Categories: Optics, Physics, Quantum mechanics, Two-level system.

Fermi’s golden rule

In quantum mechanics, Fermi’s golden rule expresses the transition rate between two states of a system, when a sinusoidal perturbation is applied at the resonance frequency ω=Eg/ℏ\omega = E_g / \hbar of the energy gap EgE_g. The main conclusion is that the rate is independent of time.

From time-dependent perturbation theory, we know that the transition probability for a particle in state ∣a⟩\Ket{a} to go to ∣b⟩\Ket{b} is as follows for a sinusoidal perturbation at frequency ω\omega:

Pab=∣Vba∣2ℏ2sin⁡2 ⁣((ωba−ω)t/2)(ωba−ω)2\begin{aligned} P_{ab} = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2} \end{aligned}

Where ωba≡(Eb−Ea)/ℏ\omega_{ba} \equiv (E_b - E_a) / \hbar. If we assume that ∣b⟩\Ket{b} irreversibly absorbs an unlimited number of particles, then we can interpret PabP_{ab} as the “amount” of the current particle that has transitioned since the last period 2πn/(ωba ⁣− ⁣ω)2 \pi n / (\omega_{ba} \!-\! \omega).

For generality, let EbE_b be the center of a state continuum with width ΔE\Delta E. In that case, PabP_{ab} must be modified as follows, where ρ(Ex)\rho(E_x) is the destination’s density of states:

Pab=∣Vba∣2ℏ2∫Eb−ΔE/2Eb+ΔE/2sin⁡2 ⁣((ωxa−ω)t/2)(ωxa−ω)2 ρ(Ex)dEx\begin{aligned} P_{ab} &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - \Delta E / 2}^{E_b + \Delta E / 2} \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x} \end{aligned}

If EbE_b is not in a continuum, then ρ(Ex)=δ(Ex−Eb)\rho(E_x) = \delta(E_x - E_b). The integrand is a sharp sinc-function around ExE_x. For large tt, it is so sharp that we can take out ρ(Ex)\rho(E_x). In that case, we also simplify the integration limits. Then we substitute x≡(ωxa ⁣− ⁣ω)/2x \equiv (\omega_{xa}\!-\!\omega) / 2 to get:

Pab≈2ℏ∣Vba∣2ρ(Eb)∫−∞∞sin⁡2(xt)x2 dx\begin{aligned} P_{ab} &\approx \frac{2}{\hbar} |V_{ba}|^2 \rho(E_b) \int_{-\infty}^\infty \frac{\sin^2(x t)}{x^2} \:dx \end{aligned}

This definite integral turns out to be π∣t∣\pi |t|, so we find, because clearly t>0t > 0:

Pab=2πℏ∣Vba∣2ρ(Eb) t\begin{aligned} P_{ab} &= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t \end{aligned}

The transition rate RabR_{ab}, i.e. the number of particles per unit time, then takes this form:

Rab=∂Pab∂t=2πℏ∣Vba∣2ρ(Eb)\begin{aligned} \boxed{ R_{ab} = \pdv{P_{ab}}{t} = \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) } \end{aligned}

Note that the tt-dependence has disappeared, and all that remains is a constant factor involving Eb=Ea ⁣+ ⁣ℏωE_b = E_a \!+\! \hbar \omega, where ω\omega is the resonance frequency.

References

  1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.