2
$\begingroup$

In QM it's taught that the frequency of a photon emitted when an electron drops from a higher-energy state to a lower-energy state is $\Delta E/h$. In E&M, it's taught that the source of an electromagnetic wave is the acceleration of charge -- and that the frequency of the wave is the same as the frequency of oscillation in the acceleration of charge. Is there a clear explanation of how the electron's motion generates the frequency when an electron drops from one energy state to a lower energy state in an atom? Of course, I realize that the electron is really a wavefunction, not a particle, but still, some component of the electron (or electron wavefunction) must oscillate at the frequency $E/h$ during the transition from the high state to the low state. What is that component?

$\endgroup$
4
  • $\begingroup$ Yes, Emilio posted an absolutely brilliant answer on exactly this point. Let me have a search for it ... $\endgroup$ – John Rennie Apr 3 '18 at 4:46
  • $\begingroup$ Aha, found it. I strongly recommend you read Emilio's answer to the question I've linked as it is the best explanation of hat is going on that I've ever seen. $\endgroup$ – John Rennie Apr 3 '18 at 4:49
  • $\begingroup$ @John Rennie Thank you for that link! Actually, it looks more than a debate than a clear answer. I wonder if the final answer might lie in observing that the frequency of the emitted photon is equal to the difference (beat) frequency between zitterbewegung frequencies of the two states . $\endgroup$ – S. McGrew Apr 4 '18 at 4:56
  • $\begingroup$ That said, Emilio's answer was indeed brilliant. $\endgroup$ – S. McGrew Apr 4 '18 at 5:03