# I 'If a photon were trapped between two perfect mirrors...'

Tags:
1. Jul 30, 2016

### peterfreed

If a photon were trapped between two perfect mirrors perpendicular to its axis of motion, and they were gradually brought together until they were touching, so that the distance between their faces was 0m, would the photon be "trapped" between the mirrors? Without space in which to move, how could its velocity be c? I am having trouble visualizing what would happen to the photon?

2. Jul 30, 2016

Staff Emeritus
It would take an infinite force to do that.

3. Jul 30, 2016

### Staff: Mentor

Photons are not what you're thinking they are, but that's a topic for the Quantum Physics forum - in fact this thread from there is our top-of-page featured thread right now.

But we can fix that problem by working with a flash of light instead a photon. Light is an electromagnetic wave, and it does indeed propagate in a vacuum at $c$. You will find it a lot easier to visualize what is happening if you think of the wave propagating back and forth between two mirrors getting ever closer - as long as there is a gap the wave can propagate at speed $c$ and when the width of the gap goes to zero you no longer have an electromagnetic wave propagating in vacuum. One way or another, the assumption of perfect reflection must break down.

[Edit: There is no conflict between this reply and Vanadium50's. Any apparent conflict just goes to show that photons aren't what you think they are].

Last edited: Jul 30, 2016
4. Jul 30, 2016

### Battlemage!

Why would it take an infinite force to bring two mirrors together? Presumably every situation where two mirrors are brought together involves light bouncing around and between them.

Of course I would assume a light wave propagates in all directions, not just along a line, so that kind of throws a monkey in the wrench for this problem, doesn't it?

5. Jul 30, 2016

### Staff: Mentor

It does, but eventually the light gets absorbed, which is to say that the "perfect reflection" idealization breaks down. If it didn't, the force would indeed become infinite; but the appearance of an infinity in a physics problem means "something eventually has to break".

A light wave emitted by an ideal spherically symmetric source does indeed propagate in all directions, but if you've ever seen a searchlight beam you'll know that there has to be more to it than that. If you google for "Huygens principle" you'll see how all real-life light rays, of any shape whatsoever, can be treated as overlapping spherical wavefronts.

6. Jul 30, 2016

Staff Emeritus
Consider an object (rubber ball, photon, it doesn't matter) bouncing back and forth between the mirrors. Every time it hits a mirror, it imparts an impulse of 2p on it. The time between collisions is 2d/v, where d is the separation. So the force needed just to hold the mirrors together at a distance d is proportional to pv/d. As d goes to zero, the force becomes infinite.

Yes, but those are physical mirrors. The problem specifies lossless, perfect mirrors. It's essentially recasting the "what happens when an irresistible force meets an immovable object" question.

7. Jul 30, 2016

### Mister T

The notion of the mirrors touching is a consideration at the macroscopic level. At a level sufficiently microscopic to consider the behavior of subatomic particles, there is no touching. The particles exert forces on each other, and as the distance between the particles approaches zero, the force required to get them closer increases beyond all bounds.

In other words, the notion of things touching each other is an illusion. My fingers don't make contact with the keys of my keyboard. They get so close that they exert a force large enough to depress them.

8. Jul 31, 2016

### tionis

But I think the poster wants to know more about what would happen to the photon than the mirrors. So suppose the mirrors are perfect or whatever, which mean the closer they get the smaller the wavelength of the photon right? So at some point the photon is small enough to escape in between the atoms of the mirror or even between its quarks?

9. Jul 31, 2016

### Staff: Mentor

As I said in #2 above... Photons aren't what you think they are. You are imagining that photons are little objects that move around in space like a ball rolls around on the floor. They aren't, and the mental picture of photons "escaping" by slipping between the atoms is very misleading.

10. Jul 31, 2016

### GeorgeDishman

The photon has a finite probability of quantum tunneling through one of the mirrors. As you bring them together, the rate of reflections rises until it is infinite at zero separation. In practice, for a finite probability, it will tunnel through after a finite number of reflections.

11. Jul 31, 2016

### Staff: Mentor

Photons don't tunnel the way the non-relativistic particles of first-year QM do; you can't just set up Schrodinger's equation and find that the wave function in the position basis is non-zero on both sides of a potential barrier. At the risk of repeating myself.... Photons aren't what nonspecialists think they are, and the references in the first two posts of this thread are a good start on what they are.

I don't think this problem can be properly analyzed without considering the interactions between the electromagnetic field and the reflecting mirrors, and I'd be surprised to hear that there is any sensible single-particle description.

