Fermi's Golden Rule

So far transitions went to a single discrete final state, giving a probability that oscillates or grows quadratically. In most real situations — an atom emitting into the continuum of photon modes, an electron scattering into a band of momentum states — the final state belongs to a continuum. Summing over that continuum converts the awkward $t^2$ growth into a clean, constant rate. The result is Fermi's golden rule.

Summing over a continuum of final states

The total probability of leaving $|i\rangle$ is obtained by summing $P_{i\to f}(t)$ over all accessible final states. For a continuum we replace the sum by an integral weighted by the density of states $\rho(E_f)$ — the number of final states per unit energy:

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

using the constant-perturbation result with $\omega_{fi}=(E_f-E_i)/\hbar$ .

The sinc-squared becomes a delta function

The factor $4\sin^2(\omega_{fi}t/2)/\omega_{fi}^2$ is the narrowing peak we met before. For large $t$ it behaves like a (scaled) Dirac delta in energy. Precisely,

\frac{\sin^2(\omega t/2)}{(\omega/2)^2} \;\xrightarrow{\;t\to\infty\;}\; 2\pi\,t\,\delta(\omega),

which, written in energy with $\omega=(E_f-E_i)/\hbar$ , gives $\delta(\omega) = \hbar\,\delta(E_f-E_i)$ . Substituting and assuming $|V_{fi}|^2$ and $\rho$ vary slowly across the narrow peak, the integral collapses:

P(t) = \frac{2\pi}{\hbar}\,|V_{fi}|^2\,\rho(E_i)\,t .

A constant rate

Because $P(t)$ is now linear in $t$ , the transition rate $W = dP/dt$ is constant:

\boxed{\;W_{i\to f} = \frac{2\pi}{\hbar}\,|V_{fi}|^2\,\rho(E_f)\;}

evaluated at the final energy $E_f = E_i$ fixed by energy conservation. This is Fermi's golden rule. For a harmonic perturbation of amplitude $\hat V$ the same derivation gives $W = \frac{2\pi}{\hbar}\,\big|\tfrac{1}{2}V_{fi}\big|^2\,\rho(E_f)$ with the resonance condition $E_f = E_i \pm \hbar\omega$ .

Reading the ingredients

$|V_{fi}|^2$ — how strongly the perturbation connects initial and final states. Vanishing matrix elements give selection rules (zero rate).
$\rho(E_f)$ — how many final states are available at the right energy. More available states means a faster transition. This is why decay into the open continuum is irreversible while transitions between two isolated levels merely oscillate.
$E_f = E_i$ (or $E_i \pm \hbar\omega$ with a drive) — energy conservation, enforced by the delta function.

Try it

Apply the golden rule. With coupling $|V_{fi}|=0.1$ , density of states $\rho(E_f)=50.0$ , and $\hbar=1$ , compute the transition rate $W = (2\pi/\hbar)\,|V_{fi}|^2\,\rho(E_f)$ and return it. (Remember to square the matrix element.)

Run your code to see the quantum state.