The adiabatic theorem

Intuition

The opposite extreme: change the Hamiltonian slowly. Then a system that starts in the $n$ -th eigenstate stays in the $n$ -th eigenstate of the Hamiltonian of each moment, following it continuously, as long as the level never meets another. A box whose wall is moved out slowly keeps its particle in the ground state, whose energy falls smoothly; an oscillator whose spring slowly stiffens keeps its quantum number, its energy growing with the frequency. The only record of the journey is a phase.

Carry a full cup of coffee across a room slowly and it arrives full: the surface tilts gently and follows your hand. Rush, and the coffee sloshes into other motions. The adiabatic theorem is the slow walk.

The same wall moved slowly from 1 to 2: the particle stays in the ground state of the box as it is at each moment, and the wavefunction widens with the box. Its energy falls as $1/L^{2}$ , to a quarter — while the sudden move of the last lesson left it in the new ground state only 36 times in 100.

Let $\hat{H}(t)$ change slowly, with the instantaneous eigenstates $\hat{H}(t)\ket{n(t)}=E_{n}(t)\ket{n(t)}$ . A system starting in is later in , up to a phase:

\begin{aligned}\ket{\psi(t)}&\approx e^{i\theta_{n}(t)}e^{i\gamma_{n}(t)}\ket{n(t)}\\ \theta_{n}(t)&=-\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')\,dt'\\ \gamma_{n}(t)&=i\int_{0}^{t}\braket{n(t')|\dot n(t')}\,dt'\end{aligned}

Properties

Slow means $\hbar\,|\bra{m}\dot{\hat H}\ket{n}|\ll(E_{m}-E_{n})^{2}$ for every other level : the change must be gentle against the gaps, and a gap must never close.

Why a slow change keeps the level

\ket{\psi(0)}=\ket{n(0)}\ \Longrightarrow\ \ket{\psi(t)}\approx e^{i(\theta_{n}+\gamma_{n})}\ket{n(t)}

Expand the state in the instantaneous eigenstates with their dynamical phases taken out. Schrödinger’s equation then changes each coefficient through the overlap of a state with its own rate of change, and through couplings to the other levels, each divided by a gap and turning at the difference of the dynamical phases. For a slow change the couplings are small and average away, and each coefficient keeps its size.

Proof steps

\ket{\psi(t)}=\sum_{n}c_{n}(t)\,e^{i\theta_{n}(t)}\ket{n(t)}

Expand in the instantaneous eigenstates, with the dynamical phases taken out.

\sum_{n}\bigl(\dot c_{n}\ket{n}+c_{n}\ket{\dot n}\bigr)\,e^{i\theta_{n}}=0

Put it into $i\hbar\,d\ket{\psi}/dt=\hat{H}\ket{\psi}$ : the terms with $E_{n}$ cancel against the derivative of .

\dot c_{m}=-\sum_{n}c_{n}\braket{m|\dot n}\,e^{i(\theta_{n}-\theta_{m})}

Project on $\bra{m}$ .

\braket{m|\dot n}=\frac{\bra{m}\dot{\hat H}\ket{n}}{E_{n}-E_{m}}\qquad(m\neq n)

Differentiate $\hat{H}\ket{n}=E_{n}\ket{n}$ in time and project on .

\dot c_{m}\approx-c_{m}\braket{m|\dot m}\ \Longrightarrow\ |c_{m}(t)|=|c_{m}(0)|

For a slow change the couplings are small and turn quickly, and are dropped; $\braket{m|\dot m}$ is imaginary, so only a phase is left.

Applications

Practice

Following the Level

Under a slow change a system that starts in the $n$ -th eigenstate stays in the $n$ -th eigenstate of the Hamiltonian of each moment. It keeps its place in the ladder of levels, not its energy.

Try it

A system starts in the second eigenstate of $\hat{H}(0)$ , and $\hat{H}$ changes slowly with no level ever meeting another. Where is it at the end?