# What Happens to the Photons in Destructive Interference?

1. Oct 3, 2011

### peter.ell

I know about the law of conservation of energy, and I understand interference effects conceptually, but when I put the two together I'm a little confused.

I know that the energy of light must simply by transformed in destructive interference...but where does it go and how?

If the interference is happening in a thin film, then the energy of the destructive interference goes to the areas of constructive interference, right? But how is this energy redistributed? Does it happen instantaneously, or at the speed of light, and if it happens at the speed of light than isn't that redistributed energy light itself...and therefore destructive interference is really just a change in the path for light so that it appears to be destructive in a certain area?

Also, what about in the case of lasers pointing at each other in such a way that there is no constructive interference to speak of? In this case, where does the energy of the light go when it destructively interferes? I can understand how the values of the electric and magnetic fields will cancel, but then what happens to the photons when thinking of light as a particle? Do they just poof out of existence?

Thank you for clearing this up for me!

2. Oct 6, 2011

### Bill_K

Laser beams are noninteracting. Their E, B fields superpose linearly, and two beams can cross without affecting one another at all. That is, if two beams intersect at a small angle, even though they cancel in the middle, they are not "destroyed", and emerge on the other side at full strength.

3. Oct 7, 2011

### yoron

What a nice question Peter. Wish I had come up with that one.

So what happens where they do interfere, as in the last example?
If we had a detector there would it register a energy?

4. Oct 7, 2011

### Ken G

In my opinion, if you set up complete destructive interference, we just call that a "wall." There are basically two ways to set up such interference-- one involves creating constructive interference back in the direction the photon came from. We call that a "mirror." The other creates no constructive interference anywhere, the photon poofs out of existence. We call that an "absorber."

5. Oct 7, 2011

### DaveC426913

Not at that point, no. But right next to it would be a band of light twice as bright as one light source. If you averaged all the light and dark areas, you would record as much energy as two light sources put out.

6. Oct 8, 2011

### Staff: Mentor

In that case the photons are never produced at all. The energy that the source would have produced as light (photon) energy manifests itself in other ways, e.g. heat.

7. Oct 8, 2011

### azaharak

Isn't conservation of energy violated for a brief instant, or do we avoid this by invoking the uncertainty principle?

If we assign the energy due to electric field a value of 1, and the same for the magnetic field.

Then send two photons head on with each other, before they intersect we have a net value of 4 for the energy.

If you send two photons head on, such that their electric fields will cancel at some intersection point, their magnetic fields should add.

So if at any other time other than the intersect, you have 4 for the energy, and 2 for when they intersect.....?

Could it be the uncertainty (ΔEΔt) principle that allows for this violation for a brief instant of time?

8. Oct 8, 2011

### Staff: Mentor

Photons are not like little bullets that you can "aim" at each other. They are not little spatially-localized bundles of electric and magnetic fields.

9. Oct 8, 2011

### azaharak

Why can't you aim a photon? Set up a source and a thin slit and a series of polarizers to reduce intensity.

But forget that, when two lasers are aimed at each other, just consider the intersection of two photons (anti parallel to each other).

...

And your saying they have no magnetic or electric fields? How do they suddenly develop when we talk about a large collection of them

10. Oct 8, 2011

### DaveC426913

If they are aimed at each other such that they are parallel, how will you detect what they are doing before they have hit a screen?

11. Oct 8, 2011

### Staff: Mentor

What happens when light goes through a thin slit?

12. Oct 8, 2011

### azaharak

Your question makes no sense to me...

Consider two photons head on with each other, as in two lasers aimed directly at each other but only considering the point of intersection of two photons.

There is no screen, i don't want to see where they are.

When they intersect (if I'm allowed to add their electric and magnetic fields), one of these components will cancel for that brief span of time which they occupy the same space.

13. Oct 8, 2011

### azaharak

You get an interference pattern, but there is no energy loss like what im talking about. Certain spots will appear bright and certain darker on your screen.

I'm not talking about a screen, and this involves many photon interference or a photon infering with itself.

The example I'm giving involves two photons head on.

The thin slit is simply acting like an aperture for what I stated before, this is getting off topic

14. Oct 8, 2011

### DaveC426913

At some point, we were talking about what would happen at the dark area on a screen.

If we're not detecting them then who's to say what they're doing?
OK, so what exactly is the problem?

Two waves on an ocean can pass each other, momentarily leaving a flat, neutral surface before they pass onward.

I think there has been some crossing of questions. What exactly is the question, and what is the problem?

15. Oct 8, 2011

### azaharak

If there was screen it would block the incoming light? We are obviously talking about two different things.

MY question which based off the the original is

I'm talking about the instant which two photons cross each other and interfere. At that point it seems as if there isn't a conservation of energy when accounting for the combined electric and magnetic fields. I'm asking if this violation is explained by the uncertainty principle.

