If one starts with $N_0$ nuclei at $t=0$, then after time $t$, the number of nuclei left un-disintegrated are given by $$N(t)=N(0)\exp(-\lambda t)\tag{1}.$$

Is there a similar statistical decay law for an assembly of atoms in a two-level system with energies $E_1$ and $E_2(>E_1)$? If at $t=0$, $N_0$ atoms are made to populate the excited state $E_2$, do we expect at a later time the population of the level $E_2$ to exponentially deplete like (1)?

If not, what type of depletion law in time do we expect?

I'm not interested in what a single atom does in presence of interaction. I know that in presence of time-dependent interactions, a single atom in a simple two-level system makes back and forth transition between the levels. My question is about the overall statistical behavior of an assembly of atoms in a two-level system prepared in the excited state at $t=0$.

  • 1
    $\begingroup$ Assuming the transitions are a Poisson process then yes. $\endgroup$
    – lemon
    Commented Jun 24, 2017 at 14:10
  • $\begingroup$ Are they Poisson Processes? $\endgroup$ Commented Jun 24, 2017 at 15:57
  • $\begingroup$ @TausifHossain Sorry. I'm not aware of what a Poisson process is though I'm aware of Poisson distribution. $\endgroup$
    – SRS
    Commented Jun 24, 2017 at 17:15
  • $\begingroup$ I think that depends on how the atoms in your assembly can interact. For example, if they're clustered within a laser cavity with an appropriate mode, they are more likely to de-excite in a sudden cascade than if they're just sitting in a vacuum. $\endgroup$ Commented Jul 10, 2017 at 17:26
  • $\begingroup$ taken from Wikipedia: "where $A _{21}$ is the Einstein coefficient for spontaneous emission, which is fixed by the intrinsic properties of the relevant atom for the two relevant energy levels." $\endgroup$
    – jim
    Commented Jul 10, 2017 at 18:35

1 Answer 1


Generically speaking, yes. This is known as Fermi's golden rule: this follows from first-order time-dependent perturbation theory (there's a reasonable derivation in Wikipedia), where you require

  • a discrete state (like an atomic excitation) in the same energy range as a continuum (like a bunch of photon states),
  • with a small coupling between the two (generically true unless you go to great lengths to increase the coupling),
  • with no memory, i.e. where you can assume that once it's gone, the excitation won't return to the initial discrete state and re-deposit its energy.

If you're really picky, these hypotheses are not completely true (as I've discussed previously in this answer) but they are excellent approximations in real-world scenarios. You can get some funny behaviours if e.g. your continuum is one-dimensional, but in the generic case of an excited atom in free space interacting with the 3D continuum of EM modes at and near the excitation wavelength, you can pretty much just assume that you have a time-independent decay rate and therefore an exponential roll-down to zero in the population.

If your two-level system is some atomic transition, then depending on the details you might struggle to get a time-dependent measurement of the decay rate (since decay lifetimes for dipole-allowed transitions are typically in the nanosecond regime), but when that happens, if you've managed to do away with any sources of inhomogeneous broadening, then an observation of a Lorentzian lineshape is a very strong indicator that the time-domain 'shape' of the emitted photons is indeed exponential.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.