5.3: Time–independent Perturbations

The results obtained in the last section can also be applied to the case where the perturbation, $\hat{V}$, is actually independent of time (strictly, ‘switched on’ at t=0).

Again, starting the system in eigenstate $|k \rangle$ of $\hat{H}_0$ we obtain,

$$c_m(t)\Delta c_m(t) = (i\hbar )^{−1} V_{mk} \int^t_0 \text{ exp}(i\omega_{mk}t) dt \\ = \frac{V_{mk}}{\hbar\omega_{mk}} [1 − \text{ exp}(i\omega_{mk}t)] \nonumber$$

for $m \neq k$, giving for the transition probability

$$p_m(t) = |\Delta c_m(t)|^2 = \frac{|V_{mk}|^2}{\hbar^2}\frac{\sin^2 (\omega_{mk}t/2)}{(\omega_{mk}/2)^2}. \nonumber$$

For sufficiently large values of $t$, the function

$$f(t, \omega_{mk}) \equiv \frac{\sin^2 (\omega_{mk}t/2)}{(\omega_{mk}/2)^2} \nonumber$$

consists essentially of a large peak, centerd on $\omega_{mk} = 0$, of height $t^2$ and width $\approx 4\pi /t$, as indicated in Fig. 4. Thus there is only a significant transition probability if $E_m \approx E_k$. That is, if $|\omega_{mk}| < 2\pi /t$.

Figure $\PageIndex{1}$: Transition probability as a function of applied harmonic perturbation frequency

Note that we are assuming that the system was prepared in some eigenstate of $\hat{H}_0$ which is not an eigenstate of $\hat{V}$: if it were, then the matrix element $V_{nm}$ would be zero and $p_m(t) = 0$. Thus although the analysis treats the perturbation as time independent, it is applied to cases where the perturbation is switched on at $t = 0$. Moreover only perturbations which are incompatible with the Hamiltonian can induce transitions.