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The Hamiltonian of an atom coupled to an EM field, both described quantum mechanically is: $$H = \frac{1}{2m}(\hat{p}-q\hat{A})^2 = \frac{\hat{p}^2}{2m} -\frac{q}{m}\hat{p}\hat{A}+\frac{q^2}{2m}\hat{A}^2$$ Under the condition of transversality ($\hat{p}$ and $\hat{A}$ are vectors). I have only seen it treated in the dipole approximation where the last term becomes a constant and so unphysical; but what is its general interpretation? Is it shifting the energy of the free photons, it doesn't look diagonal in number of occupation basis.

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This is the quantised version of the (exact) classical Hamiltonian describing a charged point particle moving in an EM field, ie the classical Hamiltonian "with hats on". Could you be a bit more specific about what you need "interpreting", or what you are confused about? Eg. A geometer would probably say something like this is the covariant derivative on a complex line bundle associated to a principal U(1) bundle. But maybe you want something a bit more meaningful...? – Mark Mitchison Nov 13 '12 at 23:13
By interpreting, I'm thinking to what kind of vertex doesn't it describe? let's say that $\hat{p}=0$ the middle term vanishes but not the last one, we have a perturbation to the Hamiltonian of the EM field, what is this effect? – Learning is a mess Nov 14 '12 at 8:42

1 Answer 1

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Neither of the interaction terms would appear were the atom not present, and you cannot simply set $\hat{p}=0$ as it is a dynamically fluctuating quantum variable. Therefore, both of the terms must describe scattering of the light from the atom. Roughly speaking, the term $\sim \hat{p}\hat{A}$ encodes inelastic scattering, while the term $\sim \hat{A}^2$ encodes elastic scattering.

In first order perturbation theory, transitions due to the first term are controlled by the matrix elements $$ \mathcal{M}_1 = \langle m, E_f| \hat{p}\hat{A} |n,E_i\rangle = \langle m|\hat{A}|n\rangle\langle E_f|\hat{p}|E_i\rangle.$$ Here, $n, m$ are the initial and final number of photons in the light field (assume for simplicity there is only one mode, obviously in a real calculation you will also have to consider different wavevectors and polarisation states of the field), while $E_{i,f}$ are the initial and final atomic eigenstates. Since the operator $\hat{A}$ is linear in annihilation/creation operators, the matrix element is zero unless $m = n\pm 1$. Meanwhile, since atomic eigenstates have definite parity, the matrix element is zero unless $|E_f\rangle \neq |E_i\rangle$ (or more precisely, it is zero unless the two states have opposite parity).

So to first order, the interaction vertex $\hat{p}\hat{A}$ describes processes in which a photon is absorbed or emitted by the atom, changing its internal state. At higher orders in perturbation theory you will see processes where the atom's internal state changes multiple times. This includes, for example, an inelastic scattering process where the atom absorbs a photon at one frequency and an electron becomes excited, then the electron drops in energy by emitting a photon at another frequency. (You could also have elastic scattering processes where the atom ends up in the same state, hence the "roughly speaking" disclaimer above.)

Matrix elements of the second interaction vertex look like $$\mathcal{M}_2 = \langle m, E_f|\hat{A}^2 |n,E_i\rangle = \langle m| \hat{A}^2 |n\rangle \langle E_f| E_i\rangle.$$ Therefore, these matrix elements vanish unless the initial and final state of the atom is the same. However, the vertex is now quadratic in annihilation/creation operators, so you will see, for example, elastic scattering processes where a photon is absorbed and then re-emitted at the same frequency (but different direction, in general). At higher orders you will start to see really interesting optical non-linearities where, for example, two photons are absorbed, and then two photons are re-emitted at different frequencies. Hopefully this explains your idea about "perturbing the Hamiltonian of the free EM field". The strong interaction between light and electrons can produce an effective interaction in the presence of matter (aka optical non-linearity) between otherwise non-interacting photons.

Finally, don't forget that the full interaction operator contains both contributions so at higher orders in perturbation theory you will also get cross terms between $\hat{p}\hat{A}$ and $\hat{A}^2$.

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