Length of wavetrain of a single photon
Ray Tomes
wave train length and amplitude profile of a single photon?
To explain what I mean by this, consider Young's two slit experiment and
the resulting interference for the case where the photon rate is very
low and we may consider that essentially all observed events are self
interference of single photons.
For any given location in the interference pattern there is a definite
difference in distance traveled for the two paths and I imagine that
there must be some difference in distance at which there is no longer
interference. This distance difference would be what I considered to be
the length of the wave train of a single photon. Of course it would be
necessary to vary slit spacing and any other variables to make certain
that the resulting value was a constant for any given wavelength of
light.
-- Ray Tomes -- http://www.kcbbs.gen.nz/users/rtomes/rt-home.htm --
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Dirk Bruere
>...
> the resulting interference for the case where the photon rate is very
> low and we may consider that essentially all observed events are self
> interference of single photons.
I thought all interference wrt photons was self interference of single
photons.
Aleksandr Timofeev
rto...@kcbbs.gen.nz (Ray Tomes) wrote:
> I am wondering whether any attempt has ever been made to measure the
> wave train length and amplitude profile of a single photon?
>
interference in an radiointerferometer with superlong base. Who has
invented the radiointerferometer with superlong base?
Aleksandr Timofeev
---
http://solar.cini.utk.edu/~russeds/unknown/astrochem/
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(Greg Weeks)
: I am wondering whether any attempt has ever been made to measure the
: wave train length and amplitude profile of a single photon?
My guess is that there is no such thing.
For massive particles (in quantum field theory), the notion of "position"
is only approximate. It cannot be made more precise than the order of an
inverse mass.
See, for example, the section on "Single Particle States" in Chapter I of
Haag's "Local Quantum Physics". Haag describes *two* position operators
for massive particles, one not quite covariant and one not quite
diagonalizable. Re massless particles, Haag states: "For massless
particles, the concept of localization is not appropriate".
Here is a partial explanation. In quantum field theory, there is no a
priori notion of position. for a single-particle state, you can try to
define Q in terms of P, eg, by:
<p|q> = exp(ipq) [Did I get the phase right?]
But this requires fixing the phases of the <p| vectors. These phases can
be fixed by obtaining all the <p| vectors by boosting a particle at rest,
ie, <p=0|. This is a bit arbitrary, but it yields *something* (known as
the Newton-Wigner position operator). However, for massless particles,
there is no particularly natural way to fix the phases (in the momentum and
helicity basis). So no even half-way natural position operator arises.
In short: Photons have momentum-helicity wave functions. But there is no
natural way to fix the phases and no natural way to obtain a position
wave-function.
Regards,
Greg
Ray Tomes
>: I am wondering whether any attempt has ever been made to measure the
>: wave train length and amplitude profile of a single photon?
we...@orpheus.dtc.hp.com ((Greg Weeks)) wrote:
>My guess is that there is no such thing.
Other people made similar replies (so I am not just getting at Greg
here). The answers have been generally unsatisfactory from my point of
view. Firstly, I am much more interested in an experimental result than
in theory. Secondly, I have specified what I mean in a way that there
is very definitely such a thing. I repeat ...
For any given location in the interference pattern there is a definite
difference in distance traveled for the two paths and I imagine that
there must be some difference in distance at which there is no longer
interference.
Specifically, I expect that if we use mirrors to make the two paths
1 kilometre different then I would be very surprised if a photon would
interact with itself. However we know that it does so when the paths
are 5 or 10 wavelengths different. Somewhere in between the
interference must fade out and I don't think that the answer is given by
the uncertainty principle. I expect that there is a definite amplitude
profile over a number of wavelengths and would like to know if this has
ever been measured experimentally.
Raymond E. Rogers
>
> Ray Tomes (rto...@kcbbs.gen.nz) wrote:
> : I am wondering whether any attempt has ever been made to measure the
> : wave train length and amplitude profile of a single photon?
>
>
> is only approximate. It cannot be made more precise than the order of an
I would think that examining the "quality" of the interference pattern
when a interferometer is set up to work with single photons would reveal
the wave packet profile. That is to say; examining the movement and
amplitude of the interference pattern as a test arm is lengthened and
shortened. I would expect (I am not a QM expert) that the wave packet
shape is just the result of the generation process and has no intrinsic
form.
Enjoy
Ray
[Moderator's note: Excess quoted text removed. -MM]
john baez
Dr Paul Kinsler <ee...@eensgi4.leeds.ac.uk> wrote:
>Start from the point of view that a photon is a single excitation
>of an EM field mode. Then the "wave train length and amplitude profile"
>is determined by the field mode, and not by the photon. For example,
>photons in EM modes of a laser cavity will be largely confined by the
>mirrors.
>
>So perhaps the relevant question is: For the case "I" am considering,
>what are the appropriate photon modes to chose?
This sounds correct to me.
By the way, I don't see the relevance of Greg Weeks' remarks
concerning localizability of photons. It's true, there's no good
"Newton-Wigner localization" for photons. This basically means that
there's no good basis of "position eigenstates" for photons. This
is closely related to the fact that Schrodinger's equation becomes
meaningless for zero-mass particles (since it involves dividing by
the mass). But position eigenstates aren't what you want to think
about if you're interested in "wave trains" - for that, you want to
use plane waves or wavelets or some other sort of basis of modes.
Ralph Frost
>
> In article <1999Jan6.1...@leeds.ac.uk>,
> Dr Paul Kinsler <ee...@eensgi4.leeds.ac.uk> wrote:
>
> >Start from the point of view that a photon is a single excitation
> >of an EM field mode. Then the "wave train length and amplitude profile"
> >is determined by the field mode, and not by the photon. For example,
> >photons in EM modes of a laser cavity will be largely confined by the
> >mirrors.
> >
> >So perhaps the relevant question is: For the case "I" am considering,
> >what are the appropriate photon modes to chose?
>
> This sounds correct to me.
Pardon this interruption but I'm wondering if you can clarify two
questions that I have which are no doubt unrelated to whatever the case
is that Dr. Kinser is considering.
1. Are you folks saying that a photon doesn't actually become/register a
photon until the EM wave train accummulates into photon-like units _AT_
the interface where it "leaks" into the detector? Kind of like an
electron being the unit of measure of the electric field "leaking",
drop-by-drop onto paper through my inkjet printer? [Please pardon my
un-quantum-like terminology.]
2. Does the EM wavetrain length/amplitude for the photon in question
include the full path of the "stuff" travelling back through the ground
cable to the source?
I realize this is way off-topic from whatever "appropriate photon
modes" are or mean. An answer on or off the NG would be greatly
appreciated.
Thanks much for all you help.
Ralph
(Greg Weeks)
: By the way, I don't see the relevance of Greg Weeks' remarks
: concerning localizability of photons. It's true, there's no good
: "Newton-Wigner localization" for photons. This basically means that
: there's no good basis of "position eigenstates" for photons. This
: is closely related to the fact that Schrodinger's equation becomes
: meaningless for zero-mass particles (since it involves dividing by
: the mass). But position eigenstates aren't what you want to think
: about if you're interested in "wave trains" - for that, you want to
: use plane waves or wavelets or some other sort of basis of modes.
Hmmm. When a single electron exhibits diffraction and interference when
scattering off a crystal -- or, if you prefer, in the two-slit experiment
-- isn't the result typically understood in terms of the electron's wave
function in the position representation? I would have figured the same for
photons. What other candidate is there for a single photon? What plane
wave? The expectation values of E and B are zero for a single photon
state. If you have a photon of definite of momentum and helicity, the
state doesn't diagonalize the field operators. Does it come close in some
sense?
Regards,
Greg
Ralph Frost
> Obviously in principle any old orthonormal set of modes will do, but
> in practise some will make more physical sense than others -- why use
> spherical modes in a cubical box?
>
Habit?
Doug Sweetser
I find this thread unsatisfying because of a piece of artwork I did on
this very topic! The title of the resulting piece was "Groups of
coherent photons behave like waves and particles". The two critical
words are "Groups" and "coherent", neither of which are central to this
discussion. I did a rough search for the word coherent in this thread
and didn't find it (neither was it in a Scientific American article a
while ago on 2-slit experiments, arg!). Incoherent light must be
disorganized. An understanding of coherence can be achieved by
comparing it to incoherence, and that was a component of the image.
There are two ways to be coherent: spatially or temporally. Lasers are
both. Think about what it means to be nicely organized in space, or in
time versus being disorganized.
I can see that contrast, but only if I look at a collection of photons.
One photon does not make a pattern. It takes groups of photons to see a
pattern. How large a group? Well, it must be smaller than its
discoherence length, since the pattern becomes random if one waits that
long. The group picture is composed of individual photons, but the data
for incoherence must be made of many photons.
Here is an ASCII rendition of "Groups of coherent photons behave like
waves and particles."
3333333333333 1320032103221 0123210123210
2222222222222 2031022101003 0123210123210
1111111111111 1032203131302 0123210123210
0000000000000 0211130221032 0123210123210
1111111111111 2033102030103 0123210123210
2222222222222 0203203200132 0123210123210
3333333333333 3012032002103 0123210123210
2222222222222 2103202221003 0123210123210
1111111111111 2102021033203 0123210123210
0000000000000 2123103202301 0123210123210
1111111111111 2103103100301 0123210123210
2222222222222 2032103221333 0123210123210
This would be a laser
0123210123210
1232101232101
2321012321012
3210123210123
2101232101232
1012321012321
0123210123210
1232101232101
2321012321012
3210123210123
2101232101232
1012321012321
A low resolution gif of the 26" x 43", IRIS print (which looks great!)
is available at
http://world.std.com/~sweetser/PopScience/photons/photons.html
I can only understand the topic of this thread by thinking of a photon
in the context of a group of photons, and thinking of coherence as
reflected by randomness. Together, it makes sense for me.
John Sidles
>>I am wondering whether any attempt has ever been made to measure the
>>wave train length and amplitude profile of a single photon?
The concept of a "single photon" is pretty slippery. For
example, consider the time evolution of a two-state electron
spin coupled to a purely classical (non-quantized) random
magnetic field field B(t).
Such classical fields can't absorb or emit quanta, can they?
After all, classical fields don't have *any* operator structure;
they're just c-numbers.
But the above expectation is physically untrue. If one
continuously observes the two state system, for example with an
interferometer -- which can be done without inducing spin flips
-- then the time evolution of the system will be *exactly* as
though quanta are being absorbed and emitted from the classical
c-number field.
Thus, under the proper condictions, even a purely classical
field can absorb and emit discrete quanta. For this reason, it
is perfectly reasonable to treat quanta as being, in essence,
artifacts of observation.
The word "artifacts" in this context is not pejorative. It
requires very careful biological or human engineering to create
quantum artifacts. Most Hamiltonians just evolve rapidly and/or
chaotically toward thermodynamic equilibrium.
For practical quantum engineers, the modern decoherence
viewpoint has the advantage of making many quantum mysteries
accessible to calculation. For example, just how long does it
take for a quantum to be absorbed? What is the chance that a
two-state system will "change its mind" partway through an
absorption process, and thus induce an error in a quantum
calculation? These are no longer regarded as philosophical
questions, rather they are becoming engineering issues of great
practical importance.
As quantum technologies come closer to practical realization,
the decoherence/engineering point of view is rapidly becoming
dominant among practicing physicists. Undergraduate textbooks,
as usual, lag about one generation behind.
Probably the greatest single need is a first-year graduate
QM textbook which devotes about 1/3 - 1/2 of its chapters
to measurement theory and practice, with particular emphasis
on noise processes.
Alastair Ward
>
>>Ray Tomes (rto...@kcbbs.gen.nz) wrote:
>>: I am wondering whether any attempt has ever been made to measure the
>>: wave train length and amplitude profile of a single photon?
>
>1 kilometre different then I would be very surprised if a photon would
>interact with itself. However we know that it does so when the paths
>are 5 or 10 wavelengths different. Somewhere in between the
>interference must fade out and I don't think that the answer is given by
>the uncertainty principle. I expect that there is a definite amplitude
>profile over a number of wavelengths and would like to know if this has
>ever been measured experimentally.
Although I don't think the idea of a wave train is really valid when
speaking of a photon I must confess to having read once in an ancient
optics book of just the sort of experiment you describe. But it was not done
on a single photon but using a beam. And I think the analysis of the experiment
was carried out in classical terms. I am really sorry I cannot remember the
name of the optics text book where I read this - it was very old - almost
pre quantum theory - a well known classic book of it's type - but blow me
if I can think of the author - he was one of the greats in the field of optics.
Sorry thats the best I can come up with - cheers
Alastair Ward.
Ralph Frost
I'm curious to discover what specific options you have in mind here and
why. Tetrahedral units stacked in a tetrahedral box? Spheres in an
octahedral box?
Or am I reading too much (or too little) into your question?
Thanks
Best regards,
Ralph Frost
ref...@dcwi.com
Bjorn Wesen
>For any given location in the interference pattern there is a definite
>difference in distance traveled for the two paths and I imagine that
>there must be some difference in distance at which there is no longer
>interference.
The temporal coherence time is the inverse of the frequency bandwidth of the
photon emitter. The photon "wavepacket" is not infinite for any light source
with a non-zero frequency bandwidth so you get a finite maximum path
difference for usage in interference experiments.
You were interested in an experiment - well you can for example do this (*):
send a mercury-arc generated lightbeam through two small circular apertures
(close to each other). You'll get an interference pattern on the other side
on your screen. Now insert a 0.5 mm thick piece of glass behind one of the
apertures, and watch the interference pattern disappear. The glass-induced
change in optical path-length was longer than the coherence length of the
light (c * the coherence time) and thus you can't get interference because
it's not the same "wavepackets" which recombine at the screen anymore.
Try the same with a laser and the interference won't disappear - the
coherence length of a laser is much longer than the above path difference.
/Bjorn W
(*) experiment from Hecht, Optics, p.518
Matt McIrvin
>Dr Paul Kinsler wrote:
>>
>> Obviously in principle any old orthonormal set of modes will do, but
>> in practise some will make more physical sense than others -- why use
>> spherical modes in a cubical box?
>
>
>I'm curious to discover what specific options you have in mind here and
>why. Tetrahedral units stacked in a tetrahedral box? Spheres in an
>octahedral box?
He probably had in mind modes that make it easy to solve differential
equations, and tend to correspond to oscillating wave solutions of
Maxwell's equations, with the given boundary conditions. I.e. products of
sines and cosines in a cubical box, or Bessel functions in a cylindrical
box, or spherical harmonics in a spherical box.
--
Matt McIrvin http://world.std.com/~mmcirvin/
Oz
<bj...@sparta.lu.se.noSPAM> writes
>The temporal coherence time is the inverse of the frequency bandwidth of the
>photon emitter. The photon "wavepacket" is not infinite for any light source
>with a non-zero frequency bandwidth so you get a finite maximum path
>difference for usage in interference experiments.
It strikes me that an awful lot of words have been issued to answer a
perfectly ordinary question. What I think the original poster wanted to
know was how long a (as in a single) *photon wavetrain was. That is
roughly equivalent to the following experiment:
1) Take a conventional double slit diffraction.
2) Send a *single*, separated, photons through it.
3) Observe diffraction.
4) Change the length of one leg of the path until no diffraction occurs.
How long is that distance?
>From which one might deduce that the photon wavetrain is less than said
distance.
This of course then begs the question of what the frequency and energy
of that photon is since if a wavetrain is finite it must be composed of
elements of differing frequency. This was discussed in s.p. some years
ago, but I am uncertain of the final conclusion.
Supplementary:
Is wavetrain length a function of wavelength?
--
Oz (who isn't here and isn't reading this ...)
Tom Roberts
> I am wondering whether any attempt has ever been made to measure the
> wave train length and amplitude profile of a single photon?
Why do you think that a single photon will generate an interference
pattern? I think it will register once on the screen. Considering
that to be an "interference pattern" is a rather large stretch in
terminology.
Asking about the properties of an "individual" photon does not really
make sense, as photons are not individually localizeable or identifiable
-- you cannot "grab onto" one and "pin it down" in any sense at all.
And the properties of a beam of photons are inherently related to the
properties of the source -- different sources will have different
coherence lengths, which are essentially what you are asking about.
So to actually obtain an interference pattern requires a beam of
independent photons, and the other responses about the coherence
length of lasers, etc. apply.
Tom Roberts tjro...@lucent.com
Raymond E. Rogers
The probabilty density function will have a wavelike/interference
distribution. True, you must run several photons, one after another if
one is truly critical, to determine the probablity distribution but each
photon has a wavelike probability spread.
Ray
Tom Roberts wrote:
> Why do you think that a single photon will generate an interference
> pattern? I think it will register once on the screen. Considering
> that to be an "interference pattern" is a rather large stretch in
> terminology.
[Moderator's note: Quoted text deleted. Please trim quoted material
yourselves, folks! -TB]
Ray Tomes
Tom Roberts <tjro...@lucent.com> wrote:
>Ray Tomes wrote:
>> I am wondering whether any attempt has ever been made to measure the
>> wave train length and amplitude profile of a single photon?
>> [two-slit interference-pattern example]
>Why do you think that a single photon will generate an interference
>pattern? I think it will register once on the screen. Considering
>that to be an "interference pattern" is a rather large stretch in
>terminology.
Yes, you are right, each photon makes a single registration.
The interference pattern comes about from the addition of many such
single registrations. Sorry if I didn't express that clearly. The
question then is "how many interference fringes are there in the
result?" as the maximum possible.
Some people have confused this with discussion of lasers where there is
a high degree of coherence and the number of fringes can be very large.
However I think that it makes sense to see the laser as many photons
that are in phase with each other. That is why I specified single
photons at a time (strictly speaking, a low photon rate) so as to remove
the question of coherence from the discussion.
>Asking about the properties of an "individual" photon does not really
>make sense, as photons are not individually localizeable or identifiable
>-- you cannot "grab onto" one and "pin it down" in any sense at all.
>And the properties of a beam of photons are inherently related to the
>properties of the source -- different sources will have different
>coherence lengths, which are essentially what you are asking about.
No it isn't, as mentioned above. Single photon events (when added
together) will have some pattern of distribution over some number of
interference fringes. You have "pinned it down" every time that a dot
appears on you photo or CCD. With a low photon rate each such dot
represents a single photon interfering with itself.
Oz <O...@upthorpe.demon.co.uk> in the post before this one has understood
the question and is the only one who seems to have done so in this
thread. Perhaps his post makes what I am asking clearer.
I did receive one piece of email from someone who also understood the
question and which stated that a 0.5 mm piece of plain glass behind one
screen is sufficient to disrupt the interference pattern. Using 0.5 mm
of glass with a R.I of say 1.5 represents a path length difference of
roughly 0.17 mm. Assuming a wavelength of 5700 A for convenience,
0.17 mm is about 300 wavelengths.
Since at least 10 fringes are visible, it narrows the range of possible
answers to ~10 to ~300 wavelengths. I just wonder if the answer will
turn out to be 137?
Bjorn Wesen
>However I think that it makes sense to see the laser as many photons
>that are in phase with each other. That is why I specified single
>photons at a time (strictly speaking, a low photon rate) so as to remove
>the question of coherence from the discussion.
"... each photon interferes only with itself. Interference between different
photons never occurs"
Dirac, Quantum Mechanics 4th edition, page 9
Which more or less removes the distinction between a beam and "single
photons", and makes it feasible to use concepts as beam coherence with as
low intensity as "single photons" as well. Some Quantum Optician could
settle this, since all this thread is just a discussion of terminology I
think :)
/Bjorn
Ale2NOSPAM
<<
Some people have confused this with discussion of lasers where there is
a high degree of coherence and the number of fringes can be very large.
However I think that it makes sense to see the laser as many photons
that are in phase with each other. That is why I specified single
photons at a time (strictly speaking, a low photon rate)
>>
But you can do the experiment you speak of with laser light, just reduce the
beam with a beam splitter. Any intensity can be sent to your interferometer.
The beam can be reduced until fast electronics can detect individual events. I
was told that the coherence length is not effected and is of order several
feet.
Would any other experts concur?
Aleksandr Timofeev
rto...@kcbbs.gen.nz (Ray Tomes) wrote:
> I am wondering whether any attempt has ever been made to measure the
> wave train length and amplitude profile of a single photon?
>
> the resulting interference for the case where the photon rate is very
> low and we may consider that essentially all observed events are self
> interference of single photons.
>
On my sight, for consideration of an offered problem from all points of
view; the most approaching measuring instrument is the microwave
interferometr with superlong basis. I would name this type of an
interferometer as an interferometer with independent registration of signals
in shoulders. Fundamentally any other interferometer by nothing differs from
an interferometer considered below.
Principles of work.
The microwave interferometr with superlong basis consists of two radio
telescopes were on a very large distance from each other. Before experiment
or after him, the nuclear hours are synchronized. Each radio telescope writes
on a videotape a transformed radiation accepted by an antenna. Simultaneously
with a signal, the scores of time received from the standard of frequency,
are written on a videotape.
After ending experiment we have two videotapes with entries of a signal
and scores of time. The "interference picture" is received after data
processing of these videotapes on the computer.
There are two graphic schemes illustrating the description:
The microwave interferometer with superlong basis. Part 1.
Block scheme.
-> radio-telescope 1
->
-> parabolic antenna tape 1 clock 1
-> \
-> \ [ microwave ]
-> \ [ receiver + ] [videotape] [hydrogen ]
-> ) )--->[analog-to-digital]--->[recorder ]<---[frequency]
-> / [ converter ] ^ ^ [standard ]
-> / | |
-> / radio-signals time-marks
-> microwave
-> radiation
-> for synchronization of atomic clocks
-> [transportable caesium]
-> [ frequency standard ]
[snip] ====================================================================
-> radio-telescope 2
->
->
-> parabolic antenna 2 tape 2 clock 2
-> \
-> \ [ microwave ]
-> \ [ receiver + ] [videotape] [hydrogen ]
-> ) )--->[analog-to-digital]--->[recorder ]<---[frequency]
-> / [ converter ] ^ ^ [standard ]
-> / | |
-> / radio-signals time-marks
->
->
->
. The microwave interferometer with superlong basis. Part 2.
