# An atom emits light. How long is the wavetrain?

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### Science Hobbyist

Apr 5, 2000, 3:00:00 AM4/5/00
to

I noticed an incorrect assertion made by somebody here
many days ago, so I'm finally bringing up the topic.

When an electron in a solitary atom experiences a
transition to a lower energy state and emits light,
typically how long is the wavetrain of EM waves in
terms of number of cycles? A few cycles? Billions of
cycles?

More than one person said this:

The wavetrain is always less than one cycle long.

Or in other words, they apparently believe that,
whenever atoms emit EM waves, the waveform looks like
a very brief pulse or "packet." This, as opposed to
an extended train of waves which continues over
many cycles.

This is wrong! This "brief packet" thinking to me
exposes their misunderstanding of QM. While it's
true that **photons** act like brief "pulses" when
they're emitted or absorbed, EM waves are not photons.
EM waves are extended entities, and are not pointlike
either in space or in time.

If working physicists believe differently, and believe
that the EM waves emitted by solitary atoms always occur
in bursts lasting less than a single cycle, I'd certainly

--
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William J. Beaty SCIENCE HOBBYIST website
bi...@eskimo.com http://www.amasci.com
EE/programmer/sci-exhibits science projects, tesla, weird science
Seattle, WA 206-781-3320 freenrg-L taoshum-L vortex-L webhead-L

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### Bilge

Apr 5, 2000, 3:00:00 AM4/5/00
to
Science Hobbyist said some stuff about

> When an electron in a solitary atom experiences a
> transition to a lower energy state and emits light,
> typically how long is the wavetrain of EM waves in
> terms of number of cycles? A few cycles? Billions of
> cycles?

Is unbounded growth a general property cluelessness or
are you special? This makes no more physical sense
single transition. However, I suppose you're going
to do that as an encore. Why dont you read one of the
books you should have left from the time that you
alledge was spent in an ee curriculum.

> they're emitted or absorbed, EM waves are not photons.
> EM waves are extended entities, and are not pointlike
> either in space or in time.

To be precise, an infimite number of waves whic extend over
all space with a frequency spectrum from 0 to infinity.

>
> If working physicists believe differently, and believe
> that the EM waves emitted by solitary atoms always occur
> in bursts lasting less than a single cycle, I'd certainly
> like to hear about it.

Make it any finite number you want to dream up. Your answer
will contain an infinite number of waves. If this is a mystery,
find the appropriate newsgroup and/or remedial math course
to explain that you cant describe something that happens in
some fixed interval any other way.

No light---->|<-- light appears here --->|--->light is definely here-->
t1 t2

Sorry. No sine wave. Now what was the wavelength of this
radiation? I'm beging to think a case can be made to simply
start out with derision, sarcasm and contempt when it's
obvious that no one could seriously continue to argue
something that they cant even comprehend the objections
against, let alone overcome them. If continuing to use the
same incorrect suppositions and not understanding why your
keep getting the same wrong result, but keep on trying to
make it fit doesnt make you wonder if perhaps there is
a problem with you approach, save the effort an write a
perl script to generate your arguments from a markov chain
that takes its input from a daily stroll around the web
sites on the internet.

### Science Hobbyist

Apr 5, 2000, 3:00:00 AM4/5/00
to
In article <8ce18p$7fc$1...@nnrp1.deja.com>,

Science Hobbyist <bbe...@microscan.com> wrote:
> I noticed an incorrect assertion made by somebody here
> many days ago, so I'm finally bringing up the topic.
>
> When an electron in a solitary atom experiences a
> transition to a lower energy state and emits light,
> typically how long is the wavetrain of EM waves in
> terms of number of cycles? A few cycles? Billions of
> cycles?

I hope the flamer crap is not too off-putting to others
here. I feel partly responsible for it, but this doesn't
mean I have any intention of going away anytime soon.

If someone can point out the errors in my previous message
in a mature and professional manner, I'll be happy to
listen (and probably raise objections!)

:)

As I've said over and over, it's the immature and
unprofessional behavior which causes me to delete any
message the moment I encounter it. For this reason a
flamer can teach me nothing, and I'll only learn from
people of integrity, no matter how correct and insightful
the information from the flamers might be.

In case anyone is wondering, yes, I've decided to make
this newsgroup a permanent home. I see that this is the
appropriate response to the present situation. Are
newsgroups full of flamers? Yes, and one reason is that
civilized people tend to abandon them in disgust. To
combat this unfortunate positive-feedback effect, we
should do the sensible thing and flock to the newsgroups,
in order to OUTNUMBER the flamers.

--

((((((((((((((((((( ( ( ( ( (O) ) ) ) ) )))))))))))))))))))

William Beaty bbe...@microscan.com
Software Engineer http://www.microscan.com
Microscan Inc., Renton, WA 425-226-5700 x1135

### Roy McCammon

Apr 5, 2000, 3:00:00 AM4/5/00
to
Science Hobbyist wrote:

> If working physicists believe differently, and believe
> that the EM waves emitted by solitary atoms always occur
> in bursts lasting less than a single cycle, I'd certainly
> like to hear about it.

