Why do we observe double slit interference?
Very cryptic
I've been thinking a bit about this. The experiment is often conducted
within the medium of air. Now, the thing about the propagation of a
photon within air is it's constantly absorbed and reemitted by the
molecules making up the air. This whole process is elastic at visible
frequencies so there's no entanglement with the internal degrees of
freedom of the molecule, but however, during each scattering, the
center of mass of the molecule gets shifted a bit. Actually, since the
emitted photon is unlikely to be reemitted in the same direction, the
momentum of the molecule would also be changed in general. Now the
wavefunction of the molecule is probably spread out enough that the
overlap between the final and initial wavefunctions of the molecule is
still significant (i.e. not orthogonal). But the catch is, the photon
gets scattered off a huge number of air molecules along its path.
Isn't that enough to cause decoherence? My question is, why doesn't
the photon get entangled with the position and momenta of sufficiently
many air molecules enough that the interference pattern vanishes?
After all, if the photon went through one slit it would get entangled
with the air molecules around that slit which are different from the
air molecules around the other slit.
Uncle Al
Very cryptic wrote:
>
> I've been thinking a bit about this. The experiment is often conducted
> within the medium of air. Now, the thing about the propagation of a
> photon within air is it's constantly absorbed and reemitted by the
> molecules making up the air.
Dead on Arrival.
http://www.quantum.univie.ac.at/research/matterwave/c60/
Crack a physics text.
[snip]
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
Nick Maclaren
In article <41237F64...@hate.spam.net>,
Uncle Al <Uncl...@hate.spam.net> wrote:
>Very cryptic wrote:
>>
>> I've been thinking a bit about this. The experiment is often conducted
>> within the medium of air. Now, the thing about the propagation of a
>> photon within air is it's constantly absorbed and reemitted by the
>> molecules making up the air.
>
>Dead on Arrival.
>
>http://www.quantum.univie.ac.at/research/matterwave/c60/
>
>Crack a physics text.
Well, some of them are pretty obscure, but I don't know any recent
ones that are deliberately encrypted against the reader. They used
to be, but that was in the heyday of the Inquisition.
I don't see that your Web reference is relevant, either. The point
isn't wave/particle duality, but why absorption and reemission
should preserve quantum coherence. That is DEFINITELY not a trivial
question, and I myself don't have a clue - despite having read a few
physics texts.
I am not sure that he is right, because my understanding was that the
probability of absorption is small in 'transparent' media, but I
could well be wrong and the actuality is as he says. Certainly, if
he is right, then there is something very special about 'elastic'
absorption and reemission.
There may well be a known answer, but it isn't quite so basic as you
imply.
Regards,
Nick Maclaren.
Ralph Hartley
Very cryptic wrote:
> the photon
> gets scattered off a huge number of air molecules along its path.
> Isn't that enough to cause decoherence? My question is, why doesn't
> the photon get entangled with the position and momenta of sufficiently
> many air molecules enough that the interference pattern vanishes?
Some materials scatter photons in the manner you describe, and some do not.
Those that don't are referred to as "transparent", while those that do are
"translucent".
I don't think I can explain why air is transparent, maybe the property can
be determined from first principles, maybe not.
Photons do not become significantly entangled with the atoms of the air
they pass through. The experiment is simple, look out the window. See
anything?
Air is not perfectly transparent. They do occasionally scatter. There, I've
explained why the sky is blue as well.
Ralph Hartley
Urs Schreiber
news:cg0e88$ket$1...@pegasus.csx.cam.ac.uk...
> >http://www.quantum.univie.ac.at/research/matterwave/c60/
> I don't see that your Web reference is relevant, either.
Uncle Al is trying to point out that even when you shoot a big fat object
like a buckminster fullerene on a screen with two slits, it still gives an
interference pattern. The decohering effects on the buckyball must be orders
of magnitude larger than for the double slit experiment with a photon - but
the interference pattern is still there.
This is pretty amazing, given that other people are using their scanning
tunneling microscopes to play classical (though minature) soccer with
buckyballs.
On the other hand, it does not completely answer the original question,
which was as asking how it can be understood that the decoherence effects in
the 2-slit experiment (under ordinary conditions, at least) are not large
enough to destroy the interference pattern. We experimentally observes that
they are not. But can anyone give a theoretical order-of-magnitude
back-of-the-envelope estimate why not? Probably this has been done somewhere
in the literature.
alistair
> Very cryptic wrote:
> >
> > I've been thinking a bit about this. The experiment is often conducted
> > within the medium of air. Now, the thing about the propagation of a
> > photon within air is it's constantly absorbed and reemitted by the
> > molecules making up the air.
>
> Dead on Arrival.
>
> http://www.quantum.univie.ac.at/research/matterwave/c60/
Alistair:
This weblink is about C60 buckyballs passing through a diffraction
grating
and showing wave-like behaviour.Since the buckyballs are moving quite
slowly -
about 210 m/s - would it be possible to fire a neutron beam through
the buckyballs at right angles to their direction of travel and to get
a neutron diffraction pattern of the buckyballs before and after they
have passed through the grating,to see how they change.Such a
procedure could yield information about how decoherence occurs.
Nick Maclaren
"Urs Schreiber" <Urs.Sc...@uni-essen.de> writes:
|> "Nick Maclaren" <nm...@cus.cam.ac.uk> schrieb im Newsbeitrag
|> news:cg0e88$ket$1...@pegasus.csx.cam.ac.uk...
|> > In article <41237F64...@hate.spam.net>,
|> > Uncle Al <Uncl...@hate.spam.net> wrote:
|>
|> > >http://www.quantum.univie.ac.at/research/matterwave/c60/
|>
|> > I don't see that your Web reference is relevant, either.
|>
|> Uncle Al is trying to point out that even when you shoot a big fat object
|> like a buckminster fullerene on a screen with two slits, it still gives an
|> interference pattern. The decohering effects on the buckyball must be orders
|> of magnitude larger than for the double slit experiment with a photon - but
|> the interference pattern is still there.
Well, yes, that is obvious.
|> This is pretty amazing, given that other people are using their scanning
|> tunneling microscopes to play classical (though minature) soccer with
|> buckyballs.
Not really. I should be amazed if it DIDN'T happen! It would be
flabberghasting if wave/particle duality applied up to a certain
size of object, and stopped doing so thereafter. I am surprised
that the wave aspect is observable, but that is a different point.
|> On the other hand, it does not completely answer the original question,
|> which was as asking how it can be understood that the decoherence effects in
|> the 2-slit experiment (under ordinary conditions, at least) are not large
|> enough to destroy the interference pattern. ...
It doesn't even ADDRESS the original question!
If you read it, you will see:
... Now, the thing about the propagation of a
photon within air is it's constantly absorbed and reemitted by the
molecules making up the air. ...
And a lot of related points, which I have snipped. It is very clear
that the question has very little to do with the slit and a great
deal to do with the preservation of coherence through absorption and
reemission by air molecules. And that I did find amazing!
However, Ralph Hartley has posted that the effect "Very cryptic" was
assuming doesn't actually happen (as I hypothesised), which explains
the situation.
If you can succeed with the double slit experiment in a translucent
medium (using his notation), then I will indeed be amazed.
Regards,
Nick Maclaren.
Urs Schreiber
> However, Ralph Hartley has posted that the effect "Very cryptic" was
> assuming doesn't actually happen (as I hypothesised), which explains
> the situation.
Sure it happens. Decoherence takes place every time your system is not
isolated but coupled to a large environment. And every realistic system is.
The only question is: "How much?"
Ralph Hartley pointed out one reason not to expect the effect to be very big
for the photon. True. But Uncle Al pointed out that even if the effect is
orders of magnitudes bigger, it would still be small enough to allow the
observation of intereference.
Just imagine, the buckyball is performing lots of internal oscillations, suffering
excitations and de-excitations of constituent atoms by interaction with the
surrounding electromagnetic field. And not too little, after all in
Zeilinger's experiment the buckyballs come out of an oven!
On the other hand, the immense decoherence induced by all these effects is
indeed being seen in the experiment. Namely, last I checked at least,
Zeilinger's team can only produce the first interference minimum in their
experiment. The more delicate higher order minima are washed away by
decoherence. Hence, roughly, the decoherence that a "hot" buckyball is
subject to is getting close to the critical value where the interference pattern
would disappear altogether.
If you find this amazing or not, it would be nice to have a theoretical
estimate for this critical value.
Danny Ross Lunsford
"Urs Schreiber" <Urs.Sc...@uni-essen.de> wrote in message news:<4124b132$1...@news.sentex.net>...
> ..they are not. But can anyone give a theoretical order-of-magnitude
> back-of-the-envelope estimate why not? Probably this has been done somewhere
> in the literature.
Near-zone vs. wave-(far)zone?
