Two-slit experiment
drspeg
experiment (showing both wave and particle properties of light). I hope
someone here might know how to respond.
Quick review of what I believe I know: Shooting a single photon (or
electron) at a wall that contains two slits will result in an interference
pattern on the wall (detector plate) beyond the holes. Blocking one hole
results in an accumulation pattern. When both holes are open, detecting
which of the two holes the electron passes through results in an
accumulation pattern (uncertainty is violated).
(1) Does obtaining an interference pattern depend at all on the timing of
the electron gun shooting each new electron? Or would the interference
pattern still obtain even if you shot one electron a day for many days (and
were able to record the impacts of electrons on the detector plate)?
(2) If, in the case where one determines through which slit each electron
passes, one were to replace the detector plate with ANOTHER wall containing
two slits, would uncertainty be restored? That is, knowing through which
slit electrons first pass should produce an accumulation pattern, but if
that pattern were displayed on a wall containing two slits (in this case the
experimenter does NOT determine through which slit the electron passes)
would an interference pattern occur?
So, in this double-two-slit experiment, determining the first "choice" made
by the pasing electron, but not the second... would there be an accumulation
pattern followed by an interference pattern?
Thanks!
Jonathan Thornburg -- remove -animal to reply
> I've been looking for an answer to two questions regarding the 2-slit
> experiment (showing both wave and particle properties of light). I hope
> someone here might know how to respond.
> (1) Does obtaining an interference pattern depend at all on the timing of
> the electron gun shooting each new electron? Or would the interference
> pattern still obtain even if you shot one electron a day for many days (and
> were able to record the impacts of electrons on the detector plate)?
Yes, you would still get an interference pattern. That is, the
(same) interference pattern is still there even at *arbitrarily low*
intensities (&& correspondingly long exposure times). I posted a
bunch of references on this to this newsgroup on 29.June.2001:
http://www.lns.cornell.edu/spr/2001-06/msg0033834.html
ciao,
--
-- "Jonathan Thornburg -- remove -animal to reply" <jth...@aei.mpg-zebra.de>
Max-Planck-Institut fuer Gravitationsphysik (Albert-Einstein-Institut),
Golm, Germany, "Old Europe" http://www.aei.mpg.de/~jthorn/home.html
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam
Uncle Al
>
> I've been looking for an answer to two questions regarding the 2-slit
> experiment (showing both wave and particle properties of light). I hope
> someone here might know how to respond.
>
> Quick review of what I believe I know: Shooting a single photon (or
> electron) at a wall that contains two slits will result in an interference
> pattern on the wall (detector plate) beyond the holes. Blocking one hole
> results in an accumulation pattern. When both holes are open, detecting
> which of the two holes the electron passes through results in an
> accumulation pattern (uncertainty is violated).
A double slit pattern is not the sum of two single slit patterns. For
one thing, the double slit maximum is directly opposite the center
barrier. You can look it up in a physics text book,
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
http://www.motionmountain.net/
> (1) Does obtaining an interference pattern depend at all on the timing of
> the electron gun shooting each new electron? Or would the interference
> pattern still obtain even if you shot one electron a day for many days (and
> were able to record the impacts of electrons on the detector plate)?
Timing is irrelevant. Each individual moiety's wavefunction passes
through both slits - photon, electron, or huge giant really big
spatially extended rigidly multiply-connected massive lump,
http://www.quantum.univie.ac.at/research/matterwave/c60/index.html
C60 diffraction
> (2) If, in the case where one determines through which slit each electron
> passes, one were to replace the detector plate with ANOTHER wall containing
> two slits, would uncertainty be restored? That is, knowing through which
> slit electrons first pass should produce an accumulation pattern, but if
> that pattern were displayed on a wall containing two slits (in this case the
> experimenter does NOT determine through which slit the electron passes)
> would an interference pattern occur?
Google
"quantum eraser" 15,400 hits
But why stop there?
http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser
> So, in this double-two-slit experiment, determining the first "choice" made
> by the pasing electron, but not the second... would there be an accumulation
> pattern followed by an interference pattern?
Here's a nice Gedankenexperiment. We take an organic molecule that
degenerately rearranges at a tremendous rate, like semibullvalene
(Ea=5.5 kcal/mol),
http://faculty.juniata.edu/reingold/rsch.html
http://en.wikipedia.org/wiki/Semibullvalene
<http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/structures/index5.html>
and we do the C60 experiment with it. However... when we thin-film
fabricate the diffraction grating we apply an alternating Peltier
heater so alternate slits are cold and hot. When a semibullvalene
wavefunction passes through the slits it has different rearrangement
rates at the different temperatures. Find a set of conditions that
dephases the wavefunction by exactly 180 degrees slit vs. slit. When
the two halves recombine on the other side... destructive
interference! Where is the molecule?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz3.pdf
Igor Khavkine
> drspeg <drs...@hotmail.com> wrote:
> > I've been looking for an answer to two questions regarding the 2-slit
> > experiment (showing both wave and particle properties of light). I hope
> > someone here might know how to respond.
> [[...]]
> > (1) Does obtaining an interference pattern depend at all on the timing of
> > the electron gun shooting each new electron? Or would the interference
> > pattern still obtain even if you shot one electron a day for many days (and
> > were able to record the impacts of electrons on the detector plate)?
>
> Yes, you would still get an interference pattern. That is, the
> (same) interference pattern is still there even at *arbitrarily low*
> intensities (&& correspondingly long exposure times). I posted a
> bunch of references on this to this newsgroup on 29.June.2001:
> http://www.lns.cornell.edu/spr/2001-06/msg0033834.html
Here are a couple of other references:
http://www.hqrd.hitachi.co.jp/em/doubleslit.cfm
http://www.hqrd.hitachi.co.jp/em/movie.cfm, number (2)
The first link describes a double slit experiment using a very low
emission electron gun. The second one shows a movie of how the
interference pattern is actually accumulated. The interference pattern
appears even when the electron gun essentially fires a single eletron
at a time.
Igor
Alf P. Steinbach
> Jonathan Thornburg -- remove -animal to reply wrote:
>> drspeg <drs...@hotmail.com> wrote:
>>> (1) Does obtaining an interference pattern depend at all on the timing of
>>> the electron gun shooting each new electron? Or would the interference
>>> pattern still obtain even if you shot one electron a day for many days (and
>>> were able to record the impacts of electrons on the detector plate)?
>> Yes, you would still get an interference pattern. That is, the
>> (same) interference pattern is still there even at *arbitrarily low*
>> intensities (&& correspondingly long exposure times). I posted a
>> bunch of references on this to this newsgroup on 29.June.2001:
>> http://www.lns.cornell.edu/spr/2001-06/msg0033834.html
>
> Here are a couple of other references:
>
> http://www.hqrd.hitachi.co.jp/em/doubleslit.cfm
> http://www.hqrd.hitachi.co.jp/em/movie.cfm, number (2)
>
> The first link describes a double slit experiment using a very low
> emission electron gun.
But are there experiments of good standing that directly rule out any
memory of past events in the detector, e.g. with hours between single
events, thermal randomization between events, or the like?
I think that that, not single particle at a time, was essentially what
the OP (drspeg <drs...@hotmail.com>) asked.
The reference list directly provided by Jonathan didn't seem to provide
any such reference, and in fact, of the four URLs listed in that
article, the three last URLs didn't work (probably moved).
And although there were many follow-ups to the posting that Jonathans's
was a response to, all of these claiming an abundance of experimental
evidence, not one provided a reference to such an experiment. The
closest was perhaps <url:
http://www.ati.ac.at/~summweb/ifm/experiment/resultsf.html#top>. But
seemingly also that that was just another statistically-one-at-a-time
experiments.
Thus, my curiosity is pickled[1]! ;-)
[1] <url: http://www.invasioncreations.com/PickleCuriosity.swf>
--
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?
bz
news:R62dnftk_8SSZQrZ...@libcom.com:
> I've been looking for an answer to two questions regarding the 2-slit
> experiment (showing both wave and particle properties of light). I hope
> someone here might know how to respond.
>
> Quick review of what I believe I know: Shooting a single photon (or
> electron) at a wall that contains two slits will result in an
> interference pattern on the wall (detector plate) beyond the holes.
Wrong. A single photon will produce a single point response somewhere on
the detector plate. The probability of hitting any one point can be
determined by accumulating data from a series of single photons(or
electrons).
A series of single photons will build up a pattern similar to that seen
when a continuous stream of photons is used.
> Blocking one hole results in an accumulation pattern.
When one hole is blocked, and a series of single photons is used, the
accumulation pattern will look similar to that seen when a continuous
stream of photons is used with a single slit.
> When both holes
> are open, detecting which of the two holes the electron passes through
> results in an accumulation pattern (uncertainty is violated).
>
> (1) Does obtaining an interference pattern depend at all on the timing
> of the electron gun shooting each new electron? Or would the
> interference pattern still obtain even if you shot one electron a day
> for many days (and were able to record the impacts of electrons on the
> detector plate)?
Appears to be rate independent. Photon(Electron) multiplyer arrays can be
used to capture and record the location of the impacts over a long period
of time.
>
> (2) If, in the case where one determines through which slit each
> electron passes, one were to replace the detector plate with ANOTHER
> wall containing two slits, would uncertainty be restored? That is,
> knowing through which slit electrons first pass should produce an
> accumulation pattern, but if that pattern were displayed on a wall
> containing two slits (in this case the experimenter does NOT determine
> through which slit the electron passes) would an interference pattern
> occur?
Perform the experiment and see.
>
> So, in this double-two-slit experiment, determining the first "choice"
> made by the pasing electron, but not the second... would there be an
> accumulation pattern followed by an interference pattern?
>
> Thanks!
>
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
Oz
>The first link describes a double slit experiment using a very low
>emission electron gun. The second one shows a movie of how the
>interference pattern is actually accumulated. The interference pattern
>appears even when the electron gun essentially fires a single eletron
>at a time.
These are all very important experiments.
1) The self-interference of both photons and electrons (and much larger
things like fullerenes). These have all been demonstrated.
2) There is also the interference of two sources each going through one
slit **when intensity is so low that the probability of finding two
'photons' in the apparatus at the same time is very low.
Now this has been demonstrated for photons. However the laser light used
for each beam (separate lasers) had extremely long coherence lengths
(times). To do this with electrons might be problematic (ie
"challenging") but would be interesting. Anybody any ideas how it could
be done?
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Arkadiusz Jadczyk
wrote:
>I've been looking for an answer to two questions regarding the 2-slit
>experiment (showing both wave and particle properties of light). I hope
>someone here might know how to respond.
>
>Quick review of what I believe I know: Shooting a single photon (or
>electron) at a wall that contains two slits will result in an interference
>pattern on the wall (detector plate) beyond the holes.
Shooting one electron produces one dot. Not a pattern.
> Blocking one hole
>results in an accumulation pattern.
Blocking one hole and shooting one electron will also produce just one
dot.
> When both holes are open, detecting
>which of the two holes the electron passes through results in an
>accumulation pattern (uncertainty is violated).
Probably when you wrote "shooting a single electron" you meant
"shooting one electron at a time, many times"?
>(1) Does obtaining an interference pattern depend at all on the timing of
>the electron gun shooting each new electron?
Generally time of the shooting has nothing to do with the pattern.
Unless the detecting device has some large relaxation time etc.
Or would the interference
>pattern still obtain even if you shot one electron a day for many days (and
>were able to record the impacts of electrons on the detector plate)?
The answer is "yes". But you should make changes to your apparatus
during the whole experiment.
>(2) If, in the case where one determines through which slit each electron
>passes, one were to replace the detector plate with ANOTHER wall containing
>two slits, would uncertainty be restored?
It is not clear to me what you have in mind. But, as far as I read your
intentions, the answer is "yes", but it would be a different
"uncertainty".
That is, knowing through which
>slit electrons first pass should produce an accumulation pattern, but if
>that pattern were displayed on a wall containing two slits (in this case the
>experimenter does NOT determine through which slit the electron passes)
>would an interference pattern occur?
The point is that there would be *very few* electrons going through your
second slits in this case. Most of them would be absorbed by the wall,
as there would be large spread of directions.
>
>So, in this double-two-slit experiment, determining the first "choice" made
>by the pasing electron, but not the second... would there be an accumulation
>pattern followed by an interference pattern?
>
>Thanks!
Here I am lost - perhaps you can make your question more precise?
ark.
--
Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--
Andreas Most
> [...]
> But are there experiments of good standing that directly rule out any
> memory of past events in the detector, e.g. with hours between single
> events, thermal randomization between events, or the like?
I remember having read about a double-slit experiment where the whole setup
was disassembled and then (after some days) reassembled between the single
events. It was even set up at different places.
After a year (or so) they could clearly see the interference pattern.
I cannot think of any memory effect in such a setup.
Unfortunately, I don't know the reference. Could have been some Japanese
people. Has anybody a clue of where to find this reference?
Andreas.
scerir
> I remember having read about a double-slit experiment
> where the whole setup was disassembled and then
> (after some days) reassembled between the single
> events.
This one?
http://leifi.physik.uni-muenchen.de/web_ph12/originalarbeiten/taylor/taylor_e.htm
In 1909 Geoffrey Ingram Taylor conducted an experiment
in which he showed that even the feeblest light source
could lead to interference fringes.
"The longest experiment took 3 months, corresponding
to the intensity of a candle more than a mile away"
It seems that this experiment led to Dirac's famous koan
"each photon then interferes only with itself".
nightlight
> Wrong. A single photon will produce a single point response somewhere
> on the detector plate. The probability of hitting any one point can be
> determined by accumulating data from a series of single photons(or
> electrons).
