M1L8d.txt

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Then let me finish this discussion with a fully
quantum mechanical description.
If you describe the detection or collision of two particles
in two modes-- so you have a mode, a mode operator,
annihilation operator, a1 and a1.
But if you are asking what happens
if I measure two particles, you annihilate two particles
with an operator, a1 and a2.
But then, because of the indistinguishability
of particles, you have to consider an exchange
term, which is a2, a1.
So whenever you detect two particles
in two different modes, your signal
is proportional to something related to a1, a2, plus a2, a1.
This extra exchange term gives you an extra factor of two
for bosons, which is exactly what appears
in the g2 function.
And, of course, if you have an exchange term for fermions,
you get zero.
And this of course leads to the antibunching or the g2 value
of zero for fermions.
However, and this is obvious, if you have a single-mode, then
the only operator to detect two particles is a1 times a1,
and this has no exchange term.
And this is the situation of the laser
and the Bose-Einstein condensate,
and that means they have a g2 function of one.
So, Cody, this is another argument
why single-mode occupation has a g2 function of one,
but I have to reconcile it with your question.
Right now, I think there may be a difference
between a canonical ensemble and grand canonical ensemble.
In a canonical ensemble, the total particle number is fixed,
but in a grand canonical it fluctuates.
And maybe this leads to additional fluctuations
for the case of single-mode chaotic light,
but I really have to think about your question.
But here's the same argument why if you occupy
only a single-mode, if you look for instance
at the Bose-Einstein condensate, you do not
have the second exchange term.
And the missing exchange term just
propagates through the equations and gives you