-drl
Very cryptic
nm...@cus.cam.ac.uk (Nick Maclaren) wrote in message news:<cg0e88$ket$1...@pegasus.csx.cam.ac.uk>...
> I am not sure that he is right, because my understanding was that the
> probability of absorption is small in 'transparent' media, but I
> could well be wrong and the actuality is as he says. Certainly, if
> he is right, then there is something very special about 'elastic'
> absorption and reemission.
>
> There may well be a known answer, but it isn't quite so basic as you
> imply.
Let me clarify a bit. When I wrote absorption, I meant absorption for
a split second followed by a reemission of another photon of the same
frequency which is elastic. Certainly, this "scattering" is
significant enough to slow down the speed of the photon appreciably
(replace air with water or glass if you wish) and "bend" its path
whenever the index of refraction changes. This is different from
inelastic absorption.
Alex Green
Ralph Hartley <har...@aic.nrl.navy.mil> wrote in message news:<cg27sh$spm$1...@ra.nrl.navy.mil>...
> Very cryptic wrote:
> > the photon
> > gets scattered off a huge number of air molecules along its path.
> > Isn't that enough to cause decoherence? My question is, why doesn't
> > the photon get entangled with the position and momenta of sufficiently
> > many air molecules enough that the interference pattern vanishes?
>
> Some materials scatter photons in the manner you describe, and some do not.
> Those that don't are referred to as "transparent", while those that do are
> "translucent".
>
Optical fibre interferometry is an interesting variation on this
theme:
http://www.irishscientist.ie/2000/contents.asp?contentxml=113s.xml&contentxsl=insight3.xsl
The coherence length of glass is measured in metres so will huge
blocks of glass be transparent but not capable of providing light as a
source for a double slit experiment? Or would the correct term for
huge lumps of glass be 'translucent' even though the decoherence may
not be due to appreciable scattering?
Peter Shor
Ralph Hartley <har...@aic.nrl.navy.mil> wrote in message news:<cg27sh$spm$1...@ra.nrl.navy.mil>...
I don't think this really answers the question. The speed of light in
air is slightly (0.03%) less than the speed of light in a vacuum. This
means that even when the photons aren't scattering, they're still
interacting with the air. But this interaction doesn't seem to affect
the double slit experiment much. Why not? Would the double slit experiment
work in water or glass, which have much higher indices of refraction but are
still pretty transparent? Why? Anyone have an answer?
Peter Shor
Douglas Natelson
Peter Shor wrote:
> I don't think this really answers the question. The speed of light in
> air is slightly (0.03%) less than the speed of light in a vacuum. This
> means that even when the photons aren't scattering, they're still
> interacting with the air. But this interaction doesn't seem to affect
> the double slit experiment much. Why not? Would the double slit experiment
> work in water or glass, which have much higher indices of refraction but are
> still pretty transparent? Why? Anyone have an answer?
My sense is that this must be because elastic interactions do
not dephase. When people first started thinking about trying
to see quantum interference effects of electrons in metals
(before 1980 or so), there was this intuitive bias that the
elastic mean free path would be the limiting length scale for
coherence -- that is, that every elastic scattering event of
an electron off of a grain boundary or impurity atom would
cause decoherence. We now know that this isn't the case.
Interactions that do not entangle the electron with dynamical
degrees of freedom of the environment do not cause decoherence.
That's why it's possible to see things like weak localization,
universal conductance fluctuations, and the Aharonov-Bohm effect
in solid state systems, even when far from the ballistic limit.
This is all discussed very nicely by people like Yoseph Imry.
It's also been discussed for photons in papers like PRL 81,
5800 (1998); PRL 92, 033903 (2004); etc.
So, my best guess is that the elastic interactions between
the light and the dielectric medium (that give rise to the
dielectric response) are (mostly) reversible, and don't
correspond to changing the internal states of the dielectric.
Now, if you had a material with weird dielectric properties
(e.g. lots of loss, resonances, etc.), then you could run into
decoherence issues.
--DN
Greg Kuperberg
In article <9b2e17b4.04082...@posting.google.com>,
Peter Shor <peter...@aol.com> wrote:
>I don't think this really answers the question. The speed of light in
>air is slightly (0.03%) less than the speed of light in a vacuum. This
>means that even when the photons aren't scattering, they're still
>interacting with the air. But this interaction doesn't seem to affect
>the double slit experiment much. Why not? Would the double slit experiment
>work in water or glass, which have much higher indices of refraction but are
>still pretty transparent? Why? Anyone have an answer?
I'm not entirely sure that this is a good explanation, but here is what I
think is the point. Because the photon has low energy, it measures very
little about the positions of the molecular dipoles that refract it,
much less their internal state, only with their presence or absence.
And because the photon has a long wavelength, it only measures the
aggregate presence of many molecules in a large volume. Thus the photon
measures almost nothing that has any uncertainty. Here "uncertainty"
means either in the narrow sense of being in a superposition, or in
the ultimately more general sense of being different from one photon
to the next. To say that the photon is measuring something is of
course equivalent to saying that its state is importing entropy.
In the visible wavelength range, you would need an astronomically high
index of refraction or a very unstable medium of propagation to invalidate
these approximations. (E.g. the two-slit experiment might not work well
in an erratic gale wind.) The two-slit experiment does of course take
a different form in the X-ray range.
It may be useful to think of the photon's evolution as a quantum
operation. For the above reasons (I think), the operation is
extremely close to sub-unitary, which is to say a single Kraus term.
--
/\ Greg Kuperberg (UC Davis)
/ \
\ / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/
\/ * All the math that's fit to e-print *
Uncle Al
>
> In article <4124b132$1...@news.sentex.net>,
> "Urs Schreiber" <Urs.Sc...@uni-essen.de> writes:
> |> > In article <41237F64...@hate.spam.net>,
> |> > Uncle Al <Uncl...@hate.spam.net> wrote:
> |>
> |> > >http://www.quantum.univie.ac.at/research/matterwave/c60/
> |>
> |> > I don't see that your Web reference is relevant, either.
> |>
> |> Uncle Al is trying to point out that even when you shoot a big fat object
> |> like a buckminster fullerene on a screen with two slits, it still gives an
> |> interference pattern. The decohering effects on the buckyball must be orders
> |> of magnitude larger than for the double slit experiment with a photon - but
> |> the interference pattern is still there.
>
> Well, yes, that is obvious.
>
> |> This is pretty amazing, given that other people are using their scanning
> |> tunneling microscopes to play classical (though minature) soccer with
> |> buckyballs.
>
> Not really. I should be amazed if it DIDN'T happen! It would be
> flabberghasting if wave/particle duality applied up to a certain
> size of object, and stopped doing so thereafter. I am surprised
> that the wave aspect is observable, but that is a different point.
>
> |> On the other hand, it does not completely answer the original question,
> |> which was as asking how it can be understood that the decoherence effects in
> |> the 2-slit experiment (under ordinary conditions, at least) are not large
> |> enough to destroy the interference pattern. ...
>
> It doesn't even ADDRESS the original question!
>
> If you read it, you will see:
>
> photon within air is it's constantly absorbed and reemitted by the
>
> And a lot of related points, which I have snipped. It is very clear
> that the question has very little to do with the slit and a great
> deal to do with the preservation of coherence through absorption and
> reemission by air molecules. And that I did find amazing!
>
> assuming doesn't actually happen (as I hypothesised), which explains
> the situation.
>
> medium (using his notation), then I will indeed be amazed.
>
> Regards,
> Nick Maclaren.
Why wouldn't a double slit experiment work in a translucent
medium? You would lose some intensity to scattering. So? When
any photon arrives at the double slit it behaves as expected.
Each slit acts as a source of radiation and they are together
coherent.
If you illuminated the slit with a frosted pane rather than a
point source of light, would you expect a diffaction pattern on
the other side? If not a diffraction pattern, what then? It
would not be sharp, but it would be there.
Jesse Mazer
>"Nick Maclaren" <nm...@cus.cam.ac.uk> schrieb im Newsbeitrag
>news:cg0e88$ket$1...@pegasus.csx.cam.ac.uk...
>
>
>>In article <41237F64...@hate.spam.net>,
>>Uncle Al <Uncl...@hate.spam.net> wrote:
>>
>>
>
>
>
>>>http://www.quantum.univie.ac.at/research/matterwave/c60/
>>>
>>>
>
>
>
>>I don't see that your Web reference is relevant, either.
>>
>>
>
>Uncle Al is trying to point out that even when you shoot a big fat object
>like a buckminster fullerene on a screen with two slits, it still gives an
>interference pattern. The decohering effects on the buckyball must be orders
>of magnitude larger than for the double slit experiment with a photon - but
>the interference pattern is still there.