I'd guess that a working physicist would say that he
didn't believe that atoms emitted EM waves at all, but
rather emitted photons, the aggregate effect of which
resembles the effect of a hypothetical EM wave.

Opinions expressed herein are my own and may not represent those of my employer.

### Roy McCammon

Apr 5, 2000, 3:00:00 AM4/5/00
to
Science Hobbyist wrote:

> So physicists typically consider photons to be real,
> and fields to be unreal?

Given some waffling room about what "real" means, yes.

QED always gives correct answers, as far as we know,
but EM does not. Photons are candidates, in a sense,
for being real, or for being representatives of
reality. EM fields are remanded to the status
of useful artifice.

> For this reason nobody can give clear answers

outcomes of experiments or calculations. But
questions about photons are difficult. Nothing
in our every day life has properties like photons.
Questions like "how long is a photon" are not just
difficult to answer, they are impossible.

> And are the chapters in physics texts about
> e-fields and b-fields perhaps misguided and
> should be removed?

No, EM fields are useful artifices.

> If it were me, I'd say that this is misguided, and
> both fields and particles should be regarded as mental
> tools, and not as "reality."

You would get substantial agreement on fields. And if you
feel that way about fields, why all the anguish over why
an atom can interact with a wave that has a wavelength
that is much longer than the atomic diameter?

> If one is more useful a tool than the
> other, that does not make it more "real,"

Yes, but EM has been disqualified as "real",
where as QED is still in the running, so to speak.

> and it should
> not prevent us from discussing the other "tools."

Certainly not, but in an open forum like this,
one should be clear that one pushing the tool
beyond where it works well, lest others be misled.
(I do not think you are trying to mislead)

> I still must ask my original question. If an atom
> behaves like a dipole antenna, then whenever an atom
> emits "light waves", should we imagine that the
> wavetrain consists of NUMEROUS cycles? Or must this
> wavetrain consist of a very brief wave packet which
> always contains less than even a single cycle? It seems
> to me that such a brief packet would have a fairly large
> distribution of frequencies. Since atomic emissions
> don't have immense spectral linewidths, then it implies
> that an atom's outgoing train of EM waves must contain
> far more than just a small fraction of a single complete
> cycle. What am I missing here?

What is present, rather than missing, is your belief
that the atom emits an EM wave. You have tossed a fact
down on the table that has no right to be there.

And now, I am afraid that I have come across a reference
that may give you encouragement. No doubt I will pay
for this, but here goes.

"Optical Electronics," Ammon Yariv (of Cal Tech),
Oxford University Press, 4th ed, 1991

chap 5: Interaction of Radiation and Atomic Systems
sec 4: The electron oscillator model of an atomic transition.

"when the frequency w of the wave is near that of
an atomic transition, the atoms acquire large dipole
moments that oscillate at w, and the total field is now
the sum of the incident field and the field radiated
by the dipoles"

He goes on to say that this method was used pre quantum
by replacing the atom with an electron oscillating in a
harmonic potential well, which is adjusted to give the
frequency and absorption width of a known transition.
An empirical "strength" factor is thrown in to match
some other characteristic, and equations are solved.
The material susceptibility is derived, which yields
a real part and an imaginary part. Apparently all this
does a pretty good job of predicting the dispersion
characteristics of the material.

But then in sec 5 he points out that the solution for
the imaginary part of the susceptibility is wrong
and that a quantum treatment gives an answer that
is right, based on experiment. The important difference
being that the QM answer allows for negative imaginary
part which allows the material to have gain and
allows lasing action.

sub

### Maarten Hoogerland

Apr 6, 2000, 3:00:00 AM4/6/00
to
Bilge wrote:
>
> Science Hobbyist said some stuff about
>
> > When an electron in a solitary atom experiences a
> > transition to a lower energy state and emits light,
> > typically how long is the wavetrain of EM waves in
> > terms of number of cycles? A few cycles? Billions of
> > cycles?
>
> Is unbounded growth a general property cluelessness or
> are you special? This makes no more physical sense
> single transition. However, I suppose you're going
> to do that as an encore. Why dont you read one of the
> books you should have left from the time that you
> alledge was spent in an ee curriculum.
>
> > they're emitted or absorbed, EM waves are not photons.
> > EM waves are extended entities, and are not pointlike
> > either in space or in time.
>
> To be precise, an infimite number of waves whic extend over
> all space with a frequency spectrum from 0 to infinity.
>
> >
> > If working physicists believe differently, and believe
> > that the EM waves emitted by solitary atoms always occur
> > in bursts lasting less than a single cycle, I'd certainly
> > like to hear about it.
>
> Make it any finite number you want to dream up. Your answer
> will contain an infinite number of waves. If this is a mystery,
> find the appropriate newsgroup and/or remedial math course
> to explain that you cant describe something that happens in
> some fixed interval any other way.
>
> No light---->|<-- light appears here --->|--->light is definely here-->
> t1 t2
>
> Sorry. No sine wave. Now what was the wavelength of this
> radiation? I'm beging to think a case can be made to simply
> start out with derision, sarcasm and contempt when it's
> obvious that no one could seriously continue to argue
> something that they cant even comprehend the objections
> against, let alone overcome them. If continuing to use the
> same incorrect suppositions and not understanding why your
> keep getting the same wrong result, but keep on trying to
> make it fit doesnt make you wonder if perhaps there is
> a problem with you approach, save the effort an write a
> perl script to generate your arguments from a markov chain
> that takes its input from a daily stroll around the web
> sites on the internet.
>