16. Oct 8, 2011

### Staff: Mentor

No, even when the system is in an eigenstate of the energy operator (so energy is certain) there is no energy loss from desctructive interference.

Don't forget that energy is proportional to the square of the field. So the added magnetic fields have an energy of 2²=4. There is no energy loss from destructive interference.

Energy is conserved, your reasoning was faulty. There is no need to invoke the uncertainty principle.

Yes, and the other will double, conserving energy.

17. Oct 8, 2011

### Cthugha

The electromagnetic fields in question are not point-like. This is in principle not different from two waves moving on a string. When they move in opposite directions and one has the peak upwards and the other has the peak downwards, would you expect conservation of energy to be violated when they meet each other? No - because there is no time when the amplitudes at each position cancellation out completely.

18. Oct 8, 2011

### azaharak

Thank you... that was an answer

Your explanation isn't a good example. I'm considering a photon, not a wave on a string. There are perpendicular electric and magnetic fields in my problem, and there is only 1 photon not an entire length L of photons interfering with each other.

Last edited: Oct 8, 2011
19. Oct 8, 2011

### Manchot

Usually, the answer is "somewhere else." For example, in a Michelson interferometer, light that destructively interferes in one arm will constructively interfere in the other. In a thin film, if a reflection destructively interferes, transmission constructively interferes. Of course, if light is being sent back to the source, emission can be inhibited, but that's easier said than done.

The previous example mentioned of firing two lasers at each other in a destructive way to inhibit lasing usually won't work: though they will cease lasing at that frequency, they'll probably just lase at a slightly different frequency (assuming the gain bandwidth is broader than the mode spacing). In that situation, you have to consider the dynamics of both lasers, and the situation can be complicated.

The simplest situation of emission being inhibited is if you put an atom in a cavity that forbids propagation of light at the atom's emission energy. One way of doing this is to put it in a photonic crystal with a photonic band gap, essentially causing all of the light that propagates outward to destructively interfere. Normally, such an atom in its excited state will spontaneously emit radiation, but in this situation, light that gets emitted will always return, so it does not radiatively emit. (Although energy can flop back and forth between the cavity and the atom, forming the basis of cavity QED.)

20. Oct 8, 2011

### Ken G

I agree with Cthugha. There is nothing particularly unusual about the "intersection" of two photons that does not appear in any classical wave description. The only difference is that for photons, you use the amplitudes to determine where, and how many, photons will appear, whereas for fields, you use the square of the amplitudes to determine the energy density in space. It is only a matter of normalization to insure that the latter gives the same answer as the former in the limit of a large number of photons. In particular, there is no unusual "cancellation of the amplitudes" when they overlap-- it's exactly the amount of cancellation you need to get energy conservation in the classical limit.

21. Oct 9, 2011

### zonde

The point is that your problem is classical. And in classical wave superposition (you can use soliton waves represented by wave packets to get closer to single photon picture) energy is conserved at all times.

So how interference of light appears if we consider light as ensemble of particles (photons)?
Interference of light have no effect on photon number (and energy) until light interacts with matter. Destructive or constructive interference is not real before this interaction i.e. it exists only as possibility of photon ensemble.
When photon ensemble interacts with matter part of photon ensemble is absorbed as heat and depending on constructive or destructive interference photon number absorbed as heat is lower or higher respectively.

22. Oct 9, 2011

### ozmbie

Using the double-slit experiment as an example, the areas of destructive and constructive interference are a result of the probability of photons arriving at those locations (zero photons = dark/destructive, many photons = bright/constructive etc). This is why the experiment can be performed using a single photon at a time and still create an interference pattern (example).

So the dark patterns aren't a result of photons canceling each other out. It's because no photons arrive there anyway.

Paul Dirac, The Principles of Quantum Mechanics, Fourth Edition, Chapter 1:

Last edited: Oct 9, 2011
23. Oct 9, 2011

### Phrak

A very good question in need of a good answer.

I set up two intersecting laser beams to cross as some acccute angle. There is an interference pattern gererated between the two. A diamond pattern of interference results.

Now I separate the source of the light from its destination into two regions with a squiggley surface, carefully drawn such that the Poynting vector, the measure of energy density transport across the surface, is zero valued everywhere on the surface.

This is what the OP was asking. How can the energy transport across a plane be zero valued according the standard argument, yet still, energy is moved. It doesn't have to be a flat plane gentlemen.

Last edited: Oct 9, 2011
24. Oct 9, 2011

### Ken G

What you are suggesting doesn't exist. There is no surface, regardless of shape, where the Poynting flux is zero, in situations like that. The interference simply works out to conserve the time-averaged energy flux through any shaped surface. But at least you are correct that the issue is essentially classical wave mechanics.

Last edited: Oct 9, 2011
25. Oct 9, 2011

### Phrak

Your thinking of an S^2T volume element. But, come to think of it, it's non-interesting.