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Even if the double slit experiment gives interesting (weird) results, it only concludes that each photon interacts with itself after passing the two slits. I have been thinking about a different experimental setup, where you have two well defined light sources (with specific wave lengths and phase) but no slits. And now to my questions: Has anyone ever done such an experiment, and will there be an interference pattern on the wall?

If the answer to the second question is "no", light can not be a true wave - it only has some wavelike properties. But if it is "yes", things become much more interesting.

If there is an interference pattern on the wall, there has to be an interference pattern even if both light sources are emitting single photons at random, but as seldom as, say, once per minute. That in turn would mean that the photons know about each other, even if they are separated in time with several seconds, and the light sources are independent (not entangled).

  • Did You read this: http://physics.stackexchange.com/questions/6234/does-a-photon-interfere-only-with-itself – Georg Mar 04 '11 at 11:14
  • I am voting to close it as an exact duplicate. The only difference is that Alexander confuses the verbs "interact" and "interfere". They're very different things. A single photon surely doesn't "interact" with itself. – Luboš Motl Mar 04 '11 at 11:34

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If the energy is large enough, photons interactively create massive particles. So yes photons do interact.

datenwolf
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    Even with low energy photons, there are virtual loops and photon photon interactions can happen with some probability: – anna v Mar 05 '11 at 07:23
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I am answering to the second part of your question, since the first has been covered two days ago. Have a look at interfering laser beams, and read up on holography, the ultimate in interfering beams.

even if both light sources are emitting single photons at random, but as seldom as, say, once per minute. This is wrong. For interference to happen the phases must be kept. Individual photons do not keep the phase with another photon a minute later, so no interference pattern will appear. In contrast to the double slit experiment, where each photon interacts with the slits, in your two beam experiment each photon from one beam has to interact with a photon from the other beam for an interference pattern to appear.

anna v
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  • I don't know whether this answer really addresses the question the OP was intending to ask. Very weak laser light (so on average there is a photon once per minute) is not the same thing as emitting single photons at random. If two lasers have stable enough phases, even if the beams are attenuated to one photon per minute, you should observe interference. – Peter Shor Mar 04 '11 at 17:21
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    @Peter Shor: The phase canonically conjugate to the photon number, and a really precise restriction on the photon number such as emitting one photon a minute would induce an uncertainty in the phase by the undcertainty principle. – Zo the Relativist Mar 04 '11 at 18:36
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    A laser beam appears because of induced deexcitation from the pumped up levels of the atoms of the crystal/gas. By construction more than one photon is needed to start the cascade. Without a cascade there are no phases. – anna v Mar 04 '11 at 19:36
  • @anna: But you can attenuate a laser beam by putting it through beam splitters to get very weak coherent light. – Peter Shor Mar 04 '11 at 23:20
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    then Jerry's comment above holds.Though I would think that logically one would need at least two photons in a beam to carry a phase information, no? The phase is with respect to the beam. See http://en.wikipedia.org/wiki/Coherent_state , where it is demonstrated that delta(theta)*delta(n)=1/2 . I was just commenting that one cannot emit a photon per minute. To define a phase one needs at least two photons. – anna v Mar 05 '11 at 05:53
  • O.K. You're right. – Peter Shor Mar 05 '11 at 13:19
  • Any lightsource (or oscillator) has a certain length and time of coherence and band width (easily calculated with the light speed). Pairs of such oscillators have a certain relative coherence. So, if You dilatate a interference experiment in time, You loose coherence due to the time of coherence. Doing such thoughts/calculations using photons in stead of waves is a bit masochistic in my eyes: "Why the easy way if there is a painful way?" – Georg Mar 05 '11 at 14:30
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It is more likely that the photons will interact with themself, let's refer to the Mach Zehnder interferometer and focusing on the last BeamSpliter(before the photodetectors) enter image description here, then suppose photon from the upper beam interact with photon the lower beam, therefore in a "casual way" there are 2 possibilities:

  1. Destructive Interference: a photon interacts with another photon then both of them destroyed without creating a new entity >>>>> However this will not fulfil the law of "conservation of energy(energy cannot be destroyed nor created it can only transform into another type of energy/entity ", therefore not possible

  2. Constructive interference: if you consider 2 "wave" then the amplitude will add up to a wave with "bigger" amplitude, however, if you consider it as particle where photon meet with photon then create 2 or more photons this seems very illogical and also doesn't follow the conservation of energy

Therefore, those two things are not possible, so the only possible explanation is that the photon interacts with itself

Lukman Hendrajaya
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