However, we can measure what the wavelength of the spontaneous emission
radiation is. That means that to a certain extent, it can be represented
with a sine wave. We know what the spread in frequency is, it is the
natural linewidth of the transition (give or take a factor of two or
pi). For a minimum uncertainty "wavepacket" delta E delta t = h so you
might think about constructing a wave packet that has looks sinusoidal,
with frequency nu, but has a gaussian intensity envelope with a width in
time of 2 pi/delta nu. Of course the total energy of the wavepacket has
to be h nu.
If we know the frequency with infinite precision, i.e. it is a delta
function, than the corresponding wave is a pure sine wave, by fourier
arguments, and we can't say anything about the
time that the photon hits the detector. Of course we always measure some
time, but if we do the experiment a large number of times, we get a flat
arrival time distribution, until t=inf.
Anyway, the wavepacket may not be "true" in the philosophical sense, but
if you use it in calculations and experiments you get all the right
answers. That's good enough for me.

Cheers,

Maarten

### Bilge

Apr 6, 2000, 3:00:00 AM4/6/00
to
Maarten Hoogerland said some stuff about

>with frequency nu, but has a gaussian intensity envelope with a width in
>time of 2 pi/delta nu. Of course the total energy of the wavepacket has
>to be h nu.

The premise explicitly did not allow a wave-packet.
So, I pointed out why a single frequency wont work.

>If we know the frequency with infinite precision, i.e. it is a delta
>function, than the corresponding wave is a pure sine wave, by fourier
>arguments, and we can't say anything about the

Why do you think I put in a diagram with a finite time
interval?

>time that the photon hits the detector. Of course we always measure some
>time, but if we do the experiment a large number of times, we get a flat
>arrival time distribution, until t=inf.

The statement was for "a single transition". I reiterate. For
a single transition, the statement he made makes no sense. If
I wanted to change the premise, I could've stated all of this.

>Anyway, the wavepacket may not be "true" in the philosophical sense, but
>if you use it in calculations and experiments you get all the right
>answers. That's good enough for me.

Im willing to take it as true in philosophical. You're
preaching to the choir.

### Bilge

Apr 6, 2000, 3:00:00 AM4/6/00
to
Science Hobbyist said some stuff about
>
> I hope the flamer crap is not too off-putting to others
> here. I feel partly responsible for it, but this doesn't
> mean I have any intention of going away anytime soon.
>
No. I'm responsible for it. If you want a serious response,
post something serious.

> flamer can teach me nothing, and I'll only learn from

No one else can or has, either. What difference does the
approach make?

> In case anyone is wondering, yes, I've decided to make
> this newsgroup a permanent home. I see that this is the
> appropriate response to the present situation. Are
> newsgroups full of flamers? Yes, and one reason is that
> civilized people tend to abandon them in disgust. To

Good. You'll provide plenty of examples of pseudoscience
and opportunity to point out why it isn't science.

### Bilge

Apr 6, 2000, 3:00:00 AM4/6/00
to
Roy McCammon said some stuff about
Re: An atom emits light. How long is the wavetrain? to usenet:

>Science Hobbyist wrote:
>
>> If working physicists believe differently, and believe
>> that the EM waves emitted by solitary atoms always occur
>> in bursts lasting less than a single cycle, I'd certainly
>> like to hear about it.
>
>I'd guess that a working physicist would say that he
>didn't believe that atoms emitted EM waves at all, but
>rather emitted photons, the aggregate effect of which
>resembles the effect of a hypothetical EM wave.
>

The physicist would say that the atom emitted a quantum
of energy which is uncertain by an amount \Delat E
that has a probability P(x) of being observed with
wavelength x. There is nothing more one can say beyond
relating the uncertainties. That's all the reality that
exists.

### Science Hobbyist

Apr 6, 2000, 3:00:00 AM4/6/00
to
In article <38EBC0CF...@mmm.com>,

rbmcc...@mmm.com (Roy McCammon) wrote:
> Science Hobbyist wrote:
>
> > If working physicists believe differently, and believe
> > that the EM waves emitted by solitary atoms always occur
> > in bursts lasting less than a single cycle, I'd certainly
> > like to hear about it.
>
> I'd guess that a working physicist would say that he
> didn't believe that atoms emitted EM waves at all, but
> rather emitted photons, the aggregate effect of which
> resembles the effect of a hypothetical EM wave.

