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If no, what is the relationship between the wave function associated with a single photon and the classical wave which describes the behavior (such as diffraction) of light (a group of photons) ?

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If no, what is the relationship between the wave function associated with a single photon and the classical wave which describes the behavior (such as diffraction) of light (a group of photons) ?

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It's also not very clear what a single photon is even supposed to be. There are different definitions of what a photon is, but you will mostly get two answers depending on whether you ask an experimentalist or a theorist. 1) A single photon is what makes the detector click exactly once and is absorbed at the same time. 2) A single photon is the state of the EM quantum field with particle number 1. Multiple photon states are defined accordingly.

These two definitions are not identical and it's not obvious if they're even compatible. If you are pedantic about the field state in 2) being an eigenstate of the particle number operator they're surely incompatible. More relaxed interpretations can bring the concepts closer together, but it's still a very difficult concept.

So asking about the field of a single photon is both difficult to answer and not even that well defined. It's however certainly not the classical EM field.

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bhobba

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It's a difficult issue.hello, is the wave function of a photon is as same as the classical wave exhibited by a group of photons ?

It has been discussed on this forum before and generates a lot of robust discussion.

At the risk of simply regurgitating that you may find the following helpful:

http://arxiv.org/ftp/quant-ph/papers/0604/0604169.pdf

Thanks

Bill

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blue_leaf77

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While coherent state is defined to be linear superposition of many number states. And if you look for the expectation value electric field in coherent states, you will find that the expectation value forms a sinusoidal function w.r.t phase. If i'm not mistaken this is why some people regards coherent state to be the closest quantum representation of light the its classical counterpart. Simply because in classical EM wave, the E field is sinusoidal in time.

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http://lanl.arxiv.org/abs/1205.1992

Sec. 8.3.3.5.

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bhobba

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Well here is one way:So how do we explain the diffraction pattern formed by emitting the photons one by one in the two-slits experiment ?

http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

The idea is the slits change the momentum of the photon.

But be aware, without going into the details it's wrong. Unfortunately physics is sometimes like that - what you learn at a less advanced level needs to be corrected later.

Thanks

Bill

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Is there any universal accepted explainations for the diffraction pattern formed by the one-by-one photon emission in two slits experiment ?But be aware, without going into the details it's wrong.

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Thanks a lot ! I will search it.Yes, it's called quantum electrodynamics and one of the most accurate mathematical models about nature every discoved :)

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Or even better, search about quantum optics, which in this case is the relevant branch of quantum electrodynamics.Thanks a lot ! I will search it.

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http://www.nature.com/nature/journal/v433/n7023/abs/nature03280.html

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The physics is precisely what Feynman describes for massive particles in the beginning of Vol. III of the Feynman Lectures. The advantage to start quantum mechanics with non-relativistic massive particles is that you can describe single particles with a wave function, which is not strictly possible for relativistic massive particles and impossible for massless particles. For photons you need quantum field theory.

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I just cited the words from the authoer Zeilinger. In the history it seems some scientists do not accepted the statistical behavior of a single particle and just thought it was the behavior exhibited by masive particles.What do you mean by "the experiments proves the random behavior not just holds for ensembles".

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I'm curious as to the distinction between equations for fields and for wavefunctions (or equivalently, solving for a field or a wavefunction). For instance, the Weyl equation is for massless spin 1/2 particles, so if you solve that in free space or in some potential, do you not get a wavefunction? Or is the distinction that it is simply artificial to consider the wavefunction in this way, as you are really considering only one mode of an oscillator in isolation?The advantage to start quantum mechanics with non-relativistic massive particles is that you can describe single particles with a wave function, which is not strictly possible for relativistic massive particles and impossible for massless particles.

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bhobba

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We have ways and means of stopping people asking certain questionsMy question is more why specifically the fact that a particle is massless is important (obviously, no rest frame exists, but I do not see a direct line from that to this issue, but i'm probably missing something stupid)

http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

Thanks

Bill

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Also, to clarify, when most people say photon, do they mean a wave packet or a plane wave? Clearly for the latter, as for any particle even in the Schrodinger case, it is non integrable and so it makes no sense to talk about position. Or is this perhaps the issue: photons are only ever plane waves and only interact by being absorbed (a photon in a 'potential' will not get trapped, except maybe in GR and then it's only some weird effective potential), thus this wave packet idea doesn't apply? I realize I am being sloppy with my terminology, but...

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bhobba

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The summery is the math which I don't entirely understand either - my knowledge of QFT isn't quite as good as I would like it. Bottom line - in QFT the position operator, hence defining a position, is problematical for massless particles. Without a position operator defining a wave-function, which is the expansion in terms of position eigenstates is impossible. Hence you can't speak of the wave-function of a photon.I don't suppose you can summarize for me? The typesetting on that page makes my eyes bleed and baby jeebus cry =P

Most people don't know QFT, so when most people speak about a photon what they speak about is wrong. Its neither a wave or wave-packet. Its an excitation of a quantum field.Also, to clarify, when most people say photon, do they mean a wave packet or a plane wave?

Thanks

Bill

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Additionally, my understanding is it is not systemic to QFT specifically. For instance, one can use the field operator to 'create' a particle 'at position x', why does this not make sense for a photon? Sure, you'd need an infinite number of wavemodes to do it, but the same is true for an electron, is it not? Also, if i have photons in a cavity, they are said to be localized surely? I'm being facetious, clearly, but you see my point.

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bhobba

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What I am getting the feeling here is this is going around in circles. I point to an advanced article a bit beyond my current understanding of QFT that explains position is not an operator for massless particles like a photon, you ask for a précis of it, I point out its technically a bit above where I am currently at with QFT, and you still want to argue about it.

If position is not an operator you cant have a wave-function in the usual sense.

Thanks

Bill

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bhobba

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There is no wave or particle - there is only a wave-function, or even more fundamentally, a state.Bhobba, isn't it also a question of scales?.

Thanks

Bill