>
But is the buckyball experiment done in open air, or in a vacuum like
with the electron version of the double-slit experiment? I thought I
read somewhere that if you try to do the electron double-slit experiment
in open air, you won't get interference on the screen regardless of
whether you measure which slit each electron went through, presumably
because of decoherence.
Jesse
Charles J. Quarra
> In article <4124b132$1...@news.sentex.net>,
> "Urs Schreiber" <Urs.Sc...@uni-essen.de> writes:
> |> "Nick Maclaren" <nm...@cus.cam.ac.uk> schrieb im Newsbeitrag
> |> news:cg0e88$ket$1...@pegasus.csx.cam.ac.uk...
> |> > In article <41237F64...@hate.spam.net>,
> |> > Uncle Al <Uncl...@hate.spam.net> wrote:
> |> This is pretty amazing, given that other people are using their scanning
> |> tunneling microscopes to play classical (though minature) soccer with
> |> buckyballs.
>
> Not really. I should be amazed if it DIDN'T happen! It would be
> flabberghasting if wave/particle duality applied up to a certain
> size of object, and stopped doing so thereafter. I am surprised
> that the wave aspect is observable, but that is a different point.
Well, this is debatable
As i commented in other post, in introductory quantum mechanics books
the relation between wavelength-momentum is used uncospicuously for
macroscopic
objects, probably for didactic purposes, however i see a problem with
that extrapolation: if one would go on and adding the momenta of the
composite objects to obtain a "body-momenta", then one should ask if
why that cant extrapolate to, for example, light beams (as the one you
got on a laser). If there is a system in which one can fulfill the
condition of coherent constructive propagation is in the laser light.
A laser pulse has a lot more total momenta than its constitutive
photons (each produced by an individual atom line) however one doesnt
see this "total momenta wavelength" at work in the physics of these
systems, at least nothing im aware.
In principle i dont see (as of yet) a compelling reason as to why
expect the net momentum of a system of particles to produce an
effective wavelength. I havent seen yet a theorem that proves that for
a N-particle wavefunction one can approximate it to something like a
quanta of net-momenta (and its corresponding wavelength) plus other
degrees of freedom. On the other hand, small composite objects like
atoms have shown to show interference during the right experimental
conditions. But we have no theory about how scales down this effect
with the particle number.
This is particularly relevant to people working on double special
relativity, were the planck wavelength is supposedly the
lowest/highest possible wavelength/momenta any _body_, being
fundamental or composite, can achieve
alistair
> On the other hand, the immense decoherence induced by all these effects is
> indeed being seen in the experiment. Namely, last I checked at least,
> Zeilinger's team can only produce the first interference minimum in their
> experiment. The more delicate higher order minima are washed away by
> decoherence. Hence, roughly, the decoherence that a "hot" buckyball is
> subject to is getting close to the critical value where the interference pattern
> would disappear altogether.
Alistair:
Suppose a cloud of 60 unbonded carbon atoms is fired simulataneously
at the slits and the cloud has photons trapped in it corresponding to
the energy a "hot" buckyball would radiate.We would expect an
interference pattern to form
because the photons ,if detected, cannot tell us which individual
carbon atom travelled through which slit.My point is this:
Decoherence probably doesn't happen for a "hot" buckyball because
it must be behaving like the cloud of unbonded carbon atoms - we have
no way of knowing which carbon atoms in the ball radiate an individual
photon. And if a radioactive carbon 14 atom was introduced into each
buckyball we could in principle by using a detector at one of the
slits find out which atom emitted an alpha particle and we would
expect the interference pattern to disappear altogether.
DrChinese
a) This experiment has been performed in a vacuum and the results are
the same. There is interference.
b) When you see a photon go in and another come out, there is no such
thing as specifically saying what happened in between unless you have
specific knowledge by observation.
c) If you did more progressively observation, then as you suspected,
there would be decoherence and the interference effect would eventually
disappear. This would still always obey the Heisenberg Uncertainty
Principle.
d) Any medium through which light travels carries a chance of an
absorption. In a transparent medium,this chance is very very low. As
absorption increases in likelihood, decoherence occurs.
In other words, there is a relationship between transparency,
decoherence and interference.
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alistair
>if the photon went through one slit it would get entangled
>with the air molecules around that slit which are different from the
>air molecules around the other slit.
Alistair:
The air molecules around the apparatus can pass through
either slit and are in a superposition themselves.
I haven't heard people here say what effect
superposed air molecules can have on superposed
photons.If the slits were made narrow enough to stop air
getting through would the interference pattern change?
Greg Kuperberg
Greg Kuperberg <gr...@conifold.math.ucdavis.edu> wrote:
>I'm not entirely sure that this is a good explanation, but here is what I
>think is the point. Because the photon has low energy, it measures very
>little about the positions of the molecular dipoles that refract it,
>much less their internal state, only with their presence or absence.
>And because the photon has a long wavelength, it only measures the
>aggregate presence of many molecules in a large volume.
Let me add that wavelength is not the only reason that a photon trajectory
measures (or entangles with) only the aggregate refractive material in
a relatively large volume. The relevant region is a tube around the
photon path. The radius of the tube is on the order of the wavelength
(I think). But the length of the tube is the full length of the path.
The arriving quantum phase of the photon does not depend on whether it
meets a slow (i.e., highly refractive) glob at the beginning, middle, or end
of the trajectory, only whether or not it meets the glob at all.
Definitely the photon measures, and hence entangles with, refraction in
the two-slit experiment. If you configure the laser as a gun that shoots
photons sequentially, and if between shots you insert glass plates of
varying composition and thickness, then the aggregate statistics will
be poor - there will be no interference pattern. However, it is not so
hard to control what the photon measures so as to introduce very little
entropy or entanglement.
Nick Maclaren
Uncle Al <Uncl...@hate.spam.net> writes:
|>
|> Why wouldn't a double slit experiment work in a translucent
|> medium? You would lose some intensity to scattering. So? When
|> any photon arrives at the double slit it behaves as expected.
|> Each slit acts as a source of radiation and they are together
|> coherent.
|>
|> If you illuminated the slit with a frosted pane rather than a
|> point source of light, would you expect a diffaction pattern on
|> the other side? If not a diffraction pattern, what then? It
|> would not be sharp, but it would be there.
I was assuming the translucent medium being between both the
illumination and the double slit and the double slit and the
'focus'.
Regards,
Nick Maclaren.
Nick Maclaren
clarification of something I posted but was moderated out as
incomprehensible :-)
All this sounds eminently testable and, if it has not been done,
it should be. Let us consider the following sequence of tests:
The double slit experiment in air
The double slit experiment in water
The double slit experiment in water, flowing in various
directions (laminar flow)
The double slit experiment in water, flowing in various
directions (turbulent flow)
The double slit experiment where the (weak) light beam
crosses a laser beam, of the same and different frequencies
I don't know which, if any, will show anomalous effects. Even
if theory does predict that, checking theory against practice
is always good. And the above would be a very reasonable series
of experiments to assign to a suitable student as a practical.
So, have those experiments ACTUALLY been done? And what did
they show?
Regards,
Nick Maclaren.
Joe Rongen
From: "Nick Maclaren" <nm...@cus.cam.ac.uk>
Sent: Wednesday, August 25, 2004 3:42 AM
Yes, some people have actually entertained themselves
already with those ideas and more....
"The Feynman Double Slit
Here we discuss one of the two major paradoxes that we use to
introduce Quantum Mechanics. It is the double slit experiment for
bullets, water waves and electrons. Although many people have
experimented with the systems to be discussed and written about
them, Richard Feynman's treatment is so clear that physicists often
call it the "Feynman" double slit."
Let the amusement start here(this should be one line):
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/DoubleSlit/DoubleSli
t.html
Regards Joe
"To us, probability is the very guide of life" -- Bishop Butler.
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Peter Shor
But it does. Two (and three!) entangled photons can be created that
behave (sort of) like they were a single particle with a wavelength
having half (a third!) the wavelength of the individual photons. I
just googled for a reference, and came up with this article from
Laser Focus World
http://lfw.pennnet.com/Articles/Article_Display.cfm?ARTICLE_ID=208663
(and learned for the first time that the three-photon experiment has
been performed. Amazing.)
> I havent seen yet a theorem that proves that for
> a N-particle wavefunction one can approximate it to something like a
> quanta of net-momenta (and its corresponding wavelength) plus other
> degrees of freedom. On the other hand, small composite objects like
> atoms have shown to show interference during the right experimental
> conditions. But we have no theory about how scales down this effect
> with the particle number.
The theory exists (it's standard quantum mechanics), but you have to
examine the details of the experiment carefully to predict how things
will behave.
Peter Shor
Nick Maclaren
In article <002101c48b0c$87cacda0$2723fea9@research>,
"Joe Rongen" <j...@alpha.to> writes:
|>
|> > All this sounds eminently testable and, if it has not been done,
|> > it should be. Let us consider the following sequence of tests:
|> >
|> >
|> > So, have those experiments ACTUALLY been done? And what did
|> > they show?
