# Are EM waves telling us the probability of finding a photon?

I feel like I've been frequently presented with an interpretation of EM waves that goes something like this:

Light is an oscillating electromagnetic field. Because changes in the electric field produce changes in the magnetic field and vice versa, it is self propagating. The electric field of light can be represented as a sine wave with a phase, frequency, and amplitude. The amplitude however, only takes on discrete values. For monochromatic light, it is an integer multiple of the "photon energy".

This seems fairly intuitive for collimated beams of light (such as from a distant source) and so the light arrives as a plane wave. That is to say, the "sine wave" moves more-or-less like a "ray of light". And if you put a sensor within that ray of light, you'll detect the light.

But I start to get confused when I think of reflections. Feynman's QED and other sources talk about reflection in terms of interference. That the light incident on a surface causes the charges in that material to oscillate, which produces new EM waves. These EM waves interfere with one another giving us the classical macroscopic perspective of the light being specularly reflected. Feynman in this context though, is talking about this in terms of the probability of detecting a photon, and so where the probabilities constructively interfere (along the line where angle of exittance equals angle of incidence), we are likely to find a photon and along all other directions we are unlikely to find a photon.

Similarly, say we have a single electron in space that is oscillating in the $$+z$$ direction. Then it would emit EM radiation out into the entire $$xy$$ plane. The intensity of this would necessarily have to decrease over distance, but we know that light is quantized. If the EM waves it was emitting were probabilities, then we should expect that for a given observer, the odds of detecting a photon decrease with distance but once we actually make a detection, the photon would have a definitive energy. This is, as far as I know, true.

It seems like both of these scenarios are pushing me towards viewing the EM wave as a representation of the probability of finding a photon, rather than the actual strength of the electric field itself. But this doesn't seem right either, as afterall charges react to light's oscillating electric field. Plus the wave doesn't just take values between 0 or 1 as a probability should, but rather has an amplitude based on the energy, and oscillates between $$-A$$ and $$A$$. Feynman refers to the square of the amplitude as the probability, but this doesn't fully solve my confusion as now the values range from $$0$$ to $$A^2$$.

What am I missing here? Talking about the wave as purely the oscillation of the electric field feels incomplete, but viewing it as a probability also seems incomplete. Afterall, the literal oscillating electric field is what produces the oscillating magnetic field, and it is what drives the oscillation of charged particles it comes across. Yet at any given point in space, you only have a probability of actually detecting a photon there, and it's energy is definitively known. So I must be misunderstanding something. I can't understand what else the wave actually is. Or are there two waves, one for the probability and one for the electric field?

Edit: I should say I have only read the first 2 chapters of Feynman's QED. Perhaps as I continue reading, my questions will be answered. I'm just finding myself very confused at the moment.

• ‘changes in the electric field produce changes in the magnetic field’ I phrase it differently. In the absence of field sources the time derivative of E is the same thing as the rotation of B, and vice versa. Commented Feb 29 at 13:07
• Feynmans path integral gives a probability distribution for the em field for the double slit. Commented Feb 29 at 16:07

Light is not a wave, that was only true in classical physics. It can not be represented (as you claim) by a sine wave with discrete values of amplitude. It can be represented as a wave functional, which is a complicated concept (a "wave function of a wave"). There are many discussions of it, e.g. this one. You will have to study Quantum Field Theory to find out the details.

As for the observer of your EM wave that becomes weak with distance: the observer's own quantum state will interact with the EM wave functional and the resulting entangled state will have a small-amplitude component where the observer's detector was triggered, and a large component where nothing was seen. All can be described with one big QM state. But due to decoherence those two components (usually) evolve in time almost independently after the interaction.

This means you can act as if one component does not exist, but in fact there are two branches of reality that have been almost completely split off from each other. That last situation is how the discreteness enters in the macroscopic world (you can call it "apparent collapse" or "many worlds" or whatever you like). It does not follow from the discreteness of eigenfunctions of operators directly, but from the fact that some of those states are "pointer states" and become independent in time. (Collapsing the wave function is then by far the most efficient way to compute things for only one of the components, but not necessarilly the truth.)

• I think what was having my confused was reading Feynman's QED, his analogy of the "clocks" was making me try to connect it to my understanding of waves from basic E&M class. Having read your comment and starting chapter 3, it's clear now this isn't the case (though, being related, i can see why the wave theory is a good classical approximation). I perhaps should have taken his multiple warnings "light is not a wave" more literally, and not tried to get his words to fit my previous exposure to the topic, which was clearly incomplete. Commented Feb 29 at 21:25
• ‘Light is not a wave, that was only true in classical physics. Etc.’ This is really bad news for astronomers, lithography and optics in general! Commented Mar 1 at 7:07
• Yes @my2cts, but sometimes they get little things back. Like the "squeezing" that is possible with a wave functional (as opposed to an exactly defined wave). Admittedly, this is a case of giving with one hand and taking away with the other, because QM creates the problem of quantum noise, and then offers a way to reduce it with a lot of effort! Anyhow, optics in general has to deal with it... Commented Mar 1 at 7:27

The electromagnetic wave gives the expectation or average value of the number of photons. The Poisson distribution with this average is what you will observe. The same is true for an electron wave.

I like to keep the math as simple as possible, but not any simpler than that. If I can be more concise let me know 😄.