Is the Wave Aspect of Light More Like a Snake or a Wave in Water?

[Ivan Seeking]

Originally posted by arcnets
(Repeat: There is no such thing as a 'single photon e.m. field.)
Now if we want to find any probability distribution that makes any sense, then we must have the photons be absorbed somewhere, meaning [del]x is finite. This means that the wave describing the process has [del]p > 0, and thus corresponds to a mixture of photons.

But you shift gears from a propogating wave function, to one of a collapsed wave function. Your statement only appears to be correct once the photon "strikes" a target. We have a completely different animal at this point.

 

[Ivan Seeking]

Originally posted by arcnets
This does not apply to particles which have mass. As you know, the bound electron states in an atom have [del]E = 0, but [del]p and [del]x both finite. [/B]

But they need not be. I can force a [del]p or [del]x equal to zero for an electron also. You consider only the bound state; then you compare this to an unbound wave-function.

 

[arcnets]

Originally posted by Ivan Seeking
Your statement only appears to be correct once the photon "strikes" a target.

That is correct. In fact, I'm trying to point out that probabiltiy statements about photons only make sense if these are detected (and not 'in dead space').
Imagine you do a photon experiment, like double-slit or anything. In doing so, you detect the photons with some detector device, say a photographic film. For high intensities, you get a distribution of the energy input per unit time on the screen which can be calculated from the e.m. field (IIRC, E(x)² + B(x)² or so).
If you dim down the source, you observe the screen being hit randomly by single photons. If you let this accumulate for a long time, you see that the probability distribution on the film converges to the one calculated from the e.m. field.
Now you could of course scale down the e.m. field so it represents one photon per unit time. But it's still a wavefunction describing a flow of photons, not a single photon.

I also have my doubts whether an e.m. field ever 'collapses'. I guess an e.m. field is ruled by Maxwell's equations, and I'm afraid these don't allow a 'collapse'.

But they need not be. I can force a p or x equal to zero for an electron also.

'Course you can. What I'm trying to say is, you can have [del]E = 0 for an electron and still get an interesting spatial distribution. But a photon must have [del]E = 0, and is then spread out all over space. Unless you detect it, see above.

You consider only the bound state; then you compare this to an unbound wave-function.

OK, bad wording by me. I should have said:
"This does not apply to particles which have mass. As you know, there are (bound or free) electron states which have [del]E = 0, but [del]p and [del]x both finite."