Harmonic perturbations

Consider a perturbation which oscillates sinusoidally in time. This is usually called a harmonic perturbation. Thus,

$$H_1(t) = V \exp( {\rm i} \omega t) + V^\dagger \exp(-{\rm i} \omega t),$$ (812)

where $V$ is, in general, a function of position, momentum, and spin operators.

Let us initiate the system in the eigenstate $\vert i\rangle$ of the unperturbed Hamiltonian, $H_0$, and switch on the harmonic perturbation at $t=0$. It follows from Eq. (787) that

$$c_n^{(1)} = \frac{-{\rm i}}{\hbar} \int_0^t \left[V_{ni} \exp({\rm i} \ome... ...agger \exp(-{\rm i} \omega t')\right]\exp( {\rm i} \omega_{ni} t') dt',$$ (813)

$$= \frac{1}{\hbar} \left(\frac{1-\exp[ {\rm i} (\omega_{ni} + \ome... ... i} (\omega_{ni}-\omega ) t]} {\omega_{ni} - \omega} V_{ni}^\dagger\right),$$

where

$$V_{ni} = \langle n\vert V\vert i\rangle,$$ (814)

$$V_{ni}^\dagger = \langle n \vert V^\dagger \vert i\rangle = \langle i\vert V\vert n\rangle^\ast.$$ (815)

This formula is analogous to Eq. (794), provided that

$$\omega_{ni} = \frac{E_n-E_i}{\hbar} \rightarrow \omega_{ni}\pm \omega.$$ (816)

Thus, it follows from the previous analysis that the transition probability $P_{i\rightarrow n}(t)=\vert c_n^{(1)}\vert^2$ is only appreciable in the limit $t\rightarrow \infty$ if

$$\omega_{ni} + \omega \simeq 0 {\rm or} E_n \simeq E_i - \hbar \omega,$$ (817)

$$[0.5ex] \omega_{ni} - \omega \simeq 0 {\rm or} E_n \simeq E_i + \hbar \omega.$$ (818)

Clearly, (817) corresponds to the first term on the right-hand side of Eq. (813), and (818) corresponds to the second term. The former term describes a process by which the system gives up energy $\hbar\omega$ to the perturbing field, whilst making a transition to a final state whose energy level is less than that of the initial state by $\hbar\omega$. This process is known as stimulated emission. The latter term describes a process by which the system gains energy $\hbar\omega$ from the perturbing field, whilst making a transition to a final state whose energy level exceeds that of the initial state by $\hbar\omega$. This process is known as absorption. In both cases, the total energy (i.e., that of the system plus the perturbing field) is conserved.

By analogy with Eq. (807),

$$w_{i\rightarrow [n]} = \left. \frac{2\pi}{\hbar} \overline{\vert V_{ni}\vert^2} \rho(E_n) \right\vert _{E_n = E_i-\hbar\omega},$$ (819)

$$w_{i\rightarrow [n]} = \left. \frac{2\pi}{\hbar} \overline{ \vert V_{ni}^\dagger\vert^2} \rho(E_n)\right\vert _{E_n = E_i+\hbar\omega}.$$ (820)

Equation (819) specifies the transition rate for stimulated emission, whereas Eq. (820) gives the transition rate for absorption. These equations are more usually written

$$w_{i\rightarrow n} = \frac{2\pi}{\hbar} \vert V_{ni}\vert^2 \delta(E_n-E_i+\hbar\omega),$$ (821)

$$w_{i\rightarrow n} = \frac{2\pi}{\hbar} \vert V_{ni}^\dagger\vert^2 \delta(E_n -E_i-\hbar\omega).$$ (822)

It is clear from Eqs. (814)-(815) that $\vert V_{ni}^\dagger\vert^2 = \vert V_{ni}\vert^2$. It follows from Eqs. (819)-(820) that

$$\frac{w_{i\rightarrow [n]}}{\rho(E_n)} = \frac{w_{n\rightarrow [i]}}{\rho(E_i)}.$$ (823)

In other words, the rate of stimulated emission, divided by the density of final states for stimulated emission, equals the rate of absorption, divided by the density of final states for absorption. This result, which expresses a fundamental symmetry between absorption and stimulated emission, is known as detailed balancing, and is very important in statistical mechanics.