Nop, that is a misconception or mischaracterization of the empirical
facts in photon experiments.
The facts are that the response of an array of photodetectors is
statistically indistingushable from droplets statistics of an array of
dripping faucets. Specifically, the best (the most narrow distribution
of triggers) that you will ever get here is a Poissonian distribution
(as long as QED vacuum is there and its vacuum fluctuations).
There isn't anything 'single-photon-like' about it that can
distinguish it from a thresholded EM field measurements (with
photo-electron current measurements results reduced via AD
conversion by the 'pulse analayzer & discriminator' electronics
to a single bit precision, 0/1 values, based on the detector
thresholds) of a classical field.
Nothing distinctly non-classical or non-local (such as the conjectured
instant collapse of a remote EM field) occurs with photo-detections
in double slit/beam splitter experiments. (Otherwise, you wouldn't
need Bell inequality violations tests as the exclusive QM prediction
capable of making such distinction). When a detector placed on
photon path A triggers, the detector at a remote path B will trigger
or not trigger entirely independently (and solely based on its local
field intensities) of what happened with the detector in path A.
In other words, the trigger statistics on A and B detectors is exactly
as if the two wave packet fragments travelled each on its own to
its detector, then triggered it or not based solely on the total
incident EM energy on a photo-cathode within the sampling
window (the 'incident field' consists of the 'signal' EM field
superposed to the ZPF or vacuum fluctuations at the cathode).
For example if you take the array of pair results (A,B) which
are (0,0), (0,1), (1,0) and (1,1) and split it into two sub-arrays
based on results on A, hence as (1,x) and (0,y) sub-arrays, you
won't find statistically smaller proportion of 1's among x values
measured on B (the so-called collapse on B when A triggers)
than among the y values measured on B.
Of course, as with dripping faucets, you can make the detectors
virtually never produce double trigger in a given sequence of sampling
time windows, but you can do exactly the same with the faucets in the
exactlly same way -- either shrink the sampling windows enough or
reduce the input intensity/water pressure enough. In both cases if you
reduce double triggers to single triggers ratio by a factor f, you will
reduce the single trigger to no-trigger ratios by at least the same
factor f, i.e. a setup that virtually never has a double trigger among
triggers, virtually never has a single trigger among non-triggers,
either. It is a trivially unsurprising phenomenon of Poissonian
distribution in the limit p -> 0.
There were a various experments (since 1950s) trying to establish
"single photon" phenomenon i.e. the trigger statistics which would be
distinguishable from that of the array of dripping faucets. It just
doesn't happen. You can read on the very latest one such experiment
claiming demonstration of "single photons" and its detalied analysis
(showing exctly how it cheats) here:
PhysicsForum thread: Photon "Wave Collapse" Experiment...
http://www.physicsforums.com/showthread.php?t=71297
The experiment paper:
1. J.J. Thorn, M.S. Neel, V.W. Donato, G.S. Bergreen,
R.E. Davies, M. Beck "Observing the quantum behavior
of light in an undergraduate laboratory"
Am. J. Phys., Vol. 72, No. 9, 1210-1219 (2004).
http://marcus.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf
The experiment Home Page
http://marcus.whitman.edu/~beckmk/QM/
> A series of single photons will build up a pattern similar
> to that seen when a continuous stream of photons is used.
Again, read or see what actually happens. Nothing statistically
distingushes the photo-detector triggers from drips of an array
of faucets (e.g. you can have interference of the water waves
above the faucets and its effects on the drop counts below).
That conjectured phenomenon (demonstration of a "single photon"
via "sub-Poissonian distribution) has been a holy grail of the
marble-photon branch of Quantum Optics since 1950s when
Hanbury Brown and Twiss experiment showed that it's not how
the EM field behaves, be it in experiments or theoretically
(clasically or in QED).
Such demonstration, experimental or theoretical (within QED
formalism), has yet to be done. You can see for yourself how
miserably the 2004 experiment, the pinnacle of nearly five
decades of pursuit, performed and marvel at the ingenuity,
the means and the lengths at which the authors went to
create the appearance that it finally worked.
If you do know of some experiments or a QED derivation (but
not some handwaved elementary/popular QM stories) that can
distinguish double slit/beam splitter photo-detection trigger
statistics from the trivial classical phenomena such as dripping
faucets in the Poissonian limit p->0, you are welcome to cite it.
Oz
>It seems that this experiment led to Dirac's famous koan
>"each photon then interferes only with itself".
Except we know this is not correct. Experiments have been done where two
separate lasers fired through two separate slits produces an
interference pattern.
Furthermore this happened even if "the probability of two photons being
in the apparatus at one time approached zero".
In any case anyone who has listened to MW radio knows that separate
radio transmitters on the same frequency can readily cause interference
patterns.
bz
news:T1db1MSc...@farmeroz.port995.com:
> scerir <sce...@libero.it> writes
>
>>It seems that this experiment led to Dirac's famous koan
>>"each photon then interferes only with itself".
>
> Except we know this is not correct. Experiments have been done where two
> separate lasers fired through two separate slits produces an
> interference pattern.
>
> Furthermore this happened even if "the probability of two photons being
> in the apparatus at one time approached zero".
>
> In any case anyone who has listened to MW radio knows that separate
> radio transmitters on the same frequency can readily cause interference
> patterns.
>
What is usually heard there is the 'beat frequency' [hetrodyne] as signals
of two slighly different frequencies (or phases) are 'combined' in the
receiver's detector.
A signal arriving over multipaths can interfer with itself as the phase of
the multipath signals varies due to changes in path length as ionospheric
conditions vary.
Of course a HUGE number of photon is involved as MW radio frequency photons
each carry very little energy.
Timo Nieminen
> scerir <sce...@libero.it> writes
>
> >It seems that this experiment led to Dirac's famous koan
> >"each photon then interferes only with itself".
>
> Except we know this is not correct. Experiments have been done where two
> separate lasers fired through two separate slits produces an
> interference pattern.
That just shows that each photon comes from both sources. A photon is the
quantum of excitation of the EM field, the EM field is the sum of the
fields individually produced by the sources, so each photon comes from
both sources.
It is an excellent demonstration that photons aren't like classical
billiard balls.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Blackbird
> scerir <sce...@libero.it> writes
>
>> It seems that this experiment led to Dirac's famous koan
>> "each photon then interferes only with itself".
>
> Except we know this is not correct. Experiments have been done where
> two separate lasers fired through two separate slits produces an
> interference pattern.
Hmm. Interesting. Do you have a reference to a paper or webpage?
> Furthermore this happened even if "the probability of two photons
> being in the apparatus at one time approached zero".
To mee, this indicates that somehow the photon must interfere with itself.
> In any case anyone who has listened to MW radio knows that separate
> radio transmitters on the same frequency can readily cause
> interference patterns.
I don't think that this is the same kind of interference. If you add up two
amplitude modulated waves of the same frequency, then obviously the result
will be garbled, but this has a different explanation than the double slit
experiment has.
bz
news:1151472082....@p79g2000cwp.googlegroups.com:
> If you do know of some experiments or a QED derivation (but
> not some handwaved elementary/popular QM stories) that can
> distinguish double slit/beam splitter photo-detection trigger
> statistics from the trivial classical phenomena such as dripping
> faucets in the Poissonian limit p->0, you are welcome to cite it.
I don't know of any that have definitely done so, but I believe the 'single
photon' lasers that are designed to only produce a single photon at a time,
have the potential of being the basis of such an experiment.
Some are based on the idea that a single excited molecule can only produce
a single 'lase' photon per incoming excitation photon.
<http://lfw.pennnet.com/Articles/Article_Display.cfm?Section=ARTCL&ARTICLE_ID=257230&VERSION_NUM=3&p=12>
The 'buzz' in secure communications is the use of an 'untappable' stream of
single photons, where the loss of even a single photon, due to attempts at
tapping into the circuit, would render the stream of encrypted data as
clearly compromised.
Of course, simply attenuating a stream of photons from a candle or laser
until the frequency of photons detected is so low as to 'assure' single
photon events, really gives no such assurance. So the availability of
inexpensive sources of true 'single photon' streams will undoubtably lead
to some interesting observations on dual slit experments.
As for producing streams of guaranteed 'single electron' or 'single He
molecules', that may or may not be easier than guaranted single photon
streams.
I expect that given a stream of true single photons, each photon will only
excite a single grain on a photographic film(or a single element on a ccd).
The PATTERN of excited grains, over time, dependent on the slit
arrangement.
Oh No
you induced me to calculate interference effects between wave functions
for different photons using qed. The same argument did not apply to
electrons, which are fermions and which are conserved in interaction.
Regards
--
Charles Francis
substitute charles for NotI to email
scerir
> >to Dirac's famous koan "Each photon then
> >interferes only with itself."
Oz writes:
> Except we know this is not correct.
> Experiments have been done where two
> separate lasers fired through
> two separate slits produces an
> interference pattern.
Yes, but this does not mean that Dirac's
koan is wrong.
Even a photon emitted by two sources (in such
a way that you cannot say which of the two sources
emitted the photon) interferes only with itself.
When the different possible photon paths,
from sources to detector, are indistinguishable,
then we have to add the corresponding amplitudes
before squaring to obtain the probability.
The second part of the koan (see below)
seems obscure (or it needs a reformulation),
since we have two-photon interference, that
is to say the interference of photons
emerging from _independent_ sources.
s.
"Each photon then interferes only with itself.
Interference between two different photons
can never occur."
- P.A.M. Dirac, Principles of Quantum Mechanics,
Clarendon, Oxford, 1930, p.15.
Oz
>On Thu, 29 Jun 2006, Oz wrote:
>
>> scerir <sce...@libero.it> writes
>>
>> >It seems that this experiment led to Dirac's famous koan
>> >"each photon then interferes only with itself".
>>
>> Except we know this is not correct. Experiments have been done where two
>> separate lasers fired through two separate slits produces an
>> interference pattern.
>
>That just shows that each photon comes from both sources.
Hmmm....
As an explanation I think Mr Occham would find this improbable.
>A photon is the
>quantum of excitation of the EM field, the EM field is the sum of the
>fields individually produced by the sources, so each photon comes from
>both sources.
A contortion IMHO.
A much more plausible explanation is that photons are completely
wavelike. One then looks to apparent quantum behaviour in the emitters
and/or detectors where of course one finds them.
All detectors I know about rely on an irreversible energy step which is
by nature quantum.
>It is an excellent demonstration that photons aren't like classical
>billiard balls.
I would agree, they are waves.
NB I must logically also conclude that massive particles are also waves
and once again the quantised behaviour is due to something else. Clearly
(to me at any rate) the 'size' of a massive particle is determined (like
the photon) by its environment (typically quantised). So an electron,
say, can have a different physical size when undisturbed in an orbital
This can in some circumstances be very large indeed when orbitals become
macroscopic, for example in conductors.
The only real problem I have with this is why masses (which one could
say is the wavelength in the time direction) for free elementary
particles is most certainly quantised. If the particles were in a box of
fixed length in the time direction then of course there would be no
problem at all. Ideally some interaction between mass and box-size
should result in only a few integral solutions that express the
particles that we see.
Clearly this is grossly oversimplified because really we should exchange
the word 'energy' for 'mass' in all the above and there are at least two
broad energy systems, that is electromagnetic and colour, and we don't
seem to have very good theoretical constructs unifying them.
nightlight
The authors of the controversial AJP 2004 paper would
have surely cited such work as 'definite' if there were
one by that time. You can read the discussion
of their experiment at the link given and see exactly
by what means they got their sub-Poissonian
distribution: they tweaked delays on the double
trigger coincidence circuit until the pulse fell
well out of the acceptance time window of their
coincidence unit (ORTEC TAC/SCA 567). With
those settings they could have as well saved
$8-10K spent on laser and PDC (BBO) crystal,
used candle light instead and got exactly
the same "mysterious" single photon collapse.
When such means are needed to demonstrate
"single photon" and its "collapse", after
five decades of attempts, odds are that the
conjectured phenomenon does not exist.
As explained in that earlier thread, this conjectured
phenomenon is simply an operational misinterpretation
of how the elements of formalism (Glauber's correlation
functions) map to the empirical facts (the photo-detection
events). You can read more on this point in an earlier
sci.physics.reasearch post:
http://groups.google.com/group/sci.physics.research/msg/7ee770990a31bd3f
with some more detail on Glauber's theory:
http://www.physicsforums.com/showpost.php?p=529314&postcount=16
http://www.physicsforums.com/showpost.php?p=535516&postcount=61
http://www.physicsforums.com/showpost.php?p=538215&postcount=73
> but I believe the 'single photon' lasers that
> are designed to only produce a single photon
> at a time, have the potential of being the
> basis of such an experiment.
The most ideal laser produces at best only the
'coherent light' which is fully described by
Maxwell's EM equations. That was known since
early 1960s, when Sudarshan proved an equivalence
theorem which established complete classicality
of such light.
> Some are based on the idea that a single excited
> molecule can only produce a single 'lase' photon
> per incoming excitation photon.
