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adiabatic section

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# The Berry Phase
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}) = -\frac{e}{\hbar} \int_{\mathbf{r}_0}^{\mathbf{r}}
\mathbf{A}(\mathbf{r}) \cdot d \mathbf{r}
$$
$g(\mathbf{r})$ is defined as a line integral, which is only well-defined if
$\mathbf{B} = \nabla \times \mathbf{A} = 0$. This is precisely the
situation we have engineered with the solenoid.
$g(\mathbf{r})$ is defined as a line integral, which is only well-defined if $\mathbf{B} = \nabla \times \mathbf{A} = 0$. This is precisely the situation we have engineered with the solenoid.
It is simple to check that this solution works:
......@@ -129,30 +135,23 @@ $$
\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}
$$
If, contrary to the figure, we take two semi-circular paths around the solenoid,
we can evaluate what the phase associated with each path is using the circle
element on the path $r d\phi$.
If, contrary to the figure, we take two semi-circular paths around the solenoid, we can evaluate what the phase associated with each path is using the circle element on the path $r d\phi$.
$$
\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}.
$$
The *difference* between the two paths is then $e \phi/\hbar$. As we saw
in Equation \ref{eq:interference}, this difference can be measured in experiment
through interference.
The *difference* between the two paths is then $e \phi/\hbar$. As we saw in the beam splitter, this difference can be measured in experiment through interference.
Some comments:
......@@ -175,7 +174,7 @@ With the solenoid present we use the same trick as we did above by setting
$\psi$ to
$$
\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}
$$
......@@ -193,7 +192,7 @@ phase:
$$
\begin{aligned}
\langle \psi| \nabla_{\mathbf{R} \rangle \psi} &= \int e^{-ig} \psi_0(\mathbf{r} -
\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 \\
......@@ -206,9 +205,26 @@ So we find that
$$
\begin{aligned}
\gamma_n &= i \oint -\frac{ie}{\hbar} \mathbf{A}(\mathbf{R}) \cdot
\gamma_n &= - i \oint \frac{ie}{\hbar} \mathbf{A}(\mathbf{R}) \cdot
d\mathbf{R} \\
&= \frac{e}{\hbar} \int \nabla \times \mathbf{A} da \\
&= \frac{e\phi}{\hbar}.
\end{aligned}
$$
## 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,
$$
\langle \psi| \nabla_{\mathbf{R}} \psi \rangle = -\frac{ie}{\hbar} \mathbf{A}(\mathbf{R}).
$$
This relation holds when $\langle \mathbf{p} \rangle = 0$, i.e. for a particle in a box.
# The Landau-Zener effect
## Adiabatic evolution of a two level system
The problem of two levels crossing is often encountered in quantum mechanics applications.
### Slow evolution
### Fast evolution
## Summary
Proof of adiabatic theorem
# 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 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 $\psi_n(t)$:
$$
\Psi(t) = \sum_n c_n(t) \psi_n(t) e^{i \theta_n(t)}.
$$
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 $\exp (-i E_n t/ \hbar)$,
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)
$$
so the $\theta$ derivative term on the left-hand side cancels with the
right-hand side of Equation \ref{eq:TDSE_superposition}.
What we are left with is
so the $\theta$ derivative term on the left-hand side cancels with the right-hand side of the previous equation. What we are left with is
$$
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}
$$
By projecting these expressions on the $\psi_m$ eigenstate (which amounts to
applying $\bra{\psi_m}$ from the left on both sides), we find that
By projecting these expressions on the $\psi_m$ eigenstate (which amounts to applying $\langle\psi_m|$ from the left on both sides), we find that
$$
\dot{c}_m = - \sum_n c_n \langle{\psi_m| \dot\psi_n} \rangle e^{i (\theta_n - \theta_m)}
$$
where we used the orthogonality of $\psi_n$ and $\psi_m$ to eliminate the sum on
the left-hand side.
where we used the orthogonality of $\psi_n$ and $\psi_m$ to eliminate the sum on the left-hand side.
In order to make progress evaluating what $\langle{\psi_m| \dot\psi_n} \rangle$ might
be, we take the derivative of the time-independent Schr\"{o}dinger equation
which we used to define the $\psi_n$ in the first place (Equation
\ref{eq:time_dep_H}).
