Photons from two sources versus one source

[Swamp Thing]

Assume that we are given a big black box with two windows. Each window emits a laser beam. All our best measurement techniques show that the two beams have the same wavelength and have near-perfect mutual coherence (or as good as the self coherence of each output beam).

Is it possible to tell whether the beams come from two different phase-locked lasers, or are split out of a single laser? Or perhaps a third case -- a small part of one beam is tapped off and amplified, then delivered to the other window?

What experiments can distinguish between these cases?

 

[DEvens]

I think it depends on your definition of "phase locked laser."

 

[Cthugha]

All photons inside a coherence volume are indistinguishable by definition. If the beams are perfectly coherent and show coherence up to arbitrary order, you cannot determine which photon originated from which source.

If you think about it, that also makes perfect sense. Even in a laser, you usually do not have one single emitter, but many atoms (or whatever is your gain medium). In the emission from an ideal laser you also cannot tell which photon originated from which atom.

 

[Swamp Thing]

Thank you.
Another question: if the two beams are caused to overlap at a detector (or multiple detectors), will the photons behave according to single-photon statistics or two-photon statistics? Will the answer to this question depend on what is inside the black box (i.e. the three cases in post #1) ?

 

[Swamp Thing]

I think it depends on your definition of "phase locked laser."

Something along these lines : http://leecenter.caltech.edu/workshop09/talks/yariv.pdf ...?
In particular, page 12-13 "Coherent power combination" and "beam steering"

 

[Cthugha]

Another question: if the two beams are caused to overlap at a detector (or multiple detectors), will the photons behave according to single-photon statistics or two-photon statistics?

Well, "single-photon statistics" and "two-photon statistics" are not fixed terms in quantum optics, so what do you mean exactly? So well defined probability distribution with different mean values? A specific behaviour like Hong-Ou-Mandel interference? I am not quite sure what you are aiming at.

 

[Swamp Thing]

Well, "single-photon statistics" and "two-photon statistics" are not fixed terms in quantum optics, so what do you mean exactly?

Perhaps my terminology was not accurate. I am referring to these two ways of getting the detection probability in an element dS of a detector, given that Beam A has amplitide |A> to be detected, and Beam B has amplitude |B> to be detected :
(i) "Single Photon"
probaility = | |A> |B> |² dS(ii) "Two Photons"
probability = | |A goes to dS1> |B goes to dS2> + |A goes to dS2> |B goes to dS1> |² dS1 dS2

 

[Cthugha]

Perhaps my terminology was not accurate. I am referring to these two ways of getting the detection probability in an element dS of a detector, given that Beam A has amplitide |A> to be detected, and Beam B has amplitude |B> to be detected :
(i) "Single Photon"
probaility = | |A> |B> |² dS(ii) "Two Photons"
probability = | |A goes to dS1> |B goes to dS2> + |A goes to dS2> |B goes to dS1> |² dS1 dS2

I see what you mean. Let me put it this way: It can be quite complicated to talk about amplitudes for beams to be detected in some certain circumstances. What you have in the general case is some initial state 1) (for example: some ensemble of atoms in the excited state) and some final state 2) (for example: one or several detectors click).

Now the general rule is to add up all the indistinguishable ways to get from 1 to 2 according to what you call two-photon statistics and then add up all the distinguishable ways to get from 1 to 2 according to what you call single photon statistics.

So if you have two beams that are mutually coherent over (ideally) infinite distance and time, you would see them behave according to two-photon statistics. However, as second- and higher-order coherent beams have the property that their joint detection function factorizes (this means that the probability to detect a photon pair is just the mean probability to detect one photon squared), this will most likely not give you interesting effects for lasers. You may get interesting effects for thermal light or single photons, though. Look up the Hong-Ou-Mandel effect or the Hanbury-Brown-Twiss effect in case you are interested.

 

[Swamp Thing]

Thank you. Actually, it was reading about HBT that got me thinking about this kind of thing.

Just a quick question to clear up my understanding reg. my original post :

If we consider the two cases where :-

(1) both beams originate from atoms in a single cavity (e.g. via a beam splitter)
versus
(2) a case where two separate laser cavities are pulled into mutual phase lock by a classical control system

-- is it that the same mathematical description applies to both "preparations"? And that description follows the "two photon" rule, but reduces (factorizes) into a simple form?

 

[Swamp Thing]

I hope it's permissible to bump up this question, but I honestly would like to know the following (that I could not clearly glean from the above discussion - perhaps my bad for not putting the questions in the clearest way):

Case 1: A beam from one cavity (one population of excited atoms) goes through a beam splitter -- its photons are then in a superposition state like
√2 |Beam A> + √2 |Beam B>
where beam A and Beam B could be characterized by position and direction, say.

So far, so good... but:

Case 2: Two separate cavities, with a classical feedback control loop imposing a coherent phase relation between them. The outputs are directed along paths A and B, just as in Case 1.
In this case, would the photons still be in a state like √2 |Beam A> + √2 |Beam B> ?

Or would it be something else, such as half the photons in state A and half in B ?

 

[Cthugha]

Case 2: Two separate cavities, with a classical feedback control loop imposing a coherent phase relation between them. The outputs are directed along paths A and B, just as in Case 1.
In this case, would the photons still be in a state like √2 |Beam A> + √2 |Beam B> ?

Or would it be something else, such as half the photons in state A and half in B ?

There is absolutely no difference if you make the beams travel the same paths in both cases. It will be the superposition state in both cases.

 

[Swamp Thing]

There is absolutely no difference if you make the beams travel the same paths in both cases. It will be the superposition state in both cases.

Thank You!