In section 4-4 the Feynman Lectures reads:

The probability that an atom will emit a photon into a particular final state is increased by the factor $(n+1)$ if there are already $n$ photons in that state.

The specific maths that I find confusing is that the probability of an atom emitting a photon is given as (4.28)

$$ P_{emission} = (n+1)|a|^2 $$

and absorption as (4.29)

$$ P_{absorption} = n|a|^2 $$

where $|a|^2$ is "the probability [an atom] would emit if no photons were present." Wouldn't these probabilties be $P_{em + ab} \gg 1$ for most typical situations where the number of photons $n$ is well over a trillion?

And is is possible to normalize this situation?

  • $\begingroup$ If $P_{emission}$ and $P_{absorption}$ are probabilities, then they are both less than one, and so will be their product. But that being said, I'm not sure what $P_{em+ab}$ represents. $\endgroup$ – garyp Nov 14 '19 at 11:41
  • $\begingroup$ I would like to offer a few comments that I hope will be useful. First, when Feynman speaks of amplitudes (and probabilities) in FLP - particularly in the introductory QM chapters - he is often speaking rather loosely of amplitudes (probabilities) per unit time. $\endgroup$ – Michael A. Gottlieb Nov 14 '19 at 11:57
  • $\begingroup$ Second, if we are talking about H and the states are 2P and 1S, the lifetime of the 2P state is only on the order of 10^-9 s, while the energy of the emitted photon is 10.2 eV. If the atom under consideration is in an unknown initial state and |a|^2 is the probability of seeing a spontaneous transition in an arbitrary 1 second period (with no photons present), then this probability must be very small, because if the atom started in the 1S state there can be no such transition, and if it started in the 2P state then in all likelihood it already made the transition before we looked at it. $\endgroup$ – Michael A. Gottlieb Nov 14 '19 at 11:59
  • $\begingroup$ Third, the ionization energy of H is 13.6 eV from the 1S state and only 3.4 eV from 2P, so if there are n photons present and n is large the atom will quickly be ionized. If we are talking about an H atom that does not get ionized, n can not be arbitrarily large. $\endgroup$ – Michael A. Gottlieb Nov 14 '19 at 11:59
  • $\begingroup$ It would be more helpful if you guys showed, for example, how to normalize what was given in the question. I sort of understand michael's arguments about how a physical system would work, but that doesn't explain that mathematics that were presented in this textbook. And yes, garyp, I do think they would be less than one, but the numbers about were also quoted as probabilities but would not be less than one unless they are normalized somehow. $\endgroup$ – guitarphish Nov 15 '19 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.