0
$\begingroup$

There is a lot of general discussion about the connection between the QM description of photons and classical EM fields. I understand one needs to solve the quantized version of Maxwell's equations, and all of this is still a little beyond me.

For now, I would like to ask a more concrete question: let's say we have a hydrogen atom, and we are able to excite it to a specific energy level such that it will always emit photons of specific $\hbar\omega_1$.

Then I would say that the closest classical description would be that of a dipole radiation, of which the real part of the electric field can be written in far-file simply as $$\bar E = \bar A(r)\cos(\omega_2 t -kr)$$

Is it true to say that the frequency of classical oscillations of the dipole is the same as that of the monochromatic photon emission? I.e., does $\omega_1=\omega_2$?

I'm trying to get some inuition on the matter. On the one hand, it is tempting to view the oscillations as the classical point of view of some stream of emissions of a given frequency. But on the other hand, the photon emissions are not necessarily time correlated, so perhaps it is the collective properties of all the emission processes that determines the macroscopic oscillation frequency.

Which of the above (if any) is the correct interpretation?

$\endgroup$
3
$\begingroup$

I recommend you the related answer to Is there oscillating charge in a hydrogen atom? by Emilio Pisanty as I think this describes what is going on very elegantly.

If you consider (using the example Emilio does) the $1s$ and $2s$ states of the hydrogen atom you find the superposition of the two states oscillates in time, and the frequency of this oscillation is exactly the frequency of the photon emitted or absorbed in transitions between the $1s$ and $2s$ states. So the classical interpretation would be that we have an oscillating electric dipole emitting the light.

The actual process would ultimately be described using quantum field theory, but QFT is not well suited to the description of bound states like a hydrogen atom. In practice we'd use ordinary non-relativistic quantum mechanics and calculate the transition probabilities using Fermi's Golden Rule. Then use QFT to calculate corrections such as the Lamb shift.

$\endgroup$
  • 2
    $\begingroup$ It's important to note, though, that the oscillating superposition in that answer is not what you get halfway through the photoemission decay of a 2p state. That is instead an entangled state of the atom and the radiation field, for which the reduced state of the atom is completely mixed, with absolutely no charge oscillations. However, the frequency of the emitted photon is still determined by the single-atom physics. $\endgroup$ – Emilio Pisanty Sep 26 '17 at 10:48
  • $\begingroup$ Great answers, thank you both! Just to be clear- what you are saying is that the single atom analog for a dipole oscillation is the time-dependence of the charge probability density, and that is situation-dependent. But nonetheless, the EM field that would be measured in a classical experiment that is the result of a single-atom monochromatic photon emission, would oscillate with the same frequency as the emitted photons? $\endgroup$ – Yoni Sep 27 '17 at 9:00

Your Answer

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

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