \mathbf{A}(\mathbf{R}) - \frac{i}{\hbar} \langle{\mathbf{p}}{\psi_0}\rangle.
\end{aligned}
$$
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,
$$
\gamma = \pm \frac{e \phi}{2\hbar}.
$$
1. **Expanding quantum well**
Consider a quantum well with hard wall boundaries with a width that
changes in time, $W=W(t)$. One of the walls moves with constant
velocity $v$ such that the well width changes from $W_1 = W(0)$ to
$W_2 = W(T)$.
- Compute the dynamical phase of this process.
- Compute explicitly the geometrical phase of the process (we know
it has to be zero), by evaluating
$$\int d\mathbf{R} \langle \psi | \nabla_\mathbf{R} \psi \rangle\,.$$
*Note: in this general formula, $\mathbf{R}$ stands for the changing
parameters. What is $\mathbf{R}$ in our case? When evaluating
the derivative, don't forget the normalization of the wave
function.*
- The dynamical phase depends on the details of the
time-evolution. Show an explicit example demonstrating that if
the well width changes in a different manner from $W_1$ to
$W_2$, the dynamical phase is different.
2. **Abrupt expansion of a quantum well**
Consider a particle in a one-dimensional, square quantum well of
width $W$ with infinitely high walls:
\begin{equation}
V(x)=\begin{cases} 0&\text{if $0< x <W$}\\
\infty&\text{else.}\end{cases}
\end{equation}
The particle is initially in the first excited state.
Then, at time $t=0$ the right wall is suddenly moved from $x=W$ to
$x=2W$.
1. Compute the time evolution of the wave function for $t>0$. To
this end, expand the wave function at time $t=0$ (for which
$\psi(x, t=0)$ is the first excited state of the original
quantum well for $0<x<W$, and $\psi(x, t=0)=0$ for $W<x<2W$) in
terms of the eigenfunctions of the quantum well with width $2W$.
Use these to express the time-evolution involving a sum over
these eigenstates.
2. Evaluate the time evolution numerically, and make a few plots
showing the wave function at different times.
3. **Recovering the adiabatic theorem for an exactly solvable problem**
The case of an infinite square well whose right wall expands at a
constant velocity $v$ can be solved exactly. A complete set of
solutions is given by (you do *not* need to check this)
$$\phi_n(x,t) = \sqrt{\frac{2}{W(t)}}
\sin \left(\frac{n \pi}{W(t)} x \right)
e^{i(m v x^2 - 2 E_n^0 a t)/2 \hbar W(t)}\,,$$ where
$W(t) = a + v t$ is the increasing width of the well, and
$E_n^0= n^2 \pi^2 \hbar^2/2 m a^2$ is the $n$-th eigenenergy of the
original well with width $a$. The $\phi_n(x, t)$ form a complete and
orthonormal set at any time $t$. Hence, one can write the general
time evolution as $$\Psi(x, t) = \sum c_n \phi_n(x,t)\,,$$ with
time-independent coefficients $c_n$.
1. Suppose a particle starts out at $t=0$ in the ground state of
the quantum well,
$$\Psi(x, t=0) = \sqrt{\frac{2}{a}} \sin \left(\frac{\pi}{a} x\right)\,.$$
Show that the expansion coefficients can be written as
$$c_n = \frac{2}{\pi} \int_0^\pi e^{i \alpha z^2} \sin(n z) \sin(z) dz\,,$$
where $\alpha = m v a/2 \pi^2 \hbar$.
2. Show that in the limit $\alpha \ll 1$ (i.e.
$e^{i\alpha z} \approx 1$) you recover the result of the
adiabatic theorem that the system stays in the ground state.
Rephrase the condition $\alpha \ll 1$ in terms of time scales:
On the one hand, there is the time to expand the well by length
$a$, $T_e = a/v$. To which other time do I have to compare to?
3. Compare the wave function that you get from a) and b) with the
wave function from the adiabatic theorem (including phases)!
From this comparison, what is the value of the geometric phase
$\gamma(t)$ that you find?