. ----------------------------------------------------------
. "Interference picture"
. ^
. |
. [videotape 1] ------> [ COMPUTER ] <---------- [videotape 2]
. ^ ^
. | |
. radio-telescope 1 <- synchronization clocks -> radio-telescope 2
. Length of basis
. |<----------------------------- {snip} ------------------------------->|
. /^\ /^\
.^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ {snip} ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
^ .| | | | | | | | | | | | | | | | | | | | | | | | .
Noise microwave radiation
Flexible possibilities of a computer interference of signals.
1. Our interferometer has the right and left shoulders. The distance
between shoulders does not influence sensitivity of an interferometer. The
sensitivity of an interferometer to a signal is determined by the worse
receiver from both radio telescopes.
- The distance between shoulders of an interferometer can be no matter
how large. (This problem is reduced to a problem of transportation of clocks
of a synchronization).
2. The addition of signals is carried out in the computer, that allows
to apply no matter how complicated algorithms of addition of signals.
- In that specific case, we can arbitrary vary delay of signals in each
from a shoulders in any direction.
Conditionality of physical concept " an Interference picture ".
Here we shall be convinced of a celebration of a principle of a causality.
The events happening on slots of an interferometer have primary significance,
all other events happening in an interferometer have the status
secondary.
Let's analyze physical concept addition of signals in an interferometer.
The radiation incident on an input of an interferometer has the following
performances:
Wavefront; Frequency band; Spectral fluence of energy;
For each frequency:
Polarization; Amplitude; Phase; Stability.
The interferometer considered by us, is an interferometer with
independent registration of signals in shoulders and the process of addition
of signals is carried out in the computer. The phrase " process of addition
of signals is carried out in the computer " allows clearly to seize essence "
concepts of an interference picture " and source of an origin of this
concept. In the given type of an interferometer there is some arbitrariness
in choice by us of the law of addition of signals from the right and left
shoulders. In our case " the kind of an interference picture " depends on
the concrete law of addition of signals selected by us. In other kinds of
interferometers geometry (physical) construction of an interferometer
determines the law of addition of signals and " a kind of an interference
picture ".
Conclusion 1. For existence of the phenomenon of an interference the
first necessary condition is the registration of a signal for two sites (two
orifices) of a wavefront, which are located on some distance from each other.
Conclusion 2. For existence of the phenomenon of an interference the
second necessary condition is the filtration of frequency band of accepted
signals. The band of a signal should decrease with a diminution of a distance
between two sites (radiation flux) of a wavefront and on the contrary. The
stability of a source of a signal plays the role also.
Conclusion 3. For existence of the phenomenon of an interference the
third necessary condition is the sufficient total energy of an accepted signal
or it is necessary to accumulate a signal.
Conclusion 4. For existence of the phenomenon of an interference the fourth
necessary condition is the existence of the adder of signals from the right
and left shoulders.
Major parameters of an interferometer.
Give is summable stated above. Major parameters of an interferometer are:
Distance between antennas (slots);
Frequency band of registered signals;
Sensitivity of an interferometer.
Sensitivity of an interferometer.
The interferometer represents the macroscopic system.
What determines sensitivity of an interferometer? The effective square of
an antenna determines a stream of energy of a signal, than more square of an
antenna by that the greater stream of energy acts on an input of the
amplifier of a signal. The first cascade of the amplifier of a signal
determines quality of the amplifier. It represents the macroscopic system. If
the macroscopic system is in a thermodynamic equilibrium, it's the state is
characterized in temperature.
This temperature characterizes own noise of the amplifier. Energia of a
signal should exceed energy of own noise of the amplifier, therefore own
noise of the amplifier set threshold energy of a signal, which determines
sensitivity of an interferometer.
The minimum energy stream discovered by an interferometer is close
to 3*10 ^ -21 wt/m ^ 2, that is equivalent to several tens photons on a
square meter per one second (this value requires multiplication on square of
an antenna).
Conclusions.
1. In a considered microwave interferometer, the properties of a rather
sparse stream of absorbed photons are registered. (What is the time of life of
"photon" absorbed by an antenna?). Interferometer has "a very large mass".
2. Varying delay of time of a signal in the certain shoulder of an
interferometer, we define stability of frequency of a source in a direction
of a falling stream. The similar experiments frequently were carried out in
optics.
3. A new information we could receive increasing a distance between
radio telescopes. We could clarify on what distance on a wavefront the
correlation disappears (if disappears). The similar experience are not known
to me. For basis equal to the Earth's diameter, the correlation reliably
exists. The resolving power of the radio interferometer approximately in
100 - 1000 times is exceeded with a resolving power of an optical telescope.
4. We shall name photons, in a stream incident on an antenna, " as free
photons ". Let's name photons absorbed by an antenna, " as bound photons ".
Whether we know a structure " of bound photons "? To me it is not clear. I
definitely know, that the structure of an electromagnetic field is determined
by boundary conditions. For this reason the structure " of free photons
" should differ from a structure " of bound photons ". The detailses of
mechanisms of a radiation and absorption of an electromagnetic field are not
known to us. On my sight the photon is simply other title for mechanisms of
a radiation and absorption of an electromagnetic field. For this reason the
PHOTONS DO NOT EXIST IN A NATURE or, if it is pleasant more to you, in
a medium of an electromagnetic field the photons will be generated as virtual
particles.
From the point of view of the Henry Poincare, we shall come to a
conclusion:
The photons are particles - ghosts, the photons are mathematical
abstraction, which allow us to calculate probability of interaction of an
electromagnetic field and substance.
Aleksandr Timofeev
---
http://solar.cini.utk.edu/~russeds/unknown/astrochem/
PS Who has invented the microwave interferometer with superlong basis?
ca31...@bestweb.net
rto...@kcbbs.gen.nz (Ray Tomes) wrote:
> I did receive one piece of email from someone who also understood the
> question and which stated that a 0.5 mm piece of plain glass behind one
> screen is sufficient to disrupt the interference pattern. Using 0.5 mm
> of glass with a R.I of say 1.5 represents a path length difference of
> roughly 0.17 mm. Assuming a wavelength of 5700 A for convenience,
> 0.17 mm is about 300 wavelengths.
>
> Since at least 10 fringes are visible, it narrows the range of possible
> answers to ~10 to ~300 wavelengths. I just wonder if the answer will
> turn out to be 137?
This sounds alittle like the old Hanbury Brown and Twiss (HBT)
argument resurfacing ? Wasn't that the case where interference
was detected between photon detectors with a wide displacement
in stellar interferometers and the quantum physicists argued
for while, that each detectors reception of a photon was
non-deterministic and so there could be no interference ?
These weren't considered entangled photons as I recall, just
photons from the same star.
If you think of the detectors as being just different parts of
the same screen then the stellar interferometer was considered
as measuring the interference between two photon's wave-functions.
Can a "quantum" can be looked at as:
[1] a wave modulated by an impulse where the impulse
is treated as an "abstract particle"
or as
[2] a particle with an associated wave-function where
the wave-function is view as an "abstract wave"
The former leads to the idea of "wavelets" in signal processing while
the later leads to "quanta" in QM ?
---
http://www.bestweb.net/~ca314159/
Ray Tomes
"Bjorn Wesen" <bj...@sparta.lu.se.noSPAM> wrote:
>"... each photon interferes only with itself. Interference between different
>photons never occurs"
>Dirac, Quantum Mechanics 4th edition, page 9
>Which more or less removes the distinction between a beam and "single
>photons", and makes it feasible to use concepts as beam coherence with as
>low intensity as "single photons" as well. Some Quantum Optician could
>settle this, since all this thread is just a discussion of terminology I
>think :)
If everything that you and Dirac say is correct then does this not lead
to the conclusion that the photons in a laser are different to the ones
in normal incoherent light? That I would find very surprising. Does it
not imply sunstructure?
Yes, the thread has been a discussion of terminology, but the question
is not about that. However if you are right then the question does not
go deep enough. Whatever, I still would like to have a narrower range
than 10-300 wavelengths for the difference in distance of the two paths
over which a single photon can interfere with itself.
Dr P. Kinsler
In sci.physics.research Bjorn Wesen <bj...@sparta.lu.se.noSPAM> wrote:
: "... each photon interferes only with itself. Interference between different
: photons never occurs"
: Dirac, Quantum Mechanics 4th edition, page 9
: Which more or less removes the distinction between a beam and "single
: photons", and makes it feasible to use concepts as beam coherence with as
: low intensity as "single photons" as well. Some Quantum Optician could
: settle this, since all this thread is just a discussion of terminology I
: think :)
If it helps any, I'm a quantum optician, so here goes:
Dirac got that one wrong[1]. Photons can interfere with themselves
or other photons. If you consider photons to be excitations of EM
field modes[2], then interfering different photons is pretty much the
same as interfering classical fields[3]. Note that the field modes
can be all sorts of things, even wavepacket modes which propagate.
Pick mode which suit your physical situation, and make the maths
part easier to do -- any orthonormal set will work. Asking "how
big is this photon?" is too ill defined -- you have to specify the
field mode the photon is (has been created) in.
[1] I've always wanted to say that. ;-)
[2] As I said earlier in this thread.
[3] ALthough creating and annihilating photons isn't the same.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Institute of Microwaves and Photonics
School of Electronic & Electrical Engineering
University of Leeds (ph) +44-113-2332089
Leeds LS2 9JT (fax)+44-113-2332032
United Kingdom. P.Ki...@elec-eng.leeds.ac.uk
WEB: http://www.elec-eng.leeds.ac.uk/staff/eenpk/P.Kinsler.html
Matt McIrvin
<bj...@sparta.lu.se.noSPAM> wrote:
>"... each photon interferes only with itself. Interference between different
>photons never occurs"
>
>Dirac, Quantum Mechanics 4th edition, page 9
>
>Which more or less removes the distinction between a beam and "single
>photons", and makes it feasible to use concepts as beam coherence with as
>low intensity as "single photons" as well. Some Quantum Optician could
>settle this, since all this thread is just a discussion of terminology I
>think :)
This quote has come up before in these discussions. I'm not sure what
Dirac meant by it, because it's impossible even in principle to tell which
photon is which in a multi-photon state-- there is no real distinction
between a photon "interfering with itself" and "interfering with another
photon."
Interference involves the state of the whole system. This state must
satisfy Bose-Einstein symmetry, and the photons in a multi-photon state
may not be considered singly without taking this into account.
Coherent light is a multi-photon state by definition, since a coherent
state is a certain kind of superposition of states with different numbers
of photons.
Usually, introductory treatments of these sorts of questions talk about
experiments in which light intensity is reduced until the expectation
value of the number of photons in flight is reduced to one or less.
However, this is not necessarily the same thing as a single-photon state,
since there could still be multi-photon components with small amplitudes,
and there will be if the light is coherent.
On the other hand, whether or not it is coherent is likely to make little
difference at these low intensities; the single-photon component is going
to dominate most phenomena, and special properties of coherence, such as
minimal uncertainty of the field observables, are not so important because
even the minimal uncertainty is going to be large compared to the field
intensity.
Greg Weeks mentioned the first thing that came to my mind as well, which
is that the infinite Compton wavelength of a single photon precludes
Newton-Wigner localization and may put limits on the applicability of the
notion of a single photon's wavetrain. John Baez translated this as
meaning that the Dirac-spike basis doesn't exist, but I think it actually
says more than that, because even for massive particles, Newton-Wigner
localization is only finitely precise.
The interesting thing would be to go further than this and ask, how well
can we do? How squashed-down can a single photon state get-- must they
have power-law tails or exponential ones? Furthermore, how do we even
answer the previous question, given that it may be hard to tell a piece of
the photon state's tail from the electromagnetic vacuum? How much
admixture of multi-photon states needs to go in before we can make
something that looks like a Gaussian lump?
I don't know the answers to any of these questions...
Alastair Ward
>
>In article <77q8lc$1c$1...@news.lth.se>, "Bjorn Wesen"
><bj...@sparta.lu.se.noSPAM> wrote:
>
>>"... each photon interferes only with itself. Interference between different
>>photons never occurs"
>>
>>Dirac, Quantum Mechanics 4th edition, page 9
snip of interesting comments
I wonder if we really ought to regard the interference pattern as an
'emergent' property in much the same way as consider a wave on water as
emergent even though the latter is classical. We can deduce almost nothing
from a flash on the screen corresponding to the detection of a single
photon. It is only after we have viewed the statistical behavior of an
ensemble of photons that we can observe the emergence of the interference
pattern. It does not matter whether we use a beam in which it might be said
that many photons are in the apparatus at the same time or whether we reduce
the intensity to such a low level that we can "talk" about one photon passing
thro at a time. What really matters I think is that we have a great many photons
before any interference is observed and so I feel we have to regard the
interference as emergent and a property of the ensemble.
Alastair Ward.
Ray Tomes
phy...@eensgi4.leeds.ac.uk (Dr P. Kinsler) wrote:
>: "... each photon interferes only with itself. Interference between different
>: photons never occurs"
>: Dirac, Quantum Mechanics 4th edition, page 9
>Dirac got that one wrong[1].
I strongly suspected so.
>Photons can interfere with themselves
>or other photons. If you consider photons to be excitations of EM
>field modes[2], then interfering different photons is pretty much the
>same as interfering classical fields[3]. Note that the field modes
>can be all sorts of things, even wavepacket modes which propagate.
>Pick mode which suit your physical situation, and make the maths
>part easier to do -- any orthonormal set will work. Asking "how
>big is this photon?" is too ill defined -- you have to specify the
>field mode the photon is (has been created) in.
I am unfamiliar with the term "field mode". What I am interested in is
for the case of an atom emitting light because an electron jumps from
one orbital state to another. Let us say hydrogen between 2nd and 1st
orbital. If there is some rule of thumb for different such cases that
would be very useful to me.
In such a case I want to know what is the envelope of the amplitude of
the emitted photon as would be measured by its ability to interfere with
itself when it travels over two different path lengths. The sort of
answer that would be useful to me is "it is a normal distribution with a
standard deviation of 88 times the wavelength" or something like that.
>[1] I've always wanted to say that. ;-)
>[2] As I said earlier in this thread.
Ah, I missed that. Unfortunately my internet provider seems to miss
some s.p.r posts which is very frustrating ... back to dejanews.
>[3] ALthough creating and annihilating photons isn't the same.
-- Ray Tomes -- http://www.kcbbs.gen.nz/users/rtomes/rt-home.htm --
Oz
<mmci...@world.std.com> writes
>
>Greg Weeks mentioned the first thing that came to my mind as well, which
>is that the infinite Compton wavelength of a single photon precludes
>Newton-Wigner localization and may put limits on the applicability of the
>notion of a single photon's wavetrain. John Baez translated this as
>meaning that the Dirac-spike basis doesn't exist, but I think it actually
>says more than that, because even for massive particles, Newton-Wigner
>localization is only finitely precise.
Um. I think you just said something important to the question. Could you
now say it again, but expressed for idiots?
>The interesting thing would be to go further than this and ask, how well
>can we do? How squashed-down can a single photon state get-- must they
>have power-law tails or exponential ones?
Yes, this would be very nice. However as a first approximation it would
be good just to have a rough handle on the length (give or take an order
or two) to get started with.
>Furthermore, how do we even
>answer the previous question, given that it may be hard to tell a piece of
>the photon state's tail from the electromagnetic vacuum? How much
>admixture of multi-photon states needs to go in before we can make
>something that looks like a Gaussian lump?
Nah, nah. Not allowed multi-photon states. Too confusing, too many
escape routes for handwaving and waffle.
>I don't know the answers to any of these questions...
Eh???
OhMyGod.
I only butted in because I can't say I have ever come across any comment
about a single photon wavetrain length. I would have thought it was
pretty 'standard' and 'well known' but I am pretty sure that I would
have remembered if I had come across it, and I don't think I have. Now
my son is just doing A-level (age 17) physics and is currently doing EM.
He hasn't asked this question, but he will. It will be a little
embarrasing if I explain that, not only do I not know, nor does
'physics' (as personified by the expert and revered posters on this
group).
As an aside that might be allowed by the Lord_High_moderator there was
an apocryphal story at Cambridge when I was an undergraduate on the
'very best' PhD topic ever. A wealthy, able and lazy postgrad did the
'single photon interfering with itself' experiment now so often quoted.
This involved setting up a series of attenuating slits in black boxes on
the roof of the Cavendish such that he could be sure only one (sunlight)
photon would ever be in the final (double slit) box at a time. He then
spent two years on a yacht in the south of france (probably doing
remarkably little physics). On his return he developed the plates and in
a month, wrote it up. The nice thing being that whatever the result, a
PhD would inevitably have to be granted due to it's importance.
I wish ...
One has the horrible (delightful) feeling that there is another,
similar, experiment going begging for the resourceful (and lazy)
postgrad.
--
Oz
ach...@my-dejanews.com
rto...@kcbbs.gen.nz (Ray Tomes) wrote:
> What I am interested in is
> for the case of an atom emitting light because an electron jumps from
> one orbital state to another. Let us say hydrogen between 2nd and 1st
> orbital. If there is some rule of thumb for different such cases that
> would be very useful to me.
>
> In such a case I want to know what is the envelope of the amplitude of
> the emitted photon as would be measured by its ability to interfere with
> itself when it travels over two different path lengths. The sort of
> answer that would be useful to me is "it is a normal distribution with a
> standard deviation of 88 times the wavelength" or something like that.
I would guess the following:
The exited state of the atom decays in a time dt. This leads to an
uncertainity in the energy of the photon of h/dt. Therefore one expects a
coherence length c dt, where c is the velocity of light. The
longer the atom can remain in its excited state, the longer is the coherence
length. E.g. for a "forbidden" transition one expects "longer wavetrains".
(I'm too lazy to do a small calculation, but I would be not astonished,
if a typical number would be a wavetrain with a length of
1/alpha=137 times the wavelength or for a "forbidden" transition
(1/alpha)^2 times the wavelength.)
This reminds me of an experiment we did in high school:
We observed some interference pattern in the light of some (Na ?) bulb.
When the bulb heated up, the interference pattern vanished. The reason
is, that the livetime dt of the excited state is determined by the time
between collisions of the atoms. With rising temperature, dt and the coherence
length gets smaller and the interference pattern vanishes.
I hope this answers your question.
Achim
Oz
a...@btinternet.com> writes
> We can deduce almost nothing
>from a flash on the screen corresponding to the detection of a single
>photon. It is only after we have viewed the statistical behavior of an
>ensemble of photons that we can observe the emergence of the interference
>pattern. It does not matter whether we use a beam in which it might be said
>that many photons are in the apparatus at the same time or whether we reduce
>the intensity to such a low level that we can "talk" about one photon passing
>thro at a time. What really matters I think is that we have a great many photons
>before any interference is observed and so I feel we have to regard the
>interference as emergent and a property of the ensemble.
Whilst this is all true, none the less it gets us nowhere further
towards answering the question. It says nothing about what would happen
(ie would a diffraction pattern form or not) if the two paths differed
by 10, 100, 1000 etc wavelengths for a SINGLE photon (to some
approximation) being in the system at any one time.
I suspect MM has bit the bullet by saying "we don't know" to this
astonishingly simple question. This suggests there is something
profoundly missing in our model of a photon.
--
Oz
[Moderator's note: Actually, I was saying *I* didn't know. The
answers to my questions should be deducible from QED if somebody
is willing to do the calculations. -MM]
Rajarshi Ray
> In sci.physics.research Bjorn Wesen <bj...@sparta.lu.se.noSPAM> wrote:
> : "... each photon interferes only with itself. Interference between different
> : photons never occurs"
>
> : Dirac, Quantum Mechanics 4th edition, page 9
>
> : Which more or less removes the distinction between a beam and "single
> : photons", and makes it feasible to use concepts as beam coherence with as
> : low intensity as "single photons" as well. Some Quantum Optician could
> : settle this, since all this thread is just a discussion of terminology I
> : think :)
>
> If it helps any, I'm a quantum optician, so here goes:
>
> Dirac got that one wrong[1]. Photons can interfere with themselves
> or other photons. If you consider photons to be excitations of EM
> field modes[2], then interfering different photons is pretty much the
> same as interfering classical fields[3]. Note that the field modes
> can be all sorts of things, even wavepacket modes which propagate.
> Pick mode which suit your physical situation, and make the maths
> part easier to do -- any orthonormal set will work. Asking "how
> big is this photon?" is too ill defined -- you have to specify the
> field mode the photon is (has been created) in.
You've got to be kidding about [1] !! It goes against the fundamental principle
of QM that DIFFERENT photons interfere with each other. Other than a single
photons amplitude interfering with itself, the only other possibility is that of
interference of scattering amplitudes like the interference of the two amplitudes
of scattering of two indistinguishable particles. That's the ONLY way two
particles can interfere with each other.
Oz
In article <782o75$hbb$1...@nnrp1.dejanews.com>, ach...@my-dejanews.com
writes
>The
>longer the atom can remain in its excited state, the longer is the coherence
>length. E.g. for a "forbidden" transition one expects "longer wavetrains".
>(I'm too lazy to do a small calculation, but I would be not astonished,
>if a typical number would be a wavetrain with a length of
>1/alpha=137 times the wavelength or for a "forbidden" transition
>(1/alpha)^2 times the wavelength.)
Hmmm. Sounds plausible. Unfortunately this means that photons carry
information about how they were made which would make 'free' photons
from different sources, but the same energy, different. I find this
disturbing. One might expect them to interact differently, in a reverse
of the above interactions for example.
Can of worms?
>This reminds me of an experiment we did in high school:
>We observed some interference pattern in the light of some (Na ?) bulb.
>When the bulb heated up, the interference pattern vanished. The reason
>is, that the livetime dt of the excited state is determined by the time
>between collisions of the atoms. With rising temperature, dt and the coherence
>length gets smaller and the interference pattern vanishes.
Hmmm. Yes. On the face of it quite convincing. I would be more convinced
if one could repeat that result without changing the temperature of the
atoms and the intensity of the lamp. In short a pointer, but one that
may easily mislead if the result was misinterpreted for teaching
purposes.
--
Oz
Oz
<ba...@galaxy.ucr.edu> writes
>(If answering
>it involves extensive computation, I'm not gonna do it - too busy!
>But if there's some way to get a rough answer without scribbling
>too much, maybe I'll give it a shot.)
The question was a simple (!) one. It takes one measure of the length of
the wavetrain of a *single* photon as the amount one can alter the path
lenths of the two legs of a simple interferometer and still see fringes
when the system is set up so that only one photon is in the equipment at
any time. (Ie the paths have different lengths).
