What does the inter-photon force look like (as mediated by electrons)? Most often we think of photons mediating a force between charged particles (e.g. electrons and positrons), and it's a standard exercise in QFT to do the computation and find that, to leading order in $\alpha$, there is a $q_1 q_2/r^2$ repulsive force between particles (which of course becomes attractive if $q_1q_2 < 0$).
But in the absence of other matter, otherwise freely propagating photons feel forces from each other, mediated by electrons and positrons (and additional contributions from all other charged particles). The point where this becomes relevant is the Schwinger limit, I suppose.
Question: what kind of form does this interaction take? My QFT is too shaky to try to do the algebra myself. Some of my first questions are "is it attractive or repulsive", but that feels like a lousy question to be asking given the fact that photons don't really localize to a point the way that electrons do.
 A: Photon-photon scattering was first calculated by Euler and Kockel in 1935. At low energy the differential cross section in the zero-momentum frame, in natural units, is
$$\frac{d\sigma}{d\Omega}=\frac{139\alpha^4}{(180\pi)^2}\frac{\omega^6}{m^8}(3+\cos^2\theta)^2$$
where $\alpha$ is the fine-structure constant, $m$ the electron mass, $\omega$ the energy of each of the photons, and $\theta$ the scattering angle. See this paper for more details.
I am not aware of any way to say whether the photons are attracting or repelling each other. Their fields are interacting and the photons are scattering as a result.
In classical physics, two particles interacting electrostatically can have the same differential cross section for scattering regardless of whether they attract or repel. For example, the differential cross section for Rutherford scattering stays the same if you change the sign of $Z_1$ or $Z_2$. The hyperbolic trajectories are similar at infinity; all that differs is what happens when the particles are close. (Does the hyperbola of one particle go in front of the other particle or behind it?) But this is meaningless in a quantum theory because there are no classical trajectories.
Since photons cannot be at rest, there cannot be a QFT calculation for photons that is analogous to the derivation of the Coulomb potential for charged particles at rest.
