In my lecture notes deriving the probability density of a two energy level atom we arive at the following equation:

$$c_f(t) = \frac{1}{2} \Omega \left[\frac{1-e^{i(\omega + \omega_0)t}}{(\omega + \omega_0)} - \frac{1-e^{i(\omega + \omega_0)t}}{(\omega - \omega_0)} \right]$$

We are considering a case with a broad band spectral energy density $\rho(\omega)$, and have assumed the state $|c_1(t=0)|^2=1$ therefore (I assume) we are including only transitions from $c_i \rightarrow c_f$ and not the reverse (for this $t\approx0$).

The notes then invoke the 'rotating wave approximation' to neglect the first term since they 'assume that the atom responds only for frequencies close to the transition frequency' i.e. $|\omega - \omega_0| \ll (\omega + \omega_0)$'. In other words they say that in the integral over $\omega$, the second term will dominate.

However in other notes describing stimulated emission and absorption (such as here) they say the first term corresponds to stimulated emission (i.e. where $E_f = E_i - \hbar \omega$, as obtained from a delta function $\delta(\omega + \omega_0)$) while the second term corresponds to stimulated absoprtion (i.e. where $E_f = E_i - \hbar \omega$, as obtained from a delta function $\delta(\omega - \omega_0)$). Therefore they ignore the first term because they just want to calculate the rate of absorption etc. This makes more sense to me as in this case where we integrate over all $\omega$ in $\rho(\omega)$ we will get a divergence in one of the terms (depending on whether $E_f$ is higher or lower than $E_i$ therefore absorption or emission is only possible etc).

My question is, how are these two different approaches equivalent? Am I correct in thinking those two terms correspond to the $E_f$ population increasing due to stimulated absorption/emission from level $E_i$ respectively?

Is it maybe because, when you account for line broadening, there are no longer divergences so you still need RWA to justify neglecting the other term after all?

Also, is it true that in general when you calculate $c_f$ accounting for transitions from emission/absorption from other levels that you will get interference terms between them in $|c_f|^2$?


1 Answer 1


In order to calculate the absorption one has to start with the atom/system in its ground state, $c_g(0)=0$ and observe how fast the probability to be found in this state decreases or how fast the probability to be found in the excited state increases: $$ w_{g\rightarrow e} = -\frac{d|c_g(t)|^2}{dt}|_{t=0} \left(= \frac{d|c_e(t)|^2}{dt}|_{t=0}\right) $$ To calculate the probability of emission one takes $c_e(0)=1$ and does the same calculation mutatis mutandis.

Both calculatuions can be unified, if we talk about the initial and final states, whatever they are, so that $$ w_{i\rightarrow f} = -\frac{d|c_i(t)|^2}{dt}|_{t=0} = \frac{d|c_f(t)|^2}{dt}|_{t=0} $$

Thus, the expression for the amplitude $c_f(t)$ given in the OP can refer either to the ground or to the excited state, and correspond to either absorption ro emission. What makes us neglect the first term is that (in practical situations) it is very small, compared to the other one, since $\omega\approx \omega_0$, whereasd the coupling strength is usually much smaller than the frequency. In other words: $$ \frac{\Omega}{\omega + \omega_0}\approx \frac{\Omega}{2\omega} \ll 1\\ \left|\frac{\Omega}{\omega - \omega_0}\right|\sim 1 $$ In which case the solution is good for describing Rabi socillations.

If we want to reduce this to the Fermi Golden rule, i.e., to a simple transition rate, the condition becomes more stringent: $$ \frac{\Omega}{\omega + \omega_0}\approx \frac{\Omega}{2\omega} \ll \left|\frac{\Omega}{\omega - \omega_0}\right|\ll 1. $$

  • $\begingroup$ Are you assuming in practical situations we are using detuned monochromatic light? Otherwise for a broad band spectrum we have a range of frequencies covering both close to and far from $\omega_0$. $\endgroup$
    – Alex Gower
    Commented Feb 23, 2021 at 23:13
  • $\begingroup$ @AlexGower yes, it is detuned monochromatic. If it were broad-band. Monochromsticity is already in your formula - you have only one frequency, the integration over the spectrum is usually done in the end. $\endgroup$
    – Roger V.
    Commented Feb 24, 2021 at 6:18
  • $\begingroup$ But won't your assumptions break down once you integrate over the spectrum? $\endgroup$
    – Alex Gower
    Commented Feb 24, 2021 at 11:20
  • $\begingroup$ Where and how one integrates over the spectrum depends on the problem. WHat we are discussing is just a little bit of math that can be used in many different contexts. $\endgroup$
    – Roger V.
    Commented Feb 24, 2021 at 11:30
  • $\begingroup$ My main problem is, how can we use the justification that you say 'What makes us neglect the first term is that (in practical situations) it is very small, compared to the other one, since ω≈ω0' when we integrate over a large range of frequencies, because surely in a large range there will be frequencies both close and far from the transition frequency? $\endgroup$
    – Alex Gower
    Commented Feb 24, 2021 at 12:57

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