OK, so that's a crude definition of 'length of wavetrain', but it ought
to give an order of magnitude of any sensible definition of 'length of
wavetrain'. I guess there are quite a lot of wavelengths in a wavetrain,
enough, say, so we know the number to within one h = energy/frequency.
Supplemental: how does this length vary with wavelength?
I would have thought this was in 'any' textbook (at least of those you
read) but it seems to have generated gales of handwaving, so I guess it
isn't. If that is the case then I guess is probably does involve
'extensive computation', probably by someone in the field. :-)
--
Oz
Raymond E. Rogers
question is a little irrelevant. Consider the two slit experiment, and
a physically very stable single photon generation apparatus. I examine
the photon densities at the detector screen and get "shape" (or length)
of the photons up to a phase term (psi* psi). Now from the theory of
differential equations I could change the two slit apparatus to produce
any pattern at the detector screen; thus I would be able to produce
photons of any "shape". The photon's shape is determined by it's
generator and history, it has no intrinsic length or shape.
Corrections welcome.
Ray
Matt McIrvin
(Dr Paul Kinsler) wrote:
>I've said previously that a "photon" is a single excitation of a
>field mode; so we add a photon to a mode by applying the creation
>operator for that mode to the quantum state of the field. How "big"
>is this photon? -- well, it is as big as the field mode to which it
>belongs.
Yes, and this is something I was not really taking into account in my
post, which implicitly assumed the plane wave modes usually used in the
formulations of quantum field theory you see in particle physics.
So, given that, one thing I find interesting is the idea of the
transformations that connect these different sets of field modes. Does a
single-photon state in one basis translate to a superposition of
multi-photon and vacuum states in another? How does this relate to what
you'd see in a particle detector? (Is it just that a photon written off
as noise in one basis becomes signal in another?)
Understand, as I ask these questions, I don't believe that my inability to
answer them is a problem with QED, it's just that these are hard and
subtle homework problems in quantum optics that I have never personally
done.
This probably somehow relates to the notion of Rindler particles, the
Unruh effect, and Hawking radiation-- there again, one is applying a
transformation between different conventional bases for the same physics,
and seeing different things... and this becomes a convienient way to
express certain real physical effects!
Oz
<rero...@gte.net> writes
>Now from the theory of
>differential equations I could change the two slit apparatus to produce
>any pattern at the detector screen; thus I would be able to produce
>photons of any "shape".
The pattern is not of interest. Merely the differential length between
the two legs that is the boundary between any diffraction pattern and
none.
>The photon's shape is determined by it's
>generator and history, it has no intrinsic length or shape.
So you suggest 'free' photons of the same energy are different?
Is there any evidence for this?
--
Oz
Oz
<mmci...@world.std.com> writes
>it's just that these are hard and
>subtle homework problems in quantum optics that I have never personally
>done.
<Oz: fearful>
>This probably somehow relates to the notion of Rindler particles, the
>Unruh effect, and Hawking radiation-- there again, one is applying a
>transformation between different conventional bases for the same physics,
>and seeing different things... and this becomes a convienient way to
>express certain real physical effects!
It does seem to be rather a lot of heavyweight processes (well, sounds
like it ..) to answer what appears to be a very simple and elementary
question. If we can't easily describe what happens to a single photon in
a simple (well, about as simple as you can get without being trivial)
interferometer then one wonders if modern particle physics is as far
advanced as one imagined.
In fact I am shocked.
--
Oz
john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>The question was a simple (!) one.
The simplest-sounding questions often have no answers until you
make them more precise - which makes them seem more complicated.
>It takes one measure of the length of
>the wavetrain of a *single* photon as the amount one can alter the path
>lenths of the two legs of a simple interferometer and still see fringes
>when the system is set up so that only one photon is in the equipment at
>any time. (Ie the paths have different lengths).
>
>OK, so that's a crude definition of 'length of wavetrain', but it ought
>to give an order of magnitude of any sensible definition of 'length of
>wavetrain'.
Okay, I think I understand this definition.
And I think you're asking: given the frequency (or if you prefer,
the wavelength) of a photon, what is the length of its wavetrain?
And the answer is: it depends on how it's prepared! You can prepare
the photon in all sorts of states with different wavefunctions; some
of these will look have zillions of wiggles in a wavetrain, others
only one wiggle.
(I use the term "wiggle" rather than "wavelength" because one very
popular official definition of wavelength applies only to *infinite*
wavetrains.)
So for someone to answer, you have to be more specific about how the
photon was prepared: by emission from an ionized atom, by sending a
laser through lots of beamsplitters, or what have you.
In short, I think you're not asking a question about photons per se,
you're asking lots of different questions about different light
sources - without saying *which* light source you're interested in.
This would explain why you're not getting satisfying answers.
Oz
In article <78atss$3n6$1...@pravda.ucr.edu>, john baez
<ba...@galaxy.ucr.edu> writes
>
>And I think you're asking: given the frequency (or if you prefer,
>the wavelength) of a photon, what is the length of its wavetrain?
I have a plausible argument that suggests they should all have the same
number of wiggles. Otherwise if I observed a boosted photon, which would
have a different wavelength depending on my boost, there would be a
different number of wiggles again making them different from those
photons emitted by me of the same wavelength.
>And the answer is: it depends on how it's prepared! You can prepare
>the photon in all sorts of states with different wavefunctions; some
>of these will look have zillions of wiggles in a wavetrain, others
>only one wiggle.
I am not convinced. At all. If they have different numbers of wiggles
then that would make photons of the same energy from different sources
different. I have a problem with this in the same way that I would have
a problem with electrons produced from different mechanisms having
different numbers of wiggles and thus being different.
>(I use the term "wiggle" rather than "wavelength" because one very
>popular official definition of wavelength applies only to *infinite*
>wavetrains.)
No, that's fine, a wiggle is good for me.
>In short, I think you're not asking a question about photons per se,
>you're asking lots of different questions about different light
>sources - without saying *which* light source you're interested in.
>This would explain why you're not getting satisfying answers.
I do not believe (but stand to be corrected with any properly plausible
argument) that this is the case for isolated 'free' photons.
--
Oz
jmf...@aol.com
sid...@u.washington.edu (John Sidles) wrote:
>>In article <368a3615...@kcbbs.gen.nz>, rto...@kcbbs.gen.nz says...
>>>I am wondering whether any attempt has ever been made to measure the
>>>wave train length and amplitude profile of a single photon?
>For practical quantum engineers, the modern decoherence
>viewpoint has the advantage of making many quantum mysteries
>accessible to calculation. For example, just how long does it
>take for a quantum to be absorbed? What is the chance that a
>two-state system will "change its mind" partway through an
>absorption process, and thus induce an error in a quantum
>calculation?
What's been bugging me is:
1. How does this "hard"ware get checked out (preventive maintenance)?
Presumably one would want the capability of "inserting" an error.
2. How does one ensure that all errors can be detected?
>These are no longer regarded as philosophical
>questions, rather they are becoming engineering issues of great
>practical importance.
I'm looking forward to it :-).
<snip documentation opportunity>
/BAH
Tom Roberts
> I have a plausible argument that suggests they should all have the same
> number of wiggles. Otherwise if I observed a boosted photon, which would
> have a different wavelength depending on my boost, there would be a
> different number of wiggles again making them different from those
> photons emitted by me of the same wavelength.
The difficulty, as John Baez pointed out, is in the preparation of the
light beams, not in the beams themselves. In your statement you are comparing
two light beams which are prepared differently -- why shouldn't they differ?
In making your claim you are implicitly assuming that all photons of a given
wavelength are identical...
But really all you can do is compare the _light_beams_ (not the photons).
The _beams_ can have different energy and momentum spreads, and this
directly affects their correlation length.
A major lesson of quantum mechanics is that one should not try
to discuss things which are not measured. In this case it is the
correlation length of the _beam_ which is measured, and only that.
Trying to turn that into properties of "individual" "photons"
is unjustified, IMHO.
> John Baez wrote:
> >And the answer is: it depends on how it's prepared! [...]
>
> I am not convinced. At all. If they have different numbers of wiggles
> then that would make photons of the same energy from different sources
> different. I have a problem with this in the same way that I would have
> a problem with electrons produced from different mechanisms having
> different numbers of wiggles and thus being different.
The point is: Electron _beams_ can differ in their energy and momentum
spreads, and this directly affects their correlation length. That is
analogous to what happens for photons in beams prepared differently. You
can really only compare _beams_, not individual photons. This remains true
even if you crank the intensity down so that on average less than one photon
is in the apparatus at a time....
> >In short, I think you're not asking a question about photons per se,
> >you're asking lots of different questions about different light
> >sources [...]
>
> I do not believe (but stand to be corrected with any properly plausible
> argument) that this is the case for isolated 'free' photons.
How do you propose to obtain these "isolated 'free' photons"? The conditions
under which you obtain them are important....
That's another hard lesson of quantum mechanics: _everything_ matters,
and you can't even look without disturbing the system you look at.
Tom Roberts tjro...@lucent.com
Oz
<tjro...@lucent.com> writes
>Oz wrote:
>> I have a plausible argument that suggests they should all have the same
>> number of wiggles. Otherwise if I observed a boosted photon, which would
>> have a different wavelength depending on my boost, there would be a
>> different number of wiggles again making them different from those
>> photons emitted by me of the same wavelength.
>
>The difficulty, as John Baez pointed out, is in the preparation of the
>light beams, not in the beams themselves. In your statement you are comparing
>two light beams which are prepared differently -- why shouldn't they differ?
I don't believe I am comparing two different light beams because there
can only be one photon in the apparatus at the time so it MUST interfere
with itself. Naively perhaps, I assume that it can only interfere with
itself if the difference in the two paths is less than it's own length
(since a photon must surely be coherent with itself). It's just a single
photon version of how you might measure the coherence length of a beam
of photons.
Of course it will require a lot of single isolated photon events to
build up a diffraction pattern, but this is not so hard since a
photographic plate, or better a CCD, can add them spacially quite
easily.
In fact I would have thought a modern undergrad optics lab with a CCD
detector would be able to do this experiment quite easily, perhaps over
a weekend. Attenuating a laser beam (easier to set the system up, I
presume) so that for a device of linear dimensions of about a meter the
CCD receives (assuming a near 100% detection rate) say under one photon
per millisecond (one could probably get away with more) should ensure
effectively only one photon in the apparatus at any one time.
>In making your claim you are implicitly assuming that all photons of a given
>wavelength are identical...
Indeed. I am perfectly happy for photon beams not to be identical (they
are not), but I am disturbed if individual photons are different. I
would like my elementary particles to all be identical, including the
photon.
>But really all you can do is compare the _light_beams_ (not the photons).
Why? Individual photons can interfere with themselves perfectly well.
>The _beams_ can have different energy and momentum spreads, and this
>directly affects their correlation length.
But individual photons should not IMHO.
> A major lesson of quantum mechanics is that one should not try
> to discuss things which are not measured.
Hence I describe the apparatus.
>In this case it is the
> correlation length of the _beam_ which is measured, and only that.
Not with one photon in the apparatus at a time.
> Trying to turn that into properties of "individual" "photons"
> is unjustified, IMHO.
So I didn't ask that.
<Compare with electrons>
>The point is: Electron _beams_ can differ in their energy and momentum
>spreads, and this directly affects their correlation length. That is
>analogous to what happens for photons in beams prepared differently. You
>can really only compare _beams_, not individual photons. This remains true
>even if you crank the intensity down so that on average less than one photon
>is in the apparatus at a time....
I haven't seen any report of work showing that elecrons can interfere
with themselves but I would be very surprised if they could not. My
argument thus extends to electrons (etc) also.
>> I do not believe (but stand to be corrected with any properly plausible
>> argument) that this is the case for isolated 'free' photons.
>
>How do you propose to obtain these "isolated 'free' photons"? The conditions
>under which you obtain them are important....
I have already said that, several times.
You can optionally use different sources and see if you get a different
wavetrain length. If you do the you have shown two photons are
different, which is rather interesting (to say the very least). I think
my argument about viewing photons at different relativistic boosts is a
reasonable first order argument that all photons (if all are identical
due to the laws of nature) should have the same number of wiggles
regardless of their energy. By boosting relative to a source I can view
the 'same' photons at different enegies and I think their shape should
(probably) stay the same.
> That's another hard lesson of quantum mechanics: _everything_ matters,
> and you can't even look without disturbing the system you look at.
Agreed.
--
Oz
john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>In article <78atss$3n6$1...@pravda.ucr.edu>, john baez
><ba...@galaxy.ucr.edu> writes
>>And I think you're asking: given the frequency (or if you prefer,
>>the wavelength) of a photon, what is the length of its wavetrain?
>I have a plausible argument that suggests they should all have the same
>number of wiggles. Otherwise if I observed a boosted photon, which would
>have a different wavelength depending on my boost, there would be a
>different number of wiggles again making them different from those
>photons emitted by me of the same wavelength.
Hmm. Suppose I have a photon in some state. Say it has a wavetrain
of length L consisting of n wiggles of length L/n. There's no need
for all the wiggles to be the same length but let's keep things simple.
Now say we boost it. The wavetrain gets Lorentz-contracted by some factor
and each wiggle gets contracted by the same factor, so now we have a
wavetrain of length L/A consisting of n wiggles of length L/nA. Great.
Now what does this have to do with your claim that every wavetrain has
the same number of wiggles? Nothing. We could have some *other* photon
with a wavetrain of length L' consisting of n' wiggles of length L'/n',
and if we boosted *that* one, we'd get a photon with a wavetrain of length
L'/A consisting of n' wiggles of length L'/n'A. Nothing in any of this
allows us to conclude that n = n', which is what you apparently want to
show.
And indeed I can actually write down a photon state consisting of
a wavetrain of more or less any length I please consisting of more
or less any number of wiggles that I desire.
>>And the answer is: it depends on how it's prepared! You can prepare
>>the photon in all sorts of states with different wavefunctions; some
>>of these will look have zillions of wiggles in a wavetrain, others
>>only one wiggle.
>
>I am not convinced.
I thought you just wanted to be informed - you didn't say you needed
to be *convinced*.
>If they have different numbers of wiggles
>then that would make photons of the same energy from different sources
>different.
Right. Exactly. Photons with the same expectation value of energy
can be in many different states. Of course, a photon in an energy
*eigenstate* - i.e., one with no uncertainty in its energy - must have
an *infinitely* long wavetrain. And it must look like a sine wave.
But in practice it's impossible to create such a state. Realistic
photons are finite-length "wave packets", and the exact shape of the
wave packet depends on the source.
>I have a problem with this in the same way that I would have
>a problem with electrons produced from different mechanisms having
>different numbers of wiggles and thus being different.
Well, I can also write down an electron state consisting of a
wavetrain of more or less any length I please, consisting of more
or less any number of wiggles that I desire. For example, if I
ignore the fact that the electron is a spinor, I can just write:
psi(x) = sin(ax) if 0 < x < L
0 otherwise
and I get a state whose wavetrain is length L with wiggles of length
2 pi / a.
john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>In article <787ep8$n57$1...@news-1.news.gte.net>, Raymond E. Rogers
><rero...@gte.net> writes
>>The photon's shape is determined by its
>>generator and history, it has no intrinsic length or shape.
>So you suggest 'free' photons of the same energy are different?
Certainly that's what the standard theory says. More precisely,
there are lots of very different 1-photon states in which the
photon has the same *expectation value of energy* - even apart
from trivial dfferences like translations and rotations.
Of course, if you demand that the photon be in an energy *eigenstate*,
so that you know its energy *exactly*, there is less room for free
play. But even so, there are lots of very different-looking states.
In a previous post I said that being an energy eigenstate forced
the photon to be a plane wave, i.e., an infinite sinusoidal wavetrain
of a given frequency. But that wasn't right: any superposition of
plane waves of the same frequency will do.
(In case you're wondering what the heck I'm talking about, I'm
dealing with the ambiguity in your phrase "same energy" by interpreting
in both possible ways: same expectation value of energy, and same
energy eigenvalue.)
>Is there any evidence for this?
If you want experimental evidence, take a photon of a given energy
and send it through a half-silvered mirror. Now you have a "split"
photon whose expected energy is the same, which is in a very
different state than the original "unsplit" one. To get even more
different states, just keep splitting and/or recombining the given
photon, sending it through lenses, refractors, etc..
Ligare Martin
: In article <782o75$hbb$1...@nnrp1.dejanews.com>, ach...@my-dejanews.com
: writes
: >The
: >longer the atom can remain in its excited state, the longer is the coherence
: >length. E.g. for a "forbidden" transition one expects "longer wavetrains".
: >(I'm too lazy to do a small calculation, but I would be not astonished,
: >if a typical number would be a wavetrain with a length of
: >1/alpha=137 times the wavelength or for a "forbidden" transition
: >(1/alpha)^2 times the wavelength.)
: Hmmm. Sounds plausible. Unfortunately this means that photons carry
: information about how they were made which would make 'free' photons
: from different sources, but the same energy, different. I find this
: disturbing. One might expect them to interact differently, in a reverse
: of the above interactions for example.
Photons do carry information about "how they were made." Consider,
for example, spontaneous emission starting from a state consisting
of an excited-state atom and the vacuum field state. This
state evolves into a linear combination of the initial state and
the multiplicity of states with a ground-state atom and one photon.
There are very many of these states with one photon because there
are essentially an infinite number of field modes that can "receive"
the excitation. The intial state is not a state of definite
energy, so it does not evolve into a state of definite energy,
and the "photon" is really a linear combination of modes with
a range of energies. The relative quantum amplitudes for the
various modes give the emitted "photon" it's character. Photons
emitted by relatively short-lived states are made up of a linear
combination of modes with a relatively broad range of frequencies
(i.e., energies), and vice versa. This is just the time-energy
uncertainty principle.
One reasonable definition of the "wave-train" of a spontaneously
emitted photon is the spatial extent of the expectation value
of the square of the electric field operator (the intensity
operator). It's not too hard to show for simple electromagnetic
cavities that spontaneous emission results in an spatial intensity
expectation value in the form of a pulse leaving the atom with a
sharply rising leading edge followed by an exponentially
falling tail which is just the spatial consequence of the
exponential time dependence of the decay. This definition
of "wave-train" does give a measure of the length by which
two photon paths can differ and still show interference.
Martin Ligare
Bucknell University Physics Department
mli...@bucknell.edu
[Moderator's note: Quoted text trimmed. -MM]
Aleksandr Timofeev
Tom Roberts <tjro...@lucent.com> wrote:
[snip]
> The difficulty, as John Baez pointed out, is in the preparation of the
> light beams, not in the beams themselves. In your statement you are comparing
> two light beams which are prepared differently -- why shouldn't they differ?
> wavelength are identical...
>
> directly affects their correlation length.
>
> correlation length of the _beam_ which is measured, and only that.
> is unjustified, IMHO.
>
>
> The point is: Electron _beams_ can differ in their energy and momentum
> spreads, and this directly affects their correlation length. That is
> analogous to what happens for photons in beams prepared differently. You
> can really only compare _beams_, not individual photons. This remains true
> even if you crank the intensity down so that on average less than one photon
> is in the apparatus at a time....
>
>
> How do you propose to obtain these "isolated 'free' photons"? The conditions
> under which you obtain them are important....
>
> and you can't even look without disturbing the system you look at.
>
If I have understood uncorrectly you, I bring deep apologies. The essence
of your reasons is reduced to refusaling a possibility of observation of
photons and therefore to refusaling of their existence. Is not it so?
AT
Michael Weiss
A most fascinating discussion, Oz! Let me see if I can make sense
of it, at least on my own terms...
The aptly named Ray *appeared* to be asking a purely experimental
question, what with half-silvered mirrors and diffraction patterns
and all that other paraphenalia that Pauli could break just by taking
a train-ride through Heidelberg --- but he wasn't, not really, because
he snuck that oh-so-innocent term "photon" into the question.
And a photon is not an experimental beast. A photon is a theoretical
construct. So we should decide first if we want a ruling on the letter
of the law, or if we'd rather take depositions from witnesses. In other
words, do we want to convert Ray's question into a truly experimental
one, or would we rather learn what theory says?
The experimentalists have spoken, though neither loudly nor long,
as is usually the case on s.p.r. For the experimental question, we replace
"photon" with "really really weak beam of light, over a long enough period
of time to get a diffraction pattern". Someone mentioned some experiment
with an ionized source, where the maximum interfereable path difference
(I *guess* this is what they mean by "coherence length") went down
as the temperature of the source went up. Ale2 seemed to think
some one had done an experiment with lasers, getting a coherence
length of several meters. Now laser beams are usually *not* weak, but
Ale2 suggested that beam-splitting would get round this problem.
Interesting to know if the experiment has actually been done, and if
so, the details.
If the hearsay testimony *does* match the direct depositions, then there
you have it --- the coherence length *does* depend on how the beam
is prepared. Depends rather strongly, it appears! Could be a fraction
of an inch, could be a few feet. So much for expectations.
If it *hasn't* been done, then get cracking, experimenters!
OK, enough from the witnesses, already! Having schmoozed with
the schmotons for lo those many months, I should understand this
stuff, Well, yes and no.
One thing I learned is just how monomaniacal QM is about superpositions.
Jb mentioned the spiral staircase of lies that is teaching. Many steps ago,
I was taught that a beam of light *could be* thought of as a barrage
of bullets. Indeed, the PSSC high school physics book (co-authored by
none other than Albert Baez!) had a picture of this, one that left an
indelible impression on my soft waxy mind. [Not that I'm saying AB was
responsible for that particular picture. The book had a few dozen
co-authors.]
Easy enough to *count* these little bullets, no? And make sure there's
just one, if you choose? And since the PSSC book said you could
*also* think of the beam of light as a splish-splash of waves
(with picture to match!), perfectly fine to ask about the total number
of splishes (or wiggles) in the wave-train, right? At least in principle,
yes?
NO! For as I have learned from Baez the Younger, to have a half-decent
shot at getting a classical-looking EM field out of photon,
you have to infuse the picture with a heavy mist of vagueness,
so you're *not at all sure* how many photons there are. In other words,
you have to take superpositions of states with 1, 2, 3, many photons.