So physicists typically consider photons to be real,
and fields to be unreal? For this reason nobody can

chapters in physics texts about e-fields and b-fields

perhaps misguided and should be removed? If it were me,

I'd say that this is misguided, and both fields and
particles should be regarded as mental tools, and not

as "reality." If one is more useful a tool than the
other, that does not make it more "real," and it should

not prevent us from discussing the other "tools."

I still must ask my original question. If an atom

behaves like a dipole antenna, then whenever an atom
emits "light waves", should we imagine that the
wavetrain consists of NUMEROUS cycles? Or must this
wavetrain consist of a very brief wave packet which
always contains less than even a single cycle? It seems
to me that such a brief packet would have a fairly large
distribution of frequencies. Since atomic emissions
don't have immense spectral linewidths, then it implies
that an atom's outgoing train of EM waves must contain
far more than just a small fraction of a single complete
cycle. What am I missing here?

--

(((((((((((((((((((( ( ( ( ( (O) ) ) ) ) )))))))))))))))))))
William J. Beaty SCIENCE HOBBYIST website
bi...@eskimo.com http://www.amasci.com
EE/programmer/sci-exhibits science projects, tesla, weird science
Seattle, WA 206-781-3320 freenrg-L taoshum-L vortex-L webhead-L

### Peter Somlo

Apr 6, 2000, 3:00:00 AM4/6/00
to
Science Hobbyist <bbe...@microscan.com> wrote:

This is the same old wave/particle duality mystery again. To my mind
the wave representation (of the phenomenon) is the frequency-domain
and the particle representation is the time-domain 'observation' of
the same thing. Think of 'some vibration', which you can view with a
Spectrum Analyzer or with a CRO. In these cases the horizontal axis
was either frequency or time respectively. If you observe an
interference pattern, you are looking in the freq. domain, if you bump
an elementary particle, you are in the time domain.
Peter
_____________________________________________________________________
Dr.Peter I Somlo FIEEE | M1: "Every coin has 3 sides - at least"
Microwave Consultant | email: so...@ieee.org
tel/fax: 61-2-9451-2478| ICQ: 1032408
Mobile(AU):041-926-3168| <http://www.zeta.org.au/~somlo/default.htm>

### Bilge

Apr 6, 2000, 3:00:00 AM4/6/00
to
Science Hobbyist said some stuff about
>
> So physicists typically consider photons to be real,
> and fields to be unreal? For this reason nobody can

Since the photon IS the field, this would be a rather
bizarre belief.

> chapters in physics texts about e-fields and b-fields

> as "reality." If one is more useful a tool than the
> other, that does not make it more "real," and it should
> not prevent us from discussing the other "tools."
>

Since the are the same tool, the only thing you
cant do is say photons and fields aren't the
same thing.

> I still must ask my original question. If an atom
> behaves like a dipole antenna, then whenever an atom
> emits "light waves", should we imagine that the
> wavetrain consists of NUMEROUS cycles?

You should imagine that what you find out from a measurement
is that you'll have a probability of measuring one of the
infinite number of different energies in that packet. Since
a wave packet isn't periodic, cycles makes no sense for
a just one of them. When the source emits lots of them,
the periodicity of your "wavetrain" represents the average
periodicity in the emitted packets.

> to me that such a brief packet would have a fairly large
> distribution of frequencies. Since atomic emissions

It does. Nothing requires the frequencies to equally probable.
The fact that the emitted photon is the result of a transition
is sufficient to garauntee the the distribution is anything
but equally probable.

> don't have immense spectral linewidths, then it implies
> that an atom's outgoing train of EM waves must contain

Whatever it emits, your notion of a wavetrain doesn't
describe it.

> far more than just a small fraction of a single complete
> cycle. What am I missing here?
>

phase space.

### Richard Herring

Apr 6, 2000, 3:00:00 AM4/6/00
to
In article <8cgu1e$faa$1...@nnrp1.deja.com>, Science Hobbyist (bbe...@microscan.com) wrote:
> In article <38EBC0CF...@mmm.com>,
> rbmcc...@mmm.com (Roy McCammon) wrote:
> > Science Hobbyist wrote:
> >
> > > If working physicists believe differently, and believe
> > > that the EM waves emitted by solitary atoms always occur
> > > in bursts lasting less than a single cycle, I'd certainly
> > > like to hear about it.
> >
> > I'd guess that a working physicist would say that he
> > didn't believe that atoms emitted EM waves at all, but
> > rather emitted photons, the aggregate effect of which
> > resembles the effect of a hypothetical EM wave.

> So physicists typically consider photons to be real,

> and fields to be unreal?

what they emit are individual photons. If you're interested in
currents, (aggregate, bulk properties of charged particles) they
produce. "Real" and "unreal" don't enter into it; it's a matter
of which description is more useful in a particular situation.

> I still must ask my original question. If an atom
> behaves like a dipole antenna,

It doesn't really. The atom emits individual photons; the dipole
emits vast clouds of coherent photons which behave like waves.

