# Two two-level atoms and form of the Hamiltonian

1. Apr 5, 2013

### McLaren Rulez

Hello,

If we look at a system of two two-level atoms interacting with light, most papers start with a Hamiltonian

$$H_{int}=(\sigma_{1}^{+}+\sigma_{2}^{+})a_{\textbf{k},\lambda} + h.c.$$

That is, we absorb a photon and lost one excitation in the atoms or vice versa. Why do we never consider terms like
$$\sigma_{1}^{+}\sigma_{2}^{+}a_{\textbf{k},\lambda}a_{\textbf{k},\lambda}$$

Here, the two photons are absorbed simultaneously and we transition directly from the ground state of both to the excited state of both atoms. I suspect that it is because this process is much less likely but how do I prove it?

2. Apr 5, 2013

### DrDu

I think it is due to the fundamental interaction in quantum electrodynamics being of the form jA where j is the charge-current density operator and A is the electromagnetic vector potential. j induces transitions between electronic states and A is linear in a and a*.
You could try to perturbatively diagonalize your hamiltonian in powers of the coupling strength $\lambda$,
$H=H_0+\lambda H_{int}$. If you do this to first order, you will get new couplings second order in lambda of the form you proposed.

3. Apr 5, 2013

### fzero

The interaction Hamiltonian appears in the full Hamiltonian with some coefficient which we can call $\epsilon$. So $H = H_0 + \epsilon H_{int}$. We can determine roughly what the value of $\epsilon$ is by considering the fundamental process that leads to $H_{int}$. This is the quantum electrodynamics interaction term

$$L_{int} = e A_\mu\bar{\psi} \gamma_\mu \psi$$

that describes the interaction of a photon, described by the quantum field $A_\mu$, with the electron, described by the field $\psi$. Of course $e$ is the electric charge, which in natural units is $e\sim 1/\sqrt{137} \sim 0.09$. Clearly $\epsilon$ is related to this value, but in order to more accurately estimate it, we would need to add details about the atomic wavefunction, etc.

Now, the 2nd order interaction that you consider is an additional correction. It is generated at 2nd order in pertubation theory in QED using the interaction $L_{int}$. It would also have some coefficient $\epsilon_2$ that is now related to $e^2$, so it is about 10% as large as the 1st order term.

On the other hand, we can also generate 2-photon processes in the 2nd-order perturbation theory of the effective theory using $H_{int}$. Including this order of correction clearly accounts for some of the lack of precision that we lost by not including the 2nd order term that you wrote down. I don't think it's easy to estimate the difference between the two methods for describing 2-photon processes.

4. Apr 6, 2013

### McLaren Rulez

Thank you for the replies. The point then, if I understand you correctly, is that the coupling strength is small and therefore higher order processes can be neglected.

Thank you!

5. Apr 6, 2013

### DrDu

Terms of the form you are interested in also arise when you project the hamiltonian for the full atomic system onto the sub-space spanned by the two electronic states you are interested in. The strength of the new couplings depends on the square of the coupling strength of H_int divided by the energetic distance of the other electronic states you are projecting out. This is known as "Löwdin projection operator method" as it was invented by Per-Olov Löwdin.