> http://lfw.pennnet.com/Articles/Article_Display.cfm?Section=ARTCL&ARTICLE_ID=257230&VERSION_NUM=3&p=12
>
All that they are creating is the Poissonian light in
the limit p->0, as they plainly say at the link you
gave. Poissonian photon statistics is indistinguishable,
even in principle, from the statistics of classical
EM wave being thresholded via AD conversion to single
bit precision, then counted as: 1 is trigger, 0 is
a non-trigger. In the classical EM model, when you
place two detectors in the two paths after the beam
splitter (or behind the two slits in the double-slit
experiment), each detector triggers based solely
on the EM energy incident on its cathode within
the sampling window and entirely independently
of whether the detector in the other path triggers
("collapses" the wave function) or not. There is no
remote collapse in the classical model, just the
local collapse via self-focusing of the local
EM field due to the resonant absorption of the
incident EM energy by the oscillating electron
cloud of the cathode. See more on this model
of photo-detection in the paper (and the more
detailed papers from the same group cited there):
V. Bykov
"Photons, photocounts and laser detection of
weak optical signals" Ann. Fond. L. de Broglie,
V 26, n. spec. 1, 115-134 (2001)
http://www.ensmp.fr/aflb/AFLB-26j/aflb26jp115.htm
The same Poissonian "glitch" plagues all PDC sources,
which are the main "non-classical" light source in
recent couple decades. Namely the laser pump
(which is a 'coherent light' source, thus it has
Poissonian photon number statistics) generates at
best the Poissonian number of PDC pairs in any
sampling window, hence one cannot observe anything
about the PDC pairs photo-counts or their
correlations that is not perfectly modeled
by classical EM field with ZPF (zero-point field)
boundary conditions. This was demonstrated in a
series of papers by T. Marshall, E. Santos
and collaborators (you can see few of these
preprints at arXiv, e.g. quant-ph/9711042,
quant-ph/0202097):
http://arxiv.org/find/quant-ph/1/AND+au:+Santos_E+au:+Marshall_T/0/1/0/all/0/1
On PDC pair state & photon statistics, see references
[16] and [17] (Teich's online papers) here:
http://www.physicsforums.com/showpost.php?p=544829&postcount=122
One needs to be careful with the wishfully loaded
terminology and imagery used in Quantum Optics and
read between the lines in papers claiming Bell inequality
violations or other "non-classicality" obtained from PDC
sources (or any other optical photon sources explored for
these purposes) -- the "non-classicality" claimed so far is
always modulo "fair sampling" conjecture, i.e. the assumption
that they can somehow subtract away the Poissonian noise.
This immovable Poissonian wall is euphemistically labeled
"detection loophole" within Bell inequality tests, since
the Poissonian distribution of photo-electrons implies
an unavoidable tradeoff between the rate of 'dark counts'
and the detector efficiency. This tradeoff then precludes
the Bell's inequality violations, even in principle, by
the actual counts. So far, after over three decades of
attempts, only the fictitious counts, the counts in which
the Poissonian statistics was willed away (via ever more
creative euphemisms), have managed to "violate" Bell's
inequality (or other classical inequalities). For more
on the present status of the experiments see preprints:
http://arxiv.org/find/quant-ph/1/AND+au:+Santos_E+abs:+bell/0/1/0/2004/0/1
> The 'buzz' in secure communications is the use of an
> 'untappable' stream of single photons, where the loss
> of even a single photon, due to attempts at tapping
> into the circuit, would render the stream of
> encrypted data as clearly compromised.
That's a marketing hype. (I hope you didn't invest your
retirement into any of that "magic" technology.) Check
at the very link you posted on what the "single photon"
in this jargon means. It is simply a highly attenuated
Poissonian source:
--- quote --
It is possible to simulate a single-photon source by
strongly attenuating light pulses that initially contain
many photons so only a single photon remains. That approach
is widely used in quantum cryptography experiments, but
is inherently very inefficient. [ed. this is a red herring;
yes, it is inefficient, but that aspect is its least problem,
since it is also a perfectly classical source] ...
..
Outlook
"The field is evolving very rapidly," says Yamamoto.
But there are formidable challenges to overcome in
developing practical single-photon emitters. The number
of time slots filled by emitted photons is at best only
a few percent. "We should improve the number to 50%,"
says Yamamoto. Even with so few time slots filled, the
probability of finding two photons in an occupied time
slot is about 1% -- too high to be acceptable. He also
stresses the need to improve waveform overlap to
virtually 100%, so photons can interfere with one
another for quantum applications.
--- end quote ---
In other words, any detections are few percent, say 5%,
while two photons occur in 1% of slots. With Poissonian
distribution the probability of k detections (or
photo-electrons) in a sampling time slot is:
P(k,A) = A^k exp(-A) / k! ... (1)
where 'A' is average detections per slot. The probability
of non-detections is then:
P(k=0,A) = exp(-A) ... (2)
hence the probability of one or more detections is:
P(k>0,A) = 1 - exp(-A) ... (3)
Take their P(k>0,A) = 0.05, which yields A = -ln(0.95) = 0.05
detections per time slot on average. With this A, you can
compute the probabilities:
P(k=0,A) = exp(-A) = 0.95 = 95%
P(k=1,A) = A*exp(-A) = 0.0487 = 4.9%
P(k>1,A) = 1 - P(k=0,A) - P(k=1,A) =
= 0.00127 = 0.1 %
P(k=2,A) = A^2 exp(-A) / 2 = 0.00125 = 0.1%
Now, they acknowledge 1% is two photon detections, which
is about ten times worse than the ideal Poissonian source
would yield for their A: P(k=1,A)=0.1%. In other words,
they haven't even reached the sharpest "single photon"
statistics allowed by classical EM, let alone got
beyond it. Even the ancient perpetuum mobile inventors
had been closer to making it over their pesky hurdle
than this.
> So the availability of inexpensive sources of true
> 'single photon' streams will undoubtably lead to
> some interesting observations on dual slit experments.
The guys you linked to aren't even close to just a
very stable classical source. I have yet to see a genuine
theoretical/QED model of a realistic process which would yield
sub-Poissonian distribution using a perfect laser (Poissonian
or super-Poissonian) pump to drive some nonlinear generator
of single photons. Merely randomly blotting out events from
a Poissonian source always yields another Poissonian source,
just a sparser one (with smaller average A). It can't yield
the sub-Poissonian (hence non-classical) statistics, since
whatever the interaction Hamiltonian does to a single
photon state (sharp Fock state), it will do exactly the
same to the coherent superposition from the laser pump
(which is a superposition of single photon states with
Poissonian distribution of photon number observable),
hence the generator will yield at best a Poissonian
photon number output.
> I expect that given a stream of true single photons,
> each photon will only excite a single grain on a
> photographic film(or a single element on a ccd).
The "single photon" is infinite in spatio-temporal
extent (e.g. infinite plane wave). Coincidence
measurements and counts correlations are operationally
vacuous for such objects (since there is no element
of the formalism with localized properties that could
map into a time-space 'coincidence window' on the
empirical side).
Note also that EM vacuum fluctuations (or ZPF in
Stochastic electrodynamics) contain 1/2 vacuum photons
per EM mode. The E field of these vacuum photons superposes
its effects on film grains or ccd elements, with the
effects of the 'signal' field, no matter how stable the
'signal' field may be.
Finally, recall that "photons" are a mathematical
artifact of perturbative QED (where photons are
quanta of free EM field, introduced perturbatively
one by one, to mediate interactions between matter fields).
There are infinitely many different bases and coordinate
axes one can pick, thus different modes, hence different
'photons', superposing into the same given state.
There are also non-perturbative formalisms, empirically
equivalent to QED (up to at least alpha^5 order,
i.e. 8+ digits of precision), which eliminate EM
from the equations altogether and express the whole
dynamics in terms of (delayed/advanced) interactions
between fermion/matter fields. There are no photons
in such formalisms. See for example Barut's "Self
Field Electrodynamics". Barut references with brief
summary were discussed in couple recent
sci.physics.research posts:
http://groups.google.com/group/sci.physics.research/msg/386f48731520d145
http://groups.google.com/group/sci.physics.research/msg/bb1225c256c34c02
Oh No
Dirac's koan is wrong. He wrote this when quantum mechanics was in its
infancy and quantum electrodynamics had hardly even been born. He was
thinking, quite naturally, of the structure of one particle Hilbert
space in which the quantum superposition, |f>+|g>, appears in the
addition of states of one particle. It makes no sense to add the state
of one particle to the state of another, because they are strictly
described by states in different Hilbert spaces. However in quantum
electrodynamics we have to think of the wave function not in terms of
the probability for where a photon is, but rather the probability for
where a photon is annihilated (detected) and we get a different result.
As a concession to ASCII I shall ignore spin and simplify the formula as
much as possible, and just show the bones of the argument. The photon
field operator is,
A(x) = |x> + <x|
where I am using ket notation for operators, so that |x> creates the
state |x>, and <x| annihilates the same state.
The photon wave function in the state |f> is
<|A(x)|f> = <x|f>
so that |<x|f>|^2 is the probability for detection of a photon at x.
Now consider a two photon state |f;g> = |f>|g> + |f>|g>. For simplicity
I am assuming no entanglement, and that the photons are distinguishable
at some time in the past (they come from different sources) so that
<f|g>=0.
When one of the photons is detected the other remains as it was, but we
don't know which one is detected and which one remains. Then the final
state is either |f> or |g>, i.e. it is |f> + |g>, so the probability for
detection at x is given by
(<f|+<g|) A(x) |f;g> = (<f|+<g|)(|x> + <x|)|f;g>)
= (<f|+<g|)(<x|f>|g> + <x|g>|f>)
where I have lost the inner product between states of different numbers
of particles (it is zero). Using <f|g> =0 and <f|f>=<g|g>=1 this comes
down to
<x|f> + <x|g>
Which shows the superposition between states of different photons.
+++++
This argument does not work for electrons. I shall ignore spin. For
electrons the field operator is
Psi(x) = |x> + <x^|
where |x> creates an electron and <x^| annihilates a positron. The
thing we detect is the current
j(x) = Psi+(x)Psi(x) = (|x^> + <x|)(|x> + <x^|)
This is "normally ordered" so that <x||x> is replaced by |x><x|,
otherwise the theory is divergent before you start.
Ignoring positrons and pair creation j boils down to
j(x) = |x><x|
which is the position operator (more strictly j^0 is charge, but here I
have lost the spin indices).
Now when an electron is detected both particles remain in the final
state, so the probability of detection at x is given by
<f;g| j(x) |f;g> = <f;g| |x><x| |g;f>
= (<f|x><g| - <g|x><f|)(<x|g>|f> - <x|f>|g>)
Again the <f|g>=0 and <f|f> = <y|y> =1 so this reduces to
= |<f|x>|^2 + |<g|x>|^2
So in this case we do not have interference, but simply the sum of two
amplitudes.
Oz
> Yes, but this does not mean that Dirac's
>koan is wrong.
hmmm...
> Even a photon emitted by two sources (in such
>a way that you cannot say which of the two sources
>emitted the photon) interferes only with itself.
Hmmm....
given the experiment.
>When the different possible photon paths,
>from sources to detector, are indistinguishable,
>then we have to add the corresponding amplitudes
>before squaring to obtain the probability.
Hmmm.... how very convenient....
> The second part of the koan (see below)
>seems obscure (or it needs a reformulation),
>since we have two-photon interference, that
>is to say the interference of photons
>emerging from _independent_ sources.
Which is precisely my point.
The below can be reformulated (I am absolutely confident mathematicians
can do this), basically by saying "take two independent sources, and
call them one source", which gives the right answer at some
philosophical cost.
"Each photon then interferes only with itself.
Interference between two different photons
can never occur."
- P.A.M. Dirac, Principles of Quantum Mechanics,
Clarendon, Oxford, 1930, p.15.
--
Oz
>Oz wrote:
>> scerir <sce...@libero.it> writes
>>
>>> It seems that this experiment led to Dirac's famous koan
>>> "each photon then interferes only with itself".
>>
>> Except we know this is not correct. Experiments have been done where
>> two separate lasers fired through two separate slits produces an
>> interference pattern.
>
>Hmm. Interesting. Do you have a reference to a paper or webpage?
It was posted here, probably on one of the "length of a photon" threads.
More likely one of the longer-standing resident experts will know
anyway.
>> Furthermore this happened even if "the probability of two photons
>> being in the apparatus at one time approached zero".
>
>To mee, this indicates that somehow the photon must interfere with itself.
Maybe, however bear in mind that each laser points at only ONE slit.
The light beams (one from each laser-slit) is only combined in the
apparatus.
single
SLIT
| <D
LASER>----------------------------<E
| <T
<E
| <C
LASER>----------------------------<T
| <O
<R
So neither laser beam goes through two slits yet a two-slit diffraction
pattern is observed.
>> In any case anyone who has listened to MW radio knows that separate
>> radio transmitters on the same frequency can readily cause
>> interference patterns.
>
>I don't think that this is the same kind of interference. If you add up two
>amplitude modulated waves of the same frequency, then obviously the result
>will be garbled, but this has a different explanation than the double slit
>experiment has.
Why? Radio waves are perfectly good EM waves just like light but with
different frequency.
Oz
>What is usually heard there is the 'beat frequency' [hetrodyne] as signals
>of two slighly different frequencies (or phases) are 'combined' in the
>receiver's detector.
Indeed.
>A signal arriving over multipaths can interfer with itself as the phase of
>the multipath signals varies due to changes in path length as ionospheric
>conditions vary.
No, I am NOT talking about that.
I am talking about two SEPARATE transmitters, which still interfere.
>Of course a HUGE number of photon is involved as MW radio frequency photons
>each carry very little energy.
So what? Its still photons from different sources interfering.
Or would you claim that the interference will reduce as amplitudes
reduce? Surely you aren't suggesting that?
Oz
>Thus spake Oz <O...@farmeroz.port995.com>
>>2) There is also the interference of two sources each going through one
>>slit **when intensity is so low that the probability of finding two
>>'photons' in the apparatus at the same time is very low.
