The principle of the Hanbury-Brown and Twiss experiment (see picture) says that the joint amplitude of probability that a particle from the source A be detected in the detector 1 and a particle from the source B in the detector 2, undergoes superposition with the joint amplitude of probability that a particle from A be detected in 2, and a particle from B in 1.
I am not sure on a couple of things when we do operatorial treatment.
Let â†A1 , â†B1 , â†A2 and â†B2 , represent the raising operators that create a particle of type A at the position of the detector 1, a particle of type B at the position of the detector 1, and so on.
If I construct the two-particle field operator (see picture)
(1) Ȃ = |pA1> âA1 |pB2> âB2 + |pB1> âB1 |pA2> âA2
and calculate , one of the terms will be
(2) .
where by x I denoted direct product.
From here on I feel confused. Which operator commutes with which? I suppose that âB1 should commute with âB2 because they act at different places. However, if the two sources are identical, âA1 and âB1 should not commute. And though, what I saw in some article is that the index of the detector is ignored, and the relation (1) is written as
(1') Ȃ = |pA1> âA |pB2> âB + |pB1> âB |pA2> âA .
Then the term in (2) becomes
(2') .
Does somebody understand this?
Well, you can treat it in a classical fashion, if you consider the electromagnetic radiation as classical wave. So, it's a bit of wave mechanics there, and Maxwell's law there, and you get a neat result. So, you can fully describe it in terms of classical optics.
Try the following articles- they'll point you out in the right direction towards a full understanding of the QMT behind it, and will make some sense on those operators above. While I love bra and ket terminology, sometimes it can be confusing, if you're not very sure of the properties of your operators.
http://journals.aps.org/pr/abstract/10.1103/PhysRev.124.1866
http://science.sciencemag.org/content/284/5412/296.full
Bernd Schmeikal, Hanbury-Brown and Twiss experiment is actually a phenomena intimately connected with interferometer. You can rationalize it in a classical fashion...
Thanks Alcides, however,
1) I work with bosons (4He atoms), not with fermions,
2) the phenomenology behind the HB&T effect is known to me,
3) two-particle interferometry is quantum, not classical. In classical wave mechanics, or in electromagnetism, we don't have two-particle or multi-particle interferometry.
I believe part of the problem is the phases: "that create a particle of type A at the position of the detector 1". The creation operator does not typically commute with the creation operators so you cannot say which detector/position the created particle came from or are located at. If the two particles are not identical (could be different particles or just same particles with different quantum numbers like |+> vs |-> spin electrons) the B and A operators will commute and you should not get interference from the identical particle phenomena. You may still get wave or superposition interference depending on your exact setup.
Dear Edward,
it's hard to understand what you are trying to say. Maybe you'll pass again over your answer and make it clearer. For instance you say
"The creation operator does not typically commute with the creation operators"
What you mean?
You also say
"If the two particles are not identical . . . the B and A operators will commute and you should not get interference from the identical particle phenomena."
I believe that you meant that one doesn't get interference from different, not from identical particles.
So, please be kind and make it clearer.
With thanks,
Sofia
Sorry, too many things at once.
Usually [a+,x] != 0 so a particle cannot be created at a position. Thus [a+A,a+B] != 0 since the particles are identical (assuming identical particles come out of A and B).
When I said:" you should not get interference from the identical particle phenomena" I mean that will not be the cause of any interference (you can still have interference from other, non-identical particle effects). Ignoring those, identical particles should interfere, while non identical particles should not.
Cheers,
Eddie
Dear Eddie,
why the creation/annihilation operators don't commute with position? These operators act on different variables. Creation/annihilation act on the number of particles, while the position operator acts on space coordinates.
So, can you explain why [â†, x] ≠ 0?
With thanks,
Sofia
I was thinking of the usual expansion of the field is into plane waves which don't have a well defined . However, I guess if you expanded the state into a localized basis that may work, but I never seen that so I don't know if that is possible.
Sorry,
Eddie
Thanks Eddie, I understand.
In fact, it is difficult to talk about measurement of observables without indicating the experimental configuration of the measurement. First of all, ↠is not a Hermitian operator, s.t. it can't be measured, what can be measured is the number operator â†â.
I saw a description of a detector that records both number of particles and positions, in experiments of the group of prof. Alain Aspect. They used a 2D array of detectors. The precision in the position measurement was quite limited because of the big linear dimension of a detector.
But, in my opinion, these experiments are quite pioneer experiments, it may be that in the future the detector linear dimension would be reduced.
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