Decoherence foiled by polarization
Neil B.
relates to precise versus careless talk of "interference", and to
decoherence: I argue we can experimentally *show* that the combined wave
trains of single photons are still distributed into two paths, despite
complete prior randomization (incoherence) of their relative phases.
IOW, we can show they *are* distinguishable from single-path photons
coming out of either channel A or B but "not both." That means, they
aren't "as if collapsed" by the decoherence (I'd call those arguments
gliboherence ;-)
What I mean: split a photon at BS1 in a Mach-Zehnder interferometer.
Have something get in the way of one or both legs (air densities, etc.)
to randomize phase as they approach recombiner BS2. The output as such,
is randomized and "shows no interference." But suppose we tag the input
legs using polarization. We can start with say |x> LP original input.
We put an optical rotator to convert the wave in leg II to |y> LP. When
they recombine in BS2, transmitted |x> and reflected |y> superpose in
channel A, and transmitted |y> and reflected |x> superpose in channel B.
We can arrange for the separated |x> output components to share the same
relative phase. We can arrange for the |y> components to be 180� out of
phase (same-to-same polarization and not within a single channel.)
By itself, that doesn't do anything. But we can recombine these beams a
second time in BS3. Now, the |x> components can constructively
interfere only out of the second channel A, "A2." The |y> components can
constructively interfere only out of the second channel B, "B2." Hence,
only |x> LP will come out of A2 and only |y> LP comes out of B2. It
doesn't matter what the phase between |x> and |y> is, because we can
show another way that two wave trains existed in transit between BS2 and
BS3. If isolated (one-path) photons had been approaching and/or exiting
BS2, each could be either reflected or transmitted there or at BS3. When
the hit BS3, they again could go either way. So after the photons were
polarization tagged, we could get either |x> or |y> in either A2 or B2.
But, we don't because of phase relations - "superposition" of vectors
despite not being "interference" in the sloppy, traditional sense.
There's the difference.
Readers may need to play around with diagrams etc. to get this, let me
know what you think. This may relate to arguments about randomized
phases being "decoherence" and the supposed implications thereof.
Androcles
"Neil B." <neil_...@caloricmail.com> wrote in message
news:nO2dnYDj19d3RsLX...@posted.widowmaker...
>I propose something, perhaps original, that is likely surprising and
>relates to precise versus careless talk of "interference", and to
>decoherence: I argue we can experimentally *show* that the combined wave
>trains of single photons are still distributed into two paths, despite
>complete prior randomization (incoherence) of their relative phases. IOW,
>we can show they *are* distinguishable from single-path photons coming out
>of either channel A or B but "not both." That means, they aren't "as if
>collapsed" by the decoherence (I'd call those arguments gliboherence ;-)
Sounds like bullshitoherence, but I'll reserve judgment and see
if you have anything of substance.
> What I mean: split a photon at BS1 in a Mach-Zehnder interferometer. Have
> something get in the way of one or both legs (air densities, etc.) to
> randomize phase as they approach recombiner BS2. The output as such, is
> randomized and "shows no interference." But suppose we tag the input legs
> using polarization. We can start with say |x> LP original input. We put
> an optical rotator to convert the wave
Sly switch from photon to wave, wasn't it?
Definitely crapoherence.
Neil B.
proves the incoming wave wasn't restricted to a single "leg" at a time.
I apparently had a misfire on the randomization issue. This point isn't
about randomization - it's about the which-way tagging. Tagging each leg
as |x>, |y> respectively should (intuitively) provide "which way
information" and spoil the chance to get "interference" later. But if we
appreciate detecting "superposition" even if it's not classic
"interference", then there's a way to show that both paths are still
active despite the apparent WWI. For example, in-phase inputs will go
out of channel A as a 45� diagonal, which we can measure and distinguish
from a mix of random |x> and |y>. Note that we could use actual LP
filters, which have a chance of absorbing the photon. One might
"intuitively" expect that passage of a split-wave through one or both
such filters would surely "collapse" it into a localized wave of the
corresponding polarization, but it doesn't - both waves continue, but
changed into that polarization.
Salmon Egg
"Neil B." <neil_...@caloricmail.com> wrote:
> I propose something, perhaps original, that is likely surprising and
> relates to precise versus careless talk of "interference", and to
> decoherence: I argue we can experimentally *show* that the combined wave
> trains of single photons are still distributed into two paths, despite
> complete prior randomization (incoherence) of their relative phases.
> IOW, we can show they *are* distinguishable from single-path photons
> coming out of either channel A or B but "not both." That means, they
> aren't "as if collapsed" by the decoherence (I'd call those arguments
> gliboherence ;-)
As soon as you associate specific wave trains with specific photons, you
indicate to me that you do not understand the quantum nature of light.
Bill
--
Private Profit; Public Poop! Avoid collateral windfall!
