What are the microscopic details of diffraction? In a recent question I asked how double-slit diffraction conserves momentum and the answer was that each photon gets a "kick" from the slits. So my question now is, what does that look like at a microscopic detail level?
I have a candidate. The lowest order loop expansion I can come up with looks like so. (Apologies. I don't have a drawing package that does electrons and photons properly. I'd be glad to be pointed at one if it exists.)
So very schematically: The photon must split into two parts, which have to start as charged particles. Otherwise they could not be interacting with the photon. These must go one on each side, otherwise there can't be any interference from the two slits. The slits provide an elastic interaction photon that provides momentum, but no change in energy.
I am presuming that the interaction photon can arise from any point on the slits, and interact with any part of the loop.
Is this what is going on? There ought to be a "well known" experiment that sees that charged loop if it is there. Say by imposing an electric field parallel to the plane of the slits.

 A: You do not state the level of your physics background in your profile, but from your question, I assume that you are not familiar with quantum mechanics.
There is no "charged loop " as  you imagine in your drawing.
The material of the slits consists of atoms and molecules, which are neutral. Atoms have charged electrons around them and those are the ones photons will interact with. BUT electrons and photons are point particles, no extension in space that you show for the "charged" particle in your drawing. Their interactions are described mathematically by quantum field theory which gives the solutions in series terms of decreasing in magnitude order. The first order diagrams and calculations   for electron photon scattering can be seen here


In the first diagram (figure 7.7a), the incident photon ($k,\varepsilon$) is absorbed by the incident electron ($p_i,s_i$) and then the electron emits a photon ( $k^\prime,\varepsilon^\prime$) into the final state. In the second diagram (figure 7.7b), the incident electron ($p_i,s_i$)

This is for a free electron. It looks a bit like your diagram but it corresponds to mathematical formulas that calculate the probability of scattering in a given direction.
You can see that from momentum conservation the directions change.
When the electron is tied up in the atoms of the slit, it becomes more complicated mathematically , but momentum conservation is a strict rule. The atom on which the  electron is bound takes up the momentum and transfers it to the lattice it belongs to.
For information here is how the double slit single photon at a time images are built up


Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The classical interference pattern appears after the accumulation of photons.
A: There is no charged loop and no extra photon.
The one and unique photon in the two slits experiment acts as a wave. It does not just "split in two parts" to go through each slit. The wave "fills" the full widths of both two slits at the same time and gets "kicks" from all four ends of the slits at the same time.
How do we know it it so ? Because of the full diffraction picture. Sure, the fact that there are two slits is the cause of the interference pattern. But the width of the entire diffraction pattern is controlled by the width of each slit.
Remember, the width of the diffraction pattern in a single slit experiment in inversely proportional to the width of the slit.
If the slits are not narrow enough, there is no visible diffraction at all. Roughly, the width of the slits must be narrow compared to their separation, otherwise the diffraction pattern is narrowerer than the interference one and one cannot see interferences.
You have to think in terms of waves. Trying to understand quantum mechanics without accepting the notion of duality leads nowhere. One has to accept the weirdness of QM, and to think differently.
A: The diffraction pattern is not formed by photons splitting up into pairs of charged particles. Unless the light has a frequency that is high enough to produce electron-positron pairs, the contribution of those virtual loops that are shown in the diagram would be negligible.
To explain the conservation of momentum here, I need to use a bit more technical language. Please bare with me.
We need to say something about the state of the light. Let's start with a single photon state, (a number state with exactly one photon) which we'll denote as $|\psi_1\rangle$. It consists of a superposition (or spectrum) of plane waves. Crudely,
$$ |\psi_1\rangle = |\mathbf{k}_a\rangle C_a + |\mathbf{k}_b\rangle C_b + |\mathbf{k}_c\rangle C_c + ... , $$
where the $\mathbf{k}$'s are the wave vectors of the plane waves and the $C$'s are complex coefficients. More accurately, we can represent it with an integral
$$ |\psi_1\rangle = \int |\mathbf{k}\rangle C(\mathbf{k}) \text{d}\mathbf{k} . $$
We can represent the screen with the two slits as a transmission function $t(\mathbf{x})$. To get the state after the screen, we need to apply an operator $\hat{T}$ that imposes this transmission function on the state. Even if the state before the screen were just a single plane wave, the transmission function would cause the state after the screen to have a spectrum of plane waves. However, the transmission function causes a loss. Therefore, the operator does not maintain the normalization. So, we would need to normalize the state afterward.
To treat the loss correctly, we can model the screen as a beamsplitter that sends the part of the state that is blocked by the screen to a different output port where we can "trace them out." That would give us a mixed state if the original state contained $n>1$ photons (a number state with exactly $n$ photons). In such cases, the interference would be lost.
Fortunately, most experiments are done with laser light, which is represented by a coherent state, instead of a number state. Coherent states remain pure states when they suffer loss. Coherent states are parameterized by spectra, similar to the way we parameterize the single photon state, but in this case, the spectrum is not in general normalized. We'll assume that input state is given by a coherent state $|\alpha\rangle$ with a spectrum $\alpha(\mathbf{k})$.
Now we can use this picture to address the issue of the momentum. The part of the light that passes through the slits is given by $|\alpha'\rangle$ where the spectrum is
$$ \alpha'(\mathbf{k}) = \int t(\mathbf{x}) 
\exp(i\mathbf{x}\cdot\mathbf{k}-i\mathbf{x}\cdot\mathbf{k}') \alpha(\mathbf{k}') \text{d}\mathbf{k}'\text{d}\mathbf{x} . $$
Note that even if $\alpha(\mathbf{k}')$ was a very narrow spectrum, the modulation with the transmission function will cause $\alpha'(\mathbf{k})$ to have a broader spectrum. On the other hand, the light that is blocked by the screen is given by $|\alpha''\rangle$
where
$$ \alpha''(\mathbf{k}) = \int [1-t(\mathbf{x})] 
\exp(i\mathbf{x}\cdot\mathbf{k}-i\mathbf{x}\cdot\mathbf{k}') \alpha(\mathbf{k}') \text{d}\mathbf{k}'\text{d}\mathbf{x} . $$
Note that
$$ \alpha'(\mathbf{k}) + \alpha''(\mathbf{k}) = \alpha(\mathbf{k}) . $$
In other words, the additional components that are present in $\alpha'(\mathbf{k})$ due to the modulation are removed by $\alpha''(\mathbf{k})$ to reproduce $\alpha(\mathbf{k})$. Moreover, for the transmission function we have $t^2(\mathbf{x})=t(\mathbf{x})$, which means that
$$ \int \alpha''(\mathbf{k}) \alpha'(\mathbf{k}) \text{d}\mathbf{k} = 0 . $$
To determine the momentum, we can compute the expectation value of the wave vector. We can do this with an operator
$$ \hat{P} = \hbar \int \mathbf{k} \hat{a}^{\dagger}(\mathbf{k}) \hat{a}(\mathbf{k}) \text{d}\mathbf{k} . $$
It then follows that
$$ \langle\hat{P}\rangle = \langle\alpha|\hat{P}|\alpha\rangle 
= \hbar \int \mathbf{k} |\alpha(\mathbf{k})|^2 \text{d}\mathbf{k} , $$
because
$$ \hat{a}(\mathbf{k})|\alpha\rangle = |\alpha\rangle \alpha(\mathbf{k}) . $$
For the light that passes through the slits, we get $\langle\hat{P}\rangle'$ and for the light that is block, we get $\langle\hat{P}\rangle''$, when we use $\alpha'(\mathbf{k})$ and $\alpha''(\mathbf{k})$ in the respective calculations. Based on the properties of these spectra, it then follows that
$$ \langle\hat{P}\rangle' + \langle\hat{P}\rangle'' = \langle\hat{P}\rangle , $$
which shows that momentum is conserved. This happens because the screen receives the momentum of the part of the state that it absorbed.
A: The photons that don't interact with the screen are the ones that form the pattern. It's the ones that do interact that transfer the momentum to the slit system.