I don't know about you, but I somehow find this far more curious than
superimposing all those plane waves to get a wave-packet. And yet
the mathematics is practically identical.
So any marriage of a single photon to a nice classical wavetrain of length
L is doomed to end in divorce. Irreconcilable differences, as they say.
So much for what I do understand --- or at least think I think I understand.
Now to throw a confusion or three into the mix.
In QM, we can take superpositions, but we can also do ordinary probability
mixtures. (That old density-matrix stuff, that's been discussed many
times on s.p.r.) When the quantum opticians talk about coherence length,
are they thinking about the superpositions, or the mixtures?
Also: maybe we don't insist that our classical wave picture have a
well-defined EM field at every point. Maybe it's enough for its *energy*
to be well-defined at every point. In other words, can we have a quantum
state of the EM field such that:
The photon number is very strongly peaked at 1 --- if you expand the
state in a basis of photon-number eigenstates, the coefficient of |1>
outshines all the other coefficients.
AND AT THE SAME TIME
The "value of E^2 + B^2 at spacetime point (x,y,z,t)"
has very low uncertainty for all (x,y,z,t) --- even though
E and B *separately* are very fuzzy indeed at all (x,y,z,t).
Oz
<ba...@galaxy.ucr.edu> writes
>In article <78ddq2$irp$1...@agate.berkeley.edu>,
>Oz <O...@upthorpe.demon.co.uk> wrote:
>
>>I have a plausible argument that suggests they should all have the same
>>number of wiggles. Otherwise if I observed a boosted photon, which would
>>have a different wavelength depending on my boost, there would be a
>>different number of wiggles again making them different from those
>>photons emitted by me of the same wavelength.
>
>the same number of wiggles? Nothing. We could have some *other* photon
>with a wavetrain of length L' consisting of n' wiggles of length L'/n',
Ah, the pickiness of a teacher! OK, I am implying a default isolated
photon in a/the 'free' state which I guess is approximated by an
isolated photon in intergalactic space. I expect them all to have the
same state and thus look alike. In my micro-experiment OK, it will have
a different state (split into two at the very least) but I hope that we
can get some plausible connection between the number of wiggles (and
maybe the waveshape) and some trival states. Just for fun, like.
>And indeed I can actually write down a photon state consisting of
>a wavetrain of more or less any length I please consisting of more
>or less any number of wiggles that I desire.
Hmmm. I would look to one stable in time containing a single isolated
photon. Can you do that? If so we get closer to the answer to my
question. However, I bet you find a way not to answer this. :-)
>>>And the answer is: it depends on how it's prepared! You can prepare
>>>the photon in all sorts of states with different wavefunctions; some
>>>of these will look have zillions of wiggles in a wavetrain, others
>>>only one wiggle.
>>
>>I am not convinced.
[See comment on state of a 'free' photon. I don't think these states are
what I describe as 'free' or 'stable in time'.]
>I thought you just wanted to be informed - you didn't say you needed
>to be *convinced*.
I am a strange character who will not believe until he is convinced and
will not be informed until he believes (the veracity of the statements).
I doubt I am the only one here with this mindset, I think you would
qualify also for example.
>>If they have different numbers of wiggles
>>then that would make photons of the same energy from different sources
>>different.
>
>Right. Exactly. Photons with the same expectation value of energy
>can be in many different states. Of course, a photon in an energy
>*eigenstate* - i.e., one with no uncertainty in its energy - must have
>an *infinitely* long wavetrain. And it must look like a sine wave.
>But in practice it's impossible to create such a state. Realistic
>photons are finite-length "wave packets", and the exact shape of the
>wave packet depends on the source.
This I don't believe for my specified 'free' photon. I am happy (indeed
might insist) that this is so, close to the emitter. However this means
that (using the single photon interferometer described elsewhere) I
could distinguish between 'free' photons of the same energy. I don't
like this.
I would expect an emitted photon to more-or-less rapidly evolve into a
standard 'free' state as it leaves the vicinity of the emitting system.
Indeed, I would expect the dynamics of emission to be such that this is
enforced. I don't think this would prevent the emitting system from
taking it's time to emit a free photon, that only requires (as far as my
very limited knowledge suggests) a longer-lived entangled emitting
state.
>>I have a problem with this in the same way that I would have
>>a problem with electrons produced from different mechanisms having
>>different numbers of wiggles and thus being different.
>
>Well, I can also write down an electron state consisting of a
>wavetrain of more or less any length I please, consisting of more
>or less any number of wiggles that I desire. For example, if I
>ignore the fact that the electron is a spinor, I can just write:
>
>psi(x) = sin(ax) if 0 < x < L
> 0 otherwise
>
>and I get a state whose wavetrain is length L with wiggles of length
>2 pi / a.
Is this stable in time, ie does not evolve into a state with a different
set of wiggles? Does this entirely describe a single isolated electron?
I hope not ......
:-)
--
Oz
Oz
<ba...@galaxy.ucr.edu> writes
>>Oz
>>So you suggest 'free' photons of the same energy are different?
>
>Certainly that's what the standard theory says. More precisely,
>there are lots of very different 1-photon states in which the
>photon has the same *expectation value of energy* - even apart
>from trivial dfferences like translations and rotations.
OK, now I'm forced to ask exactly what you mean by this but later in
this post I think you answer it. Yes, photons of the same energy can
indeed be in different states, they would have a different state in both
arms of an interferometer than zipping through intergalactic space. Yes,
I agree. However this doesn't alter my argument because a photon of the
same energy in the same environment (in this case within the two arms of
the interferometer) should be in the same state (evolving though it may
be) regardless of it's source. It thus seems to me to be a valid
question to ask a slightly modified question to whit, "what is the
apparent length of the wavetrain of a photon in an interferometer"?
Then we can argue how this information can be applied, or not, to a
photon in other situations. Intergalactic space for example.
>Of course, if you demand that the photon be in an energy *eigenstate*,
>so that you know its energy *exactly*, there is less room for free
>play.
No, this would make it a very special photon I think, and I am more
interested in a more general case. Just a common or garden photon coming
from some convenient source or other to start with.
>But even so, there are lots of very different-looking states.
>In a previous post I said that being an energy eigenstate forced
>the photon to be a plane wave, i.e., an infinite sinusoidal wavetrain
>of a given frequency. But that wasn't right: any superposition of
>plane waves of the same frequency will do.
>
>(In case you're wondering what the heck I'm talking about, I'm
>dealing with the ambiguity in your phrase "same energy" by interpreting
>in both possible ways: same expectation value of energy, and same
>energy eigenvalue.)
I am not disagreeing with your comments (regrettably). However I hope
you were impressed that I followed your dictums and set out exactly the
equipment and conditions of the experiment so you knew moderately well
what exactly we are measuring. Years ago, I wouldn't have.
>>Is there any evidence for this?
>
>If you want experimental evidence, take a photon of a given energy
>and send it through a half-silvered mirror. Now you have a "split"
>photon whose expected energy is the same, which is in a very
>different state than the original "unsplit" one. To get even more
>different states, just keep splitting and/or recombining the given
>photon, sending it through lenses, refractors, etc..
Agreed. Lets just put single photons through one double slit, maybe
allow a mirror (maybe two) and a ccd to measure the photon, hokay?
Keep it nice and simple. :-)
--
Oz
john baez
<super...@my-dejanews.com> wrote:
>In article <hnxNkLAafWr2Ew$q...@upthorpe.demon.co.uk>,
> Oz <O...@upthorpe.demon.co.uk> wrote:
>> Of course it will require a lot of single isolated photon events to
>> build up a diffraction pattern, but this is not so hard since a
>> photographic plate, or better a CCD, can add them spacially quite
>> easily.
>
>the photons as single isolated events is meaningless.
We can produce single isolated photons in the lab, we can detect single
isolated photons, and we can use quantum mechanics to predict (sometimes
probabilistically, and sometimes deterministically) what their behavior
will be, and we can do quantum interference experiments using single
photons. So I'm not sure what you mean here.
john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>In article <78edob$e7t$1...@pravda.ucr.edu>, john baez
><ba...@galaxy.ucr.edu> writes
>>>I have a plausible argument that suggests they should all have the same
>>>number of wiggles. Otherwise if I observed a boosted photon, which would
>>>have a different wavelength depending on my boost, there would be a
>>>different number of wiggles again making them different from those
>>>photons emitted by me of the same wavelength.
>>Now what does this have to do with your claim that every wavetrain has
>>the same number of wiggles? Nothing. We could have some *other* photon
>>with a wavetrain of length L' consisting of n' wiggles of length L'/n',
>Ah, the pickiness of a teacher! OK, I am implying a default isolated
>photon in a/the 'free' state which I guess is approximated by an
>isolated photon in intergalactic space. I expect them all to have the
>same state and thus look alike.
Okay, but since this is what your argument was supposed to *demonstrate*,
it's no fair to use this assumption in your argument! Yes, if all photons
of the same energy have the same number of wiggles, they all have the same
number of wiggles - that's what this argument shows.
>>And indeed I can actually write down a photon state consisting of
>>a wavetrain of more or less any length I please consisting of more
>>or less any number of wiggles that I desire.
>
>Hmmm. I would look to one stable in time containing a single isolated
>photon. Can you do that? If so we get closer to the answer to my
>question. However, I bet you find a way not to answer this. :-)
By "stable in time" do you mean that the photon's wavefunction moves
along at a constant speed while maintaining its shape? Sure, I can
write down infinitely many states like this. To keep life simple I'll
pretend the photon is a massless spin-0 particle instead of a massless
spin-1 particle: this amounts to ignoring the polarization of the photon.
I'm only doing this this reduce the mathematical clutter, and if you
force me, I'll do the actual spin-1 case - but I'm warning you, you'll
regret it!
Okay, here we go:
phi(t,x,y,z) = f(x-ct)
where f is *any function I like*, and c is the speed of light, which
I'm not setting equal to 1, just so you can see that this little wave
moves along at the speed of light while keeping its shape. Since f is
*any function I like*, the shape of this wave is *arbitrary*.
I'm actually surprised that you think the photon wavefunction has some
kind of "default" state and that all isolated photons tend to this state.
A photon is a wave. Waves come in all shapes.
>>>I am not convinced.
>
>>I thought you just wanted to be informed - you didn't say you needed
>>to be *convinced*.
>
>I am a strange character who will not believe until he is convinced and
>will not be informed until he believes (the veracity of the statements).
>I doubt I am the only one here with this mindset, I think you would
>qualify also for example.
Yes, I was just teasing you.
>I don't believe for my specified 'free' photon. I am happy (indeed
>might insist) that this is so, close to the emitter. However this means
>that (using the single photon interferometer described elsewhere) I
>could distinguish between 'free' photons of the same energy. I don't
>like this.
Well, you can call up Heaven and tell them you don't like how they designed
the universe. "If you have complaints about quantum mechanics, press 2.
I'm sorry, all our angels are busy with other calls. Please hold; we value
your opinions highly."
>I would expect an emitted photon to more-or-less rapidly evolve into a
>standard 'free' state as it leaves the vicinity of the emitting system.
I wouldn't. Just as I wouldn't expect a bunch of frictionless pendulums
given random kicks to rapidly evolve into some standard state, like all
swinging in unison or something. They will just keep swinging the way
you happened to start them up. Mathematically that's just what the
electromagnetic field is: a bunch of uncoupled harmonic oscillators.
>>Well, I can also write down an electron state consisting of a
>>wavetrain of more or less any length I please, consisting of more
>>or less any number of wiggles that I desire. For example, if I
>>ignore the fact that the electron is a spinor, I can just write:
>>
>>psi(x) = sin(ax) if 0 < x < L
>> 0 otherwise
>>
>>and I get a state whose wavetrain is length L with wiggles of length
>>2 pi / a.
>
>Is this stable in time, ie does not evolve into a state with a different
>set of wiggles? Does this entirely describe a single isolated electron?
It describes a single isolated electron. Since an electron is massive,
this wave doesn't stay the same shape - unlike the case of a photon.
The reason is that massive things can move at different speeds, while
massless things must always move at the speed of light. So unlike the
case of a photon, for the electron different Fourier components
of the above wave will move at different speeds. This leads the wave
to keep changing shape and spreading out. It doesn't settle down to
some "equilibrium shape" - it just keeps changing.
For the electron, the only way to get a wavefunction that doesn't
change shape is to start with something like this:
psi(x) = exp(iax)
In other words, an *infinite* sinusoidal wavetrain.
But of course, it's not exactly practical to create an *infinite*
wavetrain. So in practice, the wavefunction of a single isolated
electron keeps changing shape.
Raymond E. Rogers
a time! It requires reasonable resolution but the facts are there; as a
matter of fact Einstein formulated the basic theory some time ago.
My interpretation of QM says that the solution of the Schrodinger
equation (which is generally a diffuse function with wave properties)
for EM gives functions "psi" such that <psi* | psi> is the PDF
(probability density function) of finding a photon at a particular point
in space-time. The interference is the result of the PDF factor "psi"
having phase delays and such. The real interference occurs with "psi"
and we merely measure the result as <psi*|psi>.
One could say that each photon interferes with itself and we can
measure the effect of many _similiar_ photons doing it to themselves
individually to obtain a profile of "psi". It is a way to think about
the situation if one insists that our measurements are talking about
"reality" rather than the shadows of reality. This wouldn't be true to
the QM model since it is really the field described by the differential
equation with certain boundary/starting conditions (Cauchy type of
conditions) that is the reality.
Ray
super...@my-dejanews.com wrote:
> You have fallen into possibly the single biggest trap with QM. To talk about
> the photons as single isolated events is meaningless. If this were the case
> we wouldn't need QM to describe the experiment we could use statistical
> analysis. Your whole argument seems to rest on this assumption. Think about
> what we are all calling the result "interference/diffraction pattern". Where
> is the interference if every photon is completely independent from the others
> and the apparatus? As I see it there are only 2 possibilities to describe how
> we get the result. 1) Each particle somehow interferes with the others
> despite being temporally separated. 2) Each particle somehow modifies the
> apparatus in such a way as to cause the effect. Unfortunately I don't think
> we can say just what happens and we are left with having to treat the
> experiment as an "unanalyzable whole" as Bohr would say. This argument about
> particles interfering with themselves is also pretty meaningless unless
> you/we can say exactly how this happens, and I don't care much for
> reputations. NJH.
Oz
<mli...@coral.bucknell.edu> writes
>Photons
>emitted by relatively short-lived states are made up of a linear
>combination of modes with a relatively broad range of frequencies
>(i.e., energies), and vice versa.
I think you are referring to photons (plural) are you not? The
combination of modes with a broad range of frequences are in fact
photons of different energies, no?
This doesn't apply to a single photon though.
--
Oz
Oz
>In article <hnxNkLAafWr2Ew$q...@upthorpe.demon.co.uk>,
> Oz <O...@upthorpe.demon.co.uk> wrote:
>> Of course it will require a lot of single isolated photon events to
>> build up a diffraction pattern, but this is not so hard since a
>> photographic plate, or better a CCD, can add them spacially quite
>> easily.
>
>the photons as single isolated events is meaningless.
Strange. I have quite definitely seen astrophysicists detecting single
photons on ccd's. This is surely an isolated event. I notice even
Feynman himself talks about detecting individual photons using a
photomultiplier. Since it happens, I find it hard to consider it
meaningless.
>If this were the case
>we wouldn't need QM to describe the experiment we could use statistical
>analysis.
Eh? Who said you didn't need QM to describe the experiment? Certainly
not me. If you mean QM predicts exactly where the photon will strike
(OK, interact if pedantic) then of course it doesn't, actually nothing
can, but the experiment does not seek to predict a single photon, indeed
far from it.
>This argument about
>particles interfering with themselves is also pretty meaningless unless
>you/we can say exactly how this happens, and I don't care much for
>reputations. NJH.
1) Photons interfering with themselves is just the way it is, and can
hardly be meaningless if that's what the apparatus is designed to
detect. It may not be important, but that's another question.
2) If something happens that you can't say exactly how it happens then
this is definitely not meaningless. It shows a hole in one's
understanding. For such a simple system as the one described, it's a
very big hole indeed IMHO, and thus more meaningful and certainly
interesting.
3) Are you suggesting that photons cannot interfere with themselves? Are
you suggesting that a photon may have an effective wavetrainlength of
several meters no matter what the intensity is?
--
Oz
Tom Roberts
> <tjro...@lucent.com> writes
> >The difficulty, as John Baez pointed out, is in the preparation of the
> >light beams, not in the beams themselves. In your statement you are
> >comparing two light beams which are prepared differently -- why shouldn't
> > they differ?
> can only be one photon in the apparatus at the time so it MUST interfere
> with itself.
But you don't know which photon is in there, nor where in the distribution
of the beam its energy or wavelength lies. In some sense, _it_ doesn't
know, either. Elsewhere in this thread you said "Unfortunately this means
that photons carry information about how they were made" -- I would say
rather that photons carry _lack_of_knowledge_ which matches how they were
made. Like any quantum object, it was prepared with a given distribution
in wavelength, and it preserves that distribution (lack of knowledge)
when it ultimately interacts; that's why we describe them with
wavefunctions.
> Naively perhaps, I assume that it can only interfere with
> itself if the difference in the two paths is less than it's own length
> (since a photon must surely be coherent with itself). It's just a single
> photon version of how you might measure the coherence length of a beam
> of photons.
But the difficulty is in using the word "it", and the assumption that
there is only one "of them", and that "it" has a well-defined "length".
These are identical bosons and you cannot really "grab hold" of "one"
of them. The english language just doesn't cope with this very well....
You have to use a wavefunction, which is inherently a distribution, and
the correlation length is inherently related to the width of the
distribution in wavelength (among other things).
> Of course it will require a lot of single isolated photon events to
> build up a diffraction pattern, but this is not so hard since a
> photographic plate, or better a CCD, can add them spacially quite
> easily.
Right. And in this case the resulting image will certainly depend
upon how the beam was generated. This includes its coherence
length.
I don't think it is really meaningful to talk of the coherence length
of a single photon, but only of a beam. No matter what you do, you
will end up measuring properties of the beam, not of individual
photons.
Let's examine that -- consider a double-slit experiment
through which we send a single photon, and use a single-photon
detecting screen. What do we know from this single event about
the coherence length of that photon? Very little! All we know
is where it hit. We do not have enough information to both
determine its wavelength and its coherence length. There will
be a range of potential solutions for varying wavelength and
coherence length (with correlations), but no unique value
for either. And the details of how wavelength and coherence
length are correlated will depend upon how the 1-photon "beam"
was generated (and also upon where it hit, the geometry, etc.).
> In fact I would have thought a modern undergrad optics lab with a CCD
> detector would be able to do this experiment quite easily,
Sure. But it would not measure what you wish it to measure. Even with
only one photon in the apparatus at a time, the resulting image will
still depend upon properties of the initial beam.
> I am disturbed if individual photons are different. I
> would like my elementary particles to all be identical, including the
> photon.
Nature is not constrained by what you would like. Nature appears to
obey quantum mechanics (at least approximately), and that means that
the description of quantum objects requires a wavefunction, and many
of the properties of such objects are not sharp. The wavelength of
a photon is one such property, and its lack of sharpness is related to
the coherence length of the beam. That's just the way it is....
> >But really all you can do is compare the _light_beams_ (not the photons).
> Why? Individual photons can interfere with themselves perfectly well.
Sure. But they inherently come from a light beam, and the properties of
the beam will affect the interference pattern generated by the beam,
even if it is considered to be a collection of individual photons.
And as I discussed above, if you only look at a single-photon event,
you do not have enough information to determine both the wavelength
and the coherence length. There is inherently a wavefunction here, and
it implies all one can discuss is widths of distributions, correlations,
etc.
Tom Roberts tjro...@lucent.com
Bill Jefferys
>1) It is impossible to know the frequency of a single isolated photon. I
>would thus wonder if this allows it's energy to be very well defined.
Suppose I bounce a photon off of a large diffraction grating and detect it
at a particular position. To me, that seems to give me a very good idea of
what the photon's frequency was (before it was detected). The precision of
the measurement is proportional to the size of the grating.
Given this, I don't understand your comment above.
Bill
Haruspex77
worlds viewpoint, so let me try to explain the "length of the wavetrain"
from that point of view.
As a ham, I am more interested in lower energy photons, so lets say I
create one by connecting a battery between a sheet of copper and the
ground. The sheet promptly develops a charge, and the change in the
electric field over it creates a magnetic field, which in turn creates
more electric field and WHOOSH an electromagnetic ripple is propagating
itself off into space.
What Maxwell didn't know was that there were a trans-finite number of
other copies of me doing the same thing at the same time. Each of them
got a slightly different result because the amount of charge on their
copper sheet varied continuously across some range. The ripples that
went out varied in energy as a result, though each one has exactly the
profile Maxwell's equations specify.
An matching number of copies of you are standing on the other side of the
room waiting to detect the ripple of electric and magnetic fields. Each
copy sees it arrive at the same time (because the equations let you
calculate the speed), but each copy gets a different measurement of how
strong it was. Each one sees a "photon" of a different energy, some may
not see anything because the ripple was smaller than the sensitivity of
the detector.
Now the Copenhagen theory quantum physicist who is watching this
experiment describes me as preparing a photon which is tightly defined in
location but therefore undefined in energy. He calculates a
"wavefunction" which is equal to an infinite sum of sine functions of
various wavelength each with an "intensity" coefficient. This function
happens to be the eigenfunction of an operator that measures when you
will see the ripple so he can calculate the eigenvalue that tells him
when you will see it. He gets what should be an easy right answer the
hard way.
He had to do that because he didn't believe there was anything real going
on, and his professors taught him that method because it is the only way
that he can also calculate the energy of that photon. When he uses a
different operator that measures how much energy you will see, he doesn't
get just one eigenvalue he gets instead a continuous "spectrum" of them
with a weighting function. If the experiment is repeated many times,
that weighting function turns out to be an accurate measure of how often
a photon with that energy will appear.
However, none of the copies of you that detect the photon care much about
that; you see only one energy. What the physicist has calculated is how
many (or rather what proportion) of you would see energies in any
particular range. If we do repeat the experiment many times, those
proportions will be reflected in the numbers that each copy of you
collects and then the physicist can happily point to his accurate
calculations of the statistics on those numbers.