12. Jul 31, 2016

### jbriggs444

For mirrors moving towards each other, there is another effect that also makes the force become infinite. At each reflection, the [ping pong ball] increases its speed because it is bouncing from a moving mirror. The impulse is slightly greater than 2p and keeps getting bigger at every bounce. The time between collisions is slightly less than 2d/v and the v keeps getting bigger at every bounce.

Of course, things break in the limit regardless. Just somewhat sooner than might be expected.

13. Aug 1, 2016

### dragoo

"photon is small enough"
The photon is a wave. How can be small?

14. Aug 1, 2016

### Orodruin

Staff Emeritus
You can avoid this particular effect by only moving the mirrors when the light pulse is not bouncing.

15. Aug 1, 2016

### vanhees71

It must be stressed again: No picture of a photon can be more misleading than that of a "bouncing ping poing ball". It's simply wrong! A photon has not even a position. So how can you use a naive particle picture? The point is that you can't!!!

16. Aug 1, 2016

### tionis

lol We are being taken to school by the QMs experts. In response to your question: classically, photons can have wavelengths of different sizes, like for example some radio waves are several meters long. But I welcome any correction by Oro, Nugatory, or Vanhees 'cause they are all kick-a*s physicists.

17. Aug 1, 2016

### Staff: Mentor

You are right that it makes no sense to apply words like "small" and "large" to a photon, but a photon is not a wave either.

18. Aug 2, 2016

### edguy99

Dont forget your perspective here. The wavelength of a red photon is 630 nanometers, the distance where you start to see interference between photons. The average distance between molecules is only about 0.1 nanometers, some 6000X smaller.

19. Aug 2, 2016

### tech99

If we imagine a radio wave persisting between two metal plates, then there must be a resonant condition, the two surfaces forming a cavity resonator and the wave having a length corresponding to the spacing in some way. The losses at the surfaces are small. If we move the plates nearer, the radio wave does not change frequency but now finds itself in a non resonant cavity. In this case I assume that the current in the walls increases and energy of the wave is donated to electricity. It looks as though a photon will not change its frequency when the plates are moved - might it then have to give up its energy to the electrons of the mirrors?

20. Aug 3, 2016

### ZapperZ

Staff Emeritus
There is another issue that hasn't been dealt with here, and something that should have been learned from all the experiments that have been done on optical cavity.

Even for a single photon, such an entity can only "live" in a cavity that matches one of the harmonics of its "standing wave". That is why many experiments are done in such optical cavity when one tries to trap "single photons". But because of that, you simply cannot arbitrarily change the dimension of such a cavity without losing such photons that have characteristics energy and wavelength.

The classical idea of resonant cavity still applies here.

Zz.

21. Aug 4, 2016

### GeorgeDishman

Be careful how you propose to move the plates. If it is a continuous movement, there will be a Doppler effect.

22. Aug 4, 2016

### Paul Colby

There is a force on the mirrors radiation pressure. Moving the plates inward will do work on the radiation increasing the stored energy and, likewise the force due to radiation pressure.

23. Aug 4, 2016

### Paul Colby

Yes, good point. When someone says "single photon" I always think in terms of an eigenstate of the mode number operator summed over all possible modes. This actually covers a very wide set of circumstances. A single mode with a single photon excitation has nearly no information regarding the location of the energy. In a state like this the force on the mirrors would be constant in time. If the excitation is spread over many modes this is not the case.

24. Aug 4, 2016

### vanhees71

One should stress that the single-photon state $|\vec{p},\lambda \rangle=\hat{a}^{\dagger}(\vec{p},\lambda)|\Omega \rangle$ is a generalized momentum eigenstate, i.e., not renormalizable to 1 but only to a $\delta$ distribution in the sense that
$$\langle \vec{p},\lambda|\vec{p}',\lambda' \rangle = (2 \pi)^3 2 |\vec{p}| \delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda,\lambda'}.$$
So, as you said, you need a superposition
$$|\phi,\lambda \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2|\vec{p}|}} \phi(\vec{p}) \hat{a}^{\dagger}(\vec{p},\lambda) |\Omega \rangle,$$
where $\phi(\vec{p})$ is chosen such that
$$\langle \phi,\lambda|\phi,\lambda \rangle= \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3 2|\vec{p}|} |\phi(\vec{p})|^2=1.$$
This is a properly normalized true single-photon state. It's close to a single-mode photon, if $\phi(\vec{p})$ is narrowly peaked around some momentum $\vec{p}_0$.

This is all quite analogous to classical electrodynamics, where a plane-wave is also not a physical field, because it has infinite energy.