How is the wave function of a photon described in quantum field theory?

[enwa]

In the electromagnetic picture, different frequencies of light waves produce different strengths of diffraction/refraction.

In the quantum picture, a photon's energy corresponds to its frequency by the Planck constant $E = hf$.

The solution of Schrodinger's equation for a free particle (which we can model a photon with) gives a continuous spectrum of non-normalizable waves, to get a physical wave function a normalizable superposition is formed. These have both 'group' velocity $c$ and 'phase' velocity $v$.

Refractive index is related to phase velocity by $n = c/v$.

By the above, I think that (A) the wave function for a photon has wavelength $\lambda = 1/f$ as opposed to the alternative (B) the frequency of a photon is an intrinsic quantity much like spin.

Is (A) correct, and if so how is this known? But maybe I am wrong, if (B) or some alternative is true please tell me where my reasoning is wrong and how correct the correct model of a photon is known. Thanks very much.

 

[Bill_K]

(C) The wavefunction for a photon shares some of the properties you mentioned for the wavefunction for a particle. It is typically a packet formed from waves having different frequencies. Each of these frequencies corresponds to a certain energy by the usual formula, E = ħω. The photon has a probability distribution in frequency, and a corresponding probability distribution in energy. Even a "pure" spectral line from an atomic transition (e.g. the sodium D line) is not precisely sharp in frequency and energy, it has a finite spread or linewidth. Just as the atomic state it came from has a finite spread in energy.

 

[enwa]

(C) The wavefunction for a photon shares some of the properties you mentioned for the wavefunction for a particle. It is typically a packet formed from waves having different frequencies. Each of these frequencies corresponds to a certain energy by the usual formula, E = ħω. The photon has a probability distribution in frequency, and a corresponding probability distribution in energy. Even a "pure" spectral line from an atomic transition (e.g. the sodium D line) is not precisely sharp in frequency and energy, it has a finite spread or linewidth. Just as the atomic state it came from has a finite spread in energy.

Thanks very much it was fascinating to read about spectral linewidth, apparently arising from energy/time uncertainty principle.

Does anyone know of experiments or theory that back up the idea that the Schrodinger wave function of a photon has frequencies near to the "classical" frequency the light is thought of as having?

 

[Bill_K]

enwa, I should make it clear that a photon does not have a special "wavefunction" ψ in addition to its other properties. A photon is, after all, a quantum of electromagnetism. It is described by the usual E and B fields, and more importantly by the potentials A and φ. When we speak of its frequency, we are talking about oscillations in these quantities.

 

[enwa]

enwa, I should make it clear that a photon does not have a special "wavefunction" ψ in addition to its other properties. A photon is, after all, a quantum of electromagnetism. It is described by the usual E and B fields, and more importantly by the potentials A and φ. When we speak of its frequency, we are talking about oscillations in these quantities.

My understanding was that:

• In the classical picture light is made of waves in the EM field and Maxwell gave evidence for this by calculating the speed of light in terms of electric and magnetic permittivity constants.
  
• In the quantum picture light is composed of photons, which are quantum mechanical particles much like electrons in the sense that they have a Schrodinger wave function.

Based on what you said, it seems like the quantum picture I had is wrong: but I don't know what the correct quantum picture of a photon would be.

 

[EmpaDoc]

Hi enwa, the answer to this question has to go quite deeply. I'll make a try.

To describe photons we use quantum field theory. Since photons can be created and destroyed, the meaningful quantity is the wavefunction (or the state) of the whole electromagnetic field. This wavefunction can give the answer to questions like: "What is the probability of detecting N_1 photons with frequency f_1, and N_2 photons with frequency f_2 (and so on)?" It can also tell you "What is the value of the E and B fields at different points at this moment of time?". However, these descriptions are complementary, thanks to Heisenberg's uncertainty principle. It dictates that the number of photons and the phase of the E and B fields cannot be measured simultaneously. Recall that in ordinary quantum mechanics, the position and momentum of a particle cannot be measured simultaneously, but the wavefunction gives the answer to either "What is the probability to find the particle at x" or "What is the probability to find that the particle's momentum is p".

So a quantum state does not have to contain a specific number of photons. It can be a superposition of different states, one with 1 photon, one with 2 photons and so on (with specified frequencies). But it describes photons, and E and B, at the same time.

By the way, the wavefunction of a single electron is also not well defined in the relativistic case because, again, they can be created and destroyed (though only in pairs with positrons). There, again, one has to consider the state of the electron field. To be sure, even in the nonrelativistic case, if you have more than one electron, they can be entangled, and then you have to consider the wavefunction of the whole system.