# Does a photon bounce?

1. Jul 7, 2008

### worwhite

Hi,

Just a simple question. Does anyone know how a photon interacts with say, a mirror? Does it bounce off the mirror, much like a ball does, or does it get absorbed and then re-emitted?

Thanks.

2. Jul 7, 2008

### granpa

Last edited: Jul 7, 2008
3. Jul 7, 2008

### Fredrik

Staff Emeritus
"Absorbed and re-emitted" is the correct alternative, but that isn't much of an explanation. If you want to know more, check out these lectures (part 2). This must be the series of lectures that the book "QED: The strange theory of light and matter" is based on, so if you prefer a book to a lecture, buy it. (It's excellent, short and cheap, and it contains (almost) no math).

4. Jul 7, 2008

### peter0302

Ack. Absorbed and re-emitted? That's exactly what our glorious FAQ says does _not_ happen in diffraction because of the discrete energy states of atoms - but then how does it happen in reflection?

5. Jul 7, 2008

### Fredrik

Staff Emeritus
There's no relevant difference between diffraction and reflection here, so whatever the answer is for diffraction, it's definitely the same for reflection. What the answer is depends on what you feel is implied by the word "absorbed". One thing that definitely doesn't happen is this: Atom absorbs photon and jumps into a higher energy state and some time later emits the excess energy in the form of a photon.

I answered the way I did because I didn't interpret the word "absorbed" as implying all of that. (Maybe I should have. I don't know). I'll try to elaborate a bit. The question is to some extent pointless, because all we can say is that there's an "in" state that consists of a photon and an electron and an "out" state that also consists of a photon and an electron. It doesn't make sense to ask if it's the same photon, because photons are indistinguishable.

The calculation of the probability of the interaction involves a Feynman diagram that can be interpreted as a photon being absorbed and re-emitted. It wouldn't make sense to say that that's what really happened, but at least there's such a diagram present in the calculation while there are no diagrams that suggest a "bounce" interpretation. This is why I felt that "absorbed and re-emitted" is a close enough approximate description of what happens, but as always the whole truth is much more complicated.

(Some personal views that others may disagree with, especially people who like the many-worlds interpretation:) There's no theory that describes what happens. Quantum mechanics and quantum field theories can't tell us what really happens, not just because fields/particles are an approximation of something else, but because quantum mechanics isn't a model of reality. By that I mean that unlike e.g. special relativity, it isn't a description of a fictional universe that resembles our own. Quantum mechanics is "just" an algorithm that tells us how to calculate probabilities of possibilities.

6. Jul 7, 2008

### peter0302

Well, I guess you can tell us what _does not_ happen, and that is, like you said, the electron absorbing the photon, jumping to a higher energy state, jumping back down, and emitting a new photon. That, to me, is a "different photon" - I think we can all agree on that.

Taking it a step further, we know that photons always follow a geodesic in spacetime - only gravity can "bend" the path of light. So, it also follows that a photon arriving at an angle of incidence is not "bent" by any known force into the angle of reflection. Thus, again, I think we can say this is not the "same" photon.

So all that's left is that there's a new photon whose energy entirely comes from the first one. And so the question is what is the physical mechanism that causes this? Is it electromagnetism? Is it the electrons? (Would a proton / neutron plasma be invisible?) Surely we can do better than "we can't describe what happens."

7. Jul 8, 2008

### Fredrik

Staff Emeritus
It's definitely an interaction between a photon and the electrons in the mirror. The theory that tells us how to calculate the probability of a specific measurable outcome is quantum electrodynamics (QED). (I highly recommend that lecture).

8. Jul 8, 2008

### peter0302

Been watching it - it is indeed very good. I particularly was riveted by Feynman's derivation of Snell's law using the admittedly dumbed-down version of path integrals).

I understand the math more or less. And I understand also that Q-anything is not concerned with the "how" or "why" but the "what". (I've told people here that many times).

I'm not asking why do some photons reflect and why do others pass through? I understand QED is not going to tell us that - it'll only tell us the odds. What I'm interested in is once the photon DOES decide to reflect, what physical processes are we talking about? If it's not absorption and re-emission like the photoelectric effect, what is it?