>>
>>Now this has been demonstrated for photons. However the laser light used
>>for each beam (separate lasers) had extremely long coherence lengths
>>(times). To do this with electrons might be problematic (ie
>>"challenging") but would be interesting. Anybody any ideas how it could
>>be done?
>>
>It can't be done. You probably don't remember or didn't follow it, but
>you induced me to calculate interference effects between wave functions
>for different photons using qed. The same argument did not apply to
>electrons, which are fermions and which are conserved in interaction.
You have been known to make errors, later corrected (like all of us).
Personally I vote for experimental results. My question is whether
anyone has any idea how it could be done? Personally I would bet you £50
it will show the same result as the photon experiment.
The initial aim would be to produce extremely monochromatic beams of
electrons. Given that we are looking for exceedingly low beam intensity
I suspect that the problems associated with mutual repulsion of
electrons within a beam will be negligible (or am I being
oversimplistic?).
One is almost tempted to start with very low energy electrons, perhaps
separating them by time-of-flight (via a chopper) or even ballistically
as the take a parabolic path in a gravitational field (ok, VERY low
energy). From there one could pass them through two identical (but
separate) accelerators to the slit and target.
Oh No
>>>for each beam (separate lasers) had extremely long coherence lengths
>>>(times). To do this with electrons might be problematic (ie
>>>"challenging") but would be interesting. Anybody any ideas how it could
>>>be done?
>>>
>>It can't be done. You probably don't remember or didn't follow it, but
>>you induced me to calculate interference effects between wave functions
>>for different photons using qed. The same argument did not apply to
>>electrons, which are fermions and which are conserved in interaction.
>
>You have been known to make errors, later corrected (like all of us).
I've given a demonstration again, in response to scerir. It is not
difficult, and nicely illustrates that observables, such as interference
effects, rely on the properties of operators, not just states, and thus
what is wrong with the view of superposition given in Dirac's koan.
You appreciate that there is a paradox here which needed resolution.
What Dirac says is trivially true of superposition of quantum states.
Hence it also would be true if we strictly observed superposition of the
wave function. But we do not. We observe the behaviour of the position
observable, and that is subtly different. In the instance of the
electron, it boils down to the same formula. In the instance of the
photon, it does not.
>
>Personally I vote for experimental results. My question is whether
>anyone has any idea how it could be done? Personally I would bet you £50
>it will show the same result as the photon experiment.
I'll take you up on that. Will you accept this as an electronic
handshake?
>The initial aim would be to produce extremely monochromatic beams of
>electrons.
Ahem, you can't have an electron equivalent of a laser. They are
fermions and cannot all be in the same state.
> Given that we are looking for exceedingly low beam intensity
>I suspect that the problems associated with mutual repulsion of
>electrons within a beam will be negligible (or am I being
>oversimplistic?).
You are forgetting the exclusion principle, which is rather more
encompassing than charge.
>One is almost tempted to start with very low energy electrons, perhaps
>separating them by time-of-flight (via a chopper)
No you wouldn't be allowed to do that either. Remember, when they do
this for photons the stipulation that only one photon comes through per
amount of time is statistical. In one of these low energy laser thingies
any photon can come through at any time, but it is only fed so much
energy so there is only one per amount of time.
scerir
> >When the different possible photon paths,
> >from sources to detector, are indistinguishable,
> >then we have to add the corresponding amplitudes
> >before squaring to obtain the probability.
Oz writes:
> Hmmm.... how very convenient....
Try 'Quantum effects in one-photon and two-photon
interference'by L. Mandel (Rev.Mod.Phys.,vol.71,n.2,
page S274). It is online (use Scholar Google).
> > The second part of the koan (see below)
> >seems obscure (or it needs a reformulation),
> >since we have two-photon interference, that
> >is to say the interference of photons
> >emerging from _independent_ sources.
>
> Which is precisely my point.
> The below can be reformulated (I am absolutely confident mathematicians
> can do this), basically by saying "take two independent sources, and
> call them one source", which gives the right answer at some
> philosophical cost.
No, the actual meaning of interference changes here.
See Mandel's paper or, i.e.,
http://www.arxiv.org/abs/quant-ph/0603048
Regards,
s.
Timo A. Nieminen
> Timo Nieminen <ti...@physics.uq.edu.au> writes
>> On Thu, 29 Jun 2006, Oz wrote:
>>
>>> scerir <sce...@libero.it> writes
>>>
>>>> It seems that this experiment led to Dirac's famous koan
>>>> "each photon then interferes only with itself".
>>>
>>> Except we know this is not correct. Experiments have been done where two
>>> separate lasers fired through two separate slits produces an
>>> interference pattern.
>>
>> That just shows that each photon comes from both sources.
>
> Hmmm....
>
> As an explanation I think Mr Occham would find this improbable.
While William of Oakham/Occam would likely have found it improbable, due
to not knowing about either electromagnetic fields or photons, it is
not incompatible with his razor. As I wrote:
>> A photon is the
>> quantum of excitation of the EM field, the EM field is the sum of the
>> fields individually produced by the sources, so each photon comes from
>> both sources.
>
> A contortion IMHO.
> A much more plausible explanation is that photons are completely
> wavelike.
The EM field is completely wavelike. Much, even most, stuff about photons
comes from purely classical theory. However, by definition photons are
_not_ completely wavelike, being the quanta of excitation/de-excitation of
the field.
Your "much more plausible" completely ignores all of the observational,
experimental, and theoretical evidence for the existence of photons. But
yes, the original two-slit experiment did held overthrow the old
corpuscular theories of light and replace them with a wave theory of
light. But note well that the old corpuscules of light (proto-photons?)
were thought of as classical particles, and the modern photon is not a
classical particle.
> One then looks to apparent quantum behaviour in the emitters
> and/or detectors where of course one finds them.
One looks _at_ quantum behaviour in the emitters/detectors. Where else
will you find excitation/de-excitation of the EM field?
An October issue of Optics and Photonics News (iirc, in 2003) had a nice
special section on "What is a photon?". Read it. Also Lamb, "Anti-photon",
Applied Physics B from 1995.
> NB I must logically also conclude that massive particles are also waves
> and once again the quantised behaviour is due to something else.
Very de Broglie. It's been done. The "something else" is "that's the way
that nature works". If photons are waves, and massive particles are waves,
then why is the exchange of energy between matter and EM fields quantised?
Observably, it is. "Why" might be a deeper question than we can answer
(yet). Asking why hbar is non-zero and has the value we measure is like
asking why is c non-infinite with the value we measure(d).
> Clearly
> (to me at any rate) the 'size' of a massive particle is determined (like
> the photon) by its environment (typically quantised). So an electron,
> say, can have a different physical size when undisturbed in an orbital
> This can in some circumstances be very large indeed when orbitals become
> macroscopic, for example in conductors.
Don't confuse localisation with size. If a photon is "large", it should be
able to interact with and be detected by two spatially separated detectors
at the same time.
Oz
>
>Personally I vote for experimental results. My question is whether
>anyone has any idea how it could be done? Personally I would bet you £50
>it will show the same result as the photon experiment.
Bet rescinded....
It would need to be necessary to have two moving electron CLOUDS (that
is covering a large area) that emulate a laser beam. I'm fairly sure
that an electron beam does NOT have this characteristic.
FrediFizzx
news:Pine.WNT.4.64.06...@serene.st...
I don't think we actually know the answer to that "question" since
"large" photons might be radio wave photons and so far individual
detection of such photons is not possible. Do you know of any
experimental limits that we might have for this? IOW, what is the
lowest frequency at which individual detection is experimentally
possible with current technology?
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
Blackbird
> Blackbird <fa...@nospam.no> writes
[...]
>> I don't think that this is the same kind of interference. If you
>> add up two amplitude modulated waves of the same frequency, then
>> obviously the result will be garbled, but this has a different
>> explanation than the double slit experiment has.
>
> Why? Radio waves are perfectly good EM waves just like light but with
> different frequency.
Sure, so I'll try to explain my point here a little better. Say we have a
vertical receiving antenna, and two MW transmitters that transmit waves of
the same frequency, but perfectly out of phase (relatively shifted by 1/2
the wavelength) at the location of the antenna. The waves transfer energy
inducing electrons in the antenna to accelerate. According to the theory,
this energy transfer (and thus the acceleration) is quantized, hence
"photons". Now, any random electron will from time to time absorb a photon
that either accelerates it in the "up" direction, or in the "down"
direction, and since we have two sources with cancelling phases, for any
finite (and sufficiently large) interval of time, the electron will absorb
approximately as many "up" as "down" photons. The electron will thus
exhibit a random walk, and no signal will be detected. This, however, does
not mean that the *photons* interfere with eachother. Interference, as in
the double slit experiment, would mean that photons from the two different
sources cancelled each other (or more precisely, they would be more likely
to show up at another location), thus there would be no energy absorbtion by
the electron whatsoever.
Oz
>
>I've given a demonstration again, in response to scerir. It is not
>difficult, and nicely illustrates that observables, such as interference
>effects, rely on the properties of operators, not just states, and thus
>what is wrong with the view of superposition given in Dirac's koan.
Yes. Nicely done, I even mostly understood it <cough>.
>>Personally I vote for experimental results. My question is whether
>>anyone has any idea how it could be done? Personally I would bet you £50
>>it will show the same result as the photon experiment.
>
>I'll take you up on that. Will you accept this as an electronic
>handshake?
Earlier today I rescinded same....
>>The initial aim would be to produce extremely monochromatic beams of
>>electrons.
>
>Ahem, you can't have an electron equivalent of a laser. They are
>fermions and cannot all be in the same state.
Absolutely true, hence (luckily for me) no such experiment can be made.
>> Given that we are looking for exceedingly low beam intensity
>>I suspect that the problems associated with mutual repulsion of
>>electrons within a beam will be negligible (or am I being
>>oversimplistic?).
>
>You are forgetting the exclusion principle, which is rather more
>encompassing than charge.
Absolutely. Perhaps a massive boson beam would be better.
I guess that probably means they would be entangled so as to get the
long thin coherent beam. Hmm....
>>One is almost tempted to start with very low energy electrons, perhaps
>>separating them by time-of-flight (via a chopper)
>
>No you wouldn't be allowed to do that either. Remember, when they do
>this for photons the stipulation that only one photon comes through per
>amount of time is statistical. In one of these low energy laser thingies
>any photon can come through at any time, but it is only fed so much
>energy so there is only one per amount of time.
This was to ensure a very precise *energy*, but I wasn't thinking very
carefully and imagining a neutron spallation source sort of thing which
is inappropriate.
Timo A. Nieminen
> "Timo A. Nieminen" <ti...@physics.uq.edu.au> wrote:
>> On Fri, 30 Jun 2006, Oz wrote:
>
>>> Clearly (to me at any rate) the 'size' of a massive particle is
>>> determined (like the photon) by its environment (typically
>>> quantised). So an electron, say, can have a different physical size
>>> when undisturbed in an orbital This can in some circumstances be
>>> very large indeed when orbitals become macroscopic, for example in
>>> conductors.
>>
>> Don't confuse localisation with size. If a photon is "large", it
>> should be able to interact with and be detected by two spatially
>> separated detectors at the same time.
>
> I don't think we actually know the answer to that "question" since
> "large" photons might be radio wave photons and so far individual
> detection of such photons is not possible. Do you know of any
> experimental limits that we might have for this? IOW, what is the
> lowest frequency at which individual detection is experimentally
> possible with current technology?
The claim regarding electrons having different sizes depending on whether
"they are undisturbed in an orbital" reads to me as an identification of
size with spread of wavefunction. This being independent of wavelength
(but see below), low frequency isn't necessary. In fact, it's best to use
the highest frequencies available, so that the energy required for
detection is only a small fraction of the total energy. Put a gamma source
in the middle of some, preferably many, detectors. Each gamma photon can
go in any direction, the radiation field of each emission is spherically
symmetric (well, perhaps a dipole field, but spherically symmetric
averaged over many). Count, and look for coincidences. Has this been done?
I don't know offhand. You should look; if it hasn't been done, it could be
a cheap and useful entry into experimental physics for you.
But yes, there is a problem with trying to make sub-wavelength individual
photon detectors. Except for single atoms/molecules, but then you still
need an individual photon detector to detect the re-emission. IME, this is
in the near IR.
I guess that you're wondering about size of photons as it might depend on
wavelength. The above means that it's hard to answer experimentally. I
think that illuminating a group of atoms, all within a wavelength, and
only one of them absorbs and re-emits, is conclusive - the "size" of a
photon is no larger than an atom. Compton can be interpreted as saying
that the size of a photon is the size of an electron, ie zero AFAWCT.
But do consider the above gamma experiment. Data is good for you. All
theoreticians should be forced into labs at some point!
Oh No
>Absolutely. Perhaps a massive boson beam would be better. I guess that
>probably means they would be entangled so as to get the long thin
>coherent beam. Hmm....
This gets pretty confusing. Lasers are weird things, and I don't profess
to fully understand them, but I think they are pretty dependent on the
properties of photons as fundamental particles. IIRC It has been
remarked here, by those more qualified than I, that the properties of a
bose gas consisting of fundamental bosons are not the same as one
consisting of bosons which are themselves compound particles consisting
of fermions, the reason being that if compound particles were in exactly
the same state it would imply that the fermions of which they consist
would be in the same state, yielding a contradiction.