In this case your idea that the "number of wiggles is the same" is an
accurate description of what really happens, and Maxwell's equations can
get you a description of what the photons in this case look like.
But, the situation is a little different if I charge up a ball on the end
of a stick and wave it slowly back and forth. Maxwell's equations say
that I am generating a continuous chain of waves that propagate through
space toward you, but your detector (no matter how sensitive) won't see
that. Most of the time, it won't see anything at all, but occasionally
it will register the smallest ripple it can sense. Of course all those
other copies of you are seeing a similar effect, only they detect the
ripple at different times. We can describe the moment when your detector
registers as the arrival of the photon.
Our friend from Copenhagen will describe what I am doing as preparing
photons that are well defined in energy but poorly localized in time and
space. His operator that measures when you see it will give a broad
spectrum of eigenvalues that describes a statistical distribution of
probable arrival times, but his operator that measures energy will
(mostly) give a single answer. Again, if you collect enough data, his
answer will be the best one.
In this case, the photons will have widely varying numbers of ripples
depending on how long you have to wait for the next one and the frequency
with which I am waving my arm.
There are things about this second case that I don't understand, but I
dropped out of school before I got much practice in arm waving. What is
the difference between the various copies of you that determines when
their detector will register at any particular moment? It has to be
fundamental enough that it doesn't depend on the details of the detector
since the results don't change form on those details, but I am not sure
where it lies. Perhaps someone else can describe that difference.
--
Haruspex (Remove the extra x to reply)
super...@my-dejanews.com
> >In article <hnxNkLAafWr2Ew$q...@upthorpe.demon.co.uk>,
> > Oz <O...@upthorpe.demon.co.uk> wrote:
> >> build up a diffraction pattern, but this is not so hard since a
> >> photographic plate, or better a CCD, can add them spacially quite
> >> easily.
> >
> >the photons as single isolated events is meaningless.
>
> isolated photons, and we can use quantum mechanics to predict (sometimes
> probabilistically, and sometimes deterministically) what their behavior
> will be, and we can do quantum interference experiments using single
> photons. So I'm not sure what you mean here.
What I am saying is quite simple really and has little to do with any of the
mechanics of the theories or the experiments. It is a case of simple logic
and being very carefull about the language we use to describe things. This is
because our language exists in the world we live in not the quantum world
which we are often trying to describe very badly. You must surely agree that
to call each photon(by photon we mean a particle) single isolated events is
very misleading. If this where true then as I laid out an interference
pattern would have to be produced by each particle "effecting" the apparatus.
The only other alternative is that no matter how long it has been since a
particle has travelled through the interferometer it somehow leaves an
"impression" (i.e. a field is effected) that is picked up by latter photons
hence producing interference. The truth is most likely a mixture of the two,
since I cannot see how we can seperate the measuring apparatus from the
measured (some people might call this the arrogance of classical physics).
The term interference is quite unambiguous, but it seems its use in relation
to other things is not so. NJH.
Tom Roberts
> [measuring the frequency of a single photon]
> Suppose I bounce a photon off of a large diffraction grating and detect it
> at a particular position. To me, that seems to give me a very good idea of
> what the photon's frequency was (before it was detected).
It is really the angle of reflection, not the position, which gives a
measurement of the photon's wavelength. To obtain the angle you need to
know not only where it ended up, but also where on the grating it reflected
and also its original direction. But you did not measure either of these,
so you must sum over all possible paths, and that introduces inherent
uncertainties. If you attempt to limit them geometrically by using opaque
screens with small holes to constrain the geometric path, then you will
find that diffraction occuring at the apertures will limit your ability
to measure the angle DUE TO THE GRATING and thus the wavelength. If
you attempt to actually measure where the photon goes using fine-grained
photon detectors, then only one of the detectors can fire for a given
photon, and you have no wavelength measurement for a single photon at all
(once a photon is detected, it is destroyed).
In short, quantum phenomena have an intrinsic amount of "lack of knowledge"
or "uncertainty". While you can push this around somewhat by careful design
of the experiment, you cannot eliminate it. That's what Heisenberg's
uncertainty principle is all about.
> The precision of
> the measurement is proportional to the size of the grating.
Not really. The precision of the measurement is related to how well you
can measure the angle of reflection off the grating. If the size of the
grating is the limiting aperture, then increasing its size will improve
precision up to a point, but beyond that point the increasing
uncertainty in the angle will cancel out reductions in the diffraction
from the aperture of the grating's size.
Relating this back to the original question:
If we concentrate on a tiny region at the middle of a single maximum of
the interference pattern from the grating, there is no way to know whether
a photon hitting there is of the mean wavelength of the beam and
travelling down the geometric middle of the initial beam, or is of smaller
wavelength but larger angle, or of larger wavelength but smaller angle.
The parameters of the beam are inherent in determining the interference
pattern produced: not only the energy/wavelength/frequency distribution
but the angular distribution as well, and these distributions are
intrinsically correlated in any region of the interference pattern.
Tom Roberts tjro...@lucent.com
john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>Are you suggesting that photons cannot interfere with themselves?
Maybe he is, maybe not, but it doesn't really matter: as you
suspect, photons can and *do* interfere with themselves.
I'm getting so annoyed by this thread that I'm starting to read books
on quantum optics, and in _Concepts of Quantum Optics_ by P. L. Knight
and L. Allen a bunch of nice papers are reprinted, going back to
Einstein's "On the Quantum Theory of Radiation" and beyond, including
one that's especially relevant to this issue:
R. L. Pfleegor and L. Mandel, Interference effects at the single photon
level, Phys. Lett. 24A (1967), 766-767.
It's an experimental paper. I'll just quote the beginning:
------------------------------------------------------------------------
Interference effects have been demonstrated in the superposition
of two light beams from two independent lasers, under conditions
where the intensity was so low that one photon was absorbed, with
high probability, before the next one was emitted by either of the
two sources.
Since the experimental demonstration of interference effects produced
by the superposition of two independent light beams, the question has
been debated whether the effect is to be regarded as evidence for the
interference of photons from one beam with photons of the other beam.
On the face of it, the observations appeared to contradict a well-known
remark of Dirac that "... each photon interferes only with itself.
Interference between different photons never occurs."
It is the purpose of this communication to report the results of
further experiments which show that the effect cannot be reasonably
be described between photons of one beam and photons of the other.
On the contrary, it appears the Dirac's statement, in a sense, is just
as applicable to the foregoing experimental situation as to conventional
interferometry.
------------------------------------------------------------------------
Now I don't think Dirac's remark is especially well-designed to explain
quantum-theoretic interference to people who don't already understand
it. For example, he didn't bother to mention that in quantum theory
there's no real distinction between "different photons" and "the same
photon" - rendering the whole argument moot! If you don't realize
this, you're doomed to wallow in confusion about this whole issue.
However, the experiment presented in this paper is neatly summed up in
the first paragraph above (the abstract), and however one wants to *talk*
about things, one has to face up to the existence of this kind of
phenomenon. A less grandstanding example of the interference of a single
photon with itself would be to take a *single* beam of light at very low
intensity, split it with a half-silvered mirror, and recombine it -
getting interference even if the intensity is so low that there's hardly
ever more than one photon around at a time. This was done in 1909 by
G. I. Taylor, and discussed in the very first paper reprinted in _Concepts
of Quantum Optics_, namely:
G. I. Taylor, Interference fringes with feeble light, Proc. Camb. Philos.
Soc. 15 (1909), 114-115.
[Moderator's note: In this age of CCDs and desktop computers, it's a
relatively easy demonstration that is sometimes done in undergraduate
physics classes. -MM]
James Banes
>I am implying a default isolated photon in a/the 'free'
>state which I guess is approximated by an isolated photon
>in intergalactic space. I expect them all to have the same
>state and thus look alike.
The problem is your conception of a "free photon". From the
standpoint of quantum electrodynamics, the wave properties of
electromagnetic radiation are really wave properties of the
EMITTER. Feynman has a good explanation of this in his QED book,
in the section on oil films (I think). The point is that all the
potential sources of a photon have a certain (complex) amplitude
for photon emission, and this amplitude evolves in time as you
progress along the emitter's worldline. However, once a putative
photon is emitted, its phase does not advance while "in flight"
(unlike massive particles), essentially because quantum phase is
a function of the absolute spacetime interval (which, after all,
is what gives the absolute interval its physical significance),
and photons exist (give or take) on null intervals. In a sense,
the ancients who conceived of sight as something like a blind
man's incompressible cane FEELING distant objects were correct,
because our retinas actually are in "direct" contact, via null
intervals, with the sources of light. The null interval plays
the role of the incompressible cane, and the wavelike properties
we "feel" are really the advancing quantum phases of the SOURCE.
That's what people mean when they say there's less to photons
than meets the eye.
You might object to this by saying that the reception amplitude
for an individual photon advances as a function of its position,
i.e., if we had (contra-factually) encountered a particular photon
one meter further away from the source than we did, we would have
found it with a different phase, but that's not right, because the
photon we would have received one meter further away (on the same
timeslice) would have been emitted 1 light-meter earlier, carrying
the corresponding phase of the emitter at THAT point on its world-
line. When we consider different spatial locations relative to
the emitter, we have to keep clearly in mind which points they
correspond to along the worldline of the emitter.
Now, you might think you could "look at" a single photon at
different distances from the emitter (trying to show that its
phase evolves in flight) by receeding fast enough from the
emitter so that the relevant emission event remains constant.
But of course the only way to do that would be to receed at
the speed of light (i.e., along a null interval), which isn't
possible. This is basically just a variation of the 16-year-
old Einstein's thought experiment about how a "standing wave"
of light would appear to someone riding along side it. The
answer is "it wouldn't, because you can't". And the more
complete answer is that you can't because light exists on
null intervals.
Notice that if your speed of recession from the source gets
closer and closer to c, the difference between the phases of
the photons you receive gets smaller and smaller (i.e., the
"frequency" of the light gets red-shifted), and approaches
zero. This is just what you should expect based on the fact
that each photon is simply the lightlike null projection of
the emitter's phase at a point on the emitter's worldline.
Hence, if you stay on the same projection ray (i.e., null
interval), you are necessarily looking at the same phase of
the emitter, and this is true everywhere on that null ray.
Now, since the phase of a photon doesn't evolve "in flight",
it's clear that the concept of a "free photon" is meaningless,
because a photon is nothing but the communication of an emitter
event's phase to some null-separated absorber event (and vice
versa). If you think of a photon as a clap, then a "free photon"
is like clapping with no hands.
Keith Lynch
Oz <O...@upthorpe.demon.co.uk> wrote:
> 1) It is impossible to know the frequency of a single isolated photon.
Not true.
> ... there are an infinite (under caution) number of waveforms that
> can be superposed to produce a sine wave.
True, but not very interesting. It works better to regard sine waves
as fundamental, and other waves as being the sum of sine waves of
various amplitudes, frequencies, and phases. (Or of sine waves and
cosine waves of fixed phase and of various amplitudes and frequencies,
if you prefer.)
> As I understand it it is impossible to have laser action with a
> single photon.
No, but you can certainly shine your laser through several welders'
goggles, until the intensity as it enters your apparatus is down to
an average of one photon per second, or even lower. Each photon will
still have a frequency that's known to high precision: the same as the
frequency of the laser. The coherence length will be the same as it
was for the whole laser, perhaps one meter or so.
I don't think it's useful to think of this coherence length as being
the size of the photon.
> I was considering more-or-less monochromatic light would be beamed
> into the apparatus. I accept that this gives a band of probabilities
> of the likely energy of the incoming photon which reduces the
> uncertainty in it's energy and I am hoping this is compatible with
> expected uncertainty of it's waveshape (h being very small after all).
I still don't know what you mean by waveshape. Photons don't have
shapes or sizes like classical objects. They aren't little spheres
or cubes or bullet shapes. If your apparatus is receiving sunlight
filtered through neutral gray filters, all we can say is that photons
will be detected with random energies at random times between sunrise
and sunset.
Each photon detection is perfectly instantaneous in the sense that it
has always either been detected or not, and is never observed to be
partway through the process of being detected. "I just picked up half
a photon, so the other half will be trickling in over the next few
picoseconds..." It doesn't work like that.
On the other hand, you won't necessarily know exactly when it was
detected. In fact you can't know exactly.
Each photon detection is perfectly localized in the sense that it is
always absorbed by just one particle (usually an electron).
On the other hand, you won't necessarily know exactly where that one
particle was at the time. In fact you can't know exactly.
> Ah. You have experimental evidence that a photon wavepacket has a
> gaussion waveshape?
No. I said it DOESN'T necessarily have a gaussian waveshape. The
only advantage of a gaussian waveshape is that it has convenient
mathematical properties, such as being its own Fourier transform.
> I do not buy that at all. There is too much information in a
> Beethovem-shaped wavepacket for a solitary photon to carry.
Right. It's Beethoven-symphony shaped only in the sense that the
probability of detecting a photon is proportionately greater during
the louder parts of the symphony. Or, if you construct your apparatus
so as to measure the photon's frequency to very high precision at
the expense of knowing when it was detected, then the probability of
detecting a photon of a given frequency is proportionately greater
in the louder sidebands, e.g. at the carrier plus or minus 440 Hz if
the symphony is in the key of A.
> IMHO it carries energy and momentum and that's all. That's all it
> has room for.
Nitpick: It also has a polarization, a direction-of-travel, a
location-at-detection, and a time-of-detection. (Each of which can
only be known to a precision which depends on how well the other
facts about this photon are known.)
(Posted and mailed.)
--
Keith F. Lynch -- k...@clark.net -- http://www.clark.net/pub/kfl/
I always welcome replies to my e-mail, postings, and web pages, but
unsolicited bulk e-mail sent to thousands of randomly collected
addresses is not acceptable, and I do complain to the spammer's ISP.
Oz
<ba...@galaxy.ucr.edu> writes
>
>By "stable in time" do you mean that the photon's wavefunction moves
>along at a constant speed while maintaining its shape? Sure, I can
>write down infinitely many states like this. To keep life simple I'll
>pretend the photon is a massless spin-0 particle instead of a massless
>spin-1 particle: this amounts to ignoring the polarization of the photon.
>I'm only doing this this reduce the mathematical clutter, and if you
>force me, I'll do the actual spin-1 case - but I'm warning you, you'll
>regret it!
I already regret it. I wish you wouldn't blind me with science ... :-)
>Okay, here we go:
>
>phi(t,x,y,z) = f(x-ct)
>
>where f is *any function I like*, and c is the speed of light, which
>I'm not setting equal to 1, just so you can see that this little wave
>moves along at the speed of light while keeping its shape. Since f is
>*any function I like*, the shape of this wave is *arbitrary*.
Oh dear. Oh dear. Why did I ask?
OK, now (since I clearly do not have the equipment to answer this), this
looks a tad generalised. From this can you derive an expression for the
energy and momentum of this particular waveform (which must not
effectively vary with time or it's length may change) and thus set it
equal to a convenient single photon?
Because if you can't, then it doesn't seem to me to properly descibe a
single isolated photon.
>I'm actually surprised that you think the photon wavefunction has some
>kind of "default" state and that all isolated photons tend to this state.
Cos nobody has yet spotted any difference between photons other than
momentum.
>A photon is a wave. Waves come in all shapes.
Hmmm. A photon propogates as a wave, but it doesn't really seem to
interact as one. We more-or-less know what a photon in flight should
look like, it's a pair of propogating EB fields, but I don't think that
describes the field's interaction with (say) and atom at all well. It's
hard to see how a to describe how a diffuse and physically large wave
can interact with an itty bitty atom (although I assume some mutual
resonance in my personal model) in the way that it does. We are reduced
to calling it a probability function, which describes how it works but
has very limited physicality. What has that to do with a pair of crossed
EB fields, for example.
>>I don't believe for my specified 'free' photon. I am happy (indeed
>>might insist) that this is so, close to the emitter. However this means
>>that (using the single photon interferometer described elsewhere) I
>>could distinguish between 'free' photons of the same energy. I don't
>>like this.
>
>Well, you can call up Heaven and tell them you don't like how they designed
>the universe.
I do the next best thing. I call up prof Baez and tell him I don't like
the way he describes a photon (being merely mortal he needn't manage a
whole universe). :-)))
If, as you (and others) claim, photons can be different in waveshape
(and thus have different EB fields etc etc) I would expect (perhaps even
demand) that they should behave differently in at least some
interactions. I know of no hint of evidence that suggests this is true,
and indeed everyone seems to use photons as if they are completely
identical. So it seems to me that the physical evidence is more on my
side than yours.
>>I would expect an emitted photon to more-or-less rapidly evolve into a
>>standard 'free' state as it leaves the vicinity of the emitting system.
>
>I wouldn't. Just as I wouldn't expect a bunch of frictionless pendulums
>given random kicks to rapidly evolve into some standard state, like all
>swinging in unison or something. They will just keep swinging the way
>you happened to start them up. Mathematically that's just what the
>electromagnetic field is: a bunch of uncoupled harmonic oscillators.
I would agree (perhaps) were it not for the energy and momentum of the
photon. The wave must propogate so as to maintain these equal, unchanged
and to maintain it's momentum, which has a direction (usual caveats
about h apply). I suspect (but being very ignorant do not know) that
this constrains the waveform in the same way that the mass/momentum
constrains that of an electron.
>>>Well, I can also write down an electron state
But is it the electron state applicable to describing a free electron?
Sure, there are oodles of possible states, but there are not (I hope)
oodles of states for a given situation. Here, of course, I am referring
to the state (which may be a mixture) that completely describes the
particle in the environment it is in. In this case it has no external
interactions (yeah, yeah, apart from the odd virtual thingummybobbies).
>>Is this stable in time, ie does not evolve into a state with a different
>>set of wiggles? Does this entirely describe a single isolated electron?
>
>It describes a single isolated electron. Since an electron is massive,
>this wave doesn't stay the same shape - unlike the case of a photon.
>The reason is that massive things can move at different speeds, while
>massless things must always move at the speed of light. So unlike the
>case of a photon, for the electron different Fourier components
>of the above wave will move at different speeds. This leads the wave
>to keep changing shape and spreading out. It doesn't settle down to
>some "equilibrium shape" - it just keeps changing.
OK, I didn't express myself well. I suppose I mean some time-integrated
envelope, but I haven't really thought out what I mean. Perhaps we can
drop this for a bit ...
[Oz, who has fallen off the wall and is hanging from his fingertips
while he watches the beast of baez trundling towards him in his hob-
nailed boots and a threatening smile.]
--
Oz
Oz
<bi...@clyde.as.utexas.edu> writes
>At 9:09 AM +0000 1/28/99, Oz wrote:>
>>1) It is impossible to know the frequency of a single isolated photon. I
>>would thus wonder if this allows it's energy to be very well defined.
>
>at a particular position. To me, that seems to give me a very good idea of
>the measurement is proportional to the size of the grating.
But after passing therough the grating you no longer know which path it
'took', it could have gone the direct route or diffracted off the last
graticule of your very large grating (yeah, yeah, I know it did all of
them). It's position is thus very unknown indeed even if you now know
it's frequency to some sort of accuracy. However not all photons will
diffract at exactly the angle expected by the grating spacing so you can
never be sure what the energy of that particular one was. [NB h is very
small, so one can have some measure of both.]
--
Oz
Dr Paul Kinsler
> >So any marriage of a single photon to a nice classical wavetrain of length
> >L is doomed to end in divorce. Irreconcilable differences, as they say.
Umm. As I've explained in previous posts, a fair fraction of quantum
optics is based on quantising the EM field inside classical field
mode to get photons: if this isn't a marriage of a single photon to
a nice classical wavetrain of length L in at least some cases I'd
be surprised to hear it. Perhaps I've misunderstood -- can you
clarify where the "divorce proceedings" start?
Alternatively, if you see a flaw in the standard QO approach I've
described, I'd be interested to hear it.
--
------------------------------+------------------------------
Dr. Paul Kinsler
Institute of Microwaves and Photonics
University of Leeds (ph) +44-113-2332089
Leeds LS2 9JT (fax)+44-113-2332032
United Kingdom P.Ki...@ee.leeds.ac.uk
WEB: http://www.ee.leeds.ac.uk/staff/pk/P.Kinsler.html
john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>>Photons
>>emitted by relatively short-lived states are made up of a linear
>>combination of modes with a relatively broad range of frequencies
>>(i.e., energies), and vice versa.
>
>I think you are referring to photons (plural) are you not?
He's referring to a single photon. A single photon is a wave, and
we can write this wave as a superposition of sine waves. That's
what they call the "superposition principle" in quantum mechanics.
>This doesn't apply to a single photon though.
Yes it does.
Dare I ask where you are getting all your information about quantum
mechanics?
ca31...@bestweb.net
super...@my-dejanews.com wrote:
> If this where true then as I laid out an interference
> pattern would have to be produced by each particle "effecting" the apparatus.
> The only other alternative is that no matter how long it has been since a
> particle has travelled through the interferometer it somehow leaves an
> "impression" (i.e. a field is effected) that is picked up by latter photons
> hence producing interference.
"Interference" by particles does not happen.
I'm not talking about quanta here, but of "classical particles"
and sometimes quanta behave like "classical particles".
So, "classical particles", and quanta behaving like "classical particles",
do not interfere. That's what happens when you observe quanta closely,
they don't interfere (Feynman two-slit stuff).
We should probably never use the word "particle" when taking about
quanta or a quantum. It is a context-sensitive term in QM.
Unfortunately, old habits die hard.
So what happens in the other extreme ? When we don't observe these
individual quanta at all, they act like "classical waves". They
interfere completely.
And when we look at these quanta with blurry eyes, they act alittle
like "waves" and alittle like "particles" judging from the appearence
of the interference pattern (or if we fiddle with their relative
polarizations we get the same effects)
But the building up of a pattern incrementally over time has nothing to do
with a quanta being having some mysterious and lasting effect on the
apparatus. Such an incrementally generated interference patterns are
generated by computer graphics people all the time and that is a
purely symbolic process of simulation with nothing to do with quanta.
--
http://www.bestweb.net/~ca314159/
Ligare Martin
: In article <78kk6h$64r$1...@homer.bucknell.edu>, Ligare Martin
: <mli...@coral.bucknell.edu> writes
: >Photons
: >emitted by relatively short-lived states are made up of a linear
: >combination of modes with a relatively broad range of frequencies
: >(i.e., energies), and vice versa.
