How many photons are absorbed during Rabi oscillations?

Question

In my understanding, Rabi oscillations are derived using the classical approximation for the electromagnetic field. I don't get how this picture fits with a quantized EM field though. Say you excite a two level system with a coherent laser at the resonance frequency for a duration that projects the state from $ |g\rangle $ into $ \frac{1}{\sqrt 2}(|g\rangle+|e\rangle) $. How many photons are absorbed?

Answer 1

It is a superposition of absorbing one photon and not absorbing at all.

Say you shine a laser pulse onto atom. $|g\rangle$ will be the state if no photon is absorbed, and $|e\rangle$ if it is absorbed. In quantum picture this light pulse can be described by coherent state $|\alpha_1\rangle =e^{-{|\alpha_1|^2\over2}}\sum_{n=0}^{\infty}{\alpha^n\over\sqrt{n!}}|n\rangle =e^{-{|\alpha|^2\over2}}e^{\alpha\hat a^\dagger}|0\rangle ~$,

In case one photon is absorbed new state is

$|\alpha_2\rangle =e^{-{|\alpha_2|^2\over2}}\sum_{n=0}^{\infty}{\alpha^{n-1}\over\sqrt{(n-1)!}}|n-1\rangle $,

So if pulse is tuned such that there is 0.5 probability for interaction, state of the whole system will be an entangled state:

$|\alpha_1,g\rangle+|\alpha_2,e\rangle$

Catch is that coherent state is a superposition of number states so one can not distinguish between coherent state with average $n$ and $n-1$ photons, so effectivlly one gets a pure state( why this holds check out Beam splitters and Mach-Zender interferometers) :

$|g\rangle+|e\rangle$

One more remark, in case you excite atom with a Fock state you will not get a superposition but rather entangled state between photons and atom. in that case atom alone is not in superposition

Answer 2

Coherent EM field is not characterizable by photon number; it is not an eigenstate of an $a^\dagger a$ operator.