I would be flabberghasted if they hadn't. We know that the
experiment has been done (often) in air and (as you say) in hot
air over beer, but has it been done under the circumstances I
mentioned (and others)?
If not, WHY not?
While the chances of spotting anything interesting and new are
low, students are cheap, this is a cheap experiment and the
publication potential of ANY anomaly is high. Even a fairly
boring result would be publishable, if written up right.
Regards,
Nick Maclaren.
Joe Rongen
"Peter Shor" <peter...@aol.com> wrote in message
news:9b2e17b4.04082...@posting.google.com
[snip]
> But it does. Two (and three!) entangled photons can be created that
> behave (sort of) like they were a single particle with a wavelength
> having half (a third!) the wavelength of the individual photons. I
> just googled for a reference, and came up with this article from
> Laser Focus World
>
> http://lfw.pennnet.com/Articles/Article_Display.cfm?ARTICLE_ID=208663
>
> (and learned for the first time that the three-photon experiment has
> been performed. Amazing.)
>
> > I havent seen yet a theorem that proves that for
> > a N-particle wavefunction one can approximate it to something like a
> > quanta of net-momenta (and its corresponding wavelength) plus other
> > degrees of freedom. On the other hand, small composite objects like
> > atoms have shown to show interference during the right experimental
> > conditions. But we have no theory about how scales down this effect
> > with the particle number.
>
> The theory exists (it's standard quantum mechanics), but you have to
> examine the details of the experiment carefully to predict how things
> will behave.
In this August issue of "Photonics" (vol 36, issue 8) on page 16
there is a small article about "Five photon Entanglement could
benefit Quantum Communications"
"The setup which is described in the July 1 issue of "Nature"
uses two Einstein-Podolsky-Rosen entangled photon sources,
each emitting one polarized entangled photon pair, to generate
a four-photon entangled state. This combined with a single-photon
state makes it possible to observe five-photon entanglements as long
as only one photon is detected in each of the five output modes."
This research is at the University of Innsbruck, Heidelberg
University and the University of Science and Technology in China.
Regards Joe
scerir
> While the chances of spotting anything interesting and new are
> low, students are cheap, this is a cheap experiment and the
> publication potential of ANY anomaly is high. Even a fairly
> boring result would be publishable, if written up right.
A naive consideration, by a non-expert.
If I remember well there are very long interferometers,
like some old good Jamin (also 10 meter long), which is
different from the Young two-slit, of course.
As far as I remember experiments with air, other gas,
etc. have been performed, using Jamin interferometers.
Anton Zelinger Anton Zeilinger wrote (Rev.Mod.Phys., p.S-288,
1999): "The superposition of amplitudes is only valid if there
is no way to know, even in principle, which path the particle
took. It is important to realize that this does not imply
that an observer actually takes note of what happens.
It is sufficient to destroy the interference pattern,
if the path information is accessible in principle from
the experiment or even if it is dispersed in the environment
and beyond any technical possibility to be recovered, but
in principle "still out there".
Now, it seems to me that the indistinguishability principle,
as stated above, would be meaningless if there is an actual
interaction between the particle (say a photon) and the air.
s.
Joe Rongen
[snip]
> But it does. Two (and three!) entangled photons can be created that
> behave (sort of) like they were a single particle with a wavelength
> having half (a third!) the wavelength of the individual photons. I
> just googled for a reference, and came up with this article from
> Laser Focus World
>
> http://lfw.pennnet.com/Articles/Article_Display.cfm?ARTICLE_ID=208663
>
> (and learned for the first time that the three-photon experiment has
> been performed. Amazing.)
>
> > I havent seen yet a theorem that proves that for
> > a N-particle wavefunction one can approximate it to something like a
> > quanta of net-momenta (and its corresponding wavelength) plus other
> > degrees of freedom. On the other hand, small composite objects like
> > atoms have shown to show interference during the right experimental
> > conditions. But we have no theory about how scales down this effect
> > with the particle number.
>
> The theory exists (it's standard quantum mechanics), but you have to
> examine the details of the experiment carefully to predict how things
> will behave.
Nick Maclaren
In article <XKiXc.57156$1V3.1...@twister2.libero.it>,
scerir <sce...@libero.it> wrote:
>"Nick Maclaren"
>
>> While the chances of spotting anything interesting and new are
>> low, students are cheap, this is a cheap experiment and the
>> publication potential of ANY anomaly is high. Even a fairly
>> boring result would be publishable, if written up right.
>
>A naive consideration, by a non-expert.
Yes, I am a non-expert. Would you care to explain the naive?
>If I remember well there are very long interferometers,
>like some old good Jamin (also 10 meter long), which is
>different from the Young two-slit, of course.
Hmm. At a crude approximation, 10 metres of air might be similar
to 1 cm of water.
>As far as I remember experiments with air, other gas,
>etc. have been performed, using Jamin interferometers.
That is essentially what I was asking! And saying that, if it
has NOT been done, it SHOULD be done.
>Anton Zelinger Anton Zeilinger wrote (Rev.Mod.Phys., p.S-288,
>1999): "The superposition of amplitudes is only valid if there
>is no way to know, even in principle, which path the particle
>took. It is important to realize that this does not imply
>that an observer actually takes note of what happens.
Well, excluding the slightly dubious nature of the last statement
(i.e. there are differing views), yes.
>It is sufficient to destroy the interference pattern,
>if the path information is accessible in principle from
>the experiment or even if it is dispersed in the environment
>and beyond any technical possibility to be recovered, but
>in principle "still out there".
Again, yes.
>Now, it seems to me that the indistinguishability principle,
>as stated above, would be meaningless if there is an actual
>interaction between the particle (say a photon) and the air.
You are still thinking deterministically, and have just disproved
your own case!
A light beam through air does scatter slightly, and is therefore
detectable, not just in theory but in practice.
Let's use a simple probabilistic model (which is almost certainly
too naive). If there were a 99% probability of any particular
photon NOT interacting, the target pattern would be 99% as bright
as if there were no interaction. And there would be enough
scattering to detect.
You could rephrase the question that I (and, according to you and
others, many other people) have asked, as being equivalent to
saying "Exactly how does the probability of interacting with the
medium (not a good word in the laser case) vary with the conditions?"
And, as with most such questions, the information obtained from an
experiment is in direct proportion to how unexpected the result is.
Unless the experiment gets the wrong answer, of course :-)
Regards,
Nick Maclaren.
ksh95
very_c...@hotmail.com (Very cryptic) wrote in message
> But the catch is, the photon
> gets scattered off a huge number of air molecules along its path.
> Isn't that enough to cause decoherence?
So I guess the question boils down to: Why doesn't the photon
decohere?
The easiest way to think about this is classically. Think of a
hydrogen atom in its spherically symmetric S state. At some instant in
time a static electric field is present. The electric field breaks the
spherical symmetry of the atom and a new ground state is created. This
new ground state looks like a dipole (to first order) so the atom now
has an external field. If this static electric field is actually a
plane wave we can imagine the electric field strength adiabatically
changing, all the while adiabatically changing the dipole moment of
the atom and hence changing the external field of the atom. The
external field is always in phase with the oscillating dipole so
everything remains coherent. Now extend this cartoon picture to
avogadros number of atoms and we have a big coherent dance. The
frequency of this big collective state will be the frequency of the
original planewave.
Be warned that this picture is so oversimplified it's almost wrong,
but I think it catches the jist of whats going on.
Charles J. Quarra
> disposablemail...@yahoo.com.ar (Charles J. Quarra) wrote in message news:<bc979c06.04081...@posting.google.com>...
>
> > a N-particle wavefunction one can approximate it to something like a
> > quanta of net-momenta (and its corresponding wavelength) plus other
> > degrees of freedom. On the other hand, small composite objects like
> > atoms have shown to show interference during the right experimental
> > conditions. But we have no theory about how scales down this effect
> > with the particle number.
>
> The theory exists (it's standard quantum mechanics), but you have to
> examine the details of the experiment carefully to predict how things
> will behave.
>
I was aware about down-converter crystals, but i forgot about them
when i posted. Obviously they look like an example where the
net-momenta is the main transition amplitude resonance. However still
remains the problem of this behaviour is scalable to greater N. If the
answer happens to be yes, the consequences for particle physics are
enormous. That would allow the possibility of engineering quantum
accelerators of composite systems that could favour reactions of
energies of the order of the total energy of the composite object.
Franz Heymann
news:cgkac8$brc$1...@pegasus.csx.cam.ac.uk...
The Rayleigh refractometer, sometimes filled with static or flowing
water disposed of most of your worries more than a century ago.
Franz
scerir
"Nick Maclaren"
> >A naive consideration, by a non-expert.