: I think you are referring to photons (plural) are you not? The
: combination of modes with a broad range of frequences are in fact
: photons of different energies, no?
: This doesn't apply to a single photon though.
: --
: Oz
No, I am not referring to photons (plural). Consider 2
modes of a radiation field, say mode A and mode B with
different properties. (One of the properties could be
frequency.) The following state
|psi> = (|photon in mode A> + |photon in mode B>)/sqrt{2}
is a perfectly legal ONE photon state. The amplitude
for the photon to be in mode A is 1/sqrt{2} and the
amplitude for the photon to be in mode B is 1/sqrt{2},
but a measurement will always find only one photon.
This is analogous to states which are linear combinations
of spin-up and spin-down for spin one-half particles
in magnetic fields. The linear combination isn't an
energy eigenstate, but it's a perfectly good state.
Marty Ligare
Physics Department
Bucknell University
mli...@bucknell.edu
john baez
>It is impossible to know the frequency of a single isolated photon.
Why do you say this? There are lots of ways of determining the frequency
of an isolated photon! The easiest way is to let an atom slowly decay in
such a way that it *produces* a photon of a well-defined frequency.
Of course, this frequency won't be *perfectly* well-defined, since the
uncertainty principle puts a limit on how well we can know the frequency
and also when the atom decayed. But if the atom decays very slowly we
can know the photon's frequency very well. In general, the longer we
allow ourselves to do it, the better we can measure the frequency.
Oz, you seem now and then to be forgetting about wave-particle duality: a
photon has both wavelike and particle-like features! The photon is simply
a wavefunction - or more precisely, a solution of Maxwell's equations.
If we measure the frequency of a photon accurately its wavefunction must
be a long, almost perfectly sinusoidal wave. Then it seems like a wave.
If we measure its position accurately its wave must be a tall skinny spike.
Then it seems like a particle. [1]
This is true not only of photons but of every other sort of particle, of
course. You can't always get away with treating them like little tennis
balls! Nor can you always get away with treating them like nice sine waves!
They are wavefunctions, which can be of *any shape*.
>>The shape of the wave packet, in the time domain or in the frequency
>>domain, isn't necessarily a gaussian bell curve, but that's the
>>simplest case to analyze, and approximates many real-world situations.
>
>Ah. You have experimental evidence that a photon wavepacket has a
>gaussian waveshape? More info please.
Grr! He said it ISN'T NECESSARILY a Gaussian! It can be *any* shape.
But Gaussians are nice functions to deal with, mathematically, so they
serve as a good approximation to any old bump-shaped function. All he's
saying here is that photon wavefunctions often look like a bump! Which
simply means that you know roughly where the photon is!
>>And if you attenuate your radio
>>reception until you only get one photon from the orchestral broadcast,
>>that one radio-frequency photon is still best regarded as being
>>Beethoven-symphony shaped.
>
>I do not buy that at all.
Please call Heaven and ask for a refund. "If you are dissatisfied with
quantum mechanics in any way please press 3, and we will refund your purchase
and return you to a universe obeying the laws of classical physics."
-----------------------------------------------------------------------------
[1] Please, folks, don't start talking again about the lack of a Newton-
Wigner localization for massless particles; I'm using the other localization,
in which I treat the photon's wavefunction as simply a solution of Maxwell's
equations. I'm just trying to get across the idea of wave-particle duality,
without worrying about these nuances.
Anyway, this issue would be relevant if Oz said it was impossible to know
the *position* of a single photon, but not given that he's claiming it's
impossible to know the *frequency* of a single photon - regardless of what
localization we use, *that* claim is false.
Oz
<tjro...@lucent.com> writes
>Oz wrote:
>> <tjro...@lucent.com> writes
>> >The difficulty, as John Baez pointed out, is in the preparation of the
>> >light beams, not in the beams themselves. In your statement you are
>> >comparing two light beams which are prepared differently -- why shouldn't
>> > they differ?
>
>> I don't believe I am comparing two different light beams because there
>> can only be one photon in the apparatus at the time so it MUST interfere
>> with itself.
>
>But you don't know which photon is in there, nor where in the distribution
>of the beam its energy or wavelength lies. In some sense, _it_ doesn't
>know, either. Elsewhere in this thread you said "Unfortunately this means
>that photons carry information about how they were made" -- I would say
>rather that photons carry _lack_of_knowledge_ which matches how they were
>made. Like any quantum object, it was prepared with a given distribution
>in wavelength, and it preserves that distribution (lack of knowledge)
>when it ultimately interacts; that's why we describe them with
Hmmm. But is the knowledge of the preparation carried for all time with
the photon? I suspect not. Lets say we took a photon and passed it
through a grating. We will have changed the wavefunction by this process
and see a red spot at a particular angle on our screen. Now, we remove
the screen and the next photon through wings it's way to a planet going
round a sun in the andromeda galaxy. How does the observer on that
planet know it's gone through a grating? If he doesn't know, and cannot
know, then he sees a different wavefunction, surely.
Tricky stuff this ....
>> Naively perhaps, I assume that it can only interfere with
>> itself if the difference in the two paths is less than it's own length
>> (since a photon must surely be coherent with itself). It's just a single
>> photon version of how you might measure the coherence length of a beam
>> of photons.
>
>But the difficulty is in using the word "it", and the assumption that
>there is only one "of them", and that "it" has a well-defined "length".
>These are identical bosons and you cannot really "grab hold" of "one"
>of them.
But you said above that they were not identical, their wavefunction
depended on how they are made. I'm the one arguing that they are
identical bosons, remember.
>The english language just doesn't cope with this very well....
>You have to use a wavefunction, which is inherently a distribution, and
>the correlation length is inherently related to the width of the
>distribution in wavelength (among other things).
Indeed. That's why I specified the equipment. There are other possible
pieces of equipment, or even perhaps theoretical constructs, that would
give a 'measure' of the size of a photon. I suspect they will all be
rather similar unless the photon has been 'heavily processed', or
interacts significantly with other photons.
>> Of course it will require a lot of single isolated photon events to
>> build up a diffraction pattern, but this is not so hard since a
>> photographic plate, or better a CCD, can add them spacially quite
>> easily.
>
>Right. And in this case the resulting image will certainly depend
>upon how the beam was generated. This includes its coherence
>length.
I am not at all convinced. I don't think single light photons will
interact significantly over distances of (say) a meter. They will thus
be different.
>I don't think it is really meaningful to talk of the coherence length
>of a single photon, but only of a beam. No matter what you do, you
>will end up measuring properties of the beam, not of individual
>photons.
I am unconvinced.
>> In fact I would have thought a modern undergrad optics lab with a CCD
>> detector would be able to do this experiment quite easily,
>
>Sure. But it would not measure what you wish it to measure. Even with
>only one photon in the apparatus at a time, the resulting image will
>still depend upon properties of the initial beam.
Prove it by experiment then. Compare the diffraction pattern of a laser
at normal intensity with that from a sodium lamp and that from both
where the intensity is down to a few thousand (or even hundred) photons
a second. If the diffraction pattern is unchanged by the intensity and
different between the lamp and the laser then I will indeed be convinced
of your argument. If not, then it will be interesting, no?
>> I am disturbed if individual photons are different. I
>> would like my elementary particles to all be identical, including the
>> photon.
>
>Nature is not constrained by what you would like. Nature appears to
>obey quantum mechanics (at least approximately), and that means that
>the description of quantum objects requires a wavefunction, and many
>of the properties of such objects are not sharp. The wavelength of
>a photon is one such property, and its lack of sharpness is related to
>the coherence length of the beam. That's just the way it is....
I don't disagree. However I don't believe it is sensible to describe a
sequence of separated non-interacting individual photons as a 'beam'. A
beam will have interference at all points between the photons in the
beam and will have a different wavefunction to individual photons that
are not interfering with each other, surely.
NB Where are the other heavyweights? Watching and laughing? :-)
--
Oz
Matt McIrvin
>Now, since the phase of a photon doesn't evolve "in flight",
>it's clear that the concept of a "free photon" is meaningless,
>because a photon is nothing but the communication of an emitter
>event's phase to some null-separated absorber event (and vice
>versa). If you think of a photon as a clap, then a "free photon"
>is like clapping with no hands.
I don't think that this necessarily follows. As a knee-jerk
post-positivist I am reluctant to discard concepts as meaningless if they
are theoretically useful in meaningful calculations. While it's true that
in Feynman's wonderful spacetime path-integral picture of QED, a photon's
phase doesn't advance in flight, this doesn't damn the concept of a free
photon to uselessness.
For instance, consider cosmology. Cosmologists like to consider the case
of a universe that contains no massive particles, just photons. While
there is nothing to emit or absorb the photons, they *do* have
gravitational effects, and the evolution of the space-time responds to
their presence.
The reason that cosmologists do this is that the radiation-filled universe
is a good approximation to a universe that contains *mostly* radiation, as
in certain early phases of the universe's evolution in standard cosmology.
In the more realistic picture, of course, there *are* emitters and
absorbers-- but one can do good approximate calculations without them, so
I'm wary of saying that they are necessary to make the whole picture
meaningful.
There are many other cases of quantum field theorists considering "pure
gauge" theories that contain nothing but massless particles (such as QCD
without quarks, or gravity plus photons, or even quantum gravity without
anything but gravity). These aren't considered to be realistic models of
the world, but they often have instructive properties that can shed light
on more realistic models.
In the more interesting cases, the quanta, while massless, can actually
interact with one another at fundamental interaction vertices. They can
sometimes even form collective blobby things that *do* have mass, and have
other interesting properties of their own. People who study QCD often
speak of "glueballs," which are theorized blobs of gluons. There have
been some threads in this group in the recent past about Wheeler's work on
"geons," which are hypothetical gravitational/electromagnetic blobs;
Usenet archives like Alta Vista or DejaNews probably have the articles.
The fact that photons *don't* directly interact with one another is not
a consequence of their masslessness, but of their chargelessness, which is
an independent property.
--
Matt McIrvin http://world.std.com/~mmcirvin/
Oz
<ba...@galaxy.ucr.edu> writes
>Dare I ask where you are getting all your information about quantum
>mechanics?
Er, um, do I have to?
Ahem, cough, cough, well I still remember a teeny amount from college,
but I have to admit I am terribly rusty.
Oh and (brightens up) I do read some quite technical newsgroups and
sometimes I even understand bits of their posts.
--
Oz
Oz
<jba...@quanset.com> writes
>Now, since the phase of a photon doesn't evolve "in flight",
>it's clear that the concept of a "free photon" is meaningless,
Oh, sorry. I presumed that a photon could and presumably would, interact
with virtual particles in the vacuum. If it did, I would assume this
might affect it's wavefunction to some small (but possibly important)
degree. It might 'evolve' a bit, for example.
--
Oz
Bill Jefferys
>Bill Jefferys wrote:
>> [measuring the frequency of a single photon]
>> Suppose I bounce a photon off of a large diffraction grating and detect it
>> at a particular position. To me, that seems to give me a very good idea of
>It is really the angle of reflection, not the position, which gives a
>measurement of the photon's wavelength.
Yes, I know. That's why I had the foresight to put my grating into a
spectrograph (sorry, I thought that would be obvious). Angle then
translates into linear position along the detector (a CCD, for example).
>> The precision of
>> the measurement is proportional to the size of the grating.
>
>Not really. The precision of the measurement is related to how well you
>can measure the angle of reflection off the grating.
It is in my spectrograph, which has collimator and camera lens large
enough to fully illuminate the grating. It's mounted at the back end of my
telescope, which directs the light from a particular star at the entrance
slit, the size of which is adjusted appropriately. In this case the
resolving power is essentially (size of grating)/wavelength. This fully
respects all the relevant physics, including Heisenberg etc. Linear
position on the CCD correlates with frequency of photon. I am limited by
the resolving power of the spectrograph and other parameters of the
instrument (e.g., am I fully sampling the spectrum by proper choice of CCD
pixel size, f/ratio of camera, slit witdth, etc.?)
>Relating this back to the original question:
>If we concentrate on a tiny region at the middle of a single maximum of
>the interference pattern from the grating, there is no way to know whether
>a photon hitting there is of the mean wavelength of the beam and
>travelling down the geometric middle of the initial beam, or is of smaller
>wavelength but larger angle, or of larger wavelength but smaller angle.
>The parameters of the beam are inherent in determining the interference
>pattern produced: not only the energy/wavelength/frequency distribution
>but the angular distribution as well, and these distributions are
>intrinsically correlated in any region of the interference pattern.
Within the resolving power of the spectrograph, I can measure the
wavelength of the photon with great precision. Of course, I don't have
infinite resolving power in a realizable instrument, but I can, within
those constraints, "measure the frequency (wavelength) of a single
photon." Astronomers do this all the time. Again, I ask "Oz" what he means
by his comment, which I don't understand.
Bill
--
Bill Jefferys/Department of Astronomy/University of Texas/Austin, TX 78712
Email: replace 'warthog' with 'clyde' | Homepage: quasar.as.utexas.edu
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john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>In article <78r3c1$1jt$1...@pravda.ucr.edu>, john baez
><ba...@galaxy.ucr.edu> writes
>>I'm getting so annoyed by this thread that I'm starting to read books
>>on quantum optics,
>Oh dear. Oh dear. Oh dearie me. It's serious then .....
>Please don't hold me responsible, I only asked a simple little question.
You asked a simple little question so chock-full of false assumptions
that to straighten you out would take a graduate course on quantum
theory! And in the process, you have gotten people talking about some
very interesting things... some of them so interesting that I've been
forced to hit the books to figure out what the folks working on quantum
optics have to say.
Imagine some poor fellow who'd never heard a flute posted the following
question on sci.music: "How many trombones does a flute sound like?"
It would cause a tremendous argument, with various experts doing their
best to explain why this question is the wrong question to ask, various
other people doing their best to imitate the sound of a flute in ASCII,
other people waxing philosophical about the impossibility of describing
the sound of a flute, and so on. Eventually some mathematical sorts
would start proving theorems saying that the waveform of a flute cannot
be approximated well in the L^2 norm by any finite sum of trombone
waveforms... and some of these theorems would involve very interesting
questions about Fourier analysis....
Meanwhile the poor fellow would be thinking "How come nobody is giving
me a simple answer? Perhaps even the experts don't know what a flute
sounds like?"
It's a bit like that.
In short: I'm not too optimistic that you're going to walk out of this
much more informed than you walked in. The *really* good thing would
be to sit down and brush up on your classical electromagnetism, then
some quantum mechanics, and then a little quantum field theory. But
of course this is a lot of work. And unlike when I taught you general
relativity, now even *I* don't have the time for a systematic course.
So all I can is fire off thunderbolts at your more egregious errors and
hope they push you in the right direction.
In particular, I hope you spend some time pondering wave-particle
duality, the uncertainty principle, and that sort of thing.
Tom Roberts
> Hmmm. But is the knowledge of the preparation carried for all time with
> the photon?
How could it possibly "shed" such information?
> I suspect not. Lets say we took a photon and passed it
> through a grating. We will have changed the wavefunction by this process
> and see a red spot at a particular angle on our screen.
Not really. We will see a red spot at a particular distribution of angles
on the screen. It's not the same thing....
> Now, we remove
> the screen and the next photon through wings it's way to a planet going
> round a sun in the andromeda galaxy. How does the observer on that
> planet know it's gone through a grating? If he doesn't know, and cannot
> know, then he sees a different wavefunction, surely.
He will see some specific distribution. That far away the geometrical
constraints will be strong, and event the minutest diffraction will be
important, with the result that the andromeda observer will not be
able to determine accurately the energy or wavelength of the original
incident photon -- he can measure the photon which reached him, but
cannot determine the incident photon's properties due to the enormity
of diffraction effects. And, of course, he cannot observe any significant
fraction of the entire interference pattern.
> >But the difficulty is in using the word "it", and the assumption that
> >there is only one "of them", and that "it" has a well-defined "length".
> >These are identical bosons and you cannot really "grab hold" of "one"
> >of them.
> But you said above that they were not identical, their wavefunction
> depended on how they are made. I'm the one arguing that they are
> identical bosons, remember.
I meant within a given beam. They are identical within one beam, but
the method of preparation of the beam affects the properties of the beam
(obviously), and of any measurements of it, including "single-photon"
measurements.
> >Sure. But it would not measure what you wish it to measure. Even with
> >only one photon in the apparatus at a time, the resulting image will
> >still depend upon properties of the initial beam.
> Prove it by experiment then.
Such experiments have been done, with the result that the interference
pattern is essentially unchanged as intensity is varied over an enormous
range. Sorry, no reference (but I saw some in other messages of this
thread).
> However I don't believe it is sensible to describe a
> sequence of separated non-interacting individual photons as a 'beam'.
Even if they were all prepared in the same manner by the same equipment??
That _is_ what we normally mean by a "beam".
Tom Roberts tjro...@lucent.com
ach...@my-dejanews.com
ba...@galaxy.ucr.edu (john baez) wrote:
> I'm getting so annoyed by this thread that I'm starting to read books
> It's an experimental paper. I'll just quote the beginning:
>
> of two light beams from two independent lasers, under conditions
> where the intensity was so low that one photon was absorbed, with
> high probability, before the next one was emitted by either of the
> two sources.
Most of the discussion in this thread seems to be about the question
what we mean when we say "a" or "one" photon, or "two different"
photons.
This is actually a really subtle question when we start to work with
two different sources.
E.g. instead of taking two lasers in the above experiment, let us use
two excited, well separated atoms. Will their photons interfere?
The answer is no!
They will produce a state of the form:
|atom 1 excited, atom 2 in ground state> |photon 2> +
|atom 2 excited, atom 1 in ground state> |photon 1>
and there is _no_ interference of the two photons, because the
atoms "remember" where the photon was coming from (for the experts:
after "tracing out" the atoms, the photons are in a "mixed state" because
<atom 2 excited,1 in ground state | 1 excited, 2 in ground state> = 0).
One cannot approximate the state above by (|photon 2> +|photon 1>)!
Is this what Dirac had in mind when he stated that two
different photons do not interfere?
One can interpret this also as a consequence of the uncertainity
relation among the number of photons N and the phase phi of the photons.
delta N delta phi >= 2 pi
N=1 is fixed, therefore the phase is random and no
interference of the "different" photons is possible.
The story with the laser is totally different. A laser produces light
with a fixed phase and a random number of photons: a "coherent state",
a "classical" wave.
Therefore the two lasers do interfere.
(I think John Baez has explained some time ago very nicely
what a "coherent state", a "mixed state" and a "pure state" is and
hopefully why a classical EM-wave is a coherent state).
Achim
john baez
Michael Weiss <we...@spamfree.net> wrote:
>And a photon is not an experimental beast. A photon is a theoretical
>construct.
You're gonna take serious flak for this, Michael! "Theoretical construct" -
them's fighting words! We know photons! We love photons! And you're
calling them a mere "theoretical construct"???
But of course you're right. Part of the reason certain people are getting
so confused on this thread is that they secretly think a photon is, deep
down, just a little bullet - possibly a bullet with some wiggly waves
thrown in on top to jazz it up, but certainly something with a standard
shape! (Okay, so they come in different colors and sizes - the little
ones are blue, the big ones are red - but still, all basically the same!)
But of course is you ask a theorist what they mean by a photon, they'll
say: "a solution of the source-free Maxwell's equations with norm equal to
one"! The state of a photon is a unit vector in a Hilbert space! And
to explain experimental results with these critters we have to do some
fairly complicated gymnastics, especially since photons have a strong
tendency to travel in packs, being bosons. Any actual state of the
electromagnetic field is a linear combination of symmetrized tensor products
of single-photon states! Now if that ain't theoretical, I don't know what
is....
In particular, we have to carefully disentangle the *classical* superposition
principle for the classical electromagnetic field from the *quantum*
superposition principle for states of the quantum electromagnetic field.
If we don't get straight about these two sorts of linearity, we are goners.
And we can't really do this without knowing a bit about coherent states.
But this is pretty technical.... so there is endless room for confusion
until one bites the bullet [1], accepts the fact that photons are highly
sophisticated theoretical beasts, and learns the darn theory!
Which of course is why everyone should read
http://math.ucr.edu/home/baez/schmoton/
>JB mentioned the spiral staircase of lies that is teaching. Many steps ago,
>I was taught that a beam of light *could be* thought of as a barrage
>of bullets. Indeed, the PSSC high school physics book (co-authored by
>none other than Albert Baez!) had a picture of this, one that left an
>indelible impression on my soft waxy mind. [Not that I'm saying AB was
>responsible for that particular picture. The book had a few dozen
>co-authors.]
Well, my uncle *may* have been responsible for that picture, since his
specialty was optics, but he also made lots of movies about *waves*, and
I would hope some of that *wave* stuff also found its way into the PSSC
physics books. I remember him visiting me as a kid and showing me lasers,
diffraction gratings, and holographs, so I definitely got the *wave*
picture into my head as well as - and probably more than - the particle
picture.
But you're right, once these pictures get in there, they have a tendency
to stick. I still visualize protons as black and neutrons as gray.
The only solution is probably to bombard kids with a *variety* of pictures,
enough different ones so they realize that these pictures can't all be
literally true.
>In other words, can we have a quantum
>state of the EM field such that:
>
> The photon number is very strongly peaked at 1 --- if you expand the
> state in a basis of photon-number eigenstates, the coefficient of |1>
> outshines all the other coefficients.
>
> AND AT THE SAME TIME
>
> The "value of E^2 + B^2 at spacetime point (x,y,z,t)"
> has very low uncertainty for all (x,y,z,t) --- even though
> E and B *separately* are very fuzzy indeed at all (x,y,z,t).
I really doubt we can get this, but I haven't proved we can't - not yet.
Fun question. I've been reading books on quantum optics, full of fun
stuff about "nonclassical light" --- all the weird states of the
electromagnetic field that are *far* from classical. But I haven't seen
this in there.
-----------------------------------------------------------------------
[1] Pun intended, as usual.
john baez
Oz <O...@upthorpe.demon.co.uk> wrote:
>Cos nobody has yet spotted any difference between photons other than
>momentum.
STOP SAYING THAT!
