If a single photon hits a lense, will the photon follow the path given by using Snell’s law or will it simply be emitted from the surface at some random angle? Assume glass and visible light.

If it follows Snell’s law, where does the constructive interference that is required come from? The wave function of the photon has to collapse at the lense in order to be absorbed by the surface so I don’t think we can just say the photon interferes with itself.

Glass is transparent, so a photon that goes through the lens is not absorbed at all, and the photon can still interact as a wave. The path the photon takes can also be calculated by summing over all possible paths.

At the risk of putting my size 11 foot in my mouth…

As I understand it, if you are looking at wave properties of light, it will act like a wave, if you are looking at particle properties, it will act as a particle.

One experiment I heard of was the double slit interference experiment. They reduced the light intensity to the point where only individual photons would strike the double slit, yet the interference pattern appeared as usual, so a photon can apparently ‘interfere with itself’.

Since the lens acts on light ‘waves’ you treat light as a wave, regardless of how many photons are being acted on. You do not need a ‘bunch’ of photons to get wave properties.

I think RM Mentock is right. Just because the photon is affected by the lens, it doesn’t mean the wave function is collapsed there. The photon is not detected at the lens.

To elaborate on RM Mentock’s comment: One of the ways to look at the probability for a quantum particle to get from point A to point B is to assign an amplitude to each possible path that the particle can take between these points, and then to add up all of these paths to get the total propagation amplitude. (The probability is then the square of this amplitude.) In the case of a lens, the amplitudes from each path add up in such a way to produce a very large probability for the photon to turn up exactly where you’d expect it to from Snell’s law. So in this sense, the photon manages to have constructive interference with itself.

I’m probably doing a lousy job of explaining this, so I’ll also refer you to one of the masters: check out QED: The Strange Theory of Light and Matter by Richard Feynman for more information on this stuff.

First off, I’m not saying you guys are wrong because I really don’t know but I’m not convinced.

RM Mentock -
Even good optical elements have a relection of several percent of the incident light. I see no way to have a reflection if the photons aren’t being absorbed at the surface and being re-emitted, with constructive interference inside the lens maintaining the incident beam plus constructive interference outside the lens maintaining the reflected beam. Please let me know if this arguement makes sense.

MikeS -
Snell’s law maintains there will be two resulting beams from a single incident beam.

If there is a single incident photon, there cannot be two resulting photons because glass doesn’t change the color. This tells me that Snell’s law won’t work for a single photon.

The single photon will be reflected a percentage of the time equal to the percentage reflection that would normally be expected for light hitting that surface. Seems pretty simple and obvious to me.

Do have some background information that leads you this?

Consider a thin film anti-reflection coating on a PC monitor. We all know they work by having destructive interference at the film/air boundry from light reflected off the film/air boundry and the light reflected off the film/glass boundry. The steady state condition here is that they cancel and there is little or no reflected light. The transient condition as the first wavefront hits the air/film boundry is that there is a reflection that lasts until the reflection from the film/glass boundry makes it back to destructively interfere with it. If the film is thick enough(more than the wavelength), a single incident wave front will not interfere with anything to cancel the refection.

Hit send to quickly…to continue… The single photon example with the thin film doesn’t match the steady state Snell’s Law prediction. The transient response of a wavefront hitting the anti reflective coating is what got me thinking about a single photon and a lens. There may well be a transient response of lightSingle photon) hitting a lens that isn’t at all simple and obvious.

Notably, when an single atom spontaneously emits a photon, isn’t it anybody’s guess as to where it goes.

When many atoms, all next to each other, emit photons of the same wavelength, but with different phasing, Snell’s law says exactly where the photons go.

Still not saying you guys are wrong…just want to understand it without prematuring applying the results of double slit experiments.

I suppose I was assuming 100% transmission in the lens, but the “sum over paths” approach I mentioned can be modified to take reflection into account; so while there’s a probability for the go from point A to point B as before, there’s also a probability for it to go from point A to point C (where you’d say it had been reflected.) I believe that the book I mentioned talks about this as well.

A lens has to act on a single photon. All light is composed of single photons. The wave concept really only applies to “standing waves”, where a steady stream of photons is perceived as a whole, just as cloth appears to be one object.