Looking through the derivation I gave it seems to me that the critical
distinction was that photons are annihilated when they are detected, so
that the amplitude for detection at x is <|A(x)|f>, a pure number. We
can add numbers, so we can get interference effects between different
photons. Fullerenes, for example are not annihilated, and the position
operator is |x><x| which yields |x><x|f>, a state. We still cannot
superpose states of different particles, so that means we can expect
interference effects between different photons, but not interference
effects between different fullerenes.
Paul Danaher
> "Timo A. Nieminen" <ti...@physics.uq.edu.au> wrote in message
> news:Pine.WNT.4.64.06...@serene.st...
>> On Fri, 30 Jun 2006, Oz wrote:
>
>>> Clearly (to me at any rate) the 'size' of a massive particle is
>>> determined (like the photon) by its environment (typically
>>> quantised). So an electron, say, can have a different physical size
>>> when undisturbed in an orbital This can in some circumstances be
>>> very large indeed when orbitals become macroscopic, for example in
>>> conductors.
>>
>> Don't confuse localisation with size. If a photon is "large", it
>> should be able to interact with and be detected by two spatially
>> separated detectors at the same time.
>
> I don't think we actually know the answer to that "question" since
> "large" photons might be radio wave photons and so far individual
> detection of such photons is not possible. Do you know of any
> experimental limits that we might have for this? IOW, what is the
> lowest frequency at which individual detection is experimentally
> possible with current technology?
>
> FrediFizzx
This seems to bring up again the problem of the dual answers to my question
about how long it takes to absorb a photon. One answer was that the wave
function collapses instantaneously, another that it depends on the frequency
of the photon. Your "large" photon brings me back to my puzzle about when
absorption occurs - is it on the arrival of the leading edge of the wave
packet/probability density function, or at a point at which this reaches
some threshold value?
(Thank you for the further puzzles in the Diether paper you cite, with his
comment that the photon energy density volume for a radio wave could be as
big as a house or bigger.)
nightlight
> This being independent of wavelength
> (but see below), low frequency isn't necessary. In fact, it's best to use
> the highest frequencies available, so that the energy required for
> detection is only a small fraction of the total energy. Put a gamma source
> in the middle of some, preferably many, detectors. Each gamma photon can
> go in any direction, the radiation field of each emission is spherically
> symmetric (well, perhaps a dipole field, but spherically symmetric
> averaged over many). Count, and look for coincidences. Has this been done?
> I don't know offhand. You should look; if it hasn't been done, it could be
> a cheap and useful entry into experimental physics for you.
Just detecting a particle-like anticorrelation in coincidences is not
by itself proof of anything non-classical or unusual in any way. After
all, kicking a ball into a fence with large holes in it will result in
the ball ending up exclusively on one side of the fence or the other.
And the distribution of the ball's stopping places may well be
symmetrical with respect to the fence.
You also need to show the interference if you bring the far away paths
back together. With gamma rays the state may be spherically
symmetrical, but it is not a pure state. It is a spherically
symmetrical mixture (since the recoil of the source nucleus can be used
to detect the direction of the gamma photon in individual emissions).
With optical photons any recoil (such as that of a heavy half-silvered
mirror) is negligible (i.e. comparable to the Heisenberg uncertainties)
to allow measurement of whether the photon was reflected or
transmitted. In that case the separate wave packet components retain
coherence and can thus interfere when brought together. If you were to
make the half-silvered mirror/beam splitter used with optical photons
light enough to undergo recoil sufficient to find out direction of the
photon, you will automatically lose the interference pattern. There is
a continuum of tradeoffs between the sharpness (visibility) of the
interference fringes and the certainty of detection (e.g. via recoil)
which way the photon went, with the perfect interference and the
perfectly certain recoil effect on its opposite extrema.
Similarly, if you make gamma source produce coherent enough and equal
(in total energy) enough wave packet components in far away regions to
fully interfere when brought together, the coincidence measurements on
the remote packets will show the Poissonian coincidence rates (i.e.
each fragment will trigger or not trigger its detectors independently
form what the other remote detectors did i.e. the fragment A will not
suddenly vanish/collapse when the fragment B triggers its detector).
Check the cited materials from my other two posts in this thread on the
experimental and theoretical status of this conjectured wave-particle
"duality" (with collapse of the remote wave packet) -- it has never
been observed experimentally or shown to exist theoretically even in
principle (within QED; the Glauber's theory of photo-detections and
photo-correlations in his 1964 Les Houches lectures was the high point
of such efforts on the theoretical side) despite over five decades of
attempts, the handwaving in the popular and pedagogical literature
notwithstanding.
Timo A. Nieminen
> Timo A. Nieminen wrote:
>> This being independent of wavelength
>> (but see below), low frequency isn't necessary. In fact, it's best to use
>> the highest frequencies available, so that the energy required for
>> detection is only a small fraction of the total energy. Put a gamma source
>> in the middle of some, preferably many, detectors. Each gamma photon can
>> go in any direction, the radiation field of each emission is spherically
>> symmetric (well, perhaps a dipole field, but spherically symmetric
>> averaged over many). Count, and look for coincidences. Has this been done?
>> I don't know offhand. You should look; if it hasn't been done, it could be
>> a cheap and useful entry into experimental physics for you.
>
> Just detecting a particle-like anticorrelation in coincidences is not
> by itself proof of anything non-classical or unusual in any way.
The proposal is just to address the claim that the photon has a size equal
to the spread of its wavefunction, and that detection probability is
simply proportional to E. Yes, by itself, that doesn't prove anything
non-classical. For example, classical billiard ball photons will behave
this way. OTOH, I think there's already ample evidence that photons aren't
classical billiard balls (eg Hanbury Brown-Twiss).
> Similarly, if you make gamma source produce coherent enough and equal
> (in total energy) enough wave packet components in far away regions to
> fully interfere when brought together, the coincidence measurements on
> the remote packets will show the Poissonian coincidence rates (i.e.
> each fragment will trigger or not trigger its detectors independently
> form what the other remote detectors did i.e. the fragment A will not
> suddenly vanish/collapse when the fragment B triggers its detector).
Have a weak enough source so that in some time interval longer than the
time required for the detector to click and recover, you only expect one
photon to be emitted, or else there's no point to the experiment. It's not
intended as a "collapse of the wavefunction" experiment.
> Check the cited materials from my other two posts
Will do.
Oz
>Don't confuse localisation with size. If a photon is "large", it should be able
>to interact with and be detected by two spatially separated detectors at the
>same time.
In a sense it can, and in a sense it cannot.
Given exited enough detectors of course one can. Exited enough detectors
are by definition noisy and nobody can tell if two transitions are due
to the EM wave tripping two or whether one (or both) are noise.
We can only tell that on average the detector is best modelled by it
making quantum transitions. This is unsurprising given that detectors
are quantised.
Oz
>Your "large" photon brings me back to my puzzle about when
>absorption occurs - is it on the arrival of the leading edge of the wave
>packet/probability density function, or at a point at which this reaches
>some threshold value?
When it is detected, that is, when an effectively irreversible
transition has occurred. I don;t see how this can be denied.
Oz
>The claim regarding electrons having different sizes depending on whether "they
>are undisturbed in an orbital" reads to me as an identification of size with
>spread of wavefunction.
Its a very excellent description of size.
We are talking size in meters here, not energy.
>This being independent of wavelength (but see below),
>low frequency isn't necessary. In fact, it's best to use the highest frequencies
>available, so that the energy required for detection is only a small fraction of
>the total energy.
Er, yes, but such an object has a very short wavelength so is inherently
more localised (or 'smaller').
>Put a gamma source in the middle of some, preferably many,
>detectors. Each gamma photon can go in any direction, the radiation field of
>each emission is spherically symmetric (well, perhaps a dipole field, but
>spherically symmetric averaged over many).
I'm not actually completely in agreement with this statement.
On average this is so, and on average (that is, over many counts) you
get the right answer.
But the uncertainty of momentum may not be spherical because you could
measure the recoil of the source (if the gamma is energetic enough),
which would destroy the spherical symmetry.
>I guess that you're wondering about size of photons as it might depend on
>wavelength. The above means that it's hard to answer experimentally. I think
>that illuminating a group of atoms, all within a wavelength, and only one of
>them absorbs and re-emits, is conclusive - the "size" of a photon is no larger
>than an atom. Compton can be interpreted as saying that the size of a photon is
>the size of an electron, ie zero AFAWCT.
Absolutely not, as diffraction will confirm.
Remember that an aerial does NOT have to be a wavelength long to
receive. In fact its normal for low frequencies to put appropriate L & C
on a short aerial so as to tune it to the required low frequency/long
wavelength. Similarly you can see an absorbing (or emitting) atom as
tuned to its transition frequency, with the masses of the
electron/nucleus and associated electromagnetic field interaction tuning
it to the very low (in wavelength terms) transition frequency.
Just as a properly tuned small aerial can transmit a very long
wavelength signal, so a very small atom is perfectly capable of being
tuned to a very long wavelength em wave.
Clearly most transitions take very many cycles of emission before the
transition is completed.
>But do consider the above gamma experiment. Data is good for you. All
>theoreticians should be forced into labs at some point!
<shudder>
<a theoretician in a lab is probably a health hazard>
Oz
>Sure, so I'll try to explain my point here a little better. Say we have a
>vertical receiving antenna, and two MW transmitters that transmit waves of
>the same frequency, but perfectly out of phase (relatively shifted by 1/2
>the wavelength) at the location of the antenna. The waves transfer energy
>inducing electrons in the antenna to accelerate. According to the theory,
>this energy transfer (and thus the acceleration) is quantized, hence
>"photons". Now, any random electron will from time to time absorb a photon
>that either accelerates it in the "up" direction, or in the "down"
>direction, and since we have two sources with cancelling phases, for any
>finite (and sufficiently large) interval of time, the electron will absorb
>approximately as many "up" as "down" photons.
Pah! That would mean you got increased random noise, which you do NOT
get. The environment would be 'hotter'.
>The electron will thus
>exhibit a random walk, and no signal will be detected. This, however, does
>not mean that the *photons* interfere with eachother.
Its by far the simplest explanation. The em field is zero and no photons
are detected.
>Interference, as in
>the double slit experiment, would mean that photons from the two different
>sources cancelled each other (or more precisely, they would be more likely
>to show up at another location), thus there would be no energy absorbtion by
>the electron whatsoever.
Which is precisely the effect of radio waves.
Think about it.
Timo A. Nieminen
> Timo A. Nieminen <ti...@physics.uq.edu.au> writes
>
>> The claim regarding electrons having different sizes depending on whether "they
>> are undisturbed in an orbital" reads to me as an identification of size with
>> spread of wavefunction.
>
> Its a very excellent description of size.
> We are talking size in meters here, not energy.
You are saying that the "spread of a wavefunction" = "size of particle"?
>> This being independent of wavelength (but see below),
>> low frequency isn't necessary. In fact, it's best to use the highest frequencies
>> available, so that the energy required for detection is only a small fraction of
>> the total energy.
>
> Er, yes, but such an object has a very short wavelength so is inherently
> more localised (or 'smaller').
No! Not at all! What does wavelength have to do with localisation? It is
relevant to _minimum possible_ localisation, but a monochromatic plane
wave mode is _completely_ unlocalised, in both space and time,
independently of the wavelength.
>> Put a gamma source in the middle of some, preferably many,
>> detectors. Each gamma photon can go in any direction, the radiation field of
>> each emission is spherically symmetric (well, perhaps a dipole field, but
>> spherically symmetric averaged over many).
>
> I'm not actually completely in agreement with this statement.
> On average this is so, and on average (that is, over many counts) you
> get the right answer.
>
> But the uncertainty of momentum may not be spherical because you could
> measure the recoil of the source (if the gamma is energetic enough),
> which would destroy the spherical symmetry.
So don't measure the recoil. No problem!
>> I guess that you're wondering about size of photons as it might depend on
>> wavelength. The above means that it's hard to answer experimentally. I think
>> that illuminating a group of atoms, all within a wavelength, and only one of
>> them absorbs and re-emits, is conclusive - the "size" of a photon is no larger
>> than an atom. Compton can be interpreted as saying that the size of a photon is
>> the size of an electron, ie zero AFAWCT.
>
> Absolutely not, as diffraction will confirm.
No, diffraction just tells you that quantum "particles" are not classical
billiard balls. The wavefunction can have finite (ie non-zero) extent
(and, technically, extends to infinity in all directions, although the
amplitude of the wavefunction may be close enough to zero so you can
pretend it's bounded in space), but that doesn't mean that the "particle"
it prescribes the probability distribution of is as large as the
wavefunction.
Diffraction tells you about wavefunctions. Diffraction of EM waves tells
you about EM waves, not about whether energy is exchanged between EM waves
and matter in quanta.
If all the energy goes to one atom, and isn't divided among many, the
quantum is one atom in size, at most, regardless of how many atoms wide
the wave is.
> Remember that an aerial does NOT have to be a wavelength long to
> receive. In fact its normal for low frequencies to put appropriate L & C
> on a short aerial so as to tune it to the required low frequency/long
> wavelength. Similarly you can see an absorbing (or emitting) atom as
> tuned to its transition frequency, with the masses of the
> electron/nucleus and associated electromagnetic field interaction tuning
> it to the very low (in wavelength terms) transition frequency.
>
> Just as a properly tuned small aerial can transmit a very long
> wavelength signal, so a very small atom is perfectly capable of being
> tuned to a very long wavelength em wave.
>
> Clearly most transitions take very many cycles of emission before the
> transition is completed.