[Ahem. In case anyone is wondering why I'm so much ruder to Oz
than everyone else, and why Oz sometimes compares me to a mean old
"wizard", it's because we have a relationship going back a long ways,
in which I teach him physics in an efficient manner that involves lots
of abuse. He seems to thrive on it. For the adventures of "Oz and the
Wizard", wherein I taught him general relativity, see:
http://math.ucr.edu/home/baez/gr/oz1.html
You'll see what I mean. I also taught him a fair hunk of electromagnetism
once upon a time, but I haven't had time to turn this into webpages. When
I actually see Oz I am perfectly polite - this nasty wizard persona is just
an act.]
Now listen here: if you and I lie on the lawn and see a purple firework
explode in the sky, we both see purple photons coming down at us, and
since these photons have (almost) the same frequency and are going (almost)
the same direction, they have (almost) the same momentum. But there is
something very different about the photons you see and the photons I see,
namely:
THE PHOTONS YOU SEE ARE GOING INTO YOUR EYES, WHILE THE PHOTONS I SEE
ARE GOING INTO MINE!
So yes, we spot differences between photons other than momentum all
the time. In addition to momentum information there are all sorts of
other kinds of information. Whenever one photon does something that
another does not do, they differ in some sort of way - and this way is
not necessarily momentum.
It's true that if we know *everything* about the momentum, we can know
*nothing* about the position! It's also true that if we know everything
about momentum and polarization, there is nothing else left we can know!
But that doesn't mean that the only thing we can know is momentum and
polarization! You takes your pick in quantum mechanics, remember? If
you decide to know a lot about *this*, you can no longer know much about
*that*.
Subtle distinctions, eh?
>We more-or-less know what a photon in flight should look like [...]
Yes, we more-or-less know: ME MORE AND YOU LESS!
[Ahem. I'm really reverting to that mean old wizard persona. I'm
sorry, I can't help it - it's the only way I can retain my sanity while
teaching Oz quantum mechanics.]
P.S. - Yes I know there are no purple photons, just red and violet ones.
That was just a joke based on the question someone asked once here on
sci.physics.research: "How does a red photon combine with a blue photon
to form a purple one".
P.P.S. - Yes I also know that one needs to be careful talking about
the position of photons and other massless particles; that's the
bit about "no Newton-Wigner localization". However, I understand
this well enough to think it's no harm to slide this issue under the
rug, at least in the present conversation. Unlike a massive particle,
you can't say *exactly* where a photon is, but that doesn't mean you
can't have a darn good idea where it is *approximately*, and that suffices
to demolish Oz's idea that "all you can know about a photon is its momentum
(and polarization)". What I'm saying, mathematically speaking, is that
there are lots of observables that aren't in the von Neumann algebra
generated by the momentum and polarization operators.
Ale2NOSPAM
<<
No, but you can certainly shine your laser through several welders'
goggles, until the intensity as it enters your apparatus is down to
an average of one photon per second, or even lower. Each photon will
still have a frequency that's known to high precision: the same as the
frequency of the laser. The coherence length will be the same as it
was for the whole laser, perhaps one meter or so.
>>
So there are at least two ways to reduce a laser beams intensity. Attenuation,
using welders' goggles, and reducing the beam by sending part of it in another
direction with a particially silvered mirror. To very different processes. In
one case photons are "destroyed" and in the other case the photon number is
roughly the same.
But I gather from this discussion the reduced beams are the same in either case
in the sense of the coherence properties of the photons are the same? Does the
phase get "messed" up differently in either case?
Thanks for any clarification!
Keith Lynch
Oz <O...@upthorpe.demon.co.uk> wrote:
> However there is just the problem that by definition sine waves are
> infinitely long and a photon definitely isn't.
Photons don't have length any more than they have breadth, depth,
odor, or temperature. A light source has a coherence length, a
wavelength, and a color temperature, but applying these terms too
literally will lead to massive confusion. The "temperature" of the
photons in your microwave oven is hundreds of degrees below zero.
The "temperature" of the photons used to make a medical x-ray is
tens of millions of degrees. The "length" of light coming out of
the sun or a laser is... equally misleading or meaningless.
If you know the EXACT frequency of a photon then it IS "infinitely
long" in the sense that it's equally likely to be found anywhere.
This infinity is just as real, and just as useful, as the one on
the focus knob of your camera.
In another sense, every photon is a dimensionless point.
In another sense, every photon is the size of its wavelength.
It depends on what question you're really trying to answer.
> So it's of more limited use for visualising a photon.
I think the best way to visualize a photon isn't as a particle at all,
but simply as the fact that energy can be added to or removed from a
field only in units of hf, where h is Plank's constant, and f is the
frequency.
Note that I didn't say "electromagnetic field". That's because this
is true of every kind of field, and every kind of system that has a
frequency associated with it. The energy in sound waves is quantized
(the "particles" are called "phonons"). The energy in your clock's
pendulum is quantized. The energy in a spinning top is quantized.
The energy in ocean waves is quantized. I don't think it's useful
to visualize water-wave particles in the ocean, or to ask how long
they are.
> Let's take the laser beam and with a fast shutter that is open for
> 1ns and shut for 100ns. I would expect the diffraction pattern to
> be that of a laser with a coherence length of about 300mm, and thus
> different from the unshuttered pattern with a coherence length of
> 1000mm.
So would I. That's why I suggested welders goggles instead of an
intermittent fast shutter.
Note that the shutter is broadening the spectrum of the light. You've
reinvented amplitude modulation. In addition to your laser's normal
frequency, there will be sidebands present at 10 MHz above and below
the main frequency, and more weakly at 30 MHz, 50, 70, etc. (Also at
20, 40, 60, etc, if the shutter isn't time-symmetric.) These pure
unbroken sine waves interfere so as to cancel each other out 99% of
the time. When a photon is detected, it may be in the main "carrier"
frequency, or any of these sidebands.
If your detector has sufficient time resolution to determine which
instance of the shutter opening it was detecting a photon from, then
it will have insufficient frequency resolution to determine which
sideband it was in.
> Maybe not useful, but it's the question I posed to anticipate my
> dear son. He is currently not impressed by the answer. [I suspect he
> thinks we are all quite mad.]
Perhaps it's nature who is "quite mad". Or at least lots of
scientists and philosophers think so, or used to think so. But
quantum mechanics seems quite reasonable to me. Maybe that just
means I'm mad too.
Oz
<ba...@galaxy.ucr.edu> writes
[Read 'sackcloth & ashes' post first.]
>
>You asked a simple little question so chock-full of false assumptions
>that to straighten you out would take a graduate course on quantum
>theory!
Oh dear. I guess you are a bit busy then .....
>And in the process, you have gotten people talking about some
>very interesting things... some of them so interesting that I've been
>forced to hit the books to figure out what the folks working on quantum
>optics have to say.
Hmm. Not entirely wasted time I hope.
>Imagine some poor fellow who'd never heard a flute posted the following
>question on sci.music: "How many trombones does a flute sound like?"
>
>It would cause a tremendous argument,
........
Hmm. A vicious analogy, that may be irritatingly true.
>Meanwhile the poor fellow would be thinking "How come nobody is giving
>me a simple answer? Perhaps even the experts don't know what a flute
>sounds like?"
Nah. It's because my picture of a photon was probably too particle-like.
OK, it was well distributed over space (hey, I would have been quite
happy with 'lengths' of a meter or more, with supporting evidence) which
is not quite what I would have described as a 'particle', but I did want
THE WAVE to be quantised and to mostly exist in a volume of space
despite that it could be quite a large volume.
To discover that an EM wave could interact with photons not yet emitted
(but, I trust, not "in the process of being emitted") was a shock I
don't mind saying. I am somewhat at a loss to do more than indeed have
one of your -+ end-of-spacetimetime sinewaves, but I don't at the moment
much like this.
>In short: I'm not too optimistic that you're going to walk out of this
>much more informed than you walked in.
Well, I AM more informed. Also my memory has been refreshed.
>So all I can is fire off thunderbolts at your more egregious errors and
>hope they push you in the right direction.
Yes, you and others do that quite well. I am quite used to being singed.
>In particular, I hope you spend some time pondering wave-particle
>duality, the uncertainty principle, and that sort of thing.
Indeed. I will ponder away and probably ponder up a bunch more
misconceptions. Is this a good move, I ask myself?
--
Oz
Oz
<ba...@galaxy.ucr.edu> writes
>In article <$ZfqLQA9...@upthorpe.demon.co.uk>,
>Oz <O...@upthorpe.demon.co.uk> wrote:
>
>THE PHOTONS YOU SEE ARE GOING INTO YOUR EYES, WHILE THE PHOTONS I SEE
>ARE GOING INTO MINE!
>
>So yes, we spot differences between photons other than momentum all
>the time.
OK, you brought position into this, not me. I don't consider position as
a 'property'. If that were so then I would have to say that a proton
over here was different to a proton over there (flat spacetime etc etc).
If I give you a proton and you test it and say "it's a proton" I
wouldn't expect you to call it a neutron over there if it passes the
same tests.
> In addition to momentum information there are all sorts of
>other kinds of information. Whenever one photon does something that
>another does not do, they differ in some sort of way - and this way is
>not necessarily momentum.
Another example would be good here.
>It's true that if we know *everything* about the momentum, we can know
>*nothing* about the position! It's also true that if we know everything
>about momentum and polarization, there is nothing else left we can know!
>But that doesn't mean that the only thing we can know is momentum and
>polarization! You takes your pick in quantum mechanics, remember? If
>you decide to know a lot about *this*, you can no longer know much about
>*that*.
>
>Subtle distinctions, eh?
Perhaps. Some (perhaps all) of this is due to the requirements of
'measuring', ie how it interacts or is interacting with other things. I
am perfectly content to see particles as extended objects with wavelike
properties.
Supplementary:
A 'stationary' electron (ie zero momentum wrt an observer) is seen as a
very 'large' extended object cos we don't know where it is on account of
we know it's momentum exactly. Equally a relativistically travelling
observer would 'know' the electron's momentum exactly (his velocity).
Will SR predict the same distribution of knowingness by transforming the
'stationary but known' distribution into a 'with velocity and known'
wavefunction? I guess all the relevent equations are Lorenz invariant so
it will: just checking.
>>We more-or-less know what a photon in flight should look like [...]
>
>Yes, we more-or-less know: ME MORE AND YOU LESS!
Hmmm.
Yes, perhaps it's not such a splendidly brilliant thing for me to say.
>[Ahem. I'm really reverting to that mean old wizard persona. I'm
>sorry, I can't help it - it's the only way I can retain my sanity while
>teaching Oz quantum mechanics.]
The requirement is mutual, believe me ......
>P.S. - Yes I know there are no purple photons,
I am glad I read to the end. I thought you were setting me up
originally. Hmm, you probably were but thought better of the welter of
replies the group would have got.
>P.P.S. - Yes I also know that one needs to be careful talking about
>the position of photons and other massless particles; that's the
>bit about "no Newton-Wigner localization".
I'm tempted, very tempted .....
>Unlike a massive particle,
>you can't say *exactly* where a photon is, but that doesn't mean you
>can't have a darn good idea where it is *approximately*, and that suffices
>to demolish Oz's idea that "all you can know about a photon is its momentum
>(and polarization)".
Ahem. It seems you don't even know *when* it is ferchristsake! Anyhow, I
consider (I might even try to define) that (in flat empty spacetime) a
photon over here is 'the same' as a photon 'over there' if it's momentum
and polarisation are the same. Unfortunately it looks as if a photon 'in
flat empty spacetime' is a meaningless statement because it ain't 'a
photon' if there is nothing for it to interact with. Hmmm. OK
redefinition: an absorbed photon 'over here' is 'the same' as an
absorbed photon 'over there' if it interacts the same way .....
Oh, I give up.
--
Oz
Oz
<ba...@galaxy.ucr.edu> writes
>We can always write this wavefunction as a superposition
>of wavefunctions corresponding to photons with definite momentum and
>polarization - i.e., sine wave solutions of Maxwell's equations. But
>that's just another way of saying it can be of pretty much any shape
>whatsoever!
Well, if it can interact with not-yet-emitted photons as well as
already-absorbed photons this is not so unreasonable. We can always
'steal' a superposition from a not-yet-in-existance photon. Some sort of
perverted sense here.
--
Oz
super...@my-dejanews.com
ca31...@bestweb.net wrote:
> "Interference" by particles does not happen.
> I'm not talking about quanta here, but of "classical particles"
> and sometimes quanta behave like "classical particles".
> So, "classical particles", and quanta behaving like "classical particles",
> do not interfere. That's what happens when you observe quanta closely,
> they don't interfere (Feynman two-slit stuff).
> We should probably never use the word "particle" when taking about
> quanta or a quantum. It is a context-sensitive term in QM.
> Unfortunately, old habits die hard.
I agree with your comments but I feel it is worth making the comment that
theories do exist that can explain this famous experiment with particles. de
Broigle's non-linear wave mechanics was developed almost entirely to solve
this problem. Maybe JB can comment on the *mathematical* problems that exist
with such theories. The problem I see with this pilot wave stuff is it seems
to imply a fairly weak guide wave but I would think that a very powerfull
wave would be required. Does anyone on this NG have experience with solitons
and this stuff? NJH.
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
[Moderator's note: Quoted text deleted. -TB]
Oz
You are right.
I am wrong. I was ignorant of this expt.
[Oz beats his breast, rents his clothes and wears sackcloth & ashes.]
In article <78r3c1$1jt$1...@pravda.ucr.edu>, john baez
<ba...@galaxy.ucr.edu> writes
> Interference effects have been demonstrated in the superposition
> of two light beams from two independent lasers, under conditions
> where the intensity was so low that one photon was absorbed, with
> high probability, before the next one was emitted by either of the
> two sources.
Whilst mulling over photons as I dozed off last night I suddenly
remembered this, and wondered if it really said what I remember it
saying, rather than what I read it said.
It's morning here and I just checked: it did.
£")(*(&^~#~@#?+^&*"!^%)\\{[}@?&%$$** !!!!!!!!!
Good grief ..... mindbending ...... world-view-shift.
[If anyone can post/email me a copy of this expt I would be very
grateful indeed.]
As an apparently now signed up member of the cranks community:
THAT'S REALLY VERY INTERESTINGLY WEIRD.
GR was fun, a bit of a world-view change. But I really find it hard to
see how to fit this in. How can a photon, no matter how reasonably
dispersed in space, diffract with something that hasn't been either
emitted or absorbed? Mental turmoil. Someone else mentioned something
along the lines of a photon gets from A to B in some sort of phasechange
free way, the time waves we see are just us sliding through time down a
fixed waveform but that's easy to see. The expt above is going to make
any mental model that isn't so abstract as to be useless verrrrry
tricky. I'm not even sure if I can call up some virtual particles to go
back/forward in time to 'see' the other photon because I think the time
interval will be too great to be plausible.
I still don't like having an elementary particle whose members are
different, so NOW I am reduced to considering this elementary particle
as being an artifact of the observer (ie a function of the
emitting/absorbing thing) but that propogates through space in a wholly
superimposed wavefunctiony way with potentially infinitely feint long
tails unrelated to any package of energy that may eventually be
absorbed.
Hmmmm. I would like to see the detail of that experiment though. If the
emission time is comparable to the time-between-photons then all bets
are off. It won't show interaction between free photons in my view, but
long-lived entangled ones. I already am very happy with a model for that
(until the evil wiz takes that apart, that is).
PS Having gone this far, and made a fool of myself, I clearly need to
brush up on this subject a tad. Moderators permitting, just a few more
'little' questions?
--
Oz
[Moderator's note: of course, go ahead and ask more questions! - jb]
super...@my-dejanews.com
Oz <O...@upthorpe.demon.co.uk> wrote:
> PS Having gone this far, and made a fool of myself, I clearly need to
> brush up on this subject a tad. Moderators permitting, just a few more
> 'little' questions?
>
> --
> Oz
>
> [Moderator's note: of course, go ahead and ask more questions! - jb]
>
I wouldn't say you made a fool of yourself. I noticed that JB misinterpreted
one of my posts so it just goes to show how difficult it is to use words to
describe exactly what we mean. One of the problems with learning physics at
university is we tend to learn the mechanics of how things work more than the
philosophical arguements. Often the subtleties don't emerge until many years
latter. I would like to see what people think of the following steps of
reasoning. Many people when pondering questions about QM tend to get very
hung up on the question about "collapse of the wavefunction". I don't know
who is to blame for this but for me von Neumann is a big culprit. The Neumann
cut described in Bohms "Undivided Universe" for instance seems to lend weight
to the idea of the collapse being a real process. Many people seem to believe
that the wavefunction of a system (psi) is real i.e. the wavefunction of a
system prior to a measurement being performed. I cannot agree with this
viewpoint. If it was a real thing then the collapse of the wavefunction would
also be real in the physical sense. For the collapse to be real it would have
to use some energy from somewhere in much the same way as a thermodynamic
collapse. AFAIK no experiment has ever managed to demonstrate such a thing
and we are left with the collapse having no physical properties of its own.
As such it is purely a concept or step that is used since it makes the
quantum algorithm work and nothing more. How many people accept this
reasoning? or have some other ideas of their own. NJH.
ca31...@bestweb.net
super...@my-dejanews.com wrote:
> I agree with your comments but I feel it is worth making the comment that
> theories do exist that can explain this famous experiment with particles. de
> Broigle's non-linear wave mechanics was developed almost entirely to solve
> this problem. Maybe JB can comment on the *mathematical* problems that exist
> with such theories. The problem I see with this pilot wave stuff is it seems
> to imply a fairly weak guide wave but I would think that a very powerfull
> wave would be required. Does anyone on this NG have experience with solitons
> and this stuff? NJH.
I think the Young's experiment is over theorized, when simply placing
polarizers over each slit allows complete and continuous control over
the degree of "particle-ness" or "wave-ness" of the interefernce pattern
by adjusting the angle between the polarizers (photon tagging).
Bohm's pilot wave seems treatable as just a carrier wave on which "particles"
or really "highly localized impulses" from detectors are superposed.
In signal processing, only waves are considered "real" and highly
localized impulses are not called "particles" but treated as abstract
particles.
In QM, it's just the opposite, the "particles" are treated as real while
the wavefunctions are treated as abstract.
Similarly, Heisenberg's uncertainty exists in signal processing in
the form of Fourier's uncertainty, so I think the connection between
signal processing is rather strong. There seem to be many other approaches
to HUP as well: Heisenberg Uncertainty in terms of the Fourier Uncertainty[1]
or least squares analysis[2] or the Cauchy-Scwartz inequality
(Pythagorean Theorem)[3], Translation Indeterminacy[4]
Solitons exist in water waves (as does tunneling), so using this idea seems
to put you out of the context of the normal QM formalism but undoubtably
sheads some insight into QM.
[1] Digital Signal Processing in C, Reid and Passin
[2] A Least Squares analysis of the Uncertainty Principle,
John Forkosch, Am. J. Phys,39,4 Apr. 1971
[3] Funzzy Thinking, Bart Kosko
[4] On the reasons for the Indeterminacy of Translation,W. Quine
Matt McIrvin
>On 30 Jan 1999 mmci...@world.std.com (Matt McIrvin) wrote:
>>The fact that photons *don't* directly interact with one another
>>is not a consequence of their masslessness, but of their
>>chargelessness, which is an independent property.
>
>An independent property? So you're saying a massless particle could
>have charge? What would be its acceleration in an electrostatic
>field? Conversely, are you saying that massive chargeless particles
>don't interact with each other?
I didn't mean that electric charge governs the interaction of a particle
with others of the same kind. Electric charge governs the interaction of
any sort of particle *with photons*. So in the case of photons, the fact
that photons have no charge means that they don't interact with one
another (via a single "vertex" interaction, at least).
Of course, a chargeless *composite* particle can be made of charged
pieces, and the pieces could interact with photons.
There do not seem to be any massless electrically charged particles, but
there does seem to be at least one situation in nature that is similar to
that of "charged photons."
QCD, the theory of quark-gluon interactions, is in some ways very similar
to QED. There are massive quarks carrying "color charge," and there are
massless "gluons" that interact with them according to their color charge
in a manner very similar to photons and electrically charged particles.
But the gluons *also* carry color charge, so there are more interaction
vertices for gluon-gluon interactions! There is reason to believe that
the gluons can even form massive bound composites.
Miguel Carrion Alvarez
I have never posted to sci.physics.research before, but I hope I
don't botch it too badly ;-)
On Mon, 1 Feb 1999, Oz wrote:
> In article <78r3c1$1jt$1...@pravda.ucr.edu>, john baez
> <ba...@galaxy.ucr.edu> writes
> > Interference effects have been demonstrated in the superposition
> > of two light beams from two independent lasers, under conditions
> > where the intensity was so low that one photon was absorbed, with
> > high probability, before the next one was emitted by either of the
> > two sources.
>
> [...]
>
> THAT'S REALLY VERY INTERESTINGLY WEIRD.
>
> GR was fun, a bit of a world-view change. But I really find it hard to
> see how to fit this in. How can a photon, no matter how reasonably
> dispersed in space, diffract with something that hasn't been either
> emitted or absorbed? Mental turmoil. Someone else mentioned something
> along the lines of a photon gets from A to B in some sort of phasechange
> free way, the time waves we see are just us sliding through time down a
> fixed waveform but that's easy to see. The expt above is going to make
> any mental model that isn't so abstract as to be useless verrrrry
> tricky. I'm not even sure if I can call up some virtual particles to go
> back/forward in time to 'see' the other photon because I think the time
> interval will be too great to be plausible.
In a previous posting, someone mentioned the fact that emission
of radiation by an atom is not as simple as the word "emission" would have
us believe. An atom emitting radiation is not like a pitcher throwing a
ball. You have some quantum mechanical state
|excited atom>|ground field>
and end up in another state
|ground of atom>|1 photon>
(actually, because of recoil, random emission time and broadening of
energy levels due to finite lifetime, it should be an entangled state of
all kinds of states of the form
|ground of atom, -p, t>|1 photon, p>
where p tells you the direction the photon was emitted, and t is the time
of emission). The point I want to make is another, though: in the
experiment quoted by John Baez (I'd also appreciate a journal reference!)
the emitting system is not an atom but "two different lasers", but the
"field" is _the same_ field and the entanglement also includes whether the
photon was emitted by one laser or another. No wonder you can observe
interference between photons "that have not been emitted". If you avoid
thinking about photons as if they were baseballs you might see that the
experimental result is reasonable.