4. **Berry phase of spin-$\frac{1}{2}$**
Consider a spin $\frac{1}{2}$ particle in an external magnetic
field. The magnetic field is rotated slowly, so that the spin always
points in the direction of the magnetic field.
In the lecture we showed that the Berry phase over a closed path
can be expressed as
$$\gamma(T) = i \oint \langle \psi(t)|
\nabla_\mathbf{R} \psi\rangle d\mathbf{R}\,.$$
Using Stoke's theorem the Berry phase can be expressed as
$$\gamma(T) = i \int \nabla_\mathbf{R} \times \langle \psi |\nabla_\mathbf{R} \psi
\rangle d\mathbf{a}\,,$$ where the integration is over an area that is bounded
by the closed path.
Use this expression to show that the Berry phase accumulated along a
closed path is given as $$\gamma(T) = -\frac{1}{2} \Omega\,$$ where
$\Omega$ is the solid angle covered by the rotating field.
Hints:
- The spin-$\frac{1}{2}$ spinor is given as $$\psi=\begin{pmatrix}
\cos(\theta/2)\\
e^{i \phi}\sin(\theta/2)
\end{pmatrix}$$
- The set of parameters $\mathbf{R}$ here is $\theta$ and $\phi$.
For the calculation, it is advantageous to consider this as the
set of spherical coordinates (we keep the radius $r$ fixed).
Then it is easiest if you use the expressions for the gradient
and the crossproduct in spherical coordinates:
$$\nabla f = \frac{\partial f}{\partial r} \mathbf{e}_r + \frac{1}{r}
\frac{\partial f}{\partial \theta} \mathbf{e}_\theta +
\frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \mathbf{e}_\phi$$
$$\begin{split}
\nabla \times \mathbf{A} & = \frac{1}{r \sin \theta} \left(
\frac{\partial}{\partial \theta} \left(\mathbf{A}_\phi \sin \theta \right)
- \frac{\partial \mathbf{A}_\theta}{\partial \phi}
\right) \mathbf{e}_r \\
& + \frac{1}{r} \left(
\frac{1}{\sin \theta} \frac{\partial \mathbf{A}_r}{\partial \phi} -
\frac{\partial}{\partial r}\left(r \mathbf{A}_\phi \right)
\right) \mathbf{e}_\theta \\
& + \frac{1}{r} \left(
\frac{\partial}{\partial r} \left(r \mathbf{A}_\theta\right) -
\frac{\partial \mathbf{A}_r}{\partial \theta}
\right) \mathbf{e}_\phi
\end{split}$$ Here, $\mathbf{e}_{r, \theta, \phi}$ are the
local, orthogonal unit vectors for spherical coordinates.
This diff is collapsed.
# Proof of adiabatic theorem
!!! success "Expected prerequisites"
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 $\psi_n(t)$:
$$
\Psi(t) = \sum_n c_n(t) \psi_n(t) e^{i \theta_n(t)}.
$$
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
$$
i \hbar \sum_n \left( \dot{c}_n \psi_n + c_n \dot{\psi}_n \right) e^{i \theta_n} = 0 \\
\sum_n \dot{c}_n \psi_n e^{i \theta_n} = - \sum_n c_n \dot{\psi}_n e^{i \theta_n}
$$
By projecting these expressions on the $\psi_m$ eigenstate (which amounts to applying $\langle\psi_m|$ from the left on both sides), we find that
$$
\dot{c}_m = - \sum_n c_n \langle{\psi_m| \dot\psi_n} \rangle e^{i (\theta_n - \theta_m)}
$$
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,
$$
\dot{H}\psi_n + H \dot{\psi}_n = \dot{E}_n \psi_n + E_n \dot{\psi}_n
$$
Again projecting this onto the $\psi_m$ gives us:
$$
\begin{aligned}
\langle{\psi_m | \dot{H} | \psi_n }\rangle + \langle{\psi_m | {H} | \dot \psi_n }\rangle =
\dot{E}_n \langle{\psi_m| \psi_n} \rangle + E_n \langle{\psi_m| \dot\psi_n} \rangle \\
\langle{\psi_m | \dot{H} | \psi_n }\rangle + E_m\langle{\psi_m| \dot\psi_n} \rangle =
\dot{E}_n \delta_{nm} + E_n \langle{\psi_m| \dot\psi_n} \rangle \\
\end{aligned}
$$
So for $n\neq m$, we find that
$$
\langle{\psi_m| \dot\psi_n} \rangle = \frac{\langle{\psi_m | \dot{H} | \psi_n }\rangle}{E_n - E_m}.
