Transition Amplitudes

We now derive the single most important formula of the module: the first-order transition amplitude. It answers the question — if the system starts in $|i\rangle$ , how much amplitude has leaked into a different state $|f\rangle$ after the perturbation has acted?

The Dyson series, truncated

In the interaction picture the state obeys

i\hbar\,\frac{\partial}{\partial t}\,|\Psi_I(t)\rangle = \hat H'_I(t)\,|\Psi_I(t)\rangle .

Integrating both sides from $0$ to $t$ turns this into an integral equation:

|\Psi_I(t)\rangle = |\Psi_I(0)\rangle - \frac{i}{\hbar}\int_0^t \hat H'_I(t')\,|\Psi_I(t')\rangle\,dt' .

This is exact but implicit — $|\Psi_I\rangle$ appears on both sides. The perturbative idea is to iterate it, generating the Dyson series. To first order we approximate the state inside the integral by its initial value $|\Psi_I(0)\rangle = |i\rangle$ :

|\Psi_I(t)\rangle \approx |i\rangle - \frac{i}{\hbar}\int_0^t \hat H'_I(t')\,|i\rangle\,dt' .

Projecting onto the final state

The amplitude to be found in a different state $|f\rangle$ (with $f \neq i$ ) is its coefficient $c_f(t) = \langle f|\Psi_I(t)\rangle$ . Projecting the expansion above and using $\langle f|i\rangle = 0$ :

\boxed{\,c_f^{(1)}(t) = -\frac{i}{\hbar}\int_0^t \langle f|\hat H'_I(t')|i\rangle\,dt' = -\frac{i}{\hbar}\int_0^t \langle f|\hat H'(t')|i\rangle\, e^{\,i\omega_{fi}t'}\,dt'\, }

where in the last step we used $\hat H'_I(t') = e^{i\hat H_0 t'/\hbar}\hat H'(t')e^{-i\hat H_0 t'/\hbar}$ acting on the eigenstates, which produces the phase $e^{i(E_f - E_i)t'/\hbar} = e^{i\omega_{fi}t'}$ . The object $\langle f|\hat H'(t')|i\rangle \equiv H'_{fi}(t')$ is the matrix element of the perturbation between the two states.

Worked example: a constant perturbation

Suppose the perturbation is simply switched on at $t=0$ and held constant, $\hat H'(t) = \hat V$ for $t>0$ , so the matrix element $V_{fi} = \langle f|\hat V|i\rangle$ is time-independent. The integral is elementary:

c_f^{(1)}(t) = -\frac{i}{\hbar}\,V_{fi}\int_0^t e^{\,i\omega_{fi}t'}\,dt' = -\frac{i}{\hbar}\,V_{fi}\,\frac{e^{\,i\omega_{fi}t}-1}{i\omega_{fi}} = -\frac{V_{fi}}{\hbar}\,\frac{e^{\,i\omega_{fi}t}-1}{\omega_{fi}} .

Taking the magnitude and using $|e^{i\theta}-1| = 2|\sin(\theta/2)|$ :

\big|c_f^{(1)}(t)\big| = \frac{|V_{fi}|}{\hbar}\,\frac{2\,\big|\sin(\omega_{fi}t/2)\big|}{|\omega_{fi}|} .

Why the phase matters

The integrand carries the oscillating phase $e^{i\omega_{fi}t'}$ . For a constant perturbation this phase oscillates and the integral does not grow without bound — the amplitude just wiggles. The way to make the integral build up is to give the perturbation its own oscillation that cancels this phase. That is precisely the harmonic drive and resonance we study next.

Try it

Compute the magnitude of the first-order amplitude for a constant perturbation with $V_{fi}=0.5$ , $\omega_{fi}=2.0$ , $t=1.0$ , in units $\hbar=1$ . Use $|c_f^{(1)}| = (|V_{fi}|/\hbar)\,2|\sin(\omega_{fi}t/2)|/|\omega_{fi}|$ and return the number.

Run your code to see the quantum state.