> I still don't like having an elementary particle whose members are
> different, so NOW I am reduced to considering this elementary particle
> as being an artifact of the observer (ie a function of the
> emitting/absorbing thing) but that propogates through space in a wholly
> superimposed wavefunctiony way with potentially infinitely feint long
> tails unrelated to any package of energy that may eventually be
> absorbed.
Different? How can two things be different and yet be
indistinguishable? Remember that, even though your mental picture of
photons tells you that two photons of the same energy can be "different"
(because the corresponding "one-photon states" of the field can be
different), if you construct the corresponding "two-photon state", it has
to be _symmetric_ under exchange of the two photons because they are
still indistinguishable. You should re-think your phrase "an elementary
particle whose members are different", because that is not what the
experiment and the theory imply.
In a way, what theory tells you and the experiment confirms is
that, if you shine two sources of light on a detector, if you don't look
at the sources and correlate their behaviour with that of the detector,
you cannot tell the photons coming from one source from the photons coming
from the other, and that reassures you that they are indeed "equal" and
not "different"!
Now, there is a classic experiment by Handbury-Brown and Twiss
which proves that you can tell whether you're receiving light from two
different sources or from just one, and this is used to measure the
apparent diameter of stars.
I hope that helps...
Miguel
Paul Arendt
>...to have a half-decent
>shot at getting a classical-looking EM field out of photon,
>you have to infuse the picture with a heavy mist of vagueness,
>so you're *not at all sure* how many photons there are. In other words,
>you have to take superpositions of states with 1, 2, 3, many photons.
Well, yes, and...
>... In other words, can we have a quantum
>state of the EM field such that:
>
> The photon number is very strongly peaked at 1 --- if you expand the
> state in a basis of photon-number eigenstates, the coefficient of |1>
> outshines all the other coefficients.
>
> AND AT THE SAME TIME
>
> The "value of E^2 + B^2 at spacetime point (x,y,z,t)"
> has very low uncertainty for all (x,y,z,t) --- even though
> E and B *separately* are very fuzzy indeed at all (x,y,z,t).
... the answer to this is "yes" too! The state in question
IS just the state | 1 > (in the photon-number basis). The
Hamiltonian is diagonal in this photon-number basis, and the
Hamiltonian is also the space integral of (E^2 + B^2), modulo some
constants.
Er... you said "at spacetime point." Darn it! Couldn't you have
said "in this or that mode?" Oh well.
To be specific, let's quantize a mode of the free EM field in the
Coulomb gauge, where things go as smoothly as they can. As you
know, one ends up with a system equivalent to the harmonic
oscillator. What are the analogs of "q" and "p" in this case?
It turns out that they are the components of the vector potential
A^i, and its time derivative (respectively), equal to the electric
field E^i in Coulomb gauge, in each of the two directions transverse
to the direction of travel. One can examine each direction's
"harmonic oscillator" separately to describe a linearly polarized
mode, or take an appropriate mixture of the two to get a circular
mode. The magnetic field is derived from the A^i, constrained to
satisfy Maxwell's equations.
Right away one sees that measuring both E and B of a mode
simultaneously is not possible, in the usual way of applying the
Heisenberg uncertainty principle to canonically conjugate variables!
And one also can see that an eigenstate of H, no matter how many
photons are in it, always has <E> = <B> = 0 (in the same way that
the harmonic oscillator's eigenstates have <p> = <q> = 0).
To add some concreteness to the picture, let's specify a mode,
and maybe even get a picture of one type of photon in the process.
Let's examine the electromagnetic field in a box of volume V, with
periodic boundary conditions in each of 3 directions [*]. Look
at a mode which is definitely moving to the "right", with (say)
right-handed circular polarization, and with a wavelength equal
exactly to the length of the box in this direction (the fundamental
mode). Oh, and let's have the mode be a plane wave perpendicular
to the direction of travel.
All excitations of this mode are COMPLETELY delocalized in space,
so the photons (if there are any) are everywhere! Moreover, one
knows the exact value of the energy density (and therefore
E^2 + B^2) everywhere, but the fields (E and B) have a completely
unknown _phase_. But since things are the same everywhere in this
mode in this box (up to a phase), we can treat all points equally
and see what "classical" sense can be made out of things. A
one-photon mode has energy 3/2 h nu, where nu = c/L is the frequency,
and L is the wavelength (equal to the dimension of the box in the
direction of "travel", defined by the direction of phase advance
as the state evolves). So we can split this energy up equally
throughout the box, and give an energy density of 3/2 h nu/V to
all points in the box. This is the expectation value of E^2 +
B^2 in the mode, so that's what E^2 + B^2 "is" at each point! But
we can't really say how much is in E or B at any given point. Also,
keep in mind that we're talking about a single mode of the EM field;
there are always the zero-point fluctuations around from every other
possible mode (even if no photons are in other modes).
So this "photon" is a state whose electric field at any point sorta
wiggles in a circle of radius sqrt(3/4 h nu/V) (in some funky units),
in a plane transverse to the direction of motion. But the _overall_
phase is completely unknown! It's a bit like having a sine wave
(actually, helical wave) whose magnitude and wavelength are totally
known, but whose overall phase is completely unknown.
How do we get an actual _spatially_ localized photon or electric
field from this picture? If we'd like to localize the photon (in its
direction of travel), we've gotta take a superposition of many more
modes (all the harmonics of our mode, because we're looking at the
fundamental). But the contributions from each mode can all be
one-photon states! I don't know whether the phases can be adjusted
to also give a semi-non-fuzzy value of E (if all contributing states
to our localized state are one-photon eigenstates of many modes).
On the other hand, if we'd like to have some idea of what E and
B are in our mode (eliminating the overall phase unknown as much
as possible) WITHOUT localizing the "photon", we can just use our
single mode. But it's going to have a superposition of many
different photon numbers, as described in the "Schmotons" thread.
Since we're looking at a specific mode, fixing the overall phase at one
point is tantamount to knowing what the E and B fields are everywhere!
So the coherent states in this mode give you a nice crossed (but
somewhat fuzzy, thanks to Heisenberg) set of E and B fields,
propagating to the right like good little free fields do when they
satisfy Maxwell's equations.
[*] Nice way to get around the normalization problems and such, huh?
Compactify the whole spatial universe! (Physicists do it all the time
without worrying about the consequences.) This puts us in effectively
a 3-torus, S_1 x S_1 x S_1, which has nontrivial cohomology classes.
This poses a problem when showing that H (built from the A^i in
Coulomb gauge) becomes E^2 + B^2, since there's an integration by
parts involved. In fact, we have to consider the possibility of
not only photons as particle-like solutions to the quantized
Maxwell's equations in this space, but also topological solitons!
Sheesh... ask a simple question about photons, and pretty soon
some joker starts yabbering about topological solitons.
Oz
Miguel Carrion Alvarez <mcar...@eucmos.sim.ucm.es> writes
> Hello,
>
> I have never posted to sci.physics.research before, but I hope I
>don't botch it too badly ;-)
With me about you'll shine, never fear.
>On Mon, 1 Feb 1999, Oz wrote:
>
> In a previous posting, someone mentioned the fact that emission
>of radiation by an atom is not as simple as the word "emission" would have
>us believe. An atom emitting radiation is not like a pitcher throwing a
>ball. You have some quantum mechanical state
>
> |excited atom>|ground field>
>
>and end up in another state
>
> |ground of atom>|1 photon>
>
>(actually, because of recoil, random emission time and broadening of
>energy levels due to finite lifetime, it should be an entangled state of
>all kinds of states of the form
>
> |ground of atom, -p, t>|1 photon, p>
>
>where p tells you the direction the photon was emitted, and t is the time
>of emission).
Well, to be truthful the thread was getting somewhat out of control and
a pause to let the postal ripples dissipate seemed appropriate. However,
put in layman's terms are you saying (crudely) that the photon, whilst
being emitted for a very long time because of the evolution of the
states from
> |excited atom>|ground field>
to
> |ground of atom>|1 photon>
takes a long time, in effect (and I describe this crudely) the photon is
being almost simultateously emitted and absorbed throughout this
process, starting as 'mostly absorbed' but ending as 'free'?
If this is the case (or something like it) then one could imagine
(ducks) that the effect would be similar to a standing wave where the
photon wavefunction is very long but entangled with the 'still in the
process of emitting' partly excited atom. The photon could thus be very
long, perhaps approaching ct (where t is the time it takes for the
transition). If the size of the apparatus, and the intervals considered
to have 'low probability of two photons being in the apparatus
simultaneously' was comparable to this length then I am NOT surprised by
the result of this experiment. In effect the photons are not travelling
at c because of the affect of the entangled state or alternatively they
persist in a volume of space longer than you would expect if they were
'free' photons.
>If you avoid
>thinking about photons as if they were baseballs you might see that the
>experimental result is reasonable.
Perhaps. However if I could make a comment. I do NOT see photons as
'little baseballs'. I see then as extended objects of some number of
wavelengths in length (and some sort of width that correlates with their
angular uncertainty) that could be anywhere from ten to a thousand or
more (I don't know, that's why I asked) 'wiggles'. That from a LF radio
transmission could be 1000km or (much) more but would STILL dump all
it's energy in one packet of hv. I have what I consider a simple and
plausible explanation for this, but with no theoretical basis.
Definitely NOT little baseballs, anyway.
Of course this is a 'free' photon. I am perfectly happy for it to become
very highly distorted in all sorts of ways when it interacts with other
particles.
> Different? How can two things be different and yet be
>indistinguishable?
Yeah. I have a big problem with that. That's why I wanted them all the
same. This was not at all a popular picture with the good and wise here.
I got what is commonly described in the vernacular as 'shafted'.
(NB This refers to being impaled on a spear, possibly slowly, definitely
painfully.).
>You should re-think your phrase "an elementary
>particle whose members are different", because that is not what the
>experiment and the theory imply.
I confess to not to being able to make a mental picture where 'free'
photons (or more simply those in the same state) were different. I
thought perhaps if I waited a bit the light (sorry) would come.
For some strange reason I thought that all 'free' photons would be, by
definition, in the 'same state'. So, perhaps naively (ahem), I
considered that a laser beam (presumably with photons in a 'laser beam
state') that was very heavily attenuated would eventually result in the
photons being so well separated that they became in the 'free' state.
Silly me. I hoped, futilely it seemed, that it might be possible to make
an estimate of the number of (as Baez so elegantly coined) 'wiggles' in
a free photon state (hoping the effect of the measuring apparatus didn't
alter them too much). This would thus have answered a little question I
had you see.
> In a way, what theory tells you and the experiment confirms is
>that, if you shine two sources of light on a detector, if you don't look
>at the sources and correlate their behaviour with that of the detector,
>you cannot tell the photons coming from one source from the photons coming
>from the other, and that reassures you that they are indeed "equal" and
>not "different"!
I am personally very relieved you have said that. I will be even more
relieved if other expert opinion concurs.
> I hope that helps...
Well, it seems to have set me off again.
You may not be popular for doing this. :-))
--
Oz
Keith Lynch
Oz <O...@upthorpe.demon.co.uk> wrote:
> It's because my picture of a photon was probably too particle-like.
Right.
> OK, it was well distributed over space (hey, I would have been quite
> happy with 'lengths' of a meter or more, with supporting evidence) ...
I still don't know what you mean by length. If you mean uncertainty
as to the previous location of a detected photon, astronomy is always
a good place to find big numbers.
To get a sharp image (high angular resolution), it's necessary to know
the momentum of photons to a very high precision. To do that, it's
necessary to NOT be able to tell their location very closely. That's
why there are optical telescopes more than five meters in diameter.
Nobody can say where on the shiny paraboloid each of those photons hit.
But they don't have to.
And radio telescopes are far larger. I've personally visited the
world's largest, in Arecibo, Puerto Rico, where a secluded valley
has been lined with metal mesh, to construct an instrument capable
of detecting a millionth of a trillionth of a trillionth of a watt.
It's so sensitive that they wouldn't let me walk under it, lest the
harmonics of my brain waves jam the faint microwave signals that have
been in transit at the speed of light since before the earth was
formed. But its spatial resolution is only about as good as the
unaided human eye, despite the immense size, since the low total
momentum of microwave photons means that you need to be spectacularly
ignorant of photon position to get much resolution, and a valley just
doesn't cut it.
The record goes to intercontinental radio interferometers, which
manage to measure the azimuth of quasars closely enough to detect
continental drift in real time. And which do so by deliberately NOT
knowing the location of photons to within the width of our planet.
So there's your answer. Some photons are the size of the earth. And
with advancing technology, they'll get much bigger. Smaller, too.
I wish the Guinness book would cover these records instead of boring
stuff about basketball scores, swallowing goldfish, and memorizing pi.
(Posted and mailed.)
john baez
Miguel Carrion Alvarez <mcar...@eucmos.sim.ucm.es> wrote:
>The point I want to make is another, though: in the
>experiment quoted by John Baez (I'd also appreciate a journal reference!)
>the emitting system is not an atom but "two different lasers", but the
>"field" is _the same_ field and the entanglement also includes whether the
>photon was emitted by one laser or another.
The paper was:
R. L. Pfleegor and L. Mandel, Interference effects at the single photon
level, Phys. Lett. 24A (1967), 766-767.
While we're talking about weird stuff you can do with light, let me
give a few more examples. I'm getting all of these from "Concepts of
Quantum Optics" by P. L. Knight and L. Allen and "Lectures on Quantum
Optics" by W. Vogel and D.-G. Welsch.
1. Photon bunching. Actually this is not all that weird. If you
take a beam of light and repeatedly measure the intensity of the
electromagnetic field there will be fluctuations. We can model
the intensity I(t) as a random variable depending on the time t
and ask about the correlation
<I(t)I(t+T)>
as a function of the time between measurements, T. For coherent
light, i.e. a laser, this correlation is independent of the time
T. For normal light the correlation is biggest when T = 0 and
decreases as T increases.
Here is a more intuitive way of thinking about it. If you put a
photoelectric detector in a very weak laser beam and keep track
of each time a photon arrives, these times form a Poisson distribution -
the photons arrive at uncorrelated times. But if you put your detector
in a beam of normal light, the arrival times are "bunched".
2. Photon antibunching. If you are more clever, you can make
a source of light where the correlation function above is *smallest*
when T = 0 and *increases* with increasing T. Then you have "photon
antibunching". Unlike photon bunching, this has no classical analog.
3. Squeezed states. A coherent state is a quantum state approximating
a given classical state for which the uncertainties of conjugate pairs
of variables are equal and small as possible. For example, a coherent
state of a single particle is a Gaussian wavefunction such that when we take
its Fourier transform we get a Gaussian in momentum space with the same
width - so the position and momentum uncertainties are equal. But we can
reduce the uncertainty in position while increasing the uncertainty in
momentum by "squeezing" the wavefunction - so that its Fourier transform
stretches out by an equal factor. Similarly, clever quantum opticians can
now produced "squeezed states" of light in which some variables are made
less uncertain at the expense of making others more uncertain.
There are other cool forms of "nonclassical light", but I can't say
much about them yet. Maybe the quantum opticians in our midst can
wow us with far-out facts.
Oz
<par...@nmt.edu> writes
<Vicious snipping in progress>
>Er... you said "at spacetime point." Darn it! Couldn't you have
>said "in this or that mode?" Oh well.
Does a 'free' photon (perhaps taken as a single photon in a 'very large'
universe) consitiute a 'mode'?
>What are the analogs of "q" and "p" in this case?
>It turns out that they are the components of the vector potential
>A^i, and its time derivative (respectively), equal to the electric
>field E^i in Coulomb gauge,
Thank you. I didn't know that.
>To add some concreteness to the picture, let's specify a mode,
>and maybe even get a picture of one type of photon in the process.
>Let's examine the electromagnetic field in a box of volume V, with
>periodic boundary conditions in each of 3 directions [*].
Yes, this is a nice easy picture, and one which I agree with providing
the 'wavetrainlength' of the photon is large compared to the box. I'm
not sure it would be a good picture for a gamma ray in a reflecting 1m^3
box though. Oh well it would (longterm), but you see what I mean.
>All excitations of this mode are COMPLETELY delocalized in space,
>so the photons (if there are any) are everywhere!
If I fired a single gamma ray with -+known direction through a hole in
your box one after another (slowly) and arranged a hole for the exiting
gamma ray at the other end I think I would be moderately confident they
would be seen to interact along some sort of cylinder between the holes
and only very rarely elsewhere. Now blow this up to galactic size and
use an appropriate frequency to compare to the gamma ray and I would
expect the situation to remain the same, but for light (probably
microwaves) at a very much lower frequency.
<snip, axe, cut, chop>
>How do we get an actual _spatially_ localized photon or electric
>field from this picture? If we'd like to localize the photon (in its
>direction of travel),
Question: It is essential to consider a photon as an infinite sine wave
when doing these examples. I have always assumed that a sufficiently
ornery mathematician could have formulated the theory using the one-
photon mode and thus describing your
>we've gotta take a superposition of many more
>modes (all the harmonics of our mode, because we're looking at the
>fundamental). But the contributions from each mode can all be
>one-photon states!
As a basis, whereupon the nice simple infinite sinewave solution would
then become
>we've gotta take a superposition of many more
>modes
And the two be *completely equivalent* (except one may be a tad
cumbersome). One may have a somewhat different expression for e=hv, but
no matter. Of course in a coherent beam of photons I would expect the
summation of all these strange animals to deliver me the wavefunction
for a nice coherent sinewave beam as all the 'off frequencies' (so to
speak) cancelled out.
>[*] Nice way to get around the normalization problems and such, huh?
>Compactify the whole spatial universe! (Physicists do it all the time
>without worrying about the consequences.) This puts us in effectively
>a 3-torus, S_1 x S_1 x S_1, which has nontrivial cohomology classes.
>This poses a problem when showing that H (built from the A^i in
>Coulomb gauge) becomes E^2 + B^2, since there's an integration by
>parts involved. In fact, we have to consider the possibility of
>not only photons as particle-like solutions to the quantized
>Maxwell's equations in this space, but also topological solitons!
>Sheesh... ask a simple question about photons, and pretty soon
>some joker starts yabbering about topological solitons.
Er, um, isn't this of some relevence to the thread? What (in peasant's
language' is interesting about topological solitons other than the
propogate unchanged in little packets? Sorta like one might imagine a
free photon might do? I understood that one needed some non-linearity
WRT the wave component to produce a soliton which would be hard to
justify given modern theory.
--
Oz
Oz
<k...@clark.net> writes
<vicious snipping in progress>
>* The photon takes every possible path at once. Only at the points
> where there is constructive interference is it likely to be found.
Agreed, but perhaps not with equal amplitude for all paths given it has
direction.
>
>* Light is waves, and follows the same laws as sound waves or water
> waves. Energy can be removed from these waves only in packets
> of size hf. These packets are called photons. But they have no
> existence as photons until you remove them. It's like a grocery
> store where milk can exist in any quantity, but can only be removed
> from the store a quart (or some integer number of quarts) at a time.
Nice analogy with the irritating problem that when you get to the end of
the bucket you had better have *exactly* one quart left or the universe
gets annoyed. I am perfectly happy that a bunch of photons interacts
like a bunch of photons and behaves like a bunch of photons. I am asking
about a 'free' photon.
>Photons don't have serial numbers.
<Snip of statement saying all photons in the same state are identical>
OK. So:
1) All photons in the same state are idential. Good.
2) So all free photons are identical. Good.
3) Iff a 'free photon' had some sort of length/size then all free
photons would have the same length/size.
4) If I separate photons far enough they will become 'free' photons.
Now (4) this seems to be the crux of the problem.
The optics guys say that the photon, no matter how isolated, carries the
characteristics of it's origin. Since origins differ that means you
can't have 'free photons'. I find this hard to believe. I find it easier
to imagine that the act of attenuation requires that the state is
altered (after all, a state with 9 photons is presumably subtly
different from one with 10 photons). I am PERFECTLY happy that a number
of photons can describe the source, for example by the distribution of
their frequencies, even if we compare (say) a fine line from a very
distant galaxy in one direction with that from another direction even if
we swap photons between them at random.
Some say you can make no measure of the 'size' of the photon. This is
clearly incorrect. I could take some HEP interaction that is known to
generate a gamma ray. Within microseconds I could have a computer detect
the fragments and give me a measure of the direction and energy of the
resultant gamma ray. I could then (with quite a high degree of
confidence) set out a volume that would contain that gamma ray (if it
had not interacted first, but this can be minimised by experimental
conditions).
>Getting back to your intermittent shutter,
> if the paths differ by
>30.3 meters, there WILL be interference again.
I will probably only be convinced by this if the experiment was actually
done. As I suggested in another post I suspect that to have a coherence
length of 30.3m the transitions within the excited atom ought (I think)
to be very slow indeed and so the state of that photon is not (close to
the apparatus) 'free' at all but entangled with the emitting atom to a
greater or lesser extent.
>then you'd have to regard one photon as consisting of a long
>series of discontinuous lumps in this case.
Which I have no problem doing as the shutter affect the state of the
emitted photon in this case if what I say (above) is valid.
>I recommend the Feynman Lectures on Physics. Take 3, and call me in
>the morning.
Yes, I really must read it through end to end. I'll take it on my next
holiday, in two years time, family permitting.
--
Oz
Oz
<k...@clark.net> writes
>I still don't know what you mean by length. If you mean uncertainty
>as to the previous location of a detected photon, astronomy is always
>a good place to find big numbers.
No, that's not really what I mean.
>To get a sharp image (high angular resolution), it's necessary to know
>the momentum of photons to a very high precision. To do that, it's
>necessary to NOT be able to tell their location very closely. That's
>why there are optical telescopes more than five meters in diameter.
>Nobody can say where on the shiny paraboloid each of those photons hit.
Sure. I'm not sure I ever mentioned wanting to know their momentum, but
clearly that is to some extent implied if the 'size' is related to their
momentum, which it would appear to do so. The usual way round this sort
of problem is to be more explicit. I assume the 'size' is proportional
to the frequency (via my little relativistic boost argument) and thus to
the momentum. So, the question then becomes something like
what is the volume (a shape would be good) that encloses 95% (call it
chance if you will) of a photon such that the product of the relative
error of the momentum (energy will do) and the position is a minimum?
That would do me just dandy. I mean, how little do we know about it?
--
Oz