Re: Bras, kets etc.


"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote in message
news:Pine.LNX.4.50.0609150916020.421-100000@xxxxxxxxxxxx
On Thu, 14 Sep 2006, Ron Baker, Pluralitas! wrote:

"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:
On Wed, 13 Sep 2006, Ron Baker, Pluralitas! wrote:
"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:

I was thinking that natural sources would be linearly
and not circularly polarized but when I think about it
I don't have any particular justification for that idea.
And right off hand I'm not sure how one would determine
that a photon is circularly polarized.

Generally, since the usual source is uncorrelated atoms or electrons in a
solid, natural sources are unpolarised and incoherent. Mathematically, the
only difference between linear and circular polarisation is a 90 degree
phase shift between the two linear components. I'm inclined to say that
circular polarisation is the more fundamental of the two, but I'm biased
by a decade of work in optical angular momentum.

A piece of polaroid will tell you whether a source is linearly polarised,
and a piece of polaroid with a 1/4-wave plate in front of it will tell you
if a source is circularly polarised. I was at a conference where one
presenter brought in a box of beetles for show-and-tell (thanks Gorden, it
was very cool). Some of the beetles were scarabs, which on account of
helical proteins in their wing covers, reflect only one circular
polarisation of light. Scarabs, meanwhile, have eyes that can see circular
polarisation, so this lets them find their own kind to mate with much more
reliably. Look at the box of beetles, they all look pretty much the same.
Look at the box of beetles through a circular polarisation filter, and the
scarabs are obvious.

Interesting. I googled a bit and saw a little more on that.
I saw mention that it reflected circularly polarized light
when the incident light was unpolarized. That kind
of supports my bias that natural sources tend not to be
circularly polarized.


However, you can't determine if the incident photon was circularly
polarised with such a filter. All you learn is whether it made it through
the filter or not. An unpolarised photon will do so 50% of the time, and
so will a plane polarised photon.

It's not quite so simple. Magic words to google for are "Stokes
parameters" or "Stokes vector" and "measure" or "measurement". The
classical measurement exploits having lots of photons in the same state,
but the single photon quantum measurement cannot, and you won't know the
incident state without further information.

This is one of the big difference between the classical description and
the single-photon quantum description. In the former, |a|^2 + |b|^2 is
the
power before the polaroid, whereas in the latter, |a|^2 + |b|^2 = 1,
since
the photon has a 100% probability of being incident on it. Classically,
you'd have a power of |a|^2 afterwards, but quantumly, if the photon
made
it through, it still has a 100% chance of being there, while if the
photon
didn't make it through, it doesn't exist anymore, and doesn't have a
wavefunction.

Applying that to an intereferometer would seem to lead
to a contradiction. If it is as you just described then the
photon actually takes one path. How would you then
explain the interference?

OK, the above was written about a pice of polaroid, where a photon with

But the principle should apply equally to beam splitters and
polarizers.

the "wrong" measured polarisation is absorbed, so there is no

Absorbtion shouldn't be critical. There are polarizers
that reflect rather than absorb the 'wrong' polarization too, esto no?

interference, but what are are saying is the typical case that people talk
about with eg two-slit interference or in an interferometer with a
beam-splitter, where the interference pattern disappears if you put a
detector (at one slit)/(in one beam path). In "collapse of the
wavefunction" language, if you don't measure, you don't collapse, and you
get interference at the end. Measure, and no interference.

Very metaphysical.

(And 'measure' is a synonym for 'detect'.)


And that phi: you don't mean it to be random per
photon do you. If it were then a laser beam would
lose its coherence going through a polarizer.

Well, you asked about unpolarised light, not a laser beam. But if you
had
a 45 degree polarised laser beam incident on a |V> polaroid, then it
would
be 1/2, at random, but that wouldn't do anything to the coherence. All
of
the incident photons are in the same mode, the transmitted mode (if you
win the 50% coin toss) has a fixed phase cf the incident mode, so the
transmitted photons are all in the same mode.

So then phi is not random per photon?

Good question. Random over time, yes.

Hmm. How?

Random over photons, I don't know
off-hand.

How could it be random per photon? How could that
not destroy the coherence of a laser beam?

If you detect the photons at different times, then random over
time gives you random over photons. If you check out Hanbury Brown's book,
this is part of the intensity interferometer controversy.

I found this:
http://marcus.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf
In section II D and in the next to last paragraph on page 1212
it seems to describe the mathematical basis of anti-correlation.
I am not familar with the math though.

Will look as time permits, and might comment here later if appropriate.
Otherwise, I'll just read.

Yeah, I was hoping you could explain some of it.
Obviously it is not your job to give physics lessons
to every yahoo on the usenet but I figured you and
I and others would find it interesting.
The math has a similar form to the vectors/bra-kets.
They mention annihilation and creation operators
which I have heard of but don't fully understand.
(But I figure they are similar to things I've already
dealt with.) And then the colons. I wonder what
they mean.

I need to read the rest of what you sent me too.


Which photons are absorbed by the filter?

This is where philosophy comes into it. One school of thought would
say
"1/2 them, entirely determined randomly".

That is easily demonstrated to be false using a pair
of polaroid filters.

How? With one filter, half of the light makes it through. With two
filters, the light is no longer unpolarised when it reaches the second
filter, and the probility of transmission through the 2nd filter
depends
on the angle.

If it was "1/2 of them, entirely determined randomely" before
then the initial polarization of each photon did not matter.
So if they are all |H> after one filter then it still should not
matter at the second filter.

The "1/2" is beacuse the incident light is unpolarised. (It would also be
1/2 for 45 degree diagonally polarised light or circularly polarised
light or 45 degree elliptically polarised light.) After the first filter,
you have polarised light.

Maybe it is a matter of semantics. "entirely determined
randomly" seems to totally discount any incident polarization
(or lack thereof).


With unpolarised light, whether an individual photon has a polarisation
depends on your picture/interpretation of QM.

Interesting. How can an individual photon not
have a polarization if one can say they are polarized
after a polarization filter?

Again, "both polarisations
at once", "one polarisation, with 50/50 chance", or "both polarisations,
but in different universes" etc etc etc.

Point of notation: Are the orientations of |H> and |V> chosen
to be some constant experimental reference for all possible
filters and their orientations that might be in the path
or do you define a new |H> and |V> (|H1>, |H2>, etc.) at each filter?

I was thinking of |H> and |V> in a "lab frame". But you can do either,
keep the same |H> and |V> basis throughout, or use a separate one for each
waveplate/polariser. Either approach gives you Malus's law.

OK. To be clear, it should be explicit.


If the spherical is resolvable to the rectangular then
might there be filters for rectangular modes that could
be used infer the spherical?

Only if you measure phase, which the PMT does not.

Interferometry?

Yes, one can do so to get (relative) phase. But then it's more than just a
PMT.

That's doesn't cause me any worries.


--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

--
rb


.

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