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180 degree phased out photons

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Hi I have a question bugging me since yesterday.

If I combine two 180 degree phased out photons, the photons will interfere leaving nothing... By nothing I mean "not able to be detected".

My question: where is the energy of the photons ?

Is it possible to seperate the two photons ?

 

Thanks

Jacques

I'm in a rush, but do I know you? Because I was having this exact conversation with my physics teacher about 6 hours ago. Sorry, got to go, will reply tomorrow.

  • Author

No I live in Canada ! strange coincidence !

What does your teacher answered you ?

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With two lasers and a half silvered mirror.

laser-mixer.JPG

If you detected individule photons you would get the same number out as you put in. The interfearence only effects the probability of their position. So you get a pattern depending on distance from the beam splitter.

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50% is reflected and 50% goes throught. The output is two beams each composed of 50% from laser A and 50% from laser B. By moving one of the laser you can make one of the output beam vanish...

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I thinked a little bit more and I think if you make one of the output vanish the other will also. An other thing I thaught was that it is impossible to have two different photon have exactly the same axe, so after some time the two photons will split appart. If we place a detector on one output will see a concentric cercle patern of light and dark area. The question now is does any photons hitted the dark area ? A probability patern only ? Or two photons 180 out of phase hitting an electron at the same time has no effect ?

It's a probability pattern caused by the fact that there is interfearence between the two photons. So if you 'watch' two photons or two electrons passing through this experiment which is functionally equivalent to a 2 slit experiment then those particles will ALWAYS move from one side to the other, they will never 'combine' and hit one point to create nothing. But even sending single particles through you'd still get the interfearence pattern, just as if you where sending through a continuouse wave.

Wouldn't you expect bands of complete destructive interference out of this, but only in the uncollapsed wave? I think if everything were properly lined up what you'd see is a cool pattern on the wall.

The question now is does any photons hitted the dark area ? A probability patern only ? Or two photons 180 out of phase hitting an electron at the same time has no effect ?

 

Dark means no photons.

Wouldn't you expect bands of complete destructive interference out of this, but only in the uncollapsed wave? I think if everything were properly lined up what you'd see is a cool pattern on the wall.

 

Indeed, I think it's circular IIRC.

The question has been answered, and this is the conclusion I reached the other day.

 

Initially I thought about observing at a point. I realised that, at a point, the photons could cancel, just like a dark fringe. This sounded wrong though, if the two photons just cancelled, where does the energy go?

 

The answer lies in the setup. We were observing at a point, we should have been thinking about the whole system, not just a single point. If you look at the whole system then you will have fringes (dark/light). So although at a point it can be dark, this just means that the photons are somewhere else, in a light fringe.

 

From this we concluded that the photons, as such, cannot cancel. The photons will always exist, basic conservation of energy. However the probability waves of the photon, or just the light waves, can cancel. This is what you see when you look at a dark fringe.

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