Spinnor said:
Consider the microwave cavity used in a particle accelerator to accelerate particle bunches. Before a bunch of particles enters the microwave cavity can we, if only in principle, quantum mechanically describe the state of the microwave cavity as a proper sum of microwave photon number states, summing over all possible momentum?
If yes, after the particle bunch has passed through the cavity and gained energy at the expense of the energy in the cavity can we say that the new state of the microwave cavity has fewer microwave quanta?
Can we say that each particle in the accelerated bunch of particles absorbs many microwave quanta while in the cavity?
Thanks!
May I comment from an engineering perspective - sorry I should not be on this Quantum Physics section!
It seems to me that we are not dealing with radiation in the situation you describe, but the motion of charges. A cavity is an energy store, and the fields inside it may be said to be linked to the motion of the electrons in its walls. It resembles an LC circuit, where the energy flows alternately between L and C, and in a perfect case there are zero losses and no aggregate energy flow. The E and H fields in the cavity are in antiphase, unlike an EM wave in which they are in-phase, and they occur in different positions. So I would suggest that there is no radiation in the cavity, just separate E and H fields, and I cannot therefore see how photons are going to be detected.
Further than this, if the Q of the cavity is infinitely high, the bandwidth will be zero, so how can the energy within it be modulated with quantum noise and consist of photons?
I feel that the acceleration of the particles you describe is similar to an electric motor, not connected to radiation.
Now I admit that if a cavity is connected to an antenna, or shall we say a slot is cut in it, then we will see radiation. The slot radiates because the electrons in the surrounding conductor are accelerated by the fields in the cavity, and not by allowing the interior "radiation" to leak out. The slot will radiate 50% of its photons back into the cavity. And the loss of the energy radiated away by the slot will create a loss resistance, the radiation resistance, which will lower the Q of the cavity to allow the interior waves to be modulated with quantum noise, now showing the presence of photons.
Once again apologies for what I do realize must be a naive view.