--
Richard Herring | <richard...@gecm.com>

### mikegi

Apr 6, 2000, 3:00:00 AM4/6/00
to
> Make it any finite number you want to dream up. Your answer
> will contain an infinite number of waves. If this is a mystery,
> find the appropriate newsgroup and/or remedial math course
> to explain that you cant describe something that happens in
> some fixed interval any other way.

What a bunch of bullshit. You don't know the difference between a
mathematical model and what it represents. A waveform isn't composed of an
infinite number of sine waves. It simply exists and can be analyzed in
different ways: infinite series of sine waves, impulses, steps, etc...

The original poster asked a perfectly reasonable question: how many cycles
are in a single transition? There will be a average frequency in the
transition combined with an amplitude envelope. From that you should be able
to estimate the cycle count by picking a certain amplitude as the on/off
marker.

If you don't know the answer, simply say so.

Mike

### Science Hobbyist

Apr 6, 2000, 3:00:00 AM4/6/00
to
In article <8ci0ac$c5p$2...@miranda.gmrc.gecm.com>,
richard...@gecm.com wrote:

> > I still must ask my original question. If an atom
> > behaves like a dipole antenna,
>
> It doesn't really. The atom emits individual photons; the dipole
> emits vast clouds of coherent photons which behave like waves.

Individual photons don't behave like waves? So if I send
my photons through a double-slit experiment INDIVIDUALLY,
the fringe pattern which builds up will be different than
when I send them through in vast coherent clouds? This
is news to me. I was under the impression that even
individual photons have wave-nature, and therefore whenever
an atom emits a single photon, that photon has wave-nature.
Or from the fields-viewpoint, the atom has emitted a
spherewave ripple in the EM field.

--

((((((((((((((((((( ( ( ( ( (O) ) ) ) ) )))))))))))))))))))

William Beaty bbe...@microscan.com
Software Engineer http://www.microscan.com
Microscan Inc., Renton, WA 425-226-5700 x1135

### Science Hobbyist

Apr 6, 2000, 3:00:00 AM4/6/00
to
In article <38EC150C...@ieee.org>,
Roy McCammon <rbmcc...@ieee.org> wrote:

> Science Hobbyist wrote:
>
> > So physicists typically consider photons to be real,
> > and fields to be unreal?
>
> Given some waffling room about what "real" means, yes.

I did not know this. It certainly explains some of
the past failures in communication here.

I was assuming that "particles" and "fields" were
two separate ways of understanding reality. If
physicists have discovered that only particles are
real, and that fields/waves are not, then this can
act as a hidden assumption which will seriously
foul up communications with non-physicists (such
as I) who still consider fields/waves to be just
as "real" as quanta.

> QED always gives correct answers, as far as we know,
> but EM does not. Photons are candidates, in a sense,
> for being real, or for being representatives of
> reality. EM fields are remanded to the status
> of useful artifice.
>

> > For this reason nobody can give clear answers
>

> outcomes of experiments or calculations. But
> questions about photons are difficult. Nothing
> in our every day life has properties like photons.

> > If it were me, I'd say that this is misguided, and
> > both fields and particles should be regarded as mental
> > tools, and not as "reality."
>

> You would get substantial agreement on fields. And if you
> feel that way about fields, why all the anguish over why
> an atom can interact with a wave that has a wavelength
> that is much longer than the atomic diameter?

Most of the "anguish" was my attempts to be
understood, and even to be HEARD over the nasty
roar of the flamers.

And, even though I feel that way about fields, I also
feel that way about EVERYTHING. We cannot experience
reality directly, we can only experience it via our
sensory input, descriptions, and mental models. For me,
all perceptions/concepts are on a sliding scale of
usefulness and accuracy, but none of them are "real."
They are the map, not the terrain. They are the answers
to questions, but they are not the thing which creates
those answers. Something is out there, but there's no
way to get to it directly, so "reality" for me is the
stuff of Plato, not the stuff of physics.

> Certainly not, but in an open forum like this,
> one should be clear that one pushing the tool
> beyond where it works well,

I thought that this was so obvious that nobody
needed to state it. Propagating vibrations in
classical EM fields lack quanta!

> > I still must ask my original question. If an atom

> > behaves like a dipole antenna, then whenever an atom
> > emits "light waves"

> What is present, rather than missing, is your belief

> that the atom emits an EM wave. You have tossed a fact
> down on the table that has no right to be there.

Then that's a big problem, and a problem which I'd
certainly like to discuss if you have the time.

I was under the impression that the rate of photon
emission/detection in situations like the double-slit
experiment was irrelevant to the wave-nature of light,
and that single photons posess just as much wave-nature
as hoards of photons.

Or to put it another way, if we reduce the output power
of a very small radio antenna, the antenna still behaves
as if it is producing "fields" and emitting "waves" even
when the output power is down below one photon per second.

Or said like this: every time a photon hits a silver
molecule in your camera film, there WAS a wave
associated with that photon while it was still on
the fly.