Neil B.
"Salmon Egg" <Salm...@sbcglobal.net> wrote in message
news:SalmonEgg-C2761...@news60.forteinc.com...
We can associate specific wave trains with a specific photon, if only
one photon is involved. Actually we could to some extent with more, but
there are degrees of ambiguity (even then, it isn't utterly
indeterminate, IOW not the opposite of having complete specification.)
BTW, what's up with the collateral windfall?
Salmon Egg
"Neil B." <neil_...@caloricmail.com> wrote:
I do not wish argue with you. I want specifics not English teacher talk
akin to describing Shakespeare plays. Describe to me an experiment, real
or gedanken that explains your point of view.
The windfall is what Merrill Lynch (spelling?) executives for driving
their employer into ruin. There are other examples.
Neil B.
we can associate a wave train with a specific photon, if there is no
ambiguity. IOW, that's the photon that "has to be there." So if one
photon hits a beam splitter, that photon's waves are present in the two
beams, and no other.
I already made the argument and a proposed experiment. I'm not yet sure
how to simplify or make better.
p.ki...@ic.ac.uk
> Well, I have to dig into semantics to make the point. More simply, yes
> we can associate a wave train with a specific photon, if there is no
> ambiguity. IOW, that's the photon that "has to be there." So if one
> photon hits a beam splitter, that photon's waves are present in the two
> beams, and no other.
I disagree. I regard photons as discrete excitations of EM field
modes, not as some kind of thing-in-itself. Thus the choice of modes
amounts to a choice of basis for your descriptions of the system.
If each output port of the beam splitter is modelled with independent
modes O_i (the "obvious" thing to do), then the original excitation
has become two entangled excitations.
If you instead use ("combined") modes C_j that are orthonormal
combinations of the "obvious" modes, then you indeed might see a
single excitation in one of these new modes. But it'll describe the
same quantum state as the obvious mode decription did.
There is no way to tie a luggage label to a packet of energy, unless
you use that label associated with your choice of modes. But the
choice of modes is an arbitrary one (assuming each choice forms an
orthonormal set, etc).
If you know exactly how the system is going to evolve, then you might
indeed (at least in principle) construct modes that tracked that
evolution perfectly (eigenmodes!). But usually we cannot solve for
them, and so have to live with our excitation hopping between modes
(labels) and entangling itself with other modes. And if we can find
the eigenmodes, then we immediately try to solve some other more
complicated problem, so again our excitaion gets spread out.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
Neil B.
<p.ki...@ic.ac.uk> wrote in message
news:e7uvl6-...@ph-kinsle.qols.ph.ic.ac.uk...
Paul, thanks for the thoughtful reply. BTW do you remember my thread
"The problematical nature of photon spin" around ten years ago? Wow,
what an argument and IMHO never adequately resolved.
Photons: well, there are different ways of "looking at things" and
describing them, which create various pictures of the system. We can
create "individual photons" because of a process known to do so, like a
single atomic excitation-de-excitation. "One" means we can get only one
click from a surrounding photon detector (and think also, what "clicks"
mean.)
My middle-brow take only gets me so far, but note this from Wikipedia:
"A Fock state, in quantum mechanics, is any state of the Fock space with
a well-defined number of particles in each state."
I do know, some other states of light have ill-defined number of photons
and so forth. Yet in practice and accordance with particle physics, I
can use a "photon gun" to create, what only a picky person would say is
not "the emission of a single photon." If I split "one of those" with a
BS, again: only IMHO a picky person would not appreciate, having "the
wave function of a single photon" spread out into an e.g. Mach-Zehnder
interferometers. Note too, a single photon has e.g. hbar of spin, which
also can in principle be detected.
p.ki...@ic.ac.uk
ow, some other states of light have ill-defined number of photons
> and so forth. Yet in practice and accordance with particle physics, I
> can use a "photon gun" to create, what only a picky person would say is
> not "the emission of a single photon."
Again, you need to think about modes. Or, more specifically, your
photon-gun will emit a certain spatial distribution of EM amplitudes,
this spatial distribution would then match some specific photon-mode G_0
chosen by the gun's designer.
But I could represent the EM field using an incompatible set of different
modes D_i, and the gun would then appear to be emitting and entangled mess
of fractionally inhabited photon modes G_i.
An excitation only looks like a single photon _if_ the EM amplitude
distribution matches the choice of basis states (modes). You make
this kind of "sensible" mode choice inplicitly -- but it's still a
choice, and is one which would not be seen as sensible for every
purpose.
p.ki...@ic.ac.uk
> But I could represent the EM field using an incompatible set of different
> modes D_i, and the gun would then appear to be emitting and entangled mess
> of fractionally inhabited photon modes G_i.
Bah. Both instances of unnecessary notation should be the same,
e.g both D_i. Sorry.