gptejms said:
It will be nice if you can do this exercise:-show that the atom radiates in the Heisenberg picture-should be possible.
It surely is very difficult if you want the full treatment in QFT, for many reasons. The physical picture behind it is the the E-fields in the right transition mode(s) in QFT are non-zero even in vacuum (the expectation values of the squares of the amplitudes of these modes are not zero), and it is the interaction of the electon of the atom with these modes that makes that the stationary states of the atom when calculated, taking purely into account the coulomb interaction with the nucleus, are NOT stationary states anymore when taking into account the dynamical EM field (as a quantum field). As such, the atom will get entangled with the EM field, and the transition amplitude for doing so can be calculated perturbatively, but in all these calculations, approximations are made. Even this treatment is quite complicated, and I'm not aware of a "frontal attack" of the problem in full, non-perturbative QED.
Nevertheless, there ARE calculations of the rate of decay of excited atoms following the idea I outlined very roughly above, and as far as I know, they fit with observations. I don't have references handy as such, but some stuff on this must be found in Mandel and Wolf in their "bible" on quantum optics, I'm sure.
You have so far not commented on the x-part of the plane wave.
I did, in post 81.
Incoming "plane wave" for t<t0 (free hamiltonian)
At t1 < t0, the incoming wave is exp(i k x), so solving for the time evolution in the Schroedinger picture with the free hamiltonian, will give us a time dependence which is:
exp(i k x) exp(- i omega t)
up to t = t0.
Note that for the free hamiltonian, the plane wave exp(i k x) is a stationary state.
So at t = t0, the wavefunction is exp(i k x - i omega t0).
Now, the potential is switched on, but this doesn't immediately change the wavefunction which is supposed to evolve continuously.
So simply said, for t>t0, we have to solve the Schroedinger equation with the new hamiltonian, with as a starting wave function, exp(i k x - i omega t0).
But note now that this function of x is not a stationary state anymore, so this time, the time evolution will not simply add a phase factor.
A way to solve this, is to expand exp(i k x - i omega t0) in the Hermite-Gauss functions (the new stationary states), so you will find:
exp(i k x - i omega t0) = a H0(x) + b H1(x) + c H2(x) + ...
(where H0, H1, etc... are the new stationary states)
The time evolution will now do the following thing:
at t = t2 > t0, we have that the wave function equals:
psi(x) = a H0(x) exp(-i E0 t2) + b H1(x) exp(- i E1 t2) + c H2(x) exp(- i E2 t2) ...
This will of course now not take on the aspect any more of a plane wave.
cheers,
Patrick.