I don't see why the wave-nature of an atomic emission
should be ignored, and assumed to be the emission of
a particle only, and never a wave. If the "EM field"
is a useful way to describe the behavior of radio
antennas and permanent magnets, why suddenly declare
it to be "wrong" when the size of the antenna is below
a certain particular threshold? And if the emissions of
even a single atom can still produce interference
patterns (in interferometers, etc.), then how can
such things be explained in terms of single particles?

> And now, I am afraid that I have come across a reference
> that may give you encouragement. No doubt I will pay
> for this, but here goes.

Heh!

>
> "Optical Electronics," Ammon Yariv (of Cal Tech),
> Oxford University Press, 4th ed, 1991

> But then in sec 5 he points out that the solution for

> the imaginary part of the susceptibility is wrong
> and that a quantum treatment gives an answer that
> is right, based on experiment. The important difference
> being that the QM answer allows for negative imaginary
> part which allows the material to have gain and
> allows lasing action.

Is the EM solution wrong because it ignores the
quantized electron energies? Or is it wrong for
some other reason?

Whew, long message! The only response I REALLY
have interest in is the stuff directly above: the
part about EM calcs failing, while QM succeeds.

### David Green

Apr 6, 2000, 3:00:00 AM4/6/00
to
Well said, Mike.

Martin Green

"mikegi" <mik...@prestige.net> wrote in message news:sepl392...@news.supernews.com...

### mikegi

Apr 6, 2000, 3:00:00 AM4/6/00
to
> > The original poster asked a perfectly reasonable question:
> > how many cycles are in a single transition?
>
> It may be a perfectly reasonable question,
> but the question has no reasonable answer.

No. Photon theory, which is actually a macroscopic approach to interactions,
has no answer and is apparently uninterested in finding one. That's why I
believe that physics today is nothing more than engineering.

Classical em theory stumbles because it has to explain everything in detail,
both in space and time. It can't simply say that an orbiting electron is
stable and leave it at that.

Quantum theory is relieved of this burden and simply describes the end
result. Unfortunately, in doing so it runs into severe problems with
action-at-a-distance. Don't you find it strange that the many of the
originators of quantum theory (eg. Planck, Schrodinger, Einstein) consider
it a mistake and attempted to debunk it later?

Mike

### Roy McCammon

Apr 6, 2000, 3:00:00 AM4/6/00
to
Science Hobbyist wrote:
>
> In article <8ci0ac$c5p$2...@miranda.gmrc.gecm.com>,
> richard...@gecm.com wrote:
>
> > > I still must ask my original question. If an atom
> > > behaves like a dipole antenna,
> >
> > It doesn't really. The atom emits individual photons; the dipole
> > emits vast clouds of coherent photons which behave like waves.
>
> Individual photons don't behave like waves? So if I send
> my photons through a double-slit experiment INDIVIDUALLY,
> the fringe pattern which builds up will be different than
> when I send them through in vast coherent clouds?

The interference pattern is the same. The aggregate effect
of many photons whether applied one at a time or many at a
time is well modeled by an EM wave. That doesn't mean that
the photon is an EM wave or behaves like an EM wave. It
could be a hoard of little gremlins that amuse themselves
by shoving electrons around in a way that makes us think
there are EM waves.

> This is news to me. I was under the impression that even

> individual photons have wave-nature, and therefore whenever
> an atom emits a single photon, that photon has wave-nature.

It has wave nature, but not EM wave nature.

> Or from the fields-viewpoint, the atom has emitted a
> spherewave ripple in the EM field.

You can use that model in some circumstance, but in
others it will let you down.

### Roy McCammon

Apr 6, 2000, 3:00:00 AM4/6/00
to
mikegi wrote:
> Don't you find it strange that the many of the
> originators of quantum theory (eg. Planck, Schrodinger, Einstein) consider
> it a mistake and attempted to debunk it later?

When the implications became clear, Schrodinger and
Einstein opposed it for religious reasons, with the
usual results when someone tries to force science
to fit religious notions.

### mikegi

Apr 6, 2000, 3:00:00 AM4/6/00
to
> When the implications became clear, Schrodinger and
> Einstein opposed it for religious reasons, with the
> usual results when someone tries to force science
> to fit religious notions.

Normally your posts are accurate and well thought out, this one is an
exception. They opposed quantum mechanics because it has no physical basis.
Einstein's 'God does not play dice' wasn't a religious statement, it meant
that physical theories should be based on deterministic principles, not
random chance. Here's a description I found on the web:

"The success of quantum physics as a descriptive theory of large numbers of
atoms was universally accepted within the physics community. A schism
developed within the community, however, regarding the interpretation and
meaning of the theory. A number of scientists including Einstein,
Schrodinger, Planck and later, de Broglie, were unable to accept the
operation of chance within the framework of quantum mechanics. They believed
that since quantum theory could not provide a causal description of the
behaviour of individual particles that it was an incomplete theory. They
conceded that quantum mechanics was a logically consistent scheme which was
able to describe experimental results accurately. They accepted that the
limitations of measurements imposed by the uncertainty principle made a
probablistic description necessary. They believed, however, that the
uncertainty principle was only a limitation of our knowledge and that, in
actuality, the particle has both a precise position and velocity and that
its behaviour is causally determined. They could not accept the notion that
chance could actually enter into the behaviour of the physical world."