> Yes, I am a non-expert.
> Would you care to explain the naive?
No, the naive non-expert was me, not you.
> You could rephrase the question that I (and, according to you and
> others, many other people) have asked, as being equivalent to
> saying "Exactly how does the probability of interacting with the
> medium (not a good word in the laser case) vary with the conditions?"
In general the probability of interacting with
something does not destroy the interference pattern,
since there is still uncertainty about the "wich slit".
A naive experiment would be: air behind slit one,
water behind slit two!
s.
Franz Heymann
news:XwIXc.68017$OH4.1...@twister1.libero.it...
The Rayleigh refractometer splits a light beam into two paths, one of
which goes through a vacuum and the other through a gas whose
refractive index is being determined. The two beams are brought
together and allowed to produce an interference pattern. The
positions of the fringes are related to the refractive index of the
test gas.
This refractometer has also been used with one of the two light paths
immersed in water.
Franz
>
> s.
Franz Heymann
news:cgq6os$d1a$1...@hercules.btinternet.com...
Oops. Both light paths were filled with water, flowing in opposite
directions
Franz
Oz
>Now, it seems to me that the indistinguishability principle,
>as stated above, would be meaningless if there is an actual
>interaction between the particle (say a photon) and the air.
Well, of course there must be, or else we would not observe air as
having a different refractive index.
However, as one of (probably millions) who has actually done diffraction
experiments in air I don't quite see the logic. So long as the 'photon'
excites the air in both slits you still don't know which one it went
through. In point of fact it went through both so no problem arises.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use o...@farmeroz.port995.com<<
ozac...@despammed.com still functions.
Oz
Peter Shor <peter...@aol.com> writes
>air is slightly (0.03%) less than the speed of light in a vacuum. This
>means that even when the photons aren't scattering, they're still
>interacting with the air. But this interaction doesn't seem to affect
>still pretty transparent? Why? Anyone have an answer?
I don't (any longer) believe in photons as existing in a beam of light.
I think its just a maxwellian field (ok with spin), the quantum
behaviour arises because all detectors are quantised (typically
requiring jumping an energy gap).
So there never is any question of 'which slit did the photon go through'
because as a field, it can and will go through both.
That said this isn't actually very much help because all emitters and
all detectors are quantised, you never see light in any other way. So
what you are talking about when you want to understand light is really
the behaviour of an emitter and a detector. Understanding how one (or
more) complex quantised emitters affect another (or more) complex
quantised detectors is a whole different ball game. For that you need
some to very heavyweight theory and mathematics. Pretty scary very
often, too.
Because typically emitters and detectors are not very simple things.
In fact physicists delight in dreaming up ever more devious, complex and
useful ones just to make light hugely annoyed and pissed off with them.
Oz
scerir <sce...@libero.it> writes
>A naive experiment would be: air behind slit one,
>water behind slit two!
Joe Rongen
"Oz" <o...@farmeroz.port995.com> wrote in message
news:rxZV0REt...@farmeroz.port995.com...
[snip]
> My answer would be considered as cranky.
>
> I don't (any longer) believe in photons as existing in a beam of light.
> I think its just a maxwellian field (ok with spin), the quantum
> behaviour arises because all detectors are quantised (typically
> requiring jumping an energy gap).
One may have to rethink that, according to JPL-Caltech.
They have demonstrated a charge-coupled detector based
on superconducting technology that can detect individual
photons AND identify their wavelength. (Most digital cameras
do not detect the wavelength, so red, blue and green filters
are added and that in turn reduces the sensitivity.) For more
see Jan 2004 issue of "Scientific American" page 24.
Regards Joe
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.762 / Virus Database: 510 - Release Date: 9/13/04
scerir
"Oz"
> I don't (any longer) believe in photons as existing in a beam of light.
> I think its just a maxwellian field (ok with spin), the quantum
> behaviour arises because all detectors are quantised (typically
> requiring jumping an energy gap). So there never is any question of
> 'which slit did the photon go through' because as a field, it can
> and will go through both. That said this isn't actually very much
> help because all emitters and all detectors are quantised, you never
> see light in any other way.
All detectors are quantized. And N.Bohr said: "The hypothesis of
light-quanta is not able to throw light on the nature of radiation."
(in the 1922 Nobel Lecture, at that time Bohr was 'conservative'!).
But how to explain that *just* one detector 'clicks' and the other
don't? See, J.J.Thorn et al., Am.J.Phys., 72, p.1210, 2004.
[Abstract]
While the classical, wavelike behavior of light (interference
and diffraction) has been easily observed in undergraduate
laboratories for many years, explicit observation of the quantum
nature of light (i.e., photons) is much more difficult.
For example, while well-known phenomena such as the photoelectric
effect, and Compton scattering, strongly suggest the existence of
photons, they are not *definitive* proof of their existence.
[Leonard Mandel wrote the same, so: no doubt about it!]
Here we present an experiment, suitable for an undergraduate
laboratory, that unequivocally demonstrates the quantum nature of light.
Spontaneously downconverted light is incident on a *beamsplitter* and
the outputs are monitored with single-photon counting detectors
[2 detectors].
We observe a near absence of *coincidence* counts between the two
detectors - a result inconsistent with a classical wave model of light,
but consistent with a quantum [particle] description in which individual
photons are incident on the beamsplitter. More explicitly, we measured
the degree of second-order coherence between the outputs to be
g(2)(0) = 0.0177Ä…0.0026
which violates the classical inequality
g(2)(0)>=1
by 377 standard deviations.
[End of Abstract]
s.
Nick Maclaren
In article <igLHQtDk...@farmeroz.port995.com>,
Oz <o...@farmeroz.port995.com> writes:
|> scerir <sce...@libero.it> writes
|>
|> >A naive experiment would be: air behind slit one,
|> >water behind slit two!
|>
|> How is this to tell you which slit the 'photon' went through?
Time. The difference between water and air is of the order of
0.74 nS/metre. Modern computers can measure times of less than
that. You might need a previous split to get a calibration,
which might make the measurement rate rather low, but that is
not a fundamental problem.
Regards,
Nick Maclaren.
Oz
>All detectors are quantized. And N.Bohr said: "The hypothesis of
>light-quanta is not able to throw light on the nature of radiation."
>(in the 1922 Nobel Lecture, at that time Bohr was 'conservative'!).
>But how to explain that *just* one detector 'clicks' and the other
>don't? See, J.J.Thorn et al., Am.J.Phys., 72, p.1210, 2004.
> [Abstract]
>While the classical, wavelike behavior of light (interference
>and diffraction) has been easily observed in undergraduate
>laboratories for many years, explicit observation of the quantum
>nature of light (i.e., photons) is much more difficult.
Its really just a case of conservation of energy.
If the emitter is quantised it emits one quantum of radiation.
Then by definition a detector can only detect a quantum of radiation.
The precise details are called QM, or would be if there was an agreed
mechanism for 'the collapse of the wavefunction'. I have my own view on
that as you know.
>For example, while well-known phenomena such as the photoelectric
>effect, and Compton scattering, strongly suggest the existence of
>photons, they are not *definitive* proof of their existence.
>[Leonard Mandel wrote the same, so: no doubt about it!]
>Here we present an experiment, suitable for an undergraduate
>laboratory, that unequivocally demonstrates the quantum nature of light.
>Spontaneously downconverted light is incident on a *beamsplitter* and
>the outputs are monitored with single-photon counting detectors
>[2 detectors].
>We observe a near absence of *coincidence* counts between the two
>detectors - a result inconsistent with a classical wave model of light,
>but consistent with a quantum [particle] description in which individual
>photons are incident on the beamsplitter.=20
I'm not really clear about the details of this particular experiment.
You will note that my (current) worldview puts great importance on the
quantum nature of BOTH the emitter and the detector, so the precise
details are vital.
>More explicitly, we measured
>the degree of second-order coherence between the outputs to be
> g(2)(0) =3D 0.0177=B10.0026
>which violates the classical inequality
> g(2)(0)>=3D1
>by 377 standard deviations.
I would need the relevance of this explaining to me together with the
details of the experiment. I am happy (excited even) to be shown to be
wrong.
--=20
scerir
"Oz"
> > A naive experiment would be: air behind slit one,
> > water behind slit two!
> How is this to tell you which slit
> the 'photon' went through?
Assuming that the speed in air is different
from the speed in water ?
s.
"A quantum possibility is more real
than a classical possibility,
but less real than a classical reality."
- Boris Tsilerson
Oz
Joe Rongen <j...@alpha.to> writes
>One may have to rethink that, according to JPL-Caltech.
>They have demonstrated a charge-coupled detector based
>on superconducting technology that can detect individual
>photons AND identify their wavelength.
So can a leaf.
--
Oz
Nick Maclaren <nm...@cus.cam.ac.uk> writes
Of course this begs the question.
