## Motivation

In time independent perturbation theory, we saw that a time dependent perturbation $H'(t)$ can cause transitions $|\psi_a \rangle \rightarrow |\psi_b \rangle$

For this to occur, frequency $\omega$ of the perturbation had to match the spacing between those levels.

$\omega_0 = \frac{E_b - E_a}{\hbar} \quad \text{(transition frequency)}$

But what if

$\omega \ll \frac{E_b - E_a}{\hbar}$

This is a very slow perturbation, which is what we call the adiabatic limit. We expect system initially in eigen state $|\psi_a \rangle$ will stay in that state (no transitions), but eigenstate itself will slowly change in time.

$|\Psi(t) \rangle \sim |\Psi_a (t) \rangle$

Example. Harmonic oscillator with time dependence $K(t) = K_0 \cos(\omega t)$.

Suppose system starts off in ground state at time $t=0$.

Then,

$K(0) = K_0$$\Psi(x, 0) = \psi_0(x) = \left(\frac{m \Omega}{\pi \hbar}\right)^{\frac{1}{4}} e^{-m \Omega_{0} x^{2} / 2 \hbar}$$\Omega_{0}=\sqrt{\frac{k_{0}}{m}}$

If spring constant changes very slowly, $(\omega \ll \Omega_0)$ we expect the only effect is that $\Omega_0$ becomes time dependent $\Omega_0 \rightarrow \Omega(t) = \sqrt{K(t)/m}$

$\Psi(x, t)=\left(\frac{m \Omega(t)}{\pi \hbar}\right)^{\frac{1}{4}} e^{-m \Omega(t) x^{2} / 2 \hbar}$

If the time dependence is slow, the solution of the time dependent Schrodinger equation

$i \hbar \frac{d|\Psi(t)\rangle}{d t}=H(t)|\Psi(t)\rangle$

should be well approximated by a succession of eigenvalue problems.

$H(t)\left|\Psi_{n}(t)\right\rangle= E_{n}(t)\left|\Psi_{n}(t)\right\rangle$

Solving the time independent Schrodinger equation for each $t$,

$|\Psi(t)\rangle \sim\left|\Psi_{n}(t)\right\rangle$

if $|\Psi(0) \rangle=\left|\Psi_{n}(0)\right\rangle$, we have the following theorem:

If $H(t)$ varies slowly in time with respect to level spacing, the system prepared in the $n$th eigenstate $|\Psi_n(0)\rangle$ of $H(0)$ will remain in the $n$th instantaneous eigenstate $|\Psi(t)$ of $H(t)$, picking up only a phase factor.

$\boxed{|\Psi(t) \rangle = e^{i\theta_n(t)}e^{i\gamma_n(t)}|\psi_n(t)\rangle}$

We say that $\theta_n(t)$ is the dynamical phase, with $E_n(t)$ the instantaneous eigenenergies.

$\theta_n(t) = -\frac{1}{\hbar} \int_{0}^{t} d t^{\prime} E_{n}\left(t^{\prime}\right)$

and $\gamma_n(t)$ is the geometrical phase.

$\gamma_n(t) = i \int_0^t dt' \left\langle\psi_{n}\left(t^{\prime}\right)\left|\frac{\partial}{\partial t^{\prime}}\right|\psi_{n}\left(t^{\prime}\right)\right\rangle$

### Proof of the adiabatic theorem

Define $|psi_n (\theta)\rangle$ and $E_n(t)$ via the time independent schrodinger equation,

$H(t) |\psi_n \rangle = E_n(t) |\psi_n(t)\rangle$

and treat $t$ as a parameter and solve the time independent Schrodinger equation for each $t$.

