# No photon wavefunction?

1. Mar 29, 2008

### pellman

I have read it said a few times that a photon cannot be described by a wave-function but I am not far enough along in my QFT to know why. What's the story here?

And is this the same thing as saying that one cannot answer the question "What is the probability of observing the photon in volume dV?" And if so, well, isn't a diffraction pattern on a screen a statistical expression of the probability associated with the individual photons?

2. Mar 29, 2008

### pam

Where have you read that? I don't believe it.

3. Mar 30, 2008

### lbrits

I can't find the source right now, but here's an argument to mull over.

Electrodynamics without matter, that is, the pure Maxwell field, is conformally invariant in 4 dimensions. If you want to talk about the probability of finding a photon in a volume dV, then dV better have meaning. Conformal invariance means it does not.

4. Mar 30, 2008

### pellman

Last edited: Mar 30, 2008
5. Mar 30, 2008

### pellman

i'm sure there a problems with a mathematically precise definition of it. There always are.

But I'm just talking about everyday type interference effects like the two slit experiment with photons (or with a diffraction grating, if you want to be practical). Can't we talk about the probability that a dot shows up in a certain box on the screen? And can't we use a superposition argument with interference effects to predict it? Then that would be a wavefunction.

6. Mar 30, 2008

### lbrits

Well, you could look for a mathematically imprecise definition, but the point about being scale invariant is not a minor detail you can just gloss over. Lets say you want to localize a photon. It's extent will go as $$\frac{1}{r^2}$$ rather than $$\frac{1}{r^2}e^{-m |r|}$$, because the photon is massless. But the volume factor, $$dV$$ goes like $$r^2 dr \sin\theta d\theta d\phi$$, and pretty soon the photon is everywhere!

The probability of a dot showing up has to do with the probability of an event occurring (interaction with an electron), since we can only detect photons using charged particles (they don't couple to anything else). But the same electromagnetic field could interact with one electron, scatter deeper into your detector, and interact with another detector. To make sense of this, from a photon point of view, you would have to say that the number of photons isn't conserved.

But if your wavefunction is a probability amplitude for the "probability of finding a photon" somewhere, (I must stress, photons are indistinguishable; there is no such thing as "this" photon or "that" photon), then that probability is going to do funny things like become greater than 1.

With that in mind, you can talk about the probability of the fields being in a certain state, and then probabilities make sense again. But you have to look at the system as a whole. This is bigger than the problem of the photon being massless, but it serves to illustrate the point.

Last edited: Mar 30, 2008
7. Mar 30, 2008

### lightarrow

Photons do couple even with neutral particles, if they have magnetic moment.

8. Mar 30, 2008

### f95toli

Well, you can always "localize" a photon as long as it is inside a high-Q cavity (cavity-QED), at least as long as you some way of making sure that the field is in a Fock state (otherwise you can only speak of the average photon number, which can e.g. be much smaller than one even for a cavity with a non-zero field).
For a lossless cavity you can theoretically arrange for it to have exactly one photon meaning the probability of finding it there is 1.
However, any real cavity will lose energy so the "probability" (or whatever we should call it) of finding one photon inside the cavity will decay over time.

Note that in the decay time can be very long -of the order of tens of microseconds and longer- which means that there is enough time to manipulate the state of the photons. This is why cavity-QED systems are so useful for studying fundamental properties of matter and light.

9. Mar 30, 2008

### genneth

I think this is another case of the subtlety of what QFT means by particle. Remember that a wavefunction is no more or less than a representation of a vector in certain basis. If the basis is unsuitable, then the representation disappears.

As far as the conformal invariance of pure Maxwell goes, that is probably very true. But how are you going to measure anything without some matter? Almost every "paradox" in quantum-related thing comes down to forgetting that the observer has to exist...

10. Mar 31, 2008

### pellman

lbrits, do the points you make about photons not apply to, say, electrons? Where does the difference lie?

What is spreading here if it is not the wavefunction for the photon?