How do you know when a photon goes through the slits or leaves the
detector in order to get the flight time?
and that's before you even start on all the other factors.
John Gonsowski
> My answer would be considered as cranky.
>
> I don't (any longer) believe in photons as existing in a beam of light.
> I think its just a maxwellian field (ok with spin), the quantum
> behaviour arises because all detectors are quantised (typically
> requiring jumping an energy gap).
>
> So there never is any question of 'which slit did the photon go through'
> because as a field, it can and will go through both.
A field is math, math can't go through slits, don't you need some kind
of physical description for what you mean by field? I personally like
a particular many worlds view of a field where there are mutliple
world lines of particle photons going through both slits until the
environment causes a collapse and then you have only one of the
worldlines chosen.
Another question is where the photon actually is, given what Urs
Schreiber said a few months ago:
"> So a little circle doing the waltz was a photon,
This depends on where you get the U(1) gauge group from. It might come
by the Kaluza-Klein mechanism from a column of the metric tensor. In
this case the photon would indeed be a state of the closed string,
namely that where one half of the oscillators on the worldsheet is
polarized along the compact direction while the other half carries an
index in the observable dimensions."
I thinks this implies that a photon is to a compact dimension what a
graviton is to an observable dimension. So in effect the photon would
not be in the beam of light directly just indirectly via some indexing
scheme?
John
carlip...@physics.ucdavis.edu
Oz <o...@farmeroz.port995.com> wrote:
> I'm not really clear about the details of this particular experiment.
> You will note that my (current) worldview puts great importance on the
> quantum nature of BOTH the emitter and the detector, so the precise
> details are vital.
The original version of this experiment was by Grangier et al.,
Europhys. Lett. 1 (1986) 173. The idea is beautiful:
Start with a process that should, quantum mechanically, produce
a single photon. Grangier et al. used an atomic cascade, in
which an atom emits two photons, one immediately after the other,
and used the first photon as a "trigger" to open a shutter for
the second. Other versions use parametric down conversion, in
which a nonlinear optical process "splits" a photon into a pair,
one of which can be used as a trigger.
Now, send your photon through an ordinary beam splitter. If the
propagating electromagnetic field acts as a wave and not a particle,
the beam splitter should divide it in half. Set up two detectors,
one for each output of the beam splitter, and see if you ever see
simultaneous signals. You don't -- the photon always comes out one
way or the other. This is pretty much the definition of a "particle"
-- an object that can't be split by a device that does split a wave.
(If you haven't thought this through, you might be tempted to say
that the field *is* split, but that it then doesn't have the energy
to trigger both detectors. Don't go there... If the light coming
out path A doesn't have enough energy to trigger detector A, then
the light coming out path B shouldn't have the energy to trigger
detector B. You might try saying that a "half a photon" has enough
energy to trigger a detector, but only half the time. But if that
were the case, then a quarter of the time both detectors would be
triggered. Unless you want to postulate that one detector can signal
the other to say "I've got this photon, so you stop looking," you're
not going to argue your way out of the results this way.)
You can make the experiment even more vivid by adding a second step.
Repeat the experiment, but replace your two detectors with two mirrors
that recombine the output of the beam splitter. Send in one photon
at a time, and look for interference. You'll see it. So even though
your individual detectors showed that the photon went one way or the
other, when you combine the two directions you get the interference
pattern you'd expect if it had gone both ways. In particular, this
means that you can't attribute the earlier result to some property of
the beam splitter -- you can't say, "The propagating light is a wave,
but the beam splitter doesn't really split it, but directs it only one
way or the other,'' since if that were the case, you wouldn't get the
interference pattern.
Quantum mechanics really is weirder than you think.
Steve Carlip
Thomas Trotter
carlip...@physics.ucdavis.edu wrote in message news:<cicjol$koi$1...@skeeter.ucdavis.edu>...
> Oz <o...@farmeroz.port995.com> wrote:
>
> > I'm not really clear about the details of this particular experiment.
> > You will note that my (current) worldview puts great importance on the
> > quantum nature of BOTH the emitter and the detector, so the precise
> > details are vital.
>
> The original version of this experiment was by Grangier et al.,
> Europhys. Lett. 1 (1986) 173. The idea is beautiful:
>
> Start with a process that should, quantum mechanically, produce
> a single photon. Grangier et al. used an atomic cascade, in
> which an atom emits two photons, one immediately after the other,
> and used the first photon as a "trigger" to open a shutter for
> the second. Other versions use parametric down conversion, in
> which a nonlinear optical process "splits" a photon into a pair,
> one of which can be used as a trigger.
>
> Now, send your photon through an ordinary beam splitter. If the
> propagating electromagnetic field acts as a wave and not a particle,
> the beam splitter should divide it in half.
Why? Couldn't I also make the assumption that the
beamsplitter will divide the wave unevenly?
> Set up two detectors, one for each output of the beam
> splitter, and see if you ever see simultaneous signals.
> You don't -- the photon always comes out one way or the
> other. This is pretty much the definition of a "particle"
> -- an object that can't be split by a device that does
> split a wave.
This is one way of talking about it. Another way is to
say that either one detector or the other registers a
detection during any particular coincidence window. The
'particle' is the detector registration, not necessarily
something that "can't be split".
>
> (If you haven't thought this through, you might be tempted to say
> that the field *is* split, but that it then doesn't have the energy
> to trigger both detectors.
If the field is divided unevenly, even ever so slightly,
then either one detector or the other will register.
> Don't go there... If the light coming out path A doesn't
> have enough energy to trigger detector A, then
> the light coming out path B shouldn't have the energy
> to trigger detector B.
Only if the beamsplitter is dividing the wave *exactly*.
[snip
> You can make the experiment even more vivid by adding a second step.
> Repeat the experiment, but replace your two detectors with two mirrors
> that recombine the output of the beam splitter. Send in one photon
> at a time, and look for interference. You'll see it. So even though
> your individual detectors showed that the photon went one way or the
> other, when you combine the two directions you get the interference
> pattern you'd expect if it had gone both ways. In particular, this
> means that you can't attribute the earlier result to some property of
> the beam splitter -- you can't say, "The propagating light is a wave,
> but the beam splitter doesn't really split it, but directs it only one
> way or the other,'' since if that were the case, you wouldn't get the
> interference pattern.
But you can say that the beamsplitter divided the wave unevenly.
In wbich case you'll get the interference pattern on recombination.
>
> Quantum mechanics really is weirder than you think.
Maybe. But maybe it's not as weird as we might sometimes
want it to be. :-)
Arun Gupta
carlip...@physics.ucdavis.edu wrote in message
> Quantum mechanics really is weirder than you think.
Can we reverse this and say that the classical world is weirder than we think.
The Quantum Mechanical wave function would continue on its merry unitary
way but for such objects. The beam-splitter and the photon detector are both
classical objects, but one is an aid in the delocalization of the photon and the
other forces a localization or collapse of the wave function. How does one
apriori know whether a classical object will have one or the other (or both?)
properties?
-Arun
alistair
> Repeat the experiment, but replace your two detectors with two mirrors
> that recombine the output of the beam splitter. Send in one photon
> at a time, and look for interference. You'll see it. So even though
> your individual detectors showed that the photon went one way or the
> other, when you combine the two directions you get the interference
> pattern you'd expect if it had gone both ways. In particular, this
> means that you can't attribute the earlier result to some property of
> the beam splitter -- you can't say, "The propagating light is a wave,
> but the beam splitter doesn't really split it, but directs it only one
> way or the other,'' since if that were the case, you wouldn't get the
> interference pattern.
In David Bohm's version of quantum mechanics, the pilot wave would
physically guide photons to form the interference pattern.When the
detectors destroy the interference pattern, the pilot wave, on
average, is just guiding photon particles along different
trajectories, compared to the trajectories they had in the case of the
mirrors.The pilot wave is physically altered by the experimental
setup.
Oz
John Gonsowski <jcgon...@yahoo.com> writes
>
>> My answer would be considered as cranky.
>>
>> I don't (any longer) believe in photons as existing in a beam of light.
>> I think its just a maxwellian field (ok with spin), the quantum
>> behaviour arises because all detectors are quantised (typically
>> requiring jumping an energy gap).
>>
>> So there never is any question of 'which slit did the photon go through'
>> because as a field, it can and will go through both.
>
>A field is math, math can't go through slits, don't you need some kind
>of physical description for what you mean by field?
Well, I tend to use the olde fashioned meaning of a field.
I suppose I could witter on about some excitation of the vacuum.
I hate to tell you that currently I think everything (including massive
particles) are fields. Its so much simpler that way.
