# Harmonic perturbation in interaction of radiation with quantum system; making sense of approximation of the integral

In the chapter "The interaction of quantum systems with radiation" (Quantum physics book by Bransden and Joachain, 2nd edition) section 11.2 "Perturbation Theory for harmonic perturbations and transition rates" there is this integral (in equation 11.45)

$$\int_0^t\sin(\omega t'-\delta_{\omega})\exp(i\omega_{ba}t')dt'$$ where $$\omega$$ is the frequency of external radiation field and $$\omega_{ba}=\frac{E_b - E_a}{\hbar}$$.
This integral is trivially calculated to $$\frac{1}{2}\exp(-i\delta_{\omega})\left[\frac{1-\exp[i(\omega_{ba}+\omega)t]}{\omega_{ba}+\omega}\right]-\frac{1}{2}\exp(i\delta_{\omega})\left[\frac{1-\exp[i(\omega_{ba}-\omega)t]}{\omega_{ba}-\omega}\right].$$

Upto here everything is fine. The problem is how they have approximated this term for different cases as given below:

For transitions in the infrared $$\left|\omega_{ba}\right|$$ is of the order $$10^{12}-10^{14}$$ $$s^{-1}$$, and even larger in the visible and ultraviolet regions. ... The product ($$\left|\omega_{ba}\right|t$$) is much greater than unity, it follows that the first term in square brackets on the RHS is negligible unless $$\omega_{ba}\approx-\omega$$, and the second term in square brackets is negligible unless $$\omega_{ba}\approx+\omega$$.

What is happening here?

This $$\omega_{ba}t$$ term is contained in a complex exponential, which is oscillatoy in nature. So how it may matter if the term $$(\omega_{ba}\sim\omega)t\rightarrow0$$?
They said "first term in square brackets on the RHS is negligible unless $$\omega_{ba}\approx-\omega$$". But all I can see is that than the denominator will tend to zero, blowing up the whole term.
Further, if $$\omega_{ba}±\omega$$ is not close to zero, what may happen? We have an oscillatory term here, all we shall have is $$±1$$.

What are they doing here?

integrating over a slowly varying function multiplied by a very fast oscillatory function tends to zero, by Riemann-Lebesgue lemma. You can also understand this conceptually - the fast oscillating function averages the slowly varying one to zero. So the only way out of it in this case is if the oscillations cancel out, meaning that the sine function oscillates with the same frequency of the exponent, $$\omega_{ba} \simeq \pm \omega$$.

You can see that also from the explicit result of the integral you wrote. The denominator goes to infinity if $$\omega$$ is not of the order of magnitude of $$\omega_{ba}$$ and the latter is very large (i.e. goes to infinity), while the numerator is bound by absolute value to be no more than 2.

Physically speaking, we want the external EM radiation to be close to resonance of the quantum system in order to have finite transition amplitudes.

By the way, your statement that

But all I can see is that than the denominator will tend to zero, blowing up the whole term.

is not accurate. Note that the numerator also tends to zero in that case. You need to expand to leading order and then get that the result is just finite $$\lim_{\epsilon \to 0}\frac{1-e^{i\epsilon t}}{\epsilon} = -it$$ and it doesn't "blow up".

• "The denominator goes to zero if ω is not of the order of magnitude of $\omega_{ba}$ and the latter is very large, while the nominator is bound by absolute value to be no more than 2." I think there is some problem here? Jan 31, 2023 at 11:22
• thank you. I meant that the nominator is very large i.e. goes to infinity with $\omega_{ba}$, not to zero of course. I edited the response
– user275556
Jan 31, 2023 at 11:28