9. Jul 8, 2008

### Count Iblis

The photon interacts "coherently" with many electrons. The physical process is "Compton scattering", but without the electrons recoiling (the entrire mirror will recoil, or if it is fixed, the earth will aborb the momentum of the photon). Also, you have to add up all the amplitudes for the photon scaterring off of the atoms, as a result you end up with an amplitude that is strongly peaked in one direction.

10. Jul 8, 2008

### peter0302

I see, I think. So the single photon interacts with several electrons in the (say) glass molecule, which collectively generate an EM wave in a new random direction, which either may be back toward the original source (reflection), or deeper into the material (refraction), either of which, averaged over the infinite possibilities, results in the familiar laws of reflection and refraction, respectively.

Do I have it right?

11. Jul 8, 2008

### thenewmans

A quantum entangled photon is used for quantum cryptographic communication of long distances. Somehow, properties of the 2 entangled photons must be determined. I would assume some electron interaction is involved. If it’s not the same photon that pops out, it might not have those properties.

I read the QED book. (Fantastic!) Feynman shows the photon hanging around in the mirror a bit before getting re-emitted. I don’t know what to make of that.

12. Jul 8, 2008

### Fredrik

Staff Emeritus
Actually it's more correct to say that it interacts with all of the electrons in the mirror than to say that it interacts with the ones in a specific molecule. If you describe the light that's emitted towards the mirror in terms of photons (a concept from quantum electrodynamics), you shouldn't describe the reflected light as an electromagnetic wave (a concept from classical electrodynamics). The reflected photon can go in any direction, but when you add up the amplitudes of all the paths, the result should be that only paths that agree with the laws of optics are relevant. I wouldn't use the word "averaged" to describe the adding of amplitudes.

13. Jul 8, 2008

### peter0302

I accept that that is a mathematical model of deriving the correct answer, but to say that a single photon does interact with _all_ of the molecules of the mirror either requires belief in MWI, or cannot be right.

I know, I know, we're back to the double slit problem - did the photo go through both slits or just one?

What I wanted to know was what physical processes are involved in generating the new photon. And it's electron perturbation. All these other semantics about photons vs. EM waves vs. amplitudes are not what I was asking about (yet for QM scientists it's all they seem to care about).

14. Jul 8, 2008

### Count Iblis

Peter, this doesn't require belief in MWI or anything else. If you have a picture in your mind of a photon as a classical particle that has a definite velocity and position at any moment, then you are forced to add up all the paths it can take to be able to reproduce quantum mechanics. But this is indeed just a mathematical formalism.

The physical process that causes reflection of the photon off a mirror is the interaction of electrons with photons. But it is important to realize that in this case it is a coherent process and not like what you get in the compton scattering case.

15. Jul 8, 2008

### reilly

Let's suppose that the mirror surface is the x-y plane, and that light carrying images is incident along the z axis. Let's talk conductors, like silver. Classically, we characterize a conductor as something in which an electric field cannot exist, all the charges and currents occur at the conducting surface. The basic physics is first that, given the incident radiation, the transverse components of E and B are generally non-zero in the conducting surface. These fields generate forces on the "free" charges and currents on the surface, which, in turn, radiate the mirror image.

That is, Lenz's Law says that the fields generated by the induced current oppose the original fields. (Nature wants a static, uniform charge density.)Crudely speaking, this is partly due to resonant coupling between current and light. For optical frequencies, electrons will oscillate more than drift -- like alternating current -- as they are moved by the light generated forces. Again, crudely, the basic radiative atom is an oscillating charge, one that pretty much stays at home. The radiation comes in, and the electron's motion follows the light's -- the force is, say,cos(wT), or a combination of frequencies. But oscillating currents radiate. And the frequency and spatial patterns of the electrons generate the image signal; have a driven radiating resonant circuit for all practical purposes.

The classical solution can be well approximated with image charges, provided the distance between image and mirror are small -- so that lags can be neglected. The image charges are slaved to their image generating partners, with reversed charges and current -- that is, the mirror image of the original image. We see the image charges. Recall that the field of an image charge, inside a conductor, is designed to simulate the field produced by the charge distribution induced by the external charge

For a more realistic classical model, impose large but finite resistivity and a dialectric constant with both a large real and imaginary part, so that fields and currents damp out in a continuous manner away from the surface. Jackson has most, if not all, of the needed math.

