Transition amplitudes · Quantum Mechanics

Intuition

Start a system in a level of the solved Hamiltonian, switch on a perturbation, and ask for the chance of finding it in another level later. To first order the amplitude is a single time integral: the perturbation’s matrix element between the two levels, weighted by a phase that turns at their Bohr frequency. A perturbation that changes slowly compared with that frequency averages to almost nothing; one with a component at the Bohr frequency adds up. The transition probability is the squared modulus of the amplitude, and first order can be trusted while it stays small.

Pushing a child on a swing: pushes timed to the swing’s own period add up, and pushes at random largely cancel. The amplitude integral adds the pushes with the swing’s phase attached to each.

Two levels of equal energy coupled by a constant $V$ , the system starting in one: the probability of the other against $Vt/\hbar$ . Exactly it is $\sin^{2}(Vt/\hbar)$ and never passes 1; first order gives $(Vt/\hbar)^{2}$ , which agrees while both are small and then grows without bound.

The first-order transition amplitude

Expand the interaction-picture state in the eigenstates of $\hat{H}_{0}$ , start it in $\ket{i}$ , and keep the lowest order in the perturbation.

\begin{aligned}c_{f}(t)&\approx-\frac{i}{\hbar}\int_{0}^{t}V_{fi}(t')\,e^{i\omega_{fi}t'}\,dt'\\ P_{i\to f}(t)&=|c_{f}(t)|^{2},\qquad \omega_{fi}=\frac{E_{f}-E_{i}}{\hbar}\end{aligned}

Properties

The amplitudes obey exactly $i\hbar\,\dot c_{f}=\sum_{n}V_{fn}(t)\,e^{i\omega_{fn}t}c_{n}$ ; first order puts the starting values on the right.

The first-order amplitude

c_{f}(t)=-\frac{i}{\hbar}\int_{0}^{t}V_{fi}(t')\,e^{i\omega_{fi}t'}\,dt'+O(V^{2})

Write the interaction-picture state as a sum over the levels of $\hat{H}_{0}$ and project its equation on the final level. The right-hand side involves every amplitude, but to lowest order they can be replaced by their starting values, which leaves a single known term to integrate.

Proof steps

\ket{\psi_{I}(t)}=\sum_{n}c_{n}(t)\ket{n},\qquad c_{n}(0)=\delta_{ni}

Expand in the eigenstates of $\hat{H}_{0}$ ; the system starts in $\ket{i}$ .

\begin{aligned}i\hbar\,\dot c_{f}&=\sum_{n}\bra{f}\hat{V}_{I}(t)\ket{n}\,c_{n}\\ &=\sum_{n}V_{fn}(t)\,e^{i\omega_{fn}t}\,c_{n}(t)\end{aligned}

Project the equation of the interaction picture on $\bra{f}$ .

i\hbar\,\dot c_{f}\approx V_{fi}(t)\,e^{i\omega_{fi}t}

To lowest order, put $c_{n}=\delta_{ni}$ on the right.

c_{f}(t)\approx-\frac{i}{\hbar}\int_{0}^{t}V_{fi}(t')\,e^{i\omega_{fi}t'}\,dt'

Integrate from $0$ , where $c_{f}=0$ for $f\neq i$ .

Applications

Practice

The Amplitude Integral

To first order, the amplitude of a jump from level $i$ to level $f$ is the time integral of the perturbation’s matrix element between them, with a phase turning at their Bohr frequency. For two levels of equal energy the phase stays at 1.

c_{f}(t)\approx-\frac{i}{\hbar}\int_{0}^{t}V_{fi}(t')\,e^{i\omega_{fi}t'}\,dt'

Try it

A constant $V_{fi}=0.002$ eV couples two levels of equal energy. What is the first-order probability of the jump after a time with $t/\hbar=50$ eV $^{-1}$ ?