Double-Slit Interference w/ Separate Sources

[Adiater]

If you have two similar coherent sources which are separated from each other by a barrier. Now one source sends particles one by one into one slit and the other sends particles into the other in a double slit interference experiment.

Now, the photons are always undistinguishable, so they should show an interference pattern.

But, there is no way that the photon from one source has info about its path from the other slit, so there is no way that it should ever have the info of interfernce. So, it should never show interference.

 

[BvU]

Hello Adiater,

If you have two similar coherent sources

That is pretty hard to achieve. (well, with two slits you get two coherent sources...)

there is no way that the photon from one source has info about its path from the other slit, so there is no way that it should ever have the info of interfernce

'having info of' is not the issue. They will always interfere. You get a pattern if they are coherent.

Your experiment can be a setup with one source and two slits; after the slits your barrier to separate the beams and then two more slits, one on each side of the barrier. Basically these two use the center part of the central maxima of the single slit pattern as new (pretty weak) coherent sources.

the photons are always undistinguishable, so they should show an interference pattern

is not a criterion

 

[mfb]

But, there is no way that the photon from one source has info about its path from the other slit

How exactly do you make the sources coherent without some information transfer between the devices?

 

[mfb]

They only need to be self-coherent, they don't have to be phase-locked together. As long as the exposure time is shorter than the coherence time an interference pattern will be recorded - the uncontrolled phase difference between the sources just introduces a random lateral shift to the pattern.

Many random lateral shifts to the pattern summed give a uniform distribution (or, to be more precise, the sum of two double-slit patterns).

 

[Derek P]

Many random lateral shifts to the pattern summed give a uniform distribution (or, to be more precise, the sum of two double-slit patterns).

I don't think so. I think you must be overlooking the first part of my statement: "As long as the exposure time is shorter than the coherence time..." The short exposure means the pattern doesn't shift too much while you sample it.

 

[mfb]

You get a single pattern with random positions for the interference fringes each time you run the experiment. If you assume a uniform distribution for the shift you get a uniform expectation value for photon detections. There will be some correlation between the detections, sure, but that is a different statement.

 

[Adiater]

How exactly do you make the sources coherent without some information transfer between the devices?

I might be mistaken, but say you have two independent sources with same color and may be polarization and the same phase even then they should be coherent.

 

[Adiater]

But there is still a humongous conceptual problem for the OP. How can a single photon come from two physically separated lasers?

What I meant was that a beam of photons coming out of the sources.

 

[Adiater]

Edit: When I posted the question, I was thinking more on the line's of Feyman's theory that each photon simultaneously travels through infinite possible paths and thus knows the places where it is not supposed to land. This is why we see double slit interference even if we try sending one photon at a time (in theory, but we could always assume the particles are electrons which can be sent one at a time). Now, if we find a way (the barrier or maybe a box surrounding the second source and the second slit) such that there is no way for particle from one slit to know the info about the paths coming from the other slit, and thus it should never show an interference pattern.

So, @mfb are you suggesting that there would be no interference pattern observed?

 

[mfb]

and the same phase

If you require them to have the same phase they are not independent.

 

[Demystifier]

If you have two similar coherent sources which are separated from each other by a barrier. Now one source sends particles one by one into one slit and the other sends particles into the other in a double slit interference experiment.

Now, the photons are always undistinguishable, so they should show an interference pattern.

But, there is no way that the photon from one source has info about its path from the other slit, so there is no way that it should ever have the info of interfernce. So, it should never show interference.

The two sources are independent so there will be no visible interference. Let me explain it mathematically.

When you send one photon through two slits (slit A and slit B), the wave function is
$$\psi({\bf x_1})=\psi_A({\bf x_1})+\psi_B({\bf x_1})$$
If you calculate $\psi^*({\bf x_1})\psi({\bf x_1})$ you find the interference terms.

When you send one photon through slit A and the other independent photon through slit B, you actually know which photon travels through which slit, so the wave function is
$$\psi({\bf x_1,x_2})=\psi_A({\bf x_1})\psi_B({\bf x_2})$$
so there is no interference.

Now you might argue that photons are indistinguishable, so the last expression should be symmetrized with respect to ${\bf x_1}\leftrightarrow{\bf x_2}$. In principle that's true, but then one should really symmetrize the full wave function which includes not only the photons but also their sources. And then one should trace out the sources, to find the density matrix of the photons. When one does all that, one arrives at an expression equivalent to the last expression above.

 

[zonde]

I suppose that appropriate reference for OP question might be this:
https://journals.aps.org/pr/abstract/10.1103/PhysRev.159.1084

 

[Cthugha]

Now you might argue that photons are indistinguishable, so the last expression should be symmetrized with respect to ${\bf x_1}\leftrightarrow{\bf x_2}$. In principle that's true, but then one should really symmetrize the full wave function which includes not only the photons but also their sources. And then one should trace out the sources, to find the density matrix of the photons. When one does all that, one arrives at an expression equivalent to the last expression above.

I think I tend to disagree with the wording (but probably not the content). The case of distinguishable photons is clear. Fields of - for example - different polarization do not interfere and that is it. Two different independent light sources emitting into the same mode of the light field are more complicated, though. At any instant, the joint intensity distribution arising due to the two fields will necessarily contain interference terms that depend on the relative phase of these two fields. However, as this relative phase is not stable, the interference pattern will vary randomly in time. The common experimental situation is that integration times are longer than the timescale of phase randomization so just the ensemble average over all relative phases is measured, which indeed contains no interference terms anymore. However, if one keeps track of the relative phase, one can still keep the information about the interference patterns. This is what happens "naturally" in two-photon interference, where you can indeed see interference effects between very different and distinguishable light sources. Personally, I liked this demonstration of interference between a laser light field and spontaneous emission from a quantum dot very much:
https://arxiv.org/abs/1006.0820

That is from "Interference of dissimilar photon sources" by A.J. Bennett et al., Nature Physics 5, 715 (2009).