Photon Wavefunctions: Is There a Schrodinger Equation?

[space-time]

Do photons have quantum mechanical wave functions like other particles do? If so, would I use some alternate version of Schrödinger's equation when deriving said wave function? I ask this because as we know, the Schrödinger equation is as follows:

(-ħ²/2m)∇² + v(x,y,z)Ψ = EΨ

Photons however, have 0 mass, so m = 0. I'm sure the problem with this becomes obvious if you try to plug m = 0 into the equation. I can only think of one possible solution to this, and this is multiplying both sides of the equation by (-2m/ħ²). If you do that and then plug in m=0, then the equation simply becomes ∇²= 0 (which is solvable). Other than that, I don't see a solution.

That is why I ask: Is there a Schrödinger equation or wave function for photons?

 

[Orodruin]

The Schrödinger equation as it stands applies only to massive particles moving much slower than the speed of light. In order to describe particles moving at relativistic velocities you need to look at the Klein-Gordon or Dirac equations. Photons, however, are notoriously difficult to quantise properly and the appropriate framework will not be available to you until studying quantum field theory.

 

[vanhees71]

To answer the original question: No, photon's do not have a wave function. Strictly speaking the single-particle wave-function formalism (aka "first quantization") doesn't make sense for relativistic interacting particles, because particle number is not conserved under high-energy collisions. For photons, you cannot even formaly define a single-particle wave function, because photons as massless spin-1 quantum fields do not have a proper position operator at all. The only adequate description for photons known today is quantum electrodynamics, i.e., in terms of relativistic local and microcausal quantum field theory, as Orodruin already said.

 

[kith]

There are however people who think that the photon wavefunction is a useful concept even though it isn't on equal footing with the usual wavefunction. See e.g. http://arxiv.org/abs/quant-ph/0508202 for a discussion of how far the analogy goes and where the problems lay.

I've always found the "there's no position operator for the photon, so don't ask for a wavefunction" paradigm a little dissatisfying.

 

[vanhees71]

Well although it may be dissatisfying, but it's a mathematical fact. See

http://arnold-neumaier.at/physfaq/topics/position.html

and references therein.

 

[kith]

I am not dissatisfied with the mathematical fact that there is no position operator for photons. It just doesn't give much physical insight. What I am dissatisfied with is the attitude that you should stop asking questions which come from your physical intuition because of this mathematical fact.

Sure, if you want to understand how the electromagnetic interaction works, you have to learn QED sooner or later. But when I learned non-relativistic QM, I had many questions regarding how far the analogy between electrons and photons goes and what problems occur if I try to treat the photon as similar to the electron as possible. I think this is quite natural because this analogy is widely used to motivate QM. Learning the fact that there isn't a position operator for photons didn't answer my questions but increased my confusion at the time. So below are a few references which may benefit people with similar questions.

Localized single-photon wave functions in free space
KW Chan et al. - Physical review letters, 2002

The photon wave function
I Bialynicki-Birula - Coherence and Quantum Optics VII, 1996

Photon wave functions
JE Sipe - Physical Review A, 1995

 

[Demystifier]

photons as massless spin-1 quantum fields do not have a proper position operator at all.

It depends on what do you mean by "proper". A position operator can be introduced, but it is not Lorentz-covariant. This is not necessarily a problem, if you think of a position operator as an instrumentalist tool which only makes sense in a context of a specific measurement, which defines a special Lorentz frame at which the measuring apparatus is at rest.

 

[jfizzix]

In quantum optics, we often take the lab frame as being a special reference frame, and use the non-lorentz-covariant Coulomb gauge to make theoretical predictions.

In this case, you can define a Schrödinger equation to describe the evolution of the electromagnetic field, and derive the quantum state of optical fields. Then, we can define a photon wavefunction as a square-integrable function living in Hilbert space that will give you the measurement statistics of that photon interacting with a detector.

Note: You also have to normalize the photon wavefunction, because the amplitude for the photon to be in a given state with a given position distribution could be very low.

As an example, this paper:
http://arxiv-web3.library.cornell.edu/abs/1502.06996
shows how to calculate the wavefunction of a pair of entangled photons exiting a birefringent crystal.

Yes, it's a shameless plug of my own work, but you may find it helpful. Feel free to ask me any questions about it too.