• For each $t$, $\{\psi_n(t)\}$ forms a complete orthonormal basis.
• Expand solution $|Psi(t)\rangle$ of full time dependent Schrodinger equation into that basis, for each $t$.
$\Psi_{n}(t)=\psi_{n} e^{-i E_{n} t / h}$

If the Hamiltonian changes with time, then the eigenfunctions and eigenvalues are time-dependant:

$H(t) \psi_{n}(t)=E_{n}(t) \psi_{n}(t)$

But, they are at least orthonormal and complete:

$\left\langle \psi_n(t) | \psi_{m}(t)\right\rangle=\delta_{m n}$

Hence, we can take our most general state, the solution to the time dependant Schrodinger equation

$i \hbar \frac{\partial}{\partial t} \Psi(t)=H(t) \Psi(t)$

and expand into the instantaneous basis:

$\Psi(t)=\sum_{n} c_{n}(t) \psi_{n}(t) e^{i \theta_{n}(t)}$

where

$\theta_{n}(t) \equiv-\frac{1}{\hbar} \int_{0}^{t} E_{n}\left(t^{\prime}\right) d t^{\prime}$

Substitute the instantaneous basis into the time dependent Schrodinger equation:

$i \hbar \sum_{n}\left[\dot{c}_{n} \psi_{n}+c_{n} \dot{\psi}_{n}+i c_{n} \psi_{n} \dot{\theta}_{n}\right] e^{i \theta_{n}}=\sum_{n} c_{n}\left(H \psi_{n}\right) e^{i \theta_{n}}$$\sum_{n} \dot{c}_{n} \psi_{n} e^{i \theta_{n}}=-\sum_{n} c_{n} \dot{\psi}_{n} e^{i \theta_{n}}$

The inner product and orthonormality gives:

$\sum_{n} \dot{c}_{n} \delta_{m n} e^{i \theta_{n}}=-\sum_{n} c_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle e^{i \theta_{n}}$$\dot{c}_{m}(t)=-\sum_{n} c_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle e^{i\left(\theta_{n}-\theta_{m}\right)}$

Now, take the time derivative of Schrodinger equation

$\frac{\partial}{\partial t} H(t)\left|\psi_{n}(t)\right\rangle=\frac{\partial}{\partial l}\left(E_{n}(t)\left|\psi_{n}(t)\right\rangle\right)$

gives

$\dot{H} \psi_{n}+H \dot{\psi}_{n}=\dot{E}_{n} \psi_{n}+E_{n} \dot{\psi}_{n}$

Taking the inner product $\langle \psi_m (t) | \quad m\neq n$

$\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle+\left\langle\psi_{m}|H| \dot{\psi}_{n}\right\rangle=\dot{E}_{n} \delta_{m n}+E_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle$

Therefore,

$\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle=\left(E_{n}-E_{m}\right)\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle$$\dot{c}_{m}(t)=-c_{m}\left\langle\psi_{m} | \dot{\psi}_{m}\right\rangle-\sum_{n \neq m} c_{n} \frac{\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle}{E_{n}-E_{m}} e^{(-i / \hbar)\int_{0}^{t}\left[E_{n}\left(t^{\prime}\right)-E_{u t}\left(t^{\prime}\right)\right] d t^{\prime}}$

Now invoke the adiabatic approximation. Assume $\dot{H}$ is extremely small, and drop the second term. We now have:

$\dot{c}_{m}(t)=-c_{m}\left\langle\psi_{n} | \dot{\psi}_{m}\right\rangle$

the solution is:

$c_{m}(t)=c_{m}(0) e^{i \gamma_{m}(t)}$

where

$\gamma_{n}(t) \equiv i \int_{0}^{t}\left\langle\psi_{m}\left(t^{\prime}\right) | \frac{\partial}{\partial t^{\prime}} |\psi_{m}\left(t^{\prime}\right)\right\rangle d t^{\prime}$

And so, the final answer is

$\Psi_{n}(t)=e^{i \theta_{n}(t)} e^{i \gamma_{n}(t)} \psi_{n}(t)$

Which is the statement of the adiabatic theorem.
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