The sudden approximation

Intuition

If the Hamiltonian changes abruptly — in a time short against $\hbar$ over the energies involved — the state has no time to respond. Just after the change it is the same vector as just before, but it is no longer an eigenstate: spread over the eigenstates of the new Hamiltonian, it is found in each with the squared overlap. When tritium’s nucleus emits a fast electron and becomes helium-3, the atomic electron, still in hydrogen’s ground state, suddenly finds a doubly charged nucleus, and ends in the ion’s ground state only seven times in ten.

Pull a tablecloth out fast enough and the dishes stay where they were; pull it slowly and they slide along with it. The sudden approximation is the fast pull; the next lesson is the slow one.

A particle in the ground state of a box of width 1, when the right wall is moved at once to 2. Just after, the wavefunction is unchanged (before); its overlap with the new ground state is $4\sqrt2/3\pi\approx0.60$ , so the particle is found there with probability $32/9\pi^{2}\approx0.36$ .

Let the Hamiltonian jump from $\hat{H}$ to $\hat{H}'$ in a time $\delta t$ much shorter than over the energies involved. The state is unchanged by the jump, and is then expanded in the eigenstates of .

\ket{\psi(0^{+})}=\ket{\psi(0^{-})},\qquad P_{n'}=\bigl|\braket{n'|\psi(0^{-})}\bigr|^{2}

Properties

It holds when $\delta t\ll\hbar/\Delta E$ , with $\Delta E$ the spread of energies the state has under the Hamiltonians involved.
The probability of leaving the initial state during the change itself is $(\delta t\,\Delta\bar H/\hbar)^{2}$ , derived below; it vanishes for an instant change.

Leaving the initial state during a quick change

P_{\text{leave}}\approx\frac{\delta t^{2}}{\hbar^{2}}\Bigl(\bra{i}\bar H^{2}\ket{i}-\bra{i}\bar H\ket{i}^{2}\Bigr)

During a short switch the evolution operator is the identity minus a term proportional to $\delta t$ and to $\bar H$ , the Hamiltonian averaged over the switch. The part of the evolved state outside $\ket{i}$ comes from that term alone, and its squared length is $\delta t^{2}/\hbar^{2}$ times the variance of . As it vanishes: the state is left as it was.

Proof steps

\begin{aligned}\hat{U}(\delta t)&\approx\hat{1}-\frac{i}{\hbar}\,\delta t\,\bar H\\ \bar H&=\frac{1}{\delta t}\int_{0}^{\delta t}\hat{H}(t)\,dt\end{aligned}

To first order in the short time $\delta t$ .

\hat{P}_{\perp}=\hat{1}-\ket{i}\bra{i},\qquad P_{\text{leave}}=\bigl\|\hat{P}_{\perp}\hat{U}\ket{i}\bigr\|^{2}

The part of the evolved state outside $\ket{i}$ .

\hat{P}_{\perp}\hat{U}\ket{i}=-\frac{i}{\hbar}\,\delta t\,\hat{P}_{\perp}\bar H\ket{i}

The identity term lies along $\ket{i}$ and is removed.

\begin{aligned}P_{\text{leave}}&=\frac{\delta t^{2}}{\hbar^{2}}\bra{i}\bar H\hat{P}_{\perp}\bar H\ket{i}\\ &=\frac{\delta t^{2}}{\hbar^{2}}\bigl(\Delta\bar H\bigr)^{2}\end{aligned}

$\hat{P}_{\perp}$ is a projector, and $\bra{i}\bar H(\hat{1}-\ket{i}\bra{i})\bar H\ket{i}$ is the variance of .

Applications

Practice

Too Fast to Follow

A Hamiltonian that changes in a time short against $\hbar$ over the energies involved leaves the state as it was. Only afterwards does the state evolve, under the new Hamiltonian.

\ket{\psi(0^{+})}=\ket{\psi(0^{-})}

Try it

Just after a sudden change of the Hamiltonian from $\hat{H}$ to $\hat{H}'$ , what is the state?