11. Mar 31, 2008

### Demystifier

Pellman, the photon wave function does exist, but the problem with it is that, in some cases, it cannot be interpreted as a probability amplitude of photon positions. For a review, with some proposals for a solution of this problem, see
http://xxx.lanl.gov/abs/quant-ph/0609163 [Found. Phys. 37 (2007) 1563]
especially Secs. 7.1, 7.2, and 8.3.

12. Mar 31, 2008

### Demystifier

Indeed! The review mentioned in my previous post discusses it in detail in Sec. 9.

13. Mar 31, 2008

### pellman

Thanks, Demystifier. Are you the author of that paper?

14. Mar 31, 2008

### Demystifier

Yes!

15. Mar 31, 2008

### pellman

Well, then.

I'm getting maybe little off topic here, but in section 9 you point out that the quanta associated with energy raising and lowering operators cannot be identified with particles. Certainly I agree. Even the simple case in which we have equidistant energy levels in multiples of $$\omega$$, the number operator N which you discuss applied to a pure state consisting one particle of energy $$2\omega$$ and one particle of energy $$3\omega$$ would return the eigenvalue 5, correct?

But what about the density operator $$\phi(x)\phi(x)^\dag$$? Does not $$\int_V \phi(x)\phi(x)^\dag d^{3}x$$ unambigiously represent the number of particles in the volume V?

Last edited: Mar 31, 2008
16. Mar 31, 2008

### pellman

As for the question of this thread ... say no more. Demystifier points out in sec 7.2 of his linked paper that since Maxwell's equation in terms of the EM potential A are second order in both time and space, they will have the same interpretational problems that the Klein-Gordon equation has. I, for one, am well acquainted with the difficulty of identifying a pdf for spin 0 and can see where this leads.

17. Mar 31, 2008

### lbrits

Actually, I don't think this quite gets you off the hook. Because the Dirac field has the same problem, yet we are able to speak reliably of single electrons and their wave-functions etc. One is then lead to question whether or not this can be done for EM. Does it have a first-quantized approximation, in other words?

One way to think of "plain old" quantum mechanics is as being the classical limit of quantum field theory. In the sense that we refer to the classical limit ($$\hbar \to 0$$) as tree-level in Feynman diagrams, with quantum corrections appearing as loops, the tree level propagator of a particle can be written as a single-particle path integral. So in solving completely classical Green's functions for fields we obtain single-particle quantum mechanics. The probabilistic interpretation of these particles is inherited from the probabilistic nature of QFT itself. In other words, the classical limit of the QFT sports individual quanta which do not interact (interactions appear as loops), and looking at the one-quantum sector, it is described by single-particle QM.

This goes under the name of "Schwinger proper time" method, Heat kernel method or lately, "String inspired" or "worldline formalism".

Here's the crux. This can be done for a scalar field, a spinor field (by introducing world-line Grassmann numbers and world-line supersymmetry), but I don't know if it can be done for gauge fields. I would very much be interested if it can.

18. Mar 31, 2008

### peter0302

I'm confused. Why is the wave function of the photon not simply a sphere with an expanding radius defined by c*t?

19. Mar 31, 2008

### lbrits

That is a classical solution to the E and B fields (or A, in an appropriate gauge), when there is a point disturbance at some initial time. That is not the same as a wave-function, and is probably more akin to what is known as a coherent state.

The "spreading wavepacket" visual of a particle comes from finding an eigenstate of the position operator, i.e., a delta function, and then evolving it forward in time using a (particle) Hamiltonian. What Hamiltonian would you use to effect such an "expanding sphere"? The E/M Hamiltonian is that of a classical field, not a classical particle. Therefore, if you wanted to talk about the wavefunction of a photon, as a particle, you wouldn't be able to do much with it without also knowing the Hamiltonian of a photon, whatever that means.

I wonder how the situation changes, if at all, by using a Proca field instead of a photon field. In other words, generalize the E/M field so that it has a mass term in it. But now I'm rambling =)