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# 5.3: Time–independent Perturbations

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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 $$|\omega_{mk}| < 2\pi /t$$.

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.

5.3: Time–independent Perturbations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.