$$
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:
$$
\dot{c}_m = -c_m \langle{\psi_m| \dot\psi_m} \rangle.
$$
!!! summary "Adiabatic approximation"
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
$$
\Psi(t) = e^{i \theta_n(t)} e^{i\gamma_n(t)} \psi_n(t).
$$
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,
$$
\frac{\langle \psi_m | \dot H |\psi_n\rangle}{E_m - E_n}<< \frac{E_m - E_n}{\hbar}.
$$
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,
$$
\dot c_n(t) = \frac{\langle \psi_m | \dot H |\psi_n \rangle}{E_m(t) - E_n(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_m(t') - E_n(t') dt'}.
$$
The coefficient is obtained by integrating the previous equation as,
$$
\rightarrow c_n(T) = \int_{0}^T \dot c_n(t) dt = \int_{0}^T \frac{\langle \psi_m | \dot H |\psi_n\rangle}{E_m(t) - E_n(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_m(t') - E_n(t') dt'} dt.
$$
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,
$$
c_n(T) = \frac{\overline{\langle \psi_m | \dot H |\psi_n\rangle}}{\overline{E_m - E_n}} e^{-\frac{i}{\hbar} \int_0^t \overline{E_m - E_n} dt'}
$$
$$
c_n(T) = \frac{\overline{\langle \psi_m | \dot H |\psi_n\rangle}}{\overline{E_m - E_n}} \frac{i\hbar }{\overline{E_m - E_n}}\left( e^{-i \overline{E_m - E_n} T /\hbar } - 1 \right).
$$
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,
$$
|c_n(T)| \leq \frac{\hbar \overline{\langle \psi_m | \dot H |\psi_n\rangle} }{\overline{E_m - E_n}^2} << 1
$$
We recover what we first presented at this section.
## Summary
Under adiabatic evolution, the wavefunction of the system is given as,
$$
\Psi(t) = e^{i \theta_n(t)} e^{i\gamma_n(t)} \psi_n(t).
$$
This means that:
* 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.
This diff is collapsed.
# Guided exercise
In the first 30 minutes of the lecture I ask you to look at the "guided exercise below".
I have recorded solutions to all of the subproblems. The total length of the videos is
21 minutes and 16 seconds. If you have time before the lecture, you can try to solve the
problems explicitly before watching the videos. If you only have time during the lecture,
then please briefly think for each problem how you would solve it (without doing the
calculation explicitly), and test your expectations with respect to the solutions in the videos!
## Expanding quantum well
Consider a quantum well with hard wall boundaries with a width that
changes in time, $W=W(t)$. One of the walls moves with constant
velocity $v$ such that the well width changes from $W_1 = W(0)$ to
$W_2 = W(T)$ with $W_2>W_1$.
1. Write down the instantaneous eigenstates $\psi_n(t)$ and eigenenergies $E_n(t)$
of the problem. Don't forget wave function normalization!
Sketch the instantaneous eigenenergies as a function of $W(t)$.
**Solution**
<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/KnEcyJhrjnA?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
2. Compute the total dynamical phase of the process, i.e. $\theta_n(T) = -\frac{1}{\hbar} \int_0^T E_n(t) dt$.
Does this phase depend on the speed with which the wall moves?
**Solution**
<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/1xV6_l_yLFw?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
3. Compute the geometrical phase of the process, i.e. $\gamma_n(T) = i \int_0^T \langle \psi_n(t)|\frac{d}{dt} \psi_n(t)\rangle dt$.
**Hint**
<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/2Wa_uWfSAE4?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
**Solution**
<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/ZwCD0azjsWU?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
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