Re: Bras, kets etc.

On Wed, 13 Sep 2006, Ron Baker, Pluralitas! wrote:

"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:

Well, the "state" is largely how likely different measurement outcomes
are.

With that view it doesn't seem to me that it is something
physical. It is just a probability calculation tool.
But we started with the HE which is relatable to the
physical. It seems to me that logical leap has been
made without explanation.

Your description of S seems oriented to population
statistics. It doesn't seem that it can be used for single
photons.

Well, the original point was to try to use something relatable to physical
states to explain bra-ket notation. Thus, it started off classical.
Bra-ket notation works equally well classically and quantumly. But the
intepretation does differ, as you note below with the question about
anti-correlations. As the classical case is basically the high photon
number limit of the quantum case, the classical case indeed corresponds to
the population statistics. More below.

Suppose you have stream of unpolarized photons
incident on a polaroid filter. How would you describe
the photons before and after the filter?

Unpolarised implies incoherent, at least temporally. Diagonally polarised
would be (|H>+|V>)/sqrt(2) or (|H>-|V>)/sqrt(2), circularly polarised
would be (|H>+i|V>)/sqrt(2) or (|H>-i|V>)/sqrt(2). Unpolarised would be
something like (exp(ia(t))|H>+exp(ib(t))|V>)/sqrt(2) where a(t) and b(t)
are unknown functions of time - the relative phase between the two states
constantly changes, and the phase relationship during successive
measurements is uncorrelated.

Then let me revise my original statement
from "unpolarized" to "randomly linearly polarized".
"randomly linearly polarized" would be applicable
in the case of most natural sources, would it not?

Is there a difference? Actually yes, since the "instantaneous random
polarisation" could also be circular, or elliptical, but that doesn't
affect the present discussion. "Randomly linearly polarised" implies
incoherent just as much as "incoherent". The difference is that for
incoherent, when you write a|H> + b|V>, both the amplitude and phase of a
and b vary randomly, whereas for randomly linearly polarised, only their
amplitudes vary randomly, while their relative phase is constant.

But for a |H> aligned piece of polaroid, for all 5 cases there is a 50%
probability of the photon making it through.

Suppose the filter is horizontal and one particular
photon is a|H> + b|V> before the filter.
What is it after the filter?

|H>, times an unknown phase factor exp(i*phi). But only a |a|^2 chance of
it making it through.

So is that (ignoring phi for the moment)
S = 0 if blocked
= 1|H> else
or is it
S = a|H>

It would have to be a|H> in order to be consistent
with your description of QM, wouldn't it?

No. The wavefunction is normalised. There must be a 100% probability that
the system is in _some_ state. The photons that are absorbed don't get to
have a wavefunction after the piece of polaroid; they don't exist anymore.

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.

(Now that might seem to be part of a photon which
would be a problem. But if S is as I think you are
describing it then it doesn't really describe a single
photon but rather a population probability.
In which case, no problem.)

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.

Consider a beam splitter with an incident beam
of a|X> . In each of the out-going legs we
would have.... what?... (a/1.414)|X> . And if you consider population
statistics then that gives us a probability of (|a|^2)/2 of detecting
photons in either leg. That works.
But not if you consider one photon. Because if the
photom is a quantum then there
would have to be anticorrelation in detection.
I don't see that anticorrelation anywhere in the math
we've discussed so far.

Well, that's just "collapse of the wavefunction" or whatever one might
wish to call it. The state could say "50% probability of measuring A and
50% probability of measuring B", and one might classically think that that
means 25% probability of measuring both A and B and 25% probability of
measuring neither, but that isn't what we see. I can't tell you off-hand
how that anti-correlation enters the math, beyond it being in the basic
definition of a single particle wavefunction.

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.

Another school says "the
vertical photons are absorbed, and the horizontal ones make it through
(and there are equal numbers of the two)". Yet another would say "all of
the photons pass through, but only 1/2 of them in our universe, and the
others in another universe".

If S is a probabilistic description then I would be
reluctant to say that |H> and |V> are real properties
of actual photons.
Especially if trying to interpret it as physical leads
to the obviously false view that a 1|V> photon has
a 50% probability of becoming a 1|H> photon.

No, if you put |V> photons onto a vertically-oriented piece of polaroid,
they all (except for incidental reflection/absorption) make it through.
Likewise, for |H> photons onto the same polaroid, none make it through.
You did specifically ask about unpolarised light.

You can give a |V> photon a 25% chance of becoming an |H> photon, with two
pieces of polaroid. The cute thing is that the billiard-ball photon school
has no difficulty with this at all: the transmitted photons are not the
same as the incident photons, and they have a fixed polarisation
determined by the orientation of the aligned molecules in the polaroid. I
don't know how the Feynman photon-take-all-possible-paths school gets to
the answer, but I think it's the same result, since purely classical
aligned elongated particles smaller than the wavelength will change the
polarisation in the same way.

[cut]

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.

--
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
.

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