# What is the minimal Hamiltonian for Compton scattering?

Thomson scattering of photons from free electrons is the elastic limit of Compton scattering at low energies and is well-described by the following non-relativistic light-matter Hamiltonian using minimal coupling:

$$\mathcal{H}= \frac{1}{2m}(\mathbf{p}-q \mathbf{A})^2=\frac{p^2}{2m}+\frac{-q}{2m}(\mathbf{p}\cdot \mathbf{A}+\mathbf{A}\cdot \mathbf{p}) + \frac{q^2}{2m} \mathbf{A}\cdot \mathbf{A}$$

Here, the last term, $$\frac{q^2}{2m} \mathbf{A}\cdot \mathbf{A}$$, describes Thomson scattering from electrons (i.e. elastic photon scattering from free electrons). To make things more relativistically accurate, one must use the add in corrections from the Dirac equation, or even QED.

My question is, does the $$\frac{q^2}{2m} \mathbf{A}\cdot \mathbf{A}$$ term in the non-relativistic light-matter Hamiltonian permit Compton (inelastic) scattering of photons from free electrons? If it does not, what is the minimal relativistic correction to the above Hamiltonian that is needed to give rise to inelastic Compton scattering?

One needs to replace the vector potential by its operator (see here). The $$\mathbf{A}\cdot\mathbf{A}$$ term would contain products of pairs of photon operators, describing thus elastic photon scattering, e.g., $$a_\mathbf{q}^\dagger a_{\mathbf{q}'} e^{i(\mathbf{q}-\mathbf{q}')\mathbf{r}}$$ Taking the matrix element of this term between the plane wave wave functions with momenta $$\mathbf{k}, \mathbf{k'}$$ results in the momentum conservation law: $$\mathbf{k}+\mathbf{q} = \mathbf{k}'+\mathbf{q}'$$ Applying the Fermi Golden rule to calculate the scattering cross-section imposes the energy conservation: $$\frac{\hbar^2\mathbf{k}^2}{2m} + c\hbar|\mathbf{q}|= \frac{\hbar^2\mathbf{k}'^2}{2m} + c\hbar|\mathbf{q}'|$$ We are thus down to the introductory QM books description of the Compton scattering with the apparently non-zero matrix element.
The other term contains only one photon creatin/annihilation operator and cannot describe compton scattering due to the restriction imposed by the energy conservation and the momentum conservation. However, it may contribute as a higher order process, with a factor proportional to $$(\frac{e^2}{\hbar c})^2$$, just like the second term. Thus, it is not immediately obvious that we can keep one and neglect the other (even if we do not care about relativity and gauge invariance).