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Please give me your thoughts regarding the statement and questions that follow.

A light wave is a probability wave depicting where a photon is likely to be within the wave. Once the photon is found then all other possibilities are gone. Given this, is it then true to say that one light wave will only produce one photon? Could we go on to say that the number of photons emitted per second is equal to the frequency of the light?

DanielSank
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Lambda
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    What do you mean by "one light wave", exactly? Waves don't come in discrete units one could count by integers. Related: http://physics.stackexchange.com/q/90646/50583, http://physics.stackexchange.com/q/46237/50583 – ACuriousMind Mar 05 '17 at 23:13
  • A wave has a measurable length. I have measured several by physically moving an interferometer's mirror by a discrete measurable amount. Maybe you could answer my question. How many photons does a wave possess? Is there a formula? Or is the answer we just don't know? – Lambda Mar 05 '17 at 23:47
  • "A light wave is a probability wave depicting where a photon is likely to be within the wave" that's incorrect: a light wave describes the state of a field; under some circumstances the probability to reconstruct particles from fields can be derived in terms of the original state. – gented Mar 06 '17 at 08:55
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    @Lambda The amount of photons in a well defined electromagnetic wave is uncertain. If you want to have a well defined amount of photons, you can have it only in an undefined wave (superposition of various configurations). This is because the particle number operator does not commute with the annihilation operator. – mpv Mar 06 '17 at 09:18

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The wave particle dualism, which is frequently presented in popular science literature and even taught in introductory courses is a historical misconception which is resolved by the quantum field theory framework. In this framework a photon is an excitation of a (vector) field which, for each point in space and time gives the field strength $A^\mu$ (not to confuse with the physically measurable field strengths $\vec E$ and $\vec B$ which are derivatives of $A^\mu$ components).

The field $A^\mu$ has some configuration depending on the initial conditions. It is tempting to interpret it as a probability distribution for particles but afaik it is not correct to do so, it is a field, not a wave function.

If a photon is measured, it interacts with some other particle (probably an electron which is part of the measuring device). This changes the field configuration for later times. But again, I'd not interpret it as something related to the collapse of a wave function.

Photon
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  • What would be the relationship between the field strength and the probability of detecting a photon somewhere? Or what is the relationship between field strength and whatever is considered a photon? (is there anything left of the classical picture of an electromagnetic wave consisting of propagating E and B fields?). Sorry for the questions! –  Mar 05 '17 at 22:05
  • Is the photon the field source of the vector field? – Lambda Mar 05 '17 at 22:33
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    It might be better to say that a photon is an excitation of a vector field, rather than the field itself. – garyp Mar 05 '17 at 23:16
  • Thanks for your answer, Photon. I have read it several times. I can picture the vector field. I see it as many vector arrows pointing toward and away from source points. Are the source points possible photons? This vision of the wave as a vector field doesn't resolve the question as to how many photons can a wave produce. Maybe I need to just devote more time thinking about it. – Lambda Mar 06 '17 at 05:31
  • @garyp: Right, using the theorists' slang which doesn't differentiate between fields and particles is a bad idea, I edited the answer. – Photon Mar 06 '17 at 07:22
  • @HughMungus: This is how you get the E and B fields out of the A field: https://en.wikipedia.org/wiki/Electromagnetic_four-potential The probability is, as usual, dependent on the field configuration, that is, the exact form of $A^\mu(x)$. Usually people expand $A^\mu(x)$ in Fourier modes and consider photons to be single modes, that is, plane waves with an exactly specified momentum but no information about position. But I think, this is due to the fact that QFT is usually used to calculate what happens in an collider where all particles are fast enough to make this a good approximation. – Photon Mar 06 '17 at 07:31
  • @Lambda: No, the photon is not the source of the vector field, but, as garyp pointed out, the excitation of the field. As I wrote in my previous comment, usually people tend to call Fourier modes with a defined momentum photons, but there is no physical reason for it. If we want to be precise, the form of the excitation doesn't really matter, any excitation would be called a photon. – Photon Mar 06 '17 at 07:37
  • @Lambda I am suspicious that you have a picture in your head of photon "particles" that are like tiny marbles, or "bits of stuff" in some sense. An entity that has a limited spatial extent, and hence a position, and hence a velocity. The word "particle" in physics means something different than it means in everyday life. Photons are not particles in that sense. They have this in common with everyday particles: they interact with other particles at specific locations and can transfer energy and momentum in the interaction. But a photon is not a little "bullet". – garyp Mar 06 '17 at 14:48