Does it make sense to say that the quantum field of a photon is exactly proportional to the photon's electromagnetic field?

\begin{align} \bar{\Psi} = \dfrac{\bar{E}+i\bar{B}}{\sqrt{\int (E^2+B^2)dV}} \end{align}


The quantized electromagnetic field gives rise to (all) photons. There is only one (quantum) field for all photons. So asking about the quantum field of "a photon" or a "photon's electromagnetic field" doesn't seem to make sense: The photon is a quanta of the field.

  • $\begingroup$ I thought individual photons can be isolated? $\endgroup$ – linuxfreebird May 20 '14 at 21:12
  • $\begingroup$ They can, and have been isolated. That doesn't mean they are not quanta of the same electromagnetic field. Perhaps you meant to ask about the wavefunction of an isolated photon and not about fields? See: physics.stackexchange.com/questions/437/… $\endgroup$ – DrEntropy May 20 '14 at 21:16
  • $\begingroup$ Maybe, the best way to go is to try to see how we observe photons in our detectors. This is related to the 2008 Physics Nobel Prize (about spontaneous symmetry breaking). One can say that photons are Nambu-Goldstone bosons for the time-reversal symmetry breaking which appears in the interaction between the electromagnetic field and matter. Some processes are irreversible (discrete atomic level coupled to a continuum, e.g spontaneous emission) whereas Rabi oscillations is a reversible process, no energy is localised from the field in the atom over a cycle. Vacuum state conserved = no photons $\endgroup$ – JumpArtist May 20 '14 at 21:33
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    $\begingroup$ @linuxfreebird: a photon is the unit of energy exchange when the photon field exchanges energy with some other field. When you say individual photons can be isolated this is true insofar as we can isolate the events at which energy is exchanged e.g. interactions with a CCD or photomultiplier. $\endgroup$ – John Rennie May 21 '14 at 9:39

Quote from Willis Lamb, Nobel Prize in Physics, 1955.

there is no such thing as a photon.


I guess, what you meant was a Fock state. An eigenstate of the free electromagnetic field, here the first excited one associated to a well defined energy (states which then does not evolve in time) that amounts to a light quantum as said above by DrEntropy.


As an experimentalist I trust analysis by theoretical physicists and this is what I will use and quote selectively in my answer here:

When talking photons, we are talking quantum mechanics and elementary particles with their dual nature, sometimes manifesting classical particles sometimes probability waves, as as displayed by two slit experiment in this answer. The dots are the particle identification of the photon, the built up interference pattern the probability wave in space for the manifestation of a photon.

the probabilities predicted from quantum mechanics should be interpreted exactly in the same way as probabilities predicted from classical statistical physics. The only difference is that quantum mechanics implies that the "exact truth" about the system doesn't exist even in principle. However, in practice, you don't care about it because you can't know the coordinate and positions of many gas molecules in a bottle, anyway.


For a wave from a light bulb where there are different frequencies and statistically :

Imagine that they do differ and you want to calculate the average value of the electric field E⃗ at some point in space, away from the light bulb. The electric field may be rewritten as some combination of creation and annihilation operators for photons in all conceivable states, with various coefficients. And by symmetry, or because the many photons contribute randomly, you get zero. So even though we are surely imagining – and we may measure – nonzero values of the electric field away from the light bulb at some moments and locations, the statistical expectation is zero and the fluctuations of the electric field are due to the randomness of the emission processes.

For a coherent source, like a laser, where a single frequency with a width is produced coherently then the classical potential and the wave function of the photons are directly related:

I don't want to scare you by the indices but the wave function of a single photon mathematically looks like the (complexified) classical electromagnetic potential A⃗ (x,y,z), with some extra subtleties.

I think this answers that the formula you propose is wrong. It is the electromagnetic potential that enters the wave function.

If you are interested really you should read carefully the blog entry and then search further reading.


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