quantum mechanics - Interference of two non-entangled photons, calculation using tensor product of Hilbert spaces

I'm trying to calculate the interference of two non-entangled photons, like in a double-slit experiment with two photon sources, one behind each slit (follow-up on this question).

The individual particles have the wave function $\psi_i\in\mathcal{H}_i$ where $i\in\{1,2\}$. The tensor product of their wave function is

$$ \Psi(1,2) = c_1\Psi_1 + c_2\Psi_2 = c_1(\psi_1\otimes 1_2) + c_2(1_1\otimes\psi_2) $$

This is what I tried:

$$ \begin{align} |\Psi(1,2)|^2 & = \langle c_1(\psi_1\otimes\mathbf{1}) + c_2(\mathbf{1}\otimes\psi_2) | c_1(\psi_1\otimes\mathbf{1}) + c_2(\mathbf{1}\otimes\psi_2) \rangle \\ & = \sum_i|c_i|^2|\psi_i|^2 + \overline{c_1}c_2\langle\psi_1\otimes\mathbf{1} | \mathbf{1}\otimes\psi_2\rangle + \overline{c_2}c_1\langle\mathbf{1}\otimes\psi_2|\psi_1\otimes\mathbf{1}\rangle \\ & = \sum_i|c_i|^2|\psi_i|^2 + 2\,\mathfrak{Re}\left(\overline{c_1}c_2\langle\psi_1\otimes\mathbf{1} | \mathbf{1}\otimes\psi_2\rangle\right) \\ & = \sum_i|c_i|^2|\psi_i|^2 + 2\,\mathfrak{Re}\left(\overline{c_1}c_2\langle\psi_1|\mathbf{1}\rangle\langle\mathbf{1}|\psi_2\rangle\right) \end{align} $$

Is that correct so far? How to continue to show the interference?

Answer

Photons are quantum mechanical entities. This means that their interactions can be described by Feynman diagrams, which are reinterpreted as creation and annihilation operators operating on ground states with propagators in between.

It has little meaning to talk of two individual photons "interfeering" without writing down the Feynman diagrams. Photon photon interactions are four vertex at lowest order. This means a factor of ~10^-8, as the electromagnetic coupling constant is of order 10^-2.

A Feynman diagram (box diagram) for photon–photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other

The four propagators for energies below gamma energies are highly off mass shell and thus diminish the probabilities even further.

So it is very improbable that two photons will have a measurable interference.