Photons? Reflection involves billions upon billions of particles, and thus requires a statistical approach, as is done in classical E&M. Dialectric constants, resistivity, currents and charges are all averages, over small 4-volumes, with many cycles of radiation occurring with very little linear motion -- small displacements of charges. (See Jackson. ) It's possible, a good guess, I'm sure, that the most efficient way to do reflection in QM is to use the density matrix approach, applied to the Heisenberg equations of motion Should lead to the classical description.

That a photon can interact with all the electrons involved is built into the density matrix -- you can see how the photon-QM current interaction merges into the Jdot E work term.

No bouncing photons.
Regards,
Reilly Atkinson

16. Jul 9, 2008

### worwhite

Whoa, I didn't expect the replies to be so numerous. And I noticed my post is now under Quantum Physics rather than SR...apparently the mechanics of light interaction isn't simple at all...

I have another question, though.

Photons have momentum, though they have no rest mass. So imagine a tiny mirror, stationary in space. A photon hits it, then "bounces" off (regardless of whether it's a bounce or a re-emission, it simply travels in the opposite direction). Considering the conservation of momentum, does the mirror remain stationary?

If it doesn't remain stationary, thus satisfying the conservation of momentum, then the amount of kinetic energy in the system should have increased (since the mirror is now moving). Where does this increase in energy come from?

17. Jul 9, 2008

### jimgraber

I think two things happen: the mirror moves, (very slightly) and the photon is redshifted (also very slightly) as well as reflected. Energy is conserved.
Best, Jim Graber

18. Jul 9, 2008

### Count Iblis

I think that this is an extremely misleading account of how the QM computation would reproduce the reflection off a mirror. Because it suggests that the photon would change the internal state of the mirror in a very complicated way when it interacts with it. This is in fact false as can be easily demonstrated experimentally: Just do a two slit interference experiment where you place mirrors at the slits. Let's say that you reflect the light upward, so that an interference pattern appears in the ceiling.

Now, the heuristic picture painted by Reilly would suggest that when a photon interacts with the mirror in one slit, the quantum state of that mirror would change. This means that you wouldn't get any interference pattern on the ceiling as the mirrors internal state would provide you with the "which path" information. The fact that in practice you can't extract this information is irrelevant, what matters is that the two states in witch the photon takes on or the other route are orthogonal.

As I pointed out in another thread, you can actually destroy the interference of classical waves by collecting which path information of every quanta in a non destructive way. So, the classical description of light, or any other waves, like sound waves is incomplete even in the classical regime.

19. Jul 9, 2008

### Staff: Mentor

No, the mirror recoils. This principle has been tested on a small scale as a means of spacecraft propulsion. Try a Google search on "solar sail".

20. Jul 9, 2008

### reilly

Now, first of all, I've simply summarized the theory of eddy currents; those induced in conductors by external fields. Further, the image solution I outline is fully discussed in Smythe's text, Static and Dynamic Electricity. In their Electrodynamics of Continuous Media, Landau and Lifschitz give a less complete but more physically oriented discussion of the topic. The fact of radar and radio and TV give ample evidence that the subject of eddy currents is 1. important, and 2. well understood. That is, there is substantial agreement that the description of reflection I have provided is is correct, as can be seen throughout the literature.

This, of course means that QM must give the same picture at the macroscopic level.

As far as your two mirror experiment goes, you have neglected a few important details, which mitigate against your conclusion. First off, if the state of the mirror does not change when a photon is absorbed, then absolutely nothing happens. As inspection of the QED
Hamiltonian will show, whenever a photon is emitted or absorbed, the state of the charge will and must change. If it does not, then the situation is equivalent to a charge in an electric field feeling no force. If you accept your argument above about change of state implying no interference. then why is the sky blue? Why does light travel slower in water than in air?

The dynamics of your two mirror interferometer are covered in the huge literature on scattering of light or particles from multiple scattering centers -- for both classical and quantum scattering. If you dig into this, you'll find that change of state often occurs without destroying interference phenomena. Note, all of this becomes very clear with use of the Lippman-Schwinger integral equation for scattering.

As the classical theory dictates, the state of a mirror constantly changes whenever it is reflecting light.

And, with all due respect, the wave function you use, in your cited other post, is incorrect. If the two mirrors are independent, then their wave function should be a product -- or superposition of products -- not a direct sum. With the sum, you are actually describing one mirror in a superposition of states.

Can you provide any example of a known physical phenomena that confirms your claim that classical wave theory is incomplete?
Regards,
Reilly Atkinson