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I have seen some contributors (e.g. annaV) state that em-wawes are built up by coherent photons. Now imagine 2 identical thin laser beams (same intensity, same source using a beam splitter and prisms) that meet head-on. The Mie-scattering from a few microscopic particles in their common path make the beams visible. Now adjust the phase of one beam so that it is exactly opposite the other. This will make the beams more or less invisible. So what happened to the (stream of) coherent photons and their energy/momentum. Did they collide, annihilate or change into something else ? And when reajusting the phase, so that the beams are in sync again, everything is back to normal (the photons reappear)! I also wonder if the coherent photons are inseperable from the wawe contrary to what others postulate, namely that photons and em-wawes are two different worlds that are mutually exclusive (Accidental Fourier Transform).

Jens
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    How do you "adjust the phase of one beam so that it is exactly opposite the other"? If you create destructive interference between the beams, then you must, even classically, have the energy go elsewhere, usually other parts of the beam-splitter setup, See e.g this answer. For the question of photons/em waves, see http://physics.stackexchange.com/q/90646/50583, http://physics.stackexchange.com/q/46237/50583 and their linked questions. – ACuriousMind Jan 16 '17 at 23:28
  • The energy is still there but the photons gone? What exactly are you up to? – Cosmas Zachos Jan 16 '17 at 23:37
  • The energy is only manifest if the wawe "hits" something. In laser anemometry, two beams intersect and form a fringe pattern used to detemine the speed of particles passing through. The distance between the fringes increases as the intersecting angle increases and becomes infinite (in theory) when they align. As the phase of one beam changes, the fringe pattern (i.e. the center of the very wide central fringe) moves accordingly. The phase is simply adjusted by moving the reflecting prism in the direction of the beam. – Jens Jan 17 '17 at 00:08
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    If the beams meet head on it is not possible to use destructive interference to make them disappear. Instead, you will create a standing wave with "photons" traveling both left and right. – garyp Jan 17 '17 at 03:13
  • This experiment has been done , and the answer is that the power of the beams goes , photons and all, is reabsorbed in the crystals creating the laser beam. It is all one quantum mechanical system. – anna v Jan 17 '17 at 04:46
  • Here is an MIT video link which demonstrates that when destructive interference happens the beam is going back to the source: https://www.youtube.com/watch?v=RRi4dv9KgCg . – anna v Jan 17 '17 at 05:13
  • garyp: A standing wawe would require an integer number of laser wawelengths (the laser frequency does'nt change) between the reflectors. Also, considering a total round trip, the beam splitter would halve the energy of each beam, so your assumption is not correct. – Jens Jan 17 '17 at 08:46

1 Answers1

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This will make the beams more or less invisible.

Here is an MIT course video showing destructive interference using beam splitters and mirrors from one laser source.

Here is the set up, at the point where it is arranged that the beam on the left is the one returning to the source.

setup

From a larger distance interference fringes appear

interf

Here are the two beams , the one returning to the source(left) and the one coming from the source (right). The interferences are complementary.

bringing the mirror closer concentrates on the central spot

two

The instructor ( Shaoul Ezekiel) presses on the table to change the wavelengths, and force total destructive on the right,

right

and then total destructive on the left

enter image description here

Please note that I am making screen shots of the video. I advise you to watch it.

So what happened to the (stream of) coherent photons and their energy/momentum.

The photons are back in the source. It is not only the crystal (or lasing matter) of the laser that is in a quantum mechanical state of emitting and absorbing photons . Once a setup with lasers is constructed the whole is in a quantum mechanical state with the source (or sources, if there is more than one laser). The photons that go back to the crystal reduce the energy needed to excite the laser for that delta(t).

Did they collide, annihilate or change into something else ?

Optical range photons cannot collide measurably since the two photon interaction is very very small. They are in a superposition in making up the classical electromagnetic wave.

One should stress that the destructive and constructive interference patterns are local , when there is destructive interference in the outgoing beam there is constructive in the return to the source, but also there are always rings of destructive and constructive, the beam does not disappear as a whole.

anna v
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  • anna v: Interesting video, but is the path back to the laser necessary? Also this is a HeNe laser, not a crystal. To make things simpler. Suppose a mirror is placed in front of the HeNe laser to direct the beam right back into the source. Would this then cause destructive interference (at the right pathlength) and turn of the lasing completely (as your answer suggests), at least for a time? The experiment has probably been done, so maybe you can direct me. – Jens Jan 17 '17 at 09:09
  • anna v: Supposing you do turn off the lasing (my previous comment), where all the output energy is directed back. But what if a beam splitter is placed in front of the laser so only, say 1% goes directly out and reflected back so that only 1/2% enters the HeNe laser. Presumably the weakened output beam would seem to disappear, at the right pathlength/phase difference, but would this small perturbance energy be sufficient to turn off lasing and make the deflected 99% beam disappear as well? I doubt it, because we could continue reducing the direct beams power sufficiently to maintain lasing. – Jens Jan 17 '17 at 09:41
  • My answer does not suggest that the lazing stops. It keeps on going . ILasing is is a quantum mechanical process and the boundary conditions of the quantum mechanical problem affect the total solution. If the boundary conditions are such that there is total interference the beam energy as photons enters the lasing loop which in this case will be a closed loop bacward and forward to the mirrro, because no light escapes forward. – anna v Jan 17 '17 at 09:47
  • If you are asking if quantum mechanics is necessary, the answer is that lasing is a quantum mechanical process which shows synchronicity.( All emmissions of light are quantum mechanical processes). The return to the source is necessary from energy conservation. – anna v Jan 17 '17 at 09:53
  • I dont see why energy conservation is a problem. The beams just pass each other with no interaction (assuming a loss free setup) unless observed by exciting some matter in their path. If this matter is in a spot, where the fields cancel out, then no excitation occurs and no energy is lost/converted. – Jens Jan 17 '17 at 11:29
  • It is not interaction that you are seeing in the interference patterns. It is superposition. Superposition quantum mechanically happens with the wave functions psi1+psi2+...., and the square of this gives the interference patterns. The photons in the black regions go back to the source so are not seen on the screen. The energy of that part of the beam too. – anna v Jan 17 '17 at 11:36