Do Photons Emit When Electron's Energy is Measured?

[Manchot]

Hey all,

I'm currently in my second QM class, and I have a question about the emission of photons. Last semester, as we studied the wavefunction of a hydrogen electron, the professor briefly mentioned that the energy levels correspond to the observed energies of photons emitted from hydrogen atoms. In general, however, an electron will be in a linear combination of the eigenstates. Since the photon energies are only measured to be discrete, does this mean that the emission of a photon corresponds with a measurement of the electron's energy?

NB: I haven't learned QED yet, so please give me a dumbed-down answer. Thanks.

 

[Meir Achuz]

It measures what the electron's energy was before the radiation since
h\nu=E_i-E_f.

 

[vanesch]

In general, however, an electron will be in a linear combination of the eigenstates. Since the photon energies are only measured to be discrete, does this mean that the emission of a photon corresponds with a measurement of the electron's energy?

Yes! Picture it this way: the superposition of electron eigenstates, through the interaction with the EM field, will result in a superposition of the EM photon states, where each of the terms corresponds to each of the eigenstate-transitions in the hydrogen atom.
And now you measure, and see one of those photons (application of the measurement postulates).

Symbolic example:

|H-atom> = |psi3> + |psi5>
(say, superposition of the third and fifth excited state, just for the sake of it).

Interaction with EM field (from the vacuum state) gives you:

|H-atom+EM field> = a |psi0> |photon3> + b |psi0> |photon5> + c |psi1> |photon2> + ...

where the first state in each term gives you the "final" state of the H-atom, and the second state in each term gives you the EM field state (kind of photon you have).

Now you do a "photon" measurement, so, with probability |a|^2, you have the first state (ground state + photon3), with probability |b|^2 you have the second state (ground state + photon5), ...

 

[Manchot]

Ah, so the photon itself is in a superposition of eigenstates. For some reason, I never considered that possibility. Does that mean that you can determine the state of the electron by measuring the energy of the photon? Is that a form of entanglement?