>Another question is where the photon actually is, given what Urs
>Schreiber said a few months ago:
>
>"> So a little circle doing the waltz was a photon,
>This depends on where you get the U(1) gauge group from. It might come
>by the Kaluza-Klein mechanism from a column of the metric tensor. In
>this case the photon would indeed be a state of the closed string,
>namely that where one half of the oscillators on the worldsheet is
>polarized along the compact direction while the other half carries an
>index in the observable dimensions."
>
>I thinks this implies that a photon is to a compact dimension what a
>graviton is to an observable dimension. So in effect the photon would
>not be in the beam of light directly just indirectly via some indexing
>scheme?
Dunno, its all way beyond me. In my viewpoint there isn't really such a
thing as a photon, just a field. This field can be short (covering a few
fm) or long (km), high intensity or low intensity, and interacts with
charged particles just as you would expect such a field to interact.
Oz
carlip...@physics.ucdavis.edu writes
>(If you haven't thought this through, you might be tempted to say
>that the field *is* split, but that it then doesn't have the energy
>to trigger both detectors.
Absolutely. It doesn't have the energy to trigger two, but it does have
enough to trigger one. Ergo only one will be triggered, randomly.
>Don't go there... If the light coming
>out path A doesn't have enough energy to trigger detector A, then
>the light coming out path B shouldn't have the energy to trigger
>detector B.
But you are forgetting that the system photon-detector#1-detector#2 is
entangled. The end result is either
detector#2triggered+detector#1not or
detector#1triggered+detector#2not
As soon as entanglement is broken then you get one or the other state
ONLY. Its like any other entanglement experiment.
Measure spin up and the other is down...
Measure a photon and the other won't...
>You might try saying that a "half a photon" has enough
>energy to trigger a detector, but only half the time. But if that
>were the case, then a quarter of the time both detectors would be
>triggered.
Don't go for that at all.
>Unless you want to postulate that one detector can signal
>the other to say "I've got this photon, so you stop looking," you're
>not going to argue your way out of the results this way.)
In effect the latter is correct, as explained above.
--
Nick Maclaren
In article <+oNAyAF2...@farmeroz.port995.com>,
Oz <o...@farmeroz.port995.com> writes:
|>
|> >|> How is this to tell you which slit the 'photon' went through?
|> >
|> >Time. The difference between water and air is of the order of
|> >0.74 nS/metre. Modern computers can measure times of less than
|> >that. You might need a previous split to get a calibration,
|> >which might make the measurement rate rather low, but that is
|> >not a fundamental problem.
|>
|> Of course this begs the question.
|>
|> How do you know when a photon goes through the slits or leaves the
|> detector in order to get the flight time?
|>
|> and that's before you even start on all the other factors.
Hmm. I thought that I had dealt with that by the previous split,
but that won't work, because it assumes the wave model. So scratch
that idea.
You can still do it, but need some way of sending bursts of photons,
possibly by opening and closing windows, that are shorter than the
difference between your flight path. A 10 metre run gives you a
good 5 nS to play with, which is plenty for some kinds of electronic
switches. I don't know which would be appropriate, if any current
ones are.
5 nS consistency over 10 metres is not a problem.
Regards,
Nick Maclaren.
Maurice Barnhill
Nick Maclaren wrote:
You are going to run into limits from the uncertainty relation in
energy and time. Since delta-E delta-t > h-bar, the length of
the time of flight of the photon between slit and screen will
limit the accuracy with which you can determine the energy, and
E=cp= hc/wavelength will then give you an uncertainty in the
wavelength. As soon as your time measurement is accurate enough
to determine which slit the photon went through, the uncertainty
in the wavelength will kill the interference pattern.
--
Maurice Barnhill
m...@udel.edu [Use ReplyTo, not From]
[bellatlantic.net is reserved for spam only]
Department of Physics and Astronomy
University of Delaware
Newark, DE 19716
Jim Logajan
Maurice Barnhill <m...@udel.edu> wrote:
>
> Nick Maclaren wrote:
>> You can still do it, but need some way of sending bursts of photons,
>> possibly by opening and closing windows, that are shorter than the
>> difference between your flight path. A 10 metre run gives you a
>> good 5 nS to play with, which is plenty for some kinds of electronic
>> switches. I don't know which would be appropriate, if any current
>> ones are.
>>
>> 5 nS consistency over 10 metres is not a problem.
>
> energy and time. Since delta-E delta-t > h-bar, the length of
> the time of flight of the photon between slit and screen will
> limit the accuracy with which you can determine the energy, and
> E=cp= hc/wavelength will then give you an uncertainty in the
> wavelength. As soon as your time measurement is accurate enough
> to determine which slit the photon went through, the uncertainty
> in the wavelength will kill the interference pattern.
That may be an objection to the proposed scheme, but there is theoretically
no need to employ any intervening obstructions to determine the time of
flight. Consider:
If an excited atom emits a photon this way --> then the atom moves this way
<-- due to conservation of momentum. I could hook that excited (or
excitable) atom up to a small (molecular sized?) stop-watch which records
the time of emission by triggering on the recoil. On the other side of the
double slitted screen is a sheet with absorber atoms which are attached to
small stop-watches that are synced to the time on the emitter stop watch.
So it should be possible to measure time-of-flight between emission and
absorption without "touching" the photon in the intervening experimental
setup. It should then be possible to determine which slit the photon passed
through without affecting its wavelength.
John Gonsowski
Oz <o...@farmeroz.port995.com> wrote in message news:<Q$m7H6JYM...@farmeroz.port995.com>...
> Well, I tend to use the olde fashioned meaning of a field.
> I suppose I could witter on about some excitation of the vacuum.
> I hate to tell you that currently I think everything (including massive
> particles) are fields. Its so much simpler that way...
> In my viewpoint there isn't really such a
> thing as a photon, just a field. This field can be short (covering a few
> fm) or long (km), high intensity or low intensity, and interacts with
> charged particles just as you would expect such a field to interact.
Well there's definitely some kind of degrees of freedom on top of the
vacuum. You can certainly recognize a massive particle by the fields
around it. Everything as a field isn't such a bad way to think, SU(5)
GUT was doing it and Tony Smith has a nice U(8) model using D8 branes
added to his TOE. A photon could be a Planck length's worth of EM for
you.
Maurice Barnhill
Sorry, as clever as this idea is it doesn't work. The photon and
the recoiling atom are entangled, and a measurement on the atom
is equivalent to a measurement on the photon. You can't pick out
by measurement a particular recoil of the atom without also
picking out the corresponding particular wavelength of the photon.
The point here is a subtle one. The photon is not a classical
object which gets a range of momenta as a result of an attempt at
a measurement of position. Even without a measurement the photon
has a range of momenta and positions with various probabilities
[really probability amplitudes so that you can get interference].
When you make a measurement of position, you pick out that
position _and the corresponding momenta_ as the one(s) occurring
in this particular instance of your experiment. It doesn't
matter how gently you make the measurement of position, the
corresponding range of momenta is selected simultaneously.
alistair
"QED - the strange theory of light" Feynman says that the interference
pattern is affected by the reliability of the detectors!
I assume that he meant experiments have been done in which a detector
close to a slit is switched on and off so there is no guarantee of
a photon or electron that strikes the detector being registered.
Joe Rongen
"alistair" <alis...@goforit64.fsnet.co.uk> wrote in message
news:861c1b21.04092...@posting.google.com...
I think that in general "the reliability of a detector" would include
the inherit flaws in detector designs. Such as a specific dead time
between measurements, quantum efficiency, spectral response,
matching source and receiver, environmental factors, noise and fatigue.
Regards Joe
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
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juju
All that has been said here is well and good. But to answer the
original question, you have to decide how the formalism of the QM model
of superposition maps on to the real world.What is happening to the
properties (mass/energy. charge. spin. etc) of the system in question
(from photons to buckeyballs) when these systems are set into
superposition.
juju
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Jim Logajan
Maurice Barnhill <m...@udel.edu> wrote:
> Jim Logajan wrote:
>> If an excited atom emits a photon this way --> then the atom moves
>> this way <-- due to conservation of momentum. I could hook that
>> excited (or excitable) atom up to a small (molecular sized?)
>> stop-watch which records the time of emission by triggering on the
>> recoil. On the other side of the double slitted screen is a sheet
>> with absorber atoms which are attached to small stop-watches that are
>> synced to the time on the emitter stop watch. So it should be
>> possible to measure time-of-flight between emission and absorption
>> without "touching" the photon in the intervening experimental setup.
>> It should then be possible to determine which slit the photon passed
>> through without affecting its wavelength.
>
> Sorry, as clever as this idea is it doesn't work. The photon and
> the recoiling atom are entangled, and a measurement on the atom
> is equivalent to a measurement on the photon. You can't pick out
> by measurement a particular recoil of the atom without also
> picking out the corresponding particular wavelength of the photon.
That's all very nice, except it seems to me you're arguing against the
wrong class of experiments. The procedure is an attempt to measure
positions and times (which are commuting compatible observables), not
positions and momentum (which are non-commuting incompatible observables).
