# Where is the energy when two photons superimpose?

1. Sep 10, 2007

### flame_cmh

Suppose that two photons are the same and propagation in opposite direction in vacuum, the energy of this system of course is 2h*$$\nu$$.
When they coincide in position, according to the principle of superposition there is no electric and magnetic field, then the energy disappear at that time??

Last edited: Sep 10, 2007
2. Sep 16, 2007

### Claude Bile

Your proposition is flawed, as it violates the Heisenberg uncertainty principle. You can't get photons that have perfectly defined position and trajectory. Photons with a perfectly defined trajectory must have an uncertainty in their wavevector direction of zero, which implies that the wavefunction of the photon is an infinite plane wave. Since the wavefunction is infinite in extent, the uncertainty in position is infinite.

Nonetheless one could in theory add two infinite counter-propagating plane waves, in which case one gets a standing wave. You will get points (nodes) where the probability of finding a photon is exactly 0, however the total energy of the system will of course continue to be 2hv.

Claude.

3. Sep 18, 2007

### flame_cmh

Ok, I get it, and thanks!
Maybe that's why it takes so long to get a reply on the thread, cause I asked a stupid question!^_^

This question originally came out from that I read the lectures of Feynman on polarization of the light! So I think if a source shot out a unpolarized light, according to the classical interpretation of light, the electric field amplitude probabily in any direction perpendicular to the direction of propagation. And when the light arrive at a point say P, the resultant field is always zero, cause it is no reason that the oscillates of the field is perfer certain direction, so the propability of the oscillation in all direction should be a constant.
And if that's right, since the field is zero, we can't see anything. Of cause it's wrong, but I don't know where, that's what I need you to tell me.

4. Sep 18, 2007

### Claude Bile

If I understand your question correctly - For an unpolarised beam, the probability of a photon being polarised in a particular direction will be equal for all directions - but this does not imply that the resultant field ought to be zero.

Classically, the direction of oscillation is completely determined by the source. In order to generate a wave, you need to wiggle a charge - the direction you wiggle the charge will determine the polarisation of the generated wave. So you never get perfect symmetry because you need to consider how the wave is generated in the first place. A classical unpolarised beam is a statistical entity, the sum product of many-many small, polarised components.

Quantum Mechanically, a photon can exist in a superposition of polarisation states. A photon emitted via spontaneous emission usually has no preferred direction or polarisation, but this just implies that the probabilities of the emitted photon possessing a particular polarisation are equal - this is sufficient to preserve symmetry as there is still no preferred direction. In the Quantum world, one can get a truly unpolarised wave.

P.S. These questions aren't stupid - I think they are quite thought-provoking.

Claude.

5. Sep 18, 2007

### flame_cmh

I understand you, and thanks for your time!

6. Sep 18, 2007

### Loren Booda

Boundary conditions are never ideal, and thus complete destructive interference involving boundaries of finite distance and mass, gravitational (including parallelism) and E-M forces, thermal effects and material composition is impractical.

7. Sep 30, 2007

### arron

I still have something uncertain about this, and I think for this question better not use photon for instance. Discussing it with mechanical waves and classical E-M waves would be more approprite. If two waves meet together, would they be completely neutralized if their phases are opposite? And then there is no wave, and how does the energy transmit? By what? And where does it exist? How far will it utimatly come out? Is it just that the boundary is not suitable for this discussion?

8. Oct 1, 2007

### Claude Bile

Two counter-propagating waves sum to produce a standing wave. This result is well known.

Two waves can only cancel for all time if they are colinear to begin with, for example from two parallel reflecting surfaces. The effect of this is also well-known, it results in the suppression of radiation. The implication of this is that you can't emit a wave then cancel it at a later time, you can only prevent the wave from being emitted to begin with, which results in the law of conservation of energy being upheld.

Claude.

Last edited: Oct 1, 2007