If a photon can be described as an excitation in a quantum field, is this the same field for all photons, or does each photon exist in its own field?
3 Answers
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I believe it is the same field for all photons in the universe. Similarly there is one field for all up quarks in the universe, another field for all down quarks in the universe and so on for each particle type. In this way the field theory model explains the uncanny consistency of particles; a neutrino made in a supernova millions of miles away is identical (as far as we can tell) to the same species of neutrino created on Earth. This suggests that all neutrinos come from the same 'stuff', and we call this stuff the neutrino field.
I think David Tong mentions as much in this talk:
answered May 14, 2013 at 15:51
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1$\begingroup$ +1 This is good, but to be quite frank. No one knows what the correct answer to this question is. It is definitely true that it is the same type of field, but it is impossible to say for sure if it is the same field. Especially if we consider photons outside our visible horizon. $\endgroup$– JimCommented May 14, 2013 at 15:55
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1$\begingroup$ I think if someone is asking a question about photons in quantum field theory, we can implicitly assume they are referring to the standard model of particle physics. Within the standard model, there is only one photon field. $\endgroup$ Commented May 15, 2013 at 2:45
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1$\begingroup$ @Jim: It must be the same photon field. Otherwise you'd need an uncountable number of unrelated couplings to different photon fields. $\endgroup$– NeuneckCommented May 15, 2013 at 11:41
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1$\begingroup$ @Jim: It would require an infinite ammount of fine tuning. For example by your many-fields reasoning every electron could have an arbitrary charge. On the other hand, the experiments tell us that all electrons have exactly the same charge which would be an enormous coincident. Occhams razor therefore demands that we choose the most economic solution that is in accord with all observations. And that is that there is one single field for each particle type. $\endgroup$– NeuneckCommented May 16, 2013 at 7:46
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1$\begingroup$ @SirajRKhan: The field from far away would not affect our electrons. It could have an arbirary charge though, say $\sqrt{\pi}$ or $35$. If there were as many fields as there are electrons in the universe, the probability of their charges (which are unrelated, as they are unrelated fermionic fields) all being exactly equal is basically zero. Especially since no first principle sigles out charge $1$ (there are charge $\frac{1}{3}$ quarks). $\endgroup$– NeuneckCommented May 17, 2013 at 9:15
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Probability of photon emission by an atom depends on the occupation number of already existing photons of this sort. The corresponding occupation number is determined with the "boundary conditions". We cannot take the occupation number of all existing photons in the Universe, so time-space "separated" regions have effectively their own photon fields and their own occupation numbers.
On the other hand, when we observe an interference pattern, we don't care of the possible sources. To us it is crucial to have a photon at the same time at the same place whatever is its origin. Superposition happens for photons of a given sort arriving at the same time at the same place. That means the boundary conditions may be "open" for two or more sources.
answered May 14, 2013 at 22:02
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$\begingroup$ The commutation relations look something like $[a(x),a^\dagger(y)]=\delta (x-y)$ right? We don't need another label $a_i$ to denote which part of spacetime (say use a partition of unity or something) "each field" is. My impression is "one quantum field" is the more usual interpretation. $\endgroup$ Commented May 15, 2013 at 2:50
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$\begingroup$ @levitopher: How about loss of interference due to retardation of photons? Loss of coherence must be described somehow. Randomness of wave phases is a mean to "label different fields". $\endgroup$ Commented May 15, 2013 at 8:02
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$\begingroup$ @VladimirKalitvianski Would it not be correct to say that this approach of using multiple fields is more of a useful tool rather than a strict statement about the number of fields involved. Strictly speaking wouldn't we say that there is one photon field? However because the 1 field approach is too difficult (knowing the occupation number of all existing photons is too difficult) we work with the pretence of multiple fields. In this way we still get accurate answers, but via a much easier (and more possible!) method. $\endgroup$ Commented May 15, 2013 at 12:36
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$\begingroup$ @user15766: Yes, it is still possible to deal with one field, but very complicated one due to too many sources/absorbers. The boundary conditions I mentioned are approximate (simplified) solutions to the total field-matter equations. So we replace one total field with many similar ones acting each in different places of Universe (i.e., without interference). $\endgroup$ Commented May 15, 2013 at 12:41
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$\begingroup$ @VladimirKalitvianski I see, thank you for the response. I'm learning a lot on this particular web page! $\endgroup$ Commented May 15, 2013 at 12:46
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Let us consider the Hamiltonian of an electromagnetic field in a finite region of space. It is possible to find a suitable unitary transformation to change it in a form same as a Hamiltonian of a independent harmonic oscillators. A photon is a quantum of energy of such an ensemble of independent oscillators. All excitations of this ensemble will be associated with the same field.
