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I understand that technology makes it now possible to shoot single photons.

I have read this question, and the answers did not cover one thing.

Single photons: Is there a 90° offset of the electric to the magnetic component in the direction of propagation?

We know that in the answers it is correctly stated that the phenomenon can be successfully described in QM.

Although it is worth talking about this on the individual photon level, since it is now possible to shoot a single photon (it is possible now as we know). I know that that is for the double slit exp. but the single photon shoot is the technology we have and that is what it is all about.

Double Slit Experiment: How do scientists ensure that there's only one photon?

Question: If we shoot a single photon will that single photon at a certain position of the propagation axis be the excitation of both E and M fields, (and so the photon itself would have two fields coupled) or is it just the excitation for one of them and then 'changes' to being the excitation of the other one as it propagates?

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Árpád Szendrei
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Field quantization is (at least usually) done in the basis of eigenmodes. In free space, those eigenmodes are plane electromagnetic waves, which carry both $\vec{E}$ and $\vec{B}$ fields. So the short answer is, yes both fields are attached to the photon, because your whole plane wave is attached to it.

But from reading the comments it seems important to adress another topic: If you quantize your modes, physical quantities are represented as expectation values of operators. You now can formulate a photon number operator as well as field operators for $\vec{E}$ and $\vec{B}$. The problem is, that the field and number operators do not commute, so similiar as for momentum and position operators in simple QM, there is an uncertainty relation between them. You can be in a state where the photon number as well as the Intensity $|E|^2$ is well-defined (Fock States), but the field values are uncertain (and have expectation value zero). Or you are in a state where the fields are defined (Coherent States) , but the photon number is uncertain. This is probably the reason why in the comments it was stated that "you cannot talk about fields and photons at the same time".

From the way the question was asked i assume you are looking for a qualitative description. If you would like to see the math to back up my statements, i suggest https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

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The photon is a quantum mechanical point particle/entity in the table of the standard model of physics. Point means that its four vector is a function of (x,y,z,t) , it has no volume in space. It is one of the founding elements of the theory .

Question: If we shoot a single photon will that single photon at a certain position of the propagation axis be the excitation of both E and M fields, (and so the photon itself would have two fields coupled) or is it just the excitation for one of them and then 'changes' to being the excitation of the other one as it propagates?

In quantum mehanics, the point elementary particles are described by solutions of the corresponding quantum mechanical equations, Dirac for fermions, klein Gordon for bosons and a form of quantized maxwell's equation for photons.

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One usually does not discuss the wave function of a photon because the classical description of light emerges smoothly from the quantum mechanical (it does need the mathematics of quantum field theory to understand the link) and is very accurate and much more easy to use, then going to individual photon levels.

The classical electric and magnetic fields of the electromagnetic radiation are built up by a confluence of the wave functions of the zillions of photons that build it up. The fields exist in the wavefunction of the photon, i.e. before it is squared to give the probability of finding the photon with energy h*nu and an (x,y,z,t) point. It is the superposed wavefunctions that will build in confluence the classical light wave with its E and B fields.

A single photon, when detected (interacting with another particle) will only enter with the complex conjugate squared of the wave function, so the information of the electric and magnetic field is irrelevant other than its effect on the probability of the interaction happening.

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