"Einstein, who became the leader of this position, expressed the concern of
his school of thought with his often quoted remark, "I do not believe in a
God that plays dice." Einstein and the others believed in the existence of
an underlying determinism which actually guided the particles. They simply
did not accept the idea that the behaviour of particles could be governed by
chance. They wanted to know what was really happening to the particles. They
believed a theory would eventually emerge in which the hidden causality
would appear. They, therefore, considered quantum theory as incomplete and
anticipated the appearance of a fuller theory which would eventually replace
it."

IMHO, belief in QM is more akin to religion than Einstein's, etc... belief
in determinism. Again, QM is nothing more than a high-level, black-box
approach to interactions.

Mike

### Bilge

Apr 7, 2000, 3:00:00 AM4/7/00
to

>
>No. Photon theory, which is actually a macroscopic approach to interactions,
>has no answer and is apparently uninterested in finding one. That's why I
>believe that physics today is nothing more than engineering.
>

While much experimental physics may be engineering, it's a bit
different than what a professional engineer would call engineering.
For example, a physicist rarely consults the ASME Duct Tape standards
prior arguing it's merits over black tape where tensile strength
and good vacuums are paramount and where traditionally there might
be some debate over whether a 1/4-20 with 4 threads engaging is
cause to use a 1/4-28, the physics of the almost any exeriment
dictates baling wire as the fastener of choice.

>Quantum theory is relieved of this burden and simply describes the end
>result.

That is an extremely naive view of quantim mechanics. QED is the
most accurate theory ever to appear. It correctly predicts, at
the most fundamental level, the magnetic moment of the electron
to 11 decimal places (since they measure g-2, it's somewhat of
a cheat to call it 13). It comes straight from the diagrams.

No theory describes anything but the end result. Every data point
is the end result of a measurement. Quantum mechanics just recognizes
that fact and tells you when the measurement significantly changes
what it was intended to measure. If the smallest quanta were
bowling balls, pointing a flashlight at someone would be
hard to do and deadly to the people around him.

Two notable examples of great value are quantum computing and
quantum cryptography.

> Unfortunately, in doing so it runs into severe problems with
>action-at-a-distance. Don't you find it strange that the many of the

Any problems quantum mechanics runs into pale by comparison
with classical descriptions. They not only contain the same
difficulties, they contain more that qm resolves. It's
not nature's fault that you don't like the way it was done.

>originators of quantum theory (eg. Planck, Schrodinger, Einstein) consider
>it a mistake and attempted to debunk it later?

You forget that they were educated with as classical a background
as one can possibly have. The entire matrix formulation is based
on the poisson bracket formalism of classical mechanics. Now days,
by the time poisson brackets are mentioned, someone in physics
would notice their similarity to commutation relations. The wave
mechanics aspects were derived from purely classical analogs with
some sr thrown in. Schrodinger essentialy took the classical
wave eqn. made it first order in time and reinterpreted the
solutions as probabilities. the dirac and klein-gordon equation
are little but replacing the variable in einsteins relativisic
expression for energy with operators. Classical mechanics or any
non-probabilistic theory, just doesnt work.

### Bilge

Apr 7, 2000, 3:00:00 AM4/7/00
to
Science Hobbyist said some stuff about
>
> Individual photons don't behave like waves? So if I send
> my photons through a double-slit experiment INDIVIDUALLY,
> the fringe pattern which builds up will be different than
> when I send them through in vast coherent clouds? This

because the fringe pattern looks like the probability density
that describes the probability of finding a photon at a given
point.

> is news to me. I was under the impression that even
> individual photons have wave-nature, and therefore whenever

In all likelyhood because you werent interested in any thing
past the point that you thought that meant the rest of the
description qa xrbon copy of your own. Phoyoms have a nature
that makes one description preferable over another to calculate
a result. Interference is strictly a result of probabilities
that were and are interpreted convenienly in many cases in
terms of classical superposition. Just because the terminology
is retained doesn't imply its use is invariant. Out of 2 possible
conventions for assigning the neutron and proton two the isospin
up and down states, both are used. Why should something more
complicated to decide have garaunteed symantics? I'm sorry if
that isn't suitable, but no matter how much you wish it, your
constrained picture is not a candidate to standardize things.

> an atom emits a single photon, that photon has wave-nature.

As someone pointed out, why do you know what the initial
wave nature is beyond the superposed probabilities of
all possible transitions? It's exactly the same result
as determining linewidths and you cant even determine what
energy is the most probable value from a single event.
Why would there be a periodic structure to it? Periodicity
implies repetition.

> Or from the fields-viewpoint, the atom has emitted a
> spherewave ripple in the EM field.