No. The energy difference between the upper and lower levels is hf. To go
through many cycles would mean pumping out and sucking back that energy
over and over. Yet the energy is launched out at c, and you can suck it
back. Yes, for a short dipole, you put energy into the near field, and get
it back. Yes, the wavefunction of the electron oscillates (and see my
recent post in sci.physics on a content-related thread for more on this,
and some nifty references supplied by others). However, classically, you
never have "the transition is completed". Never. This is a quantum
concept. Classically, you have a damped oscillator, which never decays to
zero, but a photon is emitted, and maybe detected, and the atom is
demonstrably in the lower state.
Blackbird
> Blackbird <fa...@nospam.no> writes
>> Sure, so I'll try to explain my point here a little better. Say we
>> have a vertical receiving antenna, and two MW transmitters that
>> transmit waves of the same frequency, but perfectly out of phase
>> (relatively shifted by 1/2 the wavelength) at the location of the
>> antenna. The waves transfer energy inducing electrons in the antenna
>> to accelerate. According to the theory, this energy transfer (and
>> thus the acceleration) is quantized, hence "photons". Now, any
>> random electron will from time to time absorb a photon that either
>> accelerates it in the "up" direction, or in the "down" direction,
>> and since we have two sources with cancelling phases, for any finite
>> (and sufficiently large) interval of time, the electron will absorb
>> approximately as many "up" as "down" photons.
>
> Pah! That would mean you got increased random noise, which you do NOT
> get. The environment would be 'hotter'.
Yes, and this is exactly what I believe will happen. The energy absorbtion
should lead to a small rise in the temperature of the antenna, even though
no information will be detected.
>
>> The electron will thus
>> exhibit a random walk, and no signal will be detected. This,
>> however, does not mean that the *photons* interfere with eachother.
>
> Its by far the simplest explanation. The em field is zero and no
> photons are detected.
>
>> Interference, as in
>> the double slit experiment, would mean that photons from the two
>> different sources cancelled each other (or more precisely, they
>> would be more likely to show up at another location), thus there
>> would be no energy absorbtion by the electron whatsoever.
>
> Which is precisely the effect of radio waves.
> Think about it.
In the radio wave case, with two different transmitters, you should be able
to pick up one of the signals with a directional antenna. This is
impossible in the double slit experiment (with equipment analogous to a
directional antenna, say a focusing lens), since seen from the points of
destructive interference, both slits look dark. There is simply no energy
reaching these points.
Oz
>On Fri, 21 Jul 2006, Oz wrote:
>
>> Timo A. Nieminen <ti...@physics.uq.edu.au> writes
>>
>>> The claim regarding electrons having different sizes depending on whether
>"they
>>> are undisturbed in an orbital" reads to me as an identification of size with
>>> spread of wavefunction.
>>
>> Its a very excellent description of size.
>> We are talking size in meters here, not energy.
>
>You are saying that the "spread of a wavefunction" = "size of particle"?
Of course, its certainly the area where it interacts which is the best
description of 'size' I can think of. However I would distinguish
between the wavefunction, which can vary depending on our knowledge, and
the actual EM wave. MBR emitting atom as an example see below.
The fact that this can sometimes be absolutely HUGE (as in electrons in
metals/graphite and other things) is interesting but not counter-
evidence. As a wave the size will clearly depend on any boxes it gets
put in or other restrictions.
>>> This being independent of wavelength (but see below),
>>> low frequency isn't necessary. In fact, it's best to use the highest
>frequencies
>>> available, so that the energy required for detection is only a small fraction
>of
>>> the total energy.
>>
>> Er, yes, but such an object has a very short wavelength so is inherently
>> more localised (or 'smaller').
>
>No! Not at all! What does wavelength have to do with localisation? It is
>relevant to _minimum possible_ localisation, but a monochromatic plane
>wave mode is _completely_ unlocalised, in both space and time,
>independently of the wavelength.
See above, I agree. Why do you see that as a problem?
>>> Put a gamma source in the middle of some, preferably many,
>>> detectors. Each gamma photon can go in any direction, the radiation field of
>>> each emission is spherically symmetric (well, perhaps a dipole field, but
>>> spherically symmetric averaged over many).
>>
>> I'm not actually completely in agreement with this statement.
>> On average this is so, and on average (that is, over many counts) you
>> get the right answer.
>>
>> But the uncertainty of momentum may not be spherical because you could
>> measure the recoil of the source (if the gamma is energetic enough),
>> which would destroy the spherical symmetry.
>
>So don't measure the recoil. No problem!
Something (unless you go to extreme lengths) will interact with the
particle sooner or later, and this particle will react with others etc
etc. This means that for (taking the age-old example) a MBR emitting
atom its momentum will be fixed well before the 'photon' interacts.
>>> I guess that you're wondering about size of photons as it might depend on
>>> wavelength. The above means that it's hard to answer experimentally. I think
>>> that illuminating a group of atoms, all within a wavelength, and only one of
>>> them absorbs and re-emits, is conclusive - the "size" of a photon is no
>larger
>>> than an atom. Compton can be interpreted as saying that the size of a photon
>is
>>> the size of an electron, ie zero AFAWCT.
>>
>> Absolutely not, as diffraction will confirm.
>
>No, diffraction just tells you that quantum "particles" are not classical
>billiard balls. The wavefunction can have finite (ie non-zero) extent
>(and, technically, extends to infinity in all directions, although the
>amplitude of the wavefunction may be close enough to zero so you can
>pretend it's bounded in space), but that doesn't mean that the "particle"
>it prescribes the probability distribution of is as large as the
>wavefunction.
Indeed, see above. However I still claim that the MINIMUM (see what I
actually said) size is of the order of a wavelength.
>Diffraction tells you about wavefunctions. Diffraction of EM waves tells
>you about EM waves, not about whether energy is exchanged between EM waves
>and matter in quanta.
Of course. However if you look elsewhere someone with more knowledge
than me has described a typical detector more as I see it but in proper
language.
===From: nightlight <night...@omegapoint.com>
Newsgroups: sci.physics.research
Subject: Re: Two-slit experiment
Approved: (sci.physics.research)
Message-ID: <1151472082....@p79g2000cwp.googlegroups.com>
The facts are that the response of an array of photodetectors is
statistically indistingushable from droplets statistics of an array of
dripping faucets. Specifically, the best (the most narrow distribution
of triggers) that you will ever get here is a Poissonian distribution
(as long as QED vacuum is there and its vacuum fluctuations).
There isn't anything 'single-photon-like' about it that can
distinguish it from a thresholded EM field measurements (with
photo-electron current measurements results reduced via AD
conversion by the 'pulse analayzer & discriminator' electronics
to a single bit precision, 0/1 values, based on the detector
thresholds) of a classical field.
=======end nightlight
>If all the energy goes to one atom, and isn't divided among many,
Of course, where is the problem here?
>the
>quantum is one atom in size, at most, regardless of how many atoms wide
>the wave is.
No, you are confusing the detector and the wave.
To understand more clearly imagine the atome emitting such a huge long
wave, then play this backwards. Not a problem.
>> Remember that an aerial does NOT have to be a wavelength long to
>> receive. In fact its normal for low frequencies to put appropriate L & C
>> on a short aerial so as to tune it to the required low frequency/long
>> wavelength. Similarly you can see an absorbing (or emitting) atom as
>> tuned to its transition frequency, with the masses of the
>> electron/nucleus and associated electromagnetic field interaction tuning
>> it to the very low (in wavelength terms) transition frequency.
>>
>> Just as a properly tuned small aerial can transmit a very long
>> wavelength signal, so a very small atom is perfectly capable of being
>> tuned to a very long wavelength em wave.
>>
>> Clearly most transitions take very many cycles of emission before the
>> transition is completed.
>
>No. The energy difference between the upper and lower levels is hf. To go
>through many cycles would mean pumping out and sucking back that energy
>over and over.
This isn't so in non-linear systems.
See parametric amplifier.
Quantum mechanics is typically rather non-linear.
In any case consider the reverse of an atom emitting a photon.
>Yet the energy is launched out at c, and you can suck it
>back. Yes, for a short dipole, you put energy into the near field, and get
>it back. Yes, the wavefunction of the electron oscillates (and see my
>recent post in sci.physics on a content-related thread for more on this,
>and some nifty references supplied by others). However, classically, you
>never have "the transition is completed".
Well, of course you can have equivalent systems, see expert example
above. You cannot, however, expect to model a pure quantum example non-
quantum-mechanically. For the non-relativistic H-atom emission, though,
I believe it can be done using schroedinger only.
>Never. This is a quantum
>concept. Classically, you have a damped oscillator,
Why damped? Damping energy is going where?
>which never decays to
>zero, but a photon is emitted, and maybe detected, and the atom is
>demonstrably in the lower state.
Of course. Why do you see that as a problem? Atoms are, as I have said
repeatedly, quantum devices, they have energy levels. Thats where
quantum mechanics works well. The fact that the detectors and emitters
are quantum mechanical does not mean the EM wave is, although its hugely
convenient to consider it to be so.
Oz
>Oz wrote:
>> Blackbird <fa...@nospam.no> writes
>>> Sure, so I'll try to explain my point here a little better. Say we
>>> have a vertical receiving antenna, and two MW transmitters that
>>> transmit waves of the same frequency, but perfectly out of phase
>>> (relatively shifted by 1/2 the wavelength) at the location of the
>>> antenna. The waves transfer energy inducing electrons in the antenna
>>> to accelerate. According to the theory, this energy transfer (and
>>> thus the acceleration) is quantized, hence "photons". Now, any
>>> random electron will from time to time absorb a photon that either
>>> accelerates it in the "up" direction, or in the "down" direction,
>>> and since we have two sources with cancelling phases, for any finite
>>> (and sufficiently large) interval of time, the electron will absorb
>>> approximately as many "up" as "down" photons.
>>
>> Pah! That would mean you got increased random noise, which you do NOT
>> get. The environment would be 'hotter'.
>
>Yes, and this is exactly what I believe will happen. The energy absorbtion
>should lead to a small rise in the temperature of the antenna, even though
>no information will be detected.
No, because the field strength is demonstrably zero.
It can be measured.
>>> The electron will thus
>>> exhibit a random walk, and no signal will be detected. This,
>>> however, does not mean that the *photons* interfere with eachother.
>>
>> Its by far the simplest explanation. The em field is zero and no
>> photons are detected.
>>
>>> Interference, as in
>>> the double slit experiment, would mean that photons from the two
>>> different sources cancelled each other (or more precisely, they
>>> would be more likely to show up at another location), thus there
>>> would be no energy absorbtion by the electron whatsoever.
>>
>> Which is precisely the effect of radio waves.
>> Think about it.
>
>In the radio wave case, with two different transmitters, you should be able
>to pick up one of the signals with a directional antenna. This is
>impossible in the double slit experiment (with equipment analogous to a
>directional antenna, say a focusing lens), since seen from the points of
>destructive interference, both slits look dark. There is simply no energy
>reaching these points.
That would be an interesting example to do.
However you will then see a signal simply because there is NOT a path
from one slit to the detector which cannot see the other slit. This
would be equally true whether you used two transmitters (lasers, stable,
very) or one transmitter and reflected the radio wave (or light) to the
same location as the second slit. You certainly cannot differentiate
between self-photon interference and separate-photon interference by
this method.
In essence you CAN tell which slit the light came from because your
directional antenna detects the direction (momentum).
Ralph Hartley
> In the radio wave case, with two different transmitters, you should be able
> to pick up one of the signals with a directional antenna.
A directional antenna cannot have zero size. In order to resolve the two
transmitters, the antenna would need to be larger than the nodes in the
interference pattern.
Ralph Hartley
Blackbird
Yeas, you are right on that point. After my post, I did a
back-of-the-envelope calculation on the interference pattern from two
geostationary staellites with 3 degrees separation transmitting 2,5 cm
waves. It seems that the distance between destructive regions in the
pattern is approximately 0,5 m, and that is about the diameter you need for
a satellite dish to give good separation between two satellites that close.
Edward Green
> The EM field is completely wavelike. Much, even most, stuff about photons
> comes from purely classical theory. However, by definition photons are
> _not_ completely wavelike, being the quanta of excitation/de-excitation of
> the field.
I take it then the very simplest model of a field excited/de-excited by
"photons" would be the QHO. Or possibly this is the complete model, in
general only more complexified by additional modes, and additional
complication added by the fact that the electromagnetic field has
degrees of freedom not completely analogous with a simple mass on a
spring?
Hit, or a miss?
I also seem to recall hearing it said that the description of the
excitation of the field in terms of photons had an arbitrary element,
depending how we selected the modes.
Any truth in this?
If the photons are "quanta of excitation/de-excitation of the field",
maybe it doesn't even make sense to ask whether one photon only
interferes "with itself" as opposed to "with other photons". Surely it
is the total excitation of the field which is interfering with itself,
which is to say "arriving at a value at each event which can be
decomposed into the result of various independent processes, whose
resultants add", IOW "behaves linearly", or yet IOW "behaves like a
vector".
If we are constrained to excite/de-excite a QHO by "quanta", is it not
still the single value of psi, the wave function, which is doing the
interfering -- i.e., evolving linearly via the hamiltonian -- and not
the excitations/de-excitations?
[I seem to be taking sides with Oz, though I didn't really start out to
argue one side or the other. Such is the attractive pull of rhetoric,
even when one set out merely to grope.]
Timo A. Nieminen
> Timo A. Nieminen <ti...@physics.uq.edu.au> writes
>
>> Don't confuse localisation with size. If a photon is "large", it should be able
>> to interact with and be detected by two spatially separated detectors at the
>> same time.
>
> In a sense it can, and in a sense it cannot.
