I was following Section 5 of Ridley's book "Quantum Processes in Semiconductors" where he tries to derive transition rates for a system interacting with harmonic (cosine) potential. For simplicity lets write it in a form:

$$ \langle f|H_I|i\rangle =C e^{-i t \omega }+C e^{i t \omega }. $$

Now he says that "investigating the time dependence of the transition probability in the usual way, we find that the second term induces stimulated emission of photons and only the first term induces absorption". I never checked that and used terms for corresponding processes. Now I decided to check it explicitly and I failed to show that in time-dependent perturbation theory counter-terms coming from $e^{-i t \omega }$ and $e^{i t \omega }$ vanish. Maybe someone can help me to understand how we can use only single exponent for a process in mind. This is what I tried to do.

As it is written in Sakurai book, first order transition amplitude can be calculated using equation:

$$ c_{i\to f}(t)=-\frac{i}{\hbar }\overset{t}{\underset{0}{\int }} \langle f| H_I |i\rangle e^{i \text{$\omega $}_{\text{if}} \tau} d\tau $$

where $\omega_{\text{if}} = (E_f - E_i)/\hbar$. So I integrated this equation and got:

$$ c_{i\to f} (t)= \underbrace{\frac{C\left(1-e^{it\left(\omega_{\text{ni}}-\omega\right)}\right)}{\hbar\left(\omega_{\text{ni}}-\omega\right)}}_{c_{1}}+\underbrace{\frac{C\left(1-e^{it\left(\omega_{\text{ni}}+\omega\right)}\right)}{\hbar\left(\omega_{\text{ni}}+\omega\right)}}_{c_{2}}. $$

Now transition probability is $|c_{i\to f}(t)|=c_1c_1^* + c_2c_2^* + c_1c_2^* + c_1c_2^*$. Terms $c_1c_1^*$ and $c_2c_2^*$ are present if only single exponential is considered in perturbation and give deltas as $t\to \infty$. First delta is centered at $\omega_{if}$ and is non vanishing for absorption and second is centered at $-\omega_{if}$ and is responsible for emission. So if I could show that counter terms $ c_1c_2^*$, $c_1c_2^*$ vanish I could easily choose i.e. $C e^{-i t \omega }$ for absorption. However when I calculated these terms I got:

$$ c_1 c_2^*+c_2 c_1^*=\frac{4 C^2 (\cos (t \omega )) \left(\cos \left(t \omega _{\text{if}}\right)-\cos (t \omega )\right)}{\hbar ^2 \left(\omega -\omega _{\text{if}}\right) \left(\omega _{\text{if}}+\omega \right)}. $$

This is what I did not expect. This term is non-vanishing and in resonant conditions diverges with respect to time:

$$ \underset{\omega \to \omega _{\text{if}}}{\text{lim}}\left(c_1 c_2^*+c_2 c_1^*\right)=\frac{C^2 t \left(\sin \left(2 t \omega _{\text{if}}\right)\right)}{\hbar ^2 \omega _{\text{if}}}. $$

So I thought that maybe I made some mistakes in algebra. But this can be seen from different perspective. We can write $c_1$ and $c_2$ in terms of $\text{sinc}(x)=\sin(x)/x$:

$$ c_1 = -\frac{iCt}{\hbar}e^{-\frac{1}{2}it\left(\omega-\omega_{\text{ni}}\right)}\text{sinc}\left(t\left(\omega-\omega_{\text{ni}}\right)\right) $$

$$ c_2 = -\frac{iCt}{\hbar}e^{\frac{1}{2}it\left(\omega_{\text{ni}}+\omega\right)}\text{sinc}\left(t\left(\omega_{\text{ni}}+\omega\right)\right) $$


$$ t^{2}\text{sinc}\left(t\left(\omega_{\text{ni}}+\omega\right)\right)\text{sinc}\left(t\left(\omega-\omega_{\text{ni}}\right)\right) $$

is divergent as $t\to \infty$. Maybe I am missing something. Could someone shed a light why cross terms vanish or why they are ignored when we have real harmonic perturbation.


1 Answer 1


I found an answer to this question. Although counter terms diverge as $t\to \infty$ they are very small compared to $|c_1|^2$ and $|c_2|^2$. For example if I set $C=1, \omega_{ni}=1, \mathbf{k}\cdot \mathbf{r} =0$, $t=100\gg \frac{2 \pi }{\omega _{\text{if}}}$ and plot term dependance on $\omega$ i get

enter image description here

One can easily see that $c_2 c_1^*+c_1 c_2^*$ is relatively small. On the other hand if I take resonant condition $\omega \to \omega_{ni}$ and plot with respect to time I get:

enter image description here

So for times $t\gg \frac{2 \pi }{\omega _{\text{if}}}$ one can omit counterterms.


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