Which is wrong. Simply by stating that an atom emits a photon,
you're stuck with whatever limitations your use has imposed.
Same goes for a wave description. Your problem is you wont
stick with one long enough to figure out your casual use of
terminology and adopting what works best for the circumstances
doesn't mean there's any relationship between your personal
theory of what quantum mechanics should say and what it does.

### Bilge

Apr 7, 2000, 3:00:00 AM4/7/00
to

>

>IMHO, belief in QM is more akin to religion than Einstein's, etc... belief
>in determinism. Again, QM is nothing more than a high-level, black-box
>approach to interactions.
>

Why does every decay process work exactly the following way:

you have a particle which decays. After an interval \Delta t,
the particle either decays, or it does not. If it does not,
the probability is identical to the probability in the first
interval. Nothing at all constrains the particle from not
decaying for an arbitrarily long time, regardless. If you
put a large number of these nuclei, each one will behve this
way to give you a half-life when taken as whole. No classical
process can behave this way. It would require each particle
to determine which have decayed and communicate with the
rest. Only a probabilistic process even makes sense. That
single assumption provides all of the distinguishing features
of quantum mechanics.

### mikegi

Apr 7, 2000, 3:00:00 AM4/7/00
to
> Why does every decay process work exactly the following way:
>
> you have a particle which decays. After an interval \Delta t,
> the particle either decays, or it does not. If it does not,
> the probability is identical to the probability in the first
> interval. Nothing at all constrains the particle from not
> decaying for an arbitrarily long time, regardless. If you
> put a large number of these nuclei, each one will behve this
> way to give you a half-life when taken as whole. No classical
> process can behave this way. It would require each particle
> to determine which have decayed and communicate with the
> rest. Only a probabilistic process even makes sense. That
> single assumption provides all of the distinguishing features
> of quantum mechanics.

What causes an unstable nucleus to decay? You don't know, do you? What are
the initial conditions of the unstable nucleus and how does the process of
generating it affect them? What external stimuli is the nucleus exposed to
while you wait for it to decay? Without knowing the answers to these
questions (and many more) it seems premature to throw away a rational
explanation.

BTW, any group will interact. The question is whether the interaction is
large enough to affect the outcome.

Mike

### Richard Herring

Apr 7, 2000, 3:00:00 AM4/7/00
to
In article <8ciuap$m0e$1...@nnrp1.deja.com>, Science Hobbyist (bbe...@microscan.com) wrote:
> In article <8ci0ac$c5p$2...@miranda.gmrc.gecm.com>,
> richard...@gecm.com wrote:

> > > I still must ask my original question. If an atom
> > > behaves like a dipole antenna,
> >
> > It doesn't really. The atom emits individual photons; the dipole
> > emits vast clouds of coherent photons which behave like waves.

> Individual photons don't behave like waves?

macroscopic dipoles.

> So if I send
> my photons through a double-slit experiment INDIVIDUALLY,
> the fringe pattern which builds up will be different than
> when I send them through in vast coherent clouds?

Not at all. Where did I say that?

Each individual photon will register a single point
on your detectors. That doesn't look like the behaviour of a
wave to me. However, the probability distribution of
the locations of those points is the classical diffraction pattern.

> This is news to me.

Evidently.

> I was under the impression that even
> individual photons have wave-nature, and therefore whenever

> an atom emits a single photon, that photon has wave-nature.

First define "wave-nature". If you merel mean that the probability
distribution function for the particle has the properties
associated with a classical wave, fine. If you mean any more than
that, probably not.

> Or from the fields-viewpoint, the atom has emitted a
> spherewave ripple in the EM field.

If you surround the atom by detectors, that "spherewave ripple"
only triggers *one* of them. That's not what I would call a sphere.

--
Richard Herring | <richard...@gecm.com>

### z@z

Apr 7, 2000, 3:00:00 AM4/7/00
to
William J. Beaty wrote:

: I still must ask my original question. If an atom

: behaves like a dipole antenna, then whenever an atom
: emits "light waves", should we imagine that the
: wavetrain consists of NUMEROUS cycles? Or must this

: wavetrain consist of a very brief wave packet which
: always contains less than even a single cycle?

Your "original question" makes a lot of sense, at least according
to the epistemological point of view I adhere to.

Modern physicists often forget, that the interpretation of
photons as point-like or even virtual entities un-explains all
that what is explained by classical concepts such as concrete
electromagnetic fields. If we insist on an intuitive explanation
of photons, then we must conclude that the energy of a single
photon is present as oscillating electromagnetic fields within
a wavetrain consisting of a limited number of cycles.

The whole problem of the photon concept (in the same way as
many many other problems of Maxwell's theory and of modern
physics) is still a result of the naive materialistic premise
that instantanous actions at a distance are impossible.

The frequency of a photon is fully determined by its energy. The
assumption that an atom behaves like a dipole antenna during
emission and absorption seems untenable to me. In the Mössbauer
effect only gamma photons within