>
> Given exited enough detectors of course one can. Exited enough detectors
> are by definition noisy and nobody can tell if two transitions are due
> to the EM wave tripping two or whether one (or both) are noise.
1) Use EM waves where hf >> energy required to "click" the detectors.
2) Count for long enough (both with and without source) to get reliable
statistics.
3) Energy required to "click" detectors >> kT
.. which is the point of using a gamma source.
> We can only tell that on average the detector is best modelled by it
> making quantum transitions. This is unsurprising given that detectors
> are quantised.
No, you can do better than that, if hf >> detector-tripping-energy.
OK, this whole suggestion is for a rather simple-minded experiment aimed
at skeptics, rather than the various perhaps stronger but more obscure
"non-classical" interference experiments out there, but still, the point
remains, can a photon be detected in more than one place at once? If no,
then it has zero size; if yes, then it must be detectable in more than one
place at once.
Blackbird
out for me. Although what I have to say will be trivial to most of you, a
Google search reveals that the question about Dirac's statement and
interference from multiple sources pops up again and again, so I will
briefly write down what I've learned, hopefully to the benefit of some
ignorants like myself. (I do skip some details about coherence etc. While
important for a detailed discussion, I'm trying to make a simpler point
here.)
Dirac wrote; "Each photon then interferes only with itself. Interference
between two different photons never occurs."
(As written, the statement is taken out of its context, and when read in
context, it is obvious that Dirac intended a narrower meaning than what it
looks like, but that's not really the point here.)
And the objection is now, "Dirac must be wrong here. There are experiments
involving two different light sources that show interference, and radio
waves from two different transmitters interfere. Obviously, as single photon
can't originate from more than one source, so how on earth can we get
multiple source interference if this photon only interferes with itself?"
Let's se what Dirac is *not* saying; he is not saying that EM-waves from two
different sources don't interfere. They do, and Dirac obviously knew that.
The objection to his statement fails because of the sentence "Obviously, as
single photon can't originate from more than one source."
The source of a photon is not always well defined. When you experience
interference from two different transmitters (e.g., radio stations), it is
simply undefined from which transmitter the photon originated. It is not
that we just don't know it; it is *undefined*. According to QED and the
uncertainty principle, you *must* regard the photon as coming from both
sources, in the same way that we regard the photon as going through both
slits in Young's experiment. Try to pin down the source of the photon, and
the interference disappears. If you do the field theoric calculations (as
Walls do in the paper), it turns out that "Each photon then interferes only
with itself" is a pretty good description of the math involved, even with
two or more sources.
This also explains why two different sources will produce an interference
pattern even if the sources are of such a low intensity that the
probability of having both of them emitting a photon at approximately the
same time is neglible. As long as your experiment is set up in such a way
that it is impossible to ascribe the photon to either of the two sources
separately, the uncertainty principle says that you have essentially
recreated Young's experiment. In essence, QED doesn't distingush between
the two experiments.
Finally, let me add that we see a crude, but large-scale realization of the
uncertainty principle in our everyday lives. Two nearby broadcasting
satellites will in general interfere with eachother. Now, in this case,
interference is a bad thing, so we try to avoid it. We do that by trying to
pin down the momentum of the photon, i.e., decide what satellite it
originated from. This, however, will be at a cost; since its position is
now less defined, we need a larger satellite dish to pick it up. In fact,
to be almost certain about the photon's origin, the dish must at least span
one fringe with. And that means the interference is undetectable.
[1] Walls, D. F., 1977,"A simple field theoretic description of photon
interference," Am J Physics, 45(10)
Oz
Yes.
>2) Count for long enough (both with and without source) to get reliable
>statistics.
Yes.
>3) Energy required to "click" detectors >> kT
>
>.. which is the point of using a gamma source.
and this shows what, exactly?
Remember pretty well all transitions are effectively tuned,
which is why we can use very low intensity em waves and still get
transitions.
>> We can only tell that on average the detector is best modelled by it
>> making quantum transitions. This is unsurprising given that detectors
>> are quantised.
>
>No, you can do better than that, if hf >> detector-tripping-energy.
Yes, that's because the detector is tuned, like an aerial.
Some detectors are more tuned than others and some have complex steps
allowing a detection band.
>OK, this whole suggestion is for a rather simple-minded experiment aimed at
>skeptics, rather than the various perhaps stronger but more obscure "non-
>classical" interference experiments out there, but still, the point remains, can
>a photon be detected in more than one place at once? If no, then it has zero
>size; if yes, then it must be detectable in more than one place at once.
You are assuming a pointlike, or at least packet-of-energy-like photon
exists. If, as I have concluded, the quantisation is in the
detector/emitter and I need not (although its frequently hugely
convenient to do so) insist that light (ie an em wave) is itself
quantised.
I would argue, at least to me its the simplest solution, that the very
existence of diffraction, and its properties, show that light goes
through both slits at once. That is, its in two different places at
once. I could attenuate a laser beam so as to get only one detection per
hour and I would still get a diffraction pattern.
Your photon is of course a convenient mathematical tool. You define it
to be pointlike and then ask me to prove it isn't. Diffraction shows
that it cannot be pointlike, although using the correct mathematical
tools you can use such an animal to emulate a large wave. This is often
a most convenient methodology.
Boo
> The source of a photon is not always well defined. When you experience
> interference from two different transmitters (e.g., radio stations), it is
> simply undefined from which transmitter the photon originated. It is not
> that we just don't know it; it is *undefined*. According to QED and the
> uncertainty principle, you *must* regard the photon as coming from both
> sources, in the same way that we regard the photon as going through both
> slits in Young's experiment. Try to pin down the source of the photon, and
> the interference disappears.
Does this mean that if the photons are detected as they leave their sources (of
course I mean at the source) then they cannot interfere at the destination ?
(With the usual provisos re multiple events applying).
--
Boo
Timo A. Nieminen
> Timo A. Nieminen <ti...@physics.uq.edu.au> writes
>>
>> 1) Use EM waves where hf >> energy required to "click" the detectors.
>
> Yes.
>
>> 2) Count for long enough (both with and without source) to get reliable
>> statistics.
>
> Yes.
>
>> 3) Energy required to "click" detectors >> kT
>>
>> .. which is the point of using a gamma source.
>
> and this shows what, exactly?
>
> Remember pretty well all transitions are effectively tuned,
> which is why we can use very low intensity em waves and still get
> transitions.
[cut longer discussion of this point]
(a) If the energy is spread out evenly, as suggested by classical EM
theory, and only a small fraction of that energy is required for
detection, then the surplus of energy should allow simultaneous detection
in two locations, given classical light.
(b) I think that it's sufficient to do this with hf = detector-tripping
energy. However, some might argue that the detector magically sucks in all
the energy at the instant of detection, even from distances of up to at
least 2 times the distance to the source. Even if this is, eg, many
light-years. Given the relativity of simultaneity, instantaneous action
over an extended region is problematic, at best. hf >> detection energy
doesn't leave this weaseling-out space.
A "quantumist" would say that the detection is an inelastic collision
between a photon and a detector, so perhaps it doesn't show anything that
the Compton effect doesn't already show.
> I would argue, at least to me its the simplest solution, that the very
> existence of diffraction, and its properties, show that light goes
> through both slits at once. That is, its in two different places at
> once. I could attenuate a laser beam so as to get only one detection per
> hour and I would still get a diffraction pattern.
>
> Your photon is of course a convenient mathematical tool. You define it
> to be pointlike and then ask me to prove it isn't. Diffraction shows
> that it cannot be pointlike, although using the correct mathematical
> tools you can use such an animal to emulate a large wave. This is often
> a most convenient methodology.
Diffraction shows that _EM waves_ are not pointlike. Pointlike detection
shows that the energy, at small amplitudes is not detected as being evenly
spread over a large volume as classically predicted.
But many are happy to (a) accept that electrons are point particles, and
(b) the probability of detection of them as a function of position is
given by the wavefunction. The alternative I'm arguing against (ie
electron = wavefunction of electron, electron has a size = spread of
wavefunction) requires the electron to shrink to a pointlike size
instantly on "detection of position", even if, eg, the wavefunction
"extended" for many km (note: technically, the wavefunction extends over
all space and all time, though it might have values of zero for some
positions and some times - I think that this makes wavefunctions look even
more like convenient mathematical tools than pointlike electrons).
Replace electron with photon, wavefunction with EM wave, and ditto.
Diffraction + pointlike detection shows that light is not photons, and
that light is not a classical EM wave. Oops! Nature turned out to be
stranger than expected.
Oz
>(a) If the energy is spread out evenly, as suggested by classical EM
>theory, and only a small fraction of that energy is required for
>detection, then the surplus of energy should allow simultaneous detection
>in two locations, given classical light.
Hang on, you just cheated!
1) If only a small fraction of that energy is required for a transition
then the beam has enough energy for many transmissions, ie it contains
very many photons.
2) If it contains very many photons then its clearly unremarkable to
detect one at two different locations.
>(b) I think that it's sufficient to do this with hf = detector-tripping
>energy. However, some might argue that the detector magically sucks in all
>the energy at the instant of detection, even from distances of up to at
>least 2 times the distance to the source. Even if this is, eg, many
>light-years. Given the relativity of simultaneity, instantaneous action
>over an extended region is problematic, at best. hf >> detection energy
>doesn't leave this weaseling-out space.
1) Typically (probably invariantly) a frequency much greater (or less)
than the detection energy is not well coupled to the detector and the
energy is reflected or goes straight through (ie transparent).
2) According to ted there are some very very slow transitions seen in
emissions from rarified astronomical areas. These take seconds, even
hours and I believe he once mentioned years. These will spin a very long
photon indeed (which is why they are easily disturbed). So a transition
taking a very long time indeed can be observed.
3) The absorbing atom is entangled with the light wave. Its both
absorbing and emitting (like any aerial) so the wave could continue past
for some time. Its also worth bearing in mind that atoms are not
actually symmetrical. Even a H atom has directed orbitals (eg p, d)
which will be oriented in an em field. Every rotating H atom (let alone
more complex ones) is thus NOT identical; some will be better positioned
to absorb the incoming energy because of their rotation rate and axes.
4) The very long wavelength, low energy photon, that you propose has a
very imprecise expression of length , that is position, that is emission
and absorption times.
>A "quantumist" would say that the detection is an inelastic collision
>between a photon and a detector, so perhaps it doesn't show anything that
>the Compton effect doesn't already show.
Fine, I have no problem with using that to get simple and clear results
to problems.
>Diffraction shows that _EM waves_ are not pointlike. Pointlike detection
>shows that the energy, at small amplitudes is not detected as being evenly
>spread over a large volume as classically predicted.
I don't agree. It shows that the detector is pointlike and so (duh)
gives a point of detection. The detector is invariably quantised, that
is has energy steps.
>But many are happy to (a) accept that electrons are point particles,
They clearly are not. The orbital around any atom show not, and the
electrostatic energy would be infinite. Actually that's true of any
attempt to localise anything (even a photon), increasing the accuracy of
position decreases the accuracy of energy.
1/zero = infinity
and 1/2 of infinity=infinity.
If used ballistically its often a convenient approximation.
>(b) the probability of detection of them as a function of position is
>given by the wavefunction.
I have no problem with using this mathematical tool as a useful tool.
Which it most certainly is.
>The alternative I'm arguing against (ie
>electron = wavefunction of electron, electron has a size = spread of
>wavefunction) requires the electron to shrink to a pointlike size
>instantly on "detection of position",
Certainly not. Position is never known to infinite precision.
Usually the best we can do is atomic size, a few angstroms.
Actually we typically find the electron in orbitals of size, of a few
angstoms upwards.
[Bugger, antique units in use. Er, about a nm...]
>even if, eg, the wavefunction
>"extended" for many km (note: technically, the wavefunction extends over
>all space and all time, though it might have values of zero for some
>positions and some times - I think that this makes wavefunctions look even
>more like convenient mathematical tools than pointlike electrons).
Not at all. One could argue the wavefunction is as long as the particle
has remained undisturbed, since it propagates changes at lightspeed.
[Personally, I have a hankering for h to be related to the residual
field of the universes wavefunction. This is of course nonsense.]
>Replace electron with photon, wavefunction with EM wave, and ditto.
Indeed. However an electron is a tad more complex. It has mass.
>Diffraction + pointlike detection shows that light is not photons, and
>that light is not a classical EM wave. Oops! Nature turned out to be
>stranger than expected.
No. Diffraction shows photons are waves, and quantised pointlike
detection shows that detectors are quantised and pointlike.
Very simple really.
Timo A. Nieminen
> Timo A. Nieminen <ti...@physics.uq.edu.au> writes
>
>> (a) If the energy is spread out evenly, as suggested by classical EM
>> theory, and only a small fraction of that energy is required for
>> detection, then the surplus of energy should allow simultaneous detection
>> in two locations, given classical light.
>
> Hang on, you just cheated!
> 1) If only a small fraction of that energy is required for a transition
> then the beam has enough energy for many transmissions, ie it contains
> very many photons.
>
> 2) If it contains very many photons then its clearly unremarkable to
> detect one at two different locations.
OK, so you're saying that you expect coincidences?
Remember, the suggestion was to use a gamma source, and a detector needing
far less than the energy of the gamma ray. Yes, the point is to make sure
that the gamma ray has more than enough energy to trip at least 2
detectors.
>> (b) I think that it's sufficient to do this with hf = detector-tripping
>> energy. However, some might argue that the detector magically sucks in all
>> the energy at the instant of detection, even from distances of up to at
>> least 2 times the distance to the source. Even if this is, eg, many
>> light-years. Given the relativity of simultaneity, instantaneous action
>> over an extended region is problematic, at best. hf >> detection energy
>> doesn't leave this weaseling-out space.