In order to make progress evaluating what $\langle{\psi_m| \dot\psi_n} \rangle$ might be, we take the derivative of the time-independent Schrodinger equation which we used to define the $\psi_n$ in the first place. That is,
$$
\dot{H}\psi_n + H \dot{\psi}_n = \dot{E}_n \psi_n + E_n \dot{\psi}_n
$$
Again projecting this onto the $\psi_m$ gives us:
$$
......@@ -123,36 +116,23 @@ $$
\langle{\psi_m| \dot\psi_n} \rangle = \frac{\langle{\psi_m | \dot{H} | \psi_n }\rangle}{E_n - E_m}.
$$
When we have slow, gradual change being applied to the system, $\dot{H}$ is
small. Now we are finally ready to make the $\textbf{approximation}$ 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:
$$
\dot{c}_m = -c_m \langle{\psi_m| \dot\psi_m} \rangle.
$$
!!! 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.
> 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,
$$
\frac{\langle \psi_m | \dot H |\psi_n\rangle}{E_m - E-n}<< \frac{E_m - E_n}{\hbar}.
\frac{\langle \psi_m | \dot H |\psi_n\rangle}{E_m - E_n}<< \frac{E_m - E_n}{\hbar}.
$$
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,
$$
\dot c_n(t) = \frac{\langle \psi_m | \dot H |\psi_n \rangle}{E_m(t) - E-n(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_m(t') - E-n(t') dt'}.
\dot c_n(t) = \frac{\langle \psi_m | \dot H |\psi_n \rangle}{E_m(t) - E_n(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_m(t') - E_n(t') dt'}.
$$
The coefficient is obtained by integrating the previous equation as,
$$
\rightarrow c_n(T) = \int_{0}^T \dot c_n(t) dt = \int_{0}^T \frac{\langle \psi_m | \dot H |\psi_n\rangle}{E_m(t) - E-n(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_m(t') - E-n(t') dt'} dt.
\rightarrow c_n(T) = \int_{0}^T \dot c_n(t) dt = \int_{0}^T \frac{\langle \psi_m | \dot H |\psi_n\rangle}{E_m(t) - E_n(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_m(t') - E_n(t') dt'} dt.
$$
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.
$$
\langle \psi_m (t) | \dot H |\psi_n(t) \rangle \approx \bar{\langle \psi_m | \dot H |\psi_n\rangle}
\langle \psi_m (t) | \dot H |\psi_n(t) \rangle \approx \overline{\langle \psi_m | \dot H |\psi_n\rangle}
$$
* 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,
$$
c_n(T) = \frac{\bar{\langle \psi_m | \dot H |\psi_n\rangle}}{\bar{E_m - E-n}} e^{-\frac{i}{\hbar} \int_0^t \bar{E_m - E-n} dt'}
c_n(T) = \frac{\overline{\langle \psi_m | \dot H |\psi_n\rangle}}{\overline{E_m - E_n}} e^{-\frac{i}{\hbar} \int_0^t \overline{E_m - E_n} dt'}
$$
$$
c_n(T) = \frac{\bar{\langle \psi_m | \dot H |\psi_n\rangle}}{\bar{E_m - E-n}} \frac{i\hbar }{\bar{E_m - E-n}}\left( e^{-i \bar{E_m - E-n} T /\hbar } - 1 \right).
c_n(T) = \frac{\overline{\langle \psi_m | \dot H |\psi_n\rangle}}{\overline{E_m - E_n}} \frac{i\hbar }{\overline{E_m - E_n}}\left( e^{-i \overline{E_m - E_n} T /\hbar } - 1 \right).
$$
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,
$$
|c_n(T)| \leq \frac{\hbar \bar{\langle \psi_m | \dot H |\psi_n\rangle} }{\bar{E_m - E-n}^2} << 1
|c_n(T)| \leq \frac{\hbar \overline{\langle \psi_m | \dot H |\psi_n\rangle} }{\overline{E_m - E_n}^2} << 1
$$
Observe that from the following result 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.
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