The experiment doesn't measure the magnitude or direction of the recoil
momentum or the energy of the recoil - it just records that a recoil event
happened at particular times and places.
> The point here is a subtle one. The photon is not a classical
> object which gets a range of momenta as a result of an attempt at
> a measurement of position. Even without a measurement the photon
> has a range of momenta and positions with various probabilities
> [really probability amplitudes so that you can get interference].
> When you make a measurement of position, you pick out that
> position _and the corresponding momenta_ as the one(s) occurring
> in this particular instance of your experiment. It doesn't
> matter how gently you make the measurement of position, the
> corresponding range of momenta is selected simultaneously.
That's standard orthodoxy. So perhaps you could comment on the experimental
results in the following document where one of the conclusions reads "The
interferences appear even with knowledge of what photon goes to each
slit...":
http://www.fsp.csic.es/young/young.pdf
"Time-resolved diffraction and interference: Youngæ„€ interference with
photons of different energy as revealed by time resolution" BY N. GARCIA,
I. G. SAVELIEV and M. SHARONOV.
My QM is rusty, but it seems to me that some of the conclusions seem to be
at variance with usual QM orthodoxy.
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Maurice Barnhill
>>Jim Logajan wrote:
>>
>>>If an excited atom emits a photon this way --> then the atom moves
>>>this way <-- due to conservation of momentum. I could hook that
>>>excited (or excitable) atom up to a small (molecular sized?)
>>>stop-watch which records the time of emission by triggering on the
>>>recoil. On the other side of the double slitted screen is a sheet
>>>with absorber atoms which are attached to small stop-watches that are
>>>synced to the time on the emitter stop watch. So it should be
>>>possible to measure time-of-flight between emission and absorption
>>>without "touching" the photon in the intervening experimental setup.
>>>through without affecting its wavelength.
>>
>>the recoiling atom are entangled, and a measurement on the atom=20
>>is equivalent to a measurement on the photon. You can't pick out=20
>>by measurement a particular recoil of the atom without also=20
>>picking out the corresponding particular wavelength of the photon.
>=20
> That's all very nice, except it seems to me you're arguing against the=20
> wrong class of experiments. The procedure is an attempt to measure=20
> positions and times (which are commuting compatible observables), not=20
> positions and momentum (which are non-commuting incompatible observable=
s).=20
> The experiment doesn't measure the magnitude or direction of the recoil=
=20
> momentum or the energy of the recoil - it just records that a recoil ev=
ent=20
> happened at particular times and places.
>=20
>>The point here is a subtle one. The photon is not a classical=20
>>object which gets a range of momenta as a result of an attempt at=20
>>a measurement of position. Even without a measurement the photon=20
>>has a range of momenta and positions with various probabilities=20
>>[really probability amplitudes so that you can get interference].=20
>> When you make a measurement of position, you pick out that=20
>>position _and the corresponding momenta_ as the one(s) occurring=20
>>in this particular instance of your experiment. It doesn't=20
>>matter how gently you make the measurement of position, the=20
>>corresponding range of momenta is selected simultaneously.
MVB again: On rereading I don't like my language here very much.=20
If it isn't obvious, I have in mind using Fourier transforms of=20
a wave-packet to relate the range of positions to the range of=20
existing momenta.
>=20
>=20
> That's standard orthodoxy. So perhaps you could comment on the experime=
ntal=20
> results in the following document where one of the conclusions reads "T=
he=20
> interferences appear even with knowledge of what photon goes to each=20
> slit...":
>=20
> http://www.fsp.csic.es/young/young.pdf
> "Time-resolved diffraction and interference: Young=B4s interference wit=
h
> photons of different energy as revealed by time resolution" BY N. GARCI=
A,=20
> I. G. SAVELIEV and M. SHARONOV.
>=20
> My QM is rusty, but it seems to me that some of the conclusions seem to=
be=20
> at variance with usual QM orthodoxy.
>=20
I couldn't respond to this without working through the article,=20
which took a while. I am not an experimentalist, but the=20
experiment described in this article strikes me as something of a=20
tour-de-force. The authors use a pulsed laser to send photons=20
to a spectrometer, where the pulse is separated into its=20
component wavelengths. Two different wavelengths are fed into=20
optical fibers which run to two slits near to each other (with an=20
variable time delay in one of the fibers) and allowed to=20
interfere. The interference pattern, which is expected to be and=20
is time-dependent, is detected with a streak camera.
The authors then say that since they know that the redder photon=20
goes through one slit and the bluer one through the other, they=20
know which slit each photon went through and therefore should=20
have destroyed the interference. Therefore something is wrong=20
with many of the descriptions of the genesis of Quantum=20
Mechanical interference. I believe that this interpretation is=20
clearly wrong. Briefly, none of the photons in this experiment=20
are in states having a single, well-defined wavelength, so the=20
spectrometer does not deliver photons to one slit or the other,=20
and the slit that any given photon goes through is unknown.
Why? All of the photons have a wave-train whose duration is no=20
more than the duration of the pulse given out by the laser, and=20
persumably no longer than the natural lifetime of the excited=20
state that the photon comes from. Hence none of the photons are=20
in states of definite energy, since these states are necessarily=20
of infinite duration. Since the states do not have definite=20
energy, they do not have definite wave-length either. Hence the=20
spectrometer, while it separates the basis states of an expansion=20
of the photon's actual state in terms of states of definite=20
wavelength, does not separate photons themselves into states of=20
definite wavelength. So there is some amplitude for any given=20
photon to follow a path through the slit corresponding to a=20
redder weve-length and another amplitude for following a path=20
through the slit corresponding to a bluer wave-length. These=20
amplitudes can interfere at the location of the camera, where the=20
photon is actually detected.
Now lasers give off many photons whose phases are coherant.=20
These photons are by definition entangled and are therefore an=20
inessential complication in understanding this part of the=20
experiment. The crucial issue is a much more difficult issue: if=20
a filter was used to reduce the energy density in the optical=20
fibers enough to guarantee that there was (almost) never two=20
photons in the system at once, would there still be interference?=20
Obviously I am predicting that there would be [at least in the=20
absence of a different time delay in the two fibers].
What happened when they delayed the arrival of the wave from one=20
of the fibers? Initially, very little. Once the time delay was=20
large enough, the waves through the two slits lost coherance and=20
the interference pattern disappeared. They showed that this=20
behavior was similar to what would be expected according to a=20
theory of decoherance that I suspect was not applicable to their=20
setup. I don't know enough about their particular lasers or=20
perhaps even about pulsed lasers in general to analyze the=20
results in detail, but I'll make a guess. Perhaps someone who=20
knows more about lasers will comment on how reasonable this guess=20
is. At some time delay larger than or equal to the lifetime of=20
the originating atomic state their interference pattern would=20
have to involve interference between two entangled photons. At=20
some time less than or equal to the pulse width of the laser the=20
corresponding photons in the beam are no longer coherant and=20
therefore no longer entangled. At this point the interference=20
pattern disappears. The disappearance should somehow track the=20
coherance properties of the fields across the pulse.
I think that there is a perfectly conventional explanation of the=20
results of this experiment. Note that in agreement with my=20
earler comment about time delays as a means of labelling photons,=20
the delta-E delta-t uncertainty relation is an essential part of=20
the explanation, since that relation is what tells you that the=20
photons in a pulse are not in states of definite wavelength.
Even though I disagree with the given interpretation of the=20
experiment, I think that this is a very beautiful demonstration=20
of what quantum mechanics is all about.
--=20
Gil Fuller
Are the results the same if the double slit experiment is done in a
vacuum as when done in normal "room air"? Obviously I am but a lowly
layperson but I would appreciate an answer if anyone knows.
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Jim Logajan
>> results in the following document where one of the conclusions reads
>> each slit...":
>>
>> http://www.fsp.csic.es/young/young.pdf
>> "Time-resolved diffraction and interference: Young's interference
>> with photons of different energy as revealed by time resolution" BY N.
>> GARCIA, I. G. SAVELIEV and M. SHARONOV.
>
> I couldn't respond to this without working through the article,
I really appreciate you taking the time to review that paper.
[ Elided most of your thoughtful analysis for brevity. ]
> I think that there is a perfectly conventional explanation of the
> results of this experiment. Note that in agreement with my
> earler comment about time delays as a means of labelling photons,
> the delta-E delta-t uncertainty relation is an essential part of
> the explanation, since that relation is what tells you that the
> photons in a pulse are not in states of definite wavelength.
>
> Even though I disagree with the given interpretation of the
> experiment, I think that this is a very beautiful demonstration
> of what quantum mechanics is all about.
I think it is an intriguing paper and now have better insight into some of
the aspects I did not consider and also how to interpret the results.
Thanks again!