>
> 1) Typically (probably invariantly) a frequency much greater (or less)
> than the detection energy is not well coupled to the detector and the
> energy is reflected or goes straight through (ie transparent).
I don't know how efficient the detectors are. I don't do gamma counting.
Detector efficiency and noise will determine whether the experiment would
be practical.
At the moment, all we have is the thought experiment: will a gamma source
produce an excess of coincidences between two separated detectors if the
required energy for detection << gamma energy, as compared with detectors
with required energy > half gamma energy?
> 2) According to ted there are some very very slow transitions seen in
> emissions from rarified astronomical areas. These take seconds, even
> hours and I believe he once mentioned years. These will spin a very long
> photon indeed (which is why they are easily disturbed). So a transition
> taking a very long time indeed can be observed.
Quantumly, all that gives you is a very large uncertainty in time of
emission, and a very narrow spectral line (since the uncertainty in the
energy will be very small).
> 3) The absorbing atom is entangled with the light wave. Its both
> absorbing and emitting (like any aerial) so the wave could continue past
> for some time. Its also worth bearing in mind that atoms are not
> actually symmetrical. Even a H atom has directed orbitals (eg p, d)
> which will be oriented in an em field. Every rotating H atom (let alone
> more complex ones) is thus NOT identical; some will be better positioned
> to absorb the incoming energy because of their rotation rate and axes.
So? Yes, for a dipole emission, there will be two directions where nothing
is radiated. For any emission, there will be at least two directions where
nothing is radiated. But sources will usually - especially a source that's
much smaller than the wavelength (like an atomic nucleus compared to a
typical gamma wavelength) - radiate in a broad pattern.
> 4) The very long wavelength, low energy photon, that you propose has a
> very imprecise expression of length , that is position, that is emission
> and absorption times.
The suggested experiment uses gamma photons. Hardly very long wavelength.
[cut]
>> But many are happy to (a) accept that electrons are point particles,
>
> They clearly are not.
Many clearly are. Also, clearly not all are.
> The orbital around any atom show not,
No, this shows that the wavefunction of an electron is not a point
particle. All you are claiming above is that "wavefunction of electron" is
the same as "electron".
> and the
> electrostatic energy would be infinite.
Infinite self-energy is an old problem. If you're suggesting that the
charge distribution of an electron is spread evenly over the wavefunction,
this leads to some interesting points:
(a) If the extent of the wavefunction changes, the energy of the electron,
and its inertia change. Diffraction will change the extent of the
wavefunction. Can diffraction really cause the energy of an electron to
change as it travels in free space.
(b) What force stops the extended electron from repelling itself apart?
(The forces required were also a problem in the classical extended
electron model.)
(c) Does taking the self-energy of the extended electron into account give
the correct energy differences between different orbitals?
> Actually that's true of any
> attempt to localise anything (even a photon), increasing the accuracy of
> position decreases the accuracy of energy.
> 1/zero = infinity
> and 1/2 of infinity=infinity.
No. It's time vs energy, position vs momentum. For example, localise a
photon in a resonator. You know it's in there, so it's localised. You also
know its energy well, since you the resonant frequency of the resonator.
You end up not knowing the momentum, because it could be going in either
direction.
[cut]
>> Diffraction + pointlike detection shows that light is not photons, and
>> that light is not a classical EM wave. Oops! Nature turned out to be
>> stranger than expected.
>
> No. Diffraction shows photons are waves, and quantised pointlike
> detection shows that detectors are quantised and pointlike.
>
> Very simple really.
No, you are _assuming_ that light is photons, and that the properties of
light are the properties of photons. Diffraction clearly shows that
_light_ is a wave, not that photons are waves.
Note that I did not write that photons are pointlike. Energy can be
extracted from and added to a monochromatic classical EM wave in any
amount. A photon is the hf that, quantumly, the energy is exchanged in -
the quantum of excitation of the EM wave. Thinking of photons as point
particles can lead one astray very easily.
I've seen it written that photons are just whatever creation and
annihilation operators do in quantum optics. A fair position to take, IMO.
Paul Danaher
> Paul Danaher <paul.d...@watwinc.com> writes
>> Your "large" photon brings me back to my puzzle about when
>> absorption occurs - is it on the arrival of the leading edge of the
>> wave packet/probability density function, or at a point at which
>> this reaches some threshold value?
>
> When it is detected, that is, when an effectively irreversible
> transition has occurred. I don;t see how this can be denied.
I really don't know what you mean by "effectively irreversible transition"
here. I get a sense of a "tipping point", but no physical or mathematical
image.
Oz
Something that is thermodynamically unlikely to return to its original
state, of course.
Usually an energetic electron is freed, which then drops down many
energy states often by emitting phonons (ie heat). The reverse path then
becomes hugely improbable.
Oz
[Apologies for the delay, have been on holiday for a few days]
Timo A. Nieminen <ti...@physics.uq.edu.au> writes
>On Fri, 4 Aug 2006, Oz wrote:
>
>> Timo A. Nieminen <ti...@physics.uq.edu.au> writes
>>
>>> (a) If the energy is spread out evenly, as suggested by classical EM
>>> theory, and only a small fraction of that energy is required for
>>> detection, then the surplus of energy should allow simultaneous detection
>>> in two locations, given classical light.
>>
>> Hang on, you just cheated!
>> 1) If only a small fraction of that energy is required for a transition
>> then the beam has enough energy for many transmissions, ie it contains
>> very many photons.
>>
>> 2) If it contains very many photons then its clearly unremarkable to
>> detect one at two different locations.
>
>OK, so you're saying that you expect coincidences?
Yes, if the beam contains very many photons.
>Remember, the suggestion was to use a gamma source, and a detector needing
>far less than the energy of the gamma ray.
OK, can you give me an example detector with this characteristic.
>Yes, the point is to make sure
>that the gamma ray has more than enough energy to trip at least 2
>detectors.
Gamma rays are not ideal to shoot my argument down. They are almost
always found as a wavepacket of well defined momentum and high energy.
That is they tend to a well defined path and behave very much like a
particle.
Now if you took your single gamma, sent it through a lens that turned
the small well-define wavepacket into a spherical wave, then worked on
that it would be more interesting.
>>> (b) I think that it's sufficient to do this with hf = detector-tripping
>>> energy. However, some might argue that the detector magically sucks in all
>>> the energy at the instant of detection, even from distances of up to at
>>> least 2 times the distance to the source. Even if this is, eg, many
>>> light-years. Given the relativity of simultaneity, instantaneous action
>>> over an extended region is problematic, at best. hf >> detection energy
>>> doesn't leave this weaseling-out space.
>>
>> 1) Typically (probably invariantly) a frequency much greater (or less)
>> than the detection energy is not well coupled to the detector and the
>> energy is reflected or goes straight through (ie transparent).
>
>I don't know how efficient the detectors are. I don't do gamma counting.
>Detector efficiency and noise will determine whether the experiment would
>be practical.
That is, however, precisely my point.
Its the emitters and detectors that have the quantum behaviour.
Its thus completely inappropriate to consider them as very simple black
boxes if you wish to shoot my argument down.
>At the moment, all we have is the thought experiment: will a gamma source
>produce an excess of coincidences between two separated detectors if the
>required energy for detection << gamma energy, as compared with detectors
>with required energy > half gamma energy?
Only if you can produce an actual wave with a spherical (or well
dispersed) wavefront. Even then you will have to ensure that neither
detector is entangled with the wave (ie gamma) during the coincidental
measurement period.
>> 2) According to ted there are some very very slow transitions seen in
>> emissions from rarified astronomical areas. These take seconds, even
>> hours and I believe he once mentioned years. These will spin a very long
>> photon indeed (which is why they are easily disturbed). So a transition
>> taking a very long time indeed can be observed.
>
>Quantumly, all that gives you is a very large uncertainty in time of
>emission, and a very narrow spectral line (since the uncertainty in the
>energy will be very small).
It also gives you a very long coherence length (indeed). I would say
this was a very long photon indeed. I could do interference experiments
on it that would give good results even when the equipment was widely
spatially separated.
>> 3) The absorbing atom is entangled with the light wave. Its both
>> absorbing and emitting (like any aerial) so the wave could continue past
>> for some time. Its also worth bearing in mind that atoms are not
>> actually symmetrical. Even a H atom has directed orbitals (eg p, d)
>> which will be oriented in an em field. Every rotating H atom (let alone
>> more complex ones) is thus NOT identical; some will be better positioned
>> to absorb the incoming energy because of their rotation rate and axes.
>
>So? Yes, for a dipole emission, there will be two directions where nothing
>is radiated.
Not if the atom/dipole is rotating.
Also atoms have much more complex polarised orbitals than mere dipoles.
See p- & d- orbitals for example.
>For any emission, there will be at least two directions where
>nothing is radiated.
No, see above.
>But sources will usually - especially a source that's
>much smaller than the wavelength (like an atomic nucleus compared to a
>typical gamma wavelength) - radiate in a broad pattern.
You might think so, but atoms are much more complex than that.
Typically there will be a change of momentum so that the atom recoils on
emitting a photon. In this case you cannot have a spherical wavefront.
NB I think direct transitions between spherical s-orbitals are
prohibited on momentum grounds. So a spherical wavefront cannot be
emitted from an atomic transition.
>> 4) The very long wavelength, low energy photon, that you propose has a
>> very imprecise expression of length , that is position, that is emission
>> and absorption times.
>
>The suggested experiment uses gamma photons. Hardly very long wavelength.
Indeed. Usually very incoherent and a tight wavepacket of well defined
momentum.
>>> But many are happy to (a) accept that electrons are point particles,
>>
>> They clearly are not.
>
>Many clearly are. Also, clearly not all are.
No electrons are point particles.
Its impossible on energy grounds alone.
>> The orbital around any atom show not,
>
>No, this shows that the wavefunction of an electron is not a point
>particle. All you are claiming above is that "wavefunction of electron" is
>the same as "electron".
In many cases, yes.
The electron here is a wave in an energy box.
Its actually a different thing to a free electron, or an electron in
another energy box.
>> and the
>> electrostatic energy would be infinite.
>
>Infinite self-energy is an old problem.
It still exists however. Then you can explain how to get a precise
position with a finite energy (see uncertainty...)
>If you're suggesting that the
>charge distribution of an electron is spread evenly over the wavefunction,
>this leads to some interesting points:
>
>(a) If the extent of the wavefunction changes, the energy of the electron,
>and its inertia change. Diffraction will change the extent of the
>wavefunction. Can diffraction really cause the energy of an electron to
>change as it travels in free space.
Of course, why not?
Then entangled particles become a lot less strange for one thing.
>(b) What force stops the extended electron from repelling itself apart?
>(The forces required were also a problem in the classical extended
>electron model.)
Ah. If we knew that then we would also know why its charge is precisely
e and its mass e_m. We would probably be very close to knowing the
theory of everything. My working hypothesis is that its a soliton of
some sort.
>(c) Does taking the self-energy of the extended electron into account give
>the correct energy differences between different orbitals?
I'm sure it will, its a solution after all.
>> Actually that's true of any
>> attempt to localise anything (even a photon), increasing the accuracy of
>> position decreases the accuracy of energy.
>> 1/zero = infinity
>> and 1/2 of infinity=infinity.
>
>No. It's time vs energy, position vs momentum. For example, localise a
>photon in a resonator. You know it's in there, so it's localised. You also
>know its energy well, since you the resonant frequency of the resonator.
>You end up not knowing the momentum, because it could be going in either
>direction.
So if you know its position precisely, then it has infinite momentum.
Are you suggesting that this is not effectively saying it has infinite
energy.
In your own example the resonator volume tends to zero so its wavelength
tends to zero so its energy ......
>>> Diffraction + pointlike detection shows that light is not photons, and
>>> that light is not a classical EM wave. Oops! Nature turned out to be
>>> stranger than expected.
>>
>> No. Diffraction shows photons are waves, and quantised pointlike
>> detection shows that detectors are quantised and pointlike.
>>
>> Very simple really.
>
>No, you are _assuming_ that light is photons,
I don;t see where.
Detectors are almost always (always?) quantised (eg see atom energy
levels).
>and that the properties of
>light are the properties of photons. Diffraction clearly shows that
>_light_ is a wave, not that photons are waves.
OK. I am happy that you have an unknown called a photon.
Personally I don;t think there is anything I can't handwavingly explain
using my wave model combined with energy steps (ie quantisd). You can
get quantised energy steps entirely from waves a la schroedinger.
>Note that I did not write that photons are pointlike. Energy can be
>extracted from and added to a monochromatic classical EM wave in any
>amount. A photon is the hf that, quantumly, the energy is exchanged in -
>the quantum of excitation of the EM wave. Thinking of photons as point
>particles can lead one astray very easily.
That's fine. It give the right answers conveniently and easily.
I'm all for that.
Unfortunately you now have no physical description of a photon but
merely state that its convenient to take it as a wave in some
circumstances and a particle (or quanta) in others. That's fine, you get
the answers as above. No problem.
I'm saying that I can have everything as a wave and still 'explain'
quantised situations. I'm delighted you are taking the effort to shoot
me down, until you do, it allows me to strengthen my mental model.
>I've seen it written that photons are just whatever creation and
>annihilation operators do in quantum optics. A fair position to take, IMO.
Of course. A mathematical and unphysical method that is ideally suited
to getting the right answers conveniently and easily by sweeping
unknowns under the carpet. No problem with that as a method. It works
will billiard balls most successfully!