Transition Probabilities

The amplitude $c_f^{(1)}(t)$ is complex and not directly observable. By the Born rule, the physically measurable quantity is its squared magnitude — the probability of finding the system in state $|f\rangle$ .

From amplitude to probability

The first-order transition probability is

P_{i\to f}(t) = \big|c_f^{(1)}(t)\big|^2 .

For the constant perturbation $\hat H' = \hat V$ switched on at $t=0$ , we found

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

so squaring gives

\boxed{\;P_{i\to f}(t) = \frac{|V_{fi}|^2}{\hbar^2}\,\frac{4\sin^2(\omega_{fi}t/2)}{\omega_{fi}^2}\;}

where $V_{fi}=\langle f|\hat V|i\rangle$ and $\omega_{fi}=(E_f-E_i)/\hbar$ .

Reading the formula

Two factors control the probability:

The matrix element $|V_{fi}|^2$ . If the perturbation cannot connect $|i\rangle$ and $|f\rangle$ — that is, $\langle f|\hat V|i\rangle = 0$ — then $P_{i\to f}=0$ to this order. This is the origin of selection rules, studied later in the module.
The energy-mismatch factor $4\sin^2(\omega_{fi}t/2)/\omega_{fi}^2$ . This is sharply peaked at $\omega_{fi}\to 0$ , i.e. when the initial and final states are nearly degenerate. Away from that, the probability oscillates but stays small.

The small-energy-gap limit

Near $\omega_{fi}=0$ the factor has a finite limit. Using $\sin x \approx x$ ,

\frac{4\sin^2(\omega_{fi}t/2)}{\omega_{fi}^2} \;\xrightarrow{\;\omega_{fi}\to 0\;}\; t^2 ,

so for transitions to (near-)degenerate states the probability grows quadratically in time:

P_{i\to f}(t) \;\approx\; \frac{|V_{fi}|^2}{\hbar^2}\,t^2 .

A useful shape: the sinc function

Writing $\Omega = \omega_{fi}$ , the energy-mismatch factor is $t^2\,\mathrm{sinc}^2(\Omega t/2)$ with $\mathrm{sinc}(x)=\sin x / x$ . As $t$ grows the central peak at $\Omega=0$ becomes taller (height $t^2$ ) and narrower (width $\sim 2\pi/t$ ). This narrowing is the seed of energy conservation: only final states with $E_f \approx E_i$ accumulate appreciable probability, and the longer the perturbation acts, the more strictly that condition is enforced — a time–energy uncertainty relation $\Delta E\,t \sim \hbar$ .

Try it

For a constant perturbation with $V_{fi}=0.3$ , $\omega_{fi}=1.5$ , $t=2.0$ , and $\hbar=1$ , compute $P_{i\to f}(t) = (|V_{fi}|^2/\hbar^2)\,4\sin^2(\omega_{fi}t/2)/\omega_{fi}^2$ and return it. (Check that your answer lies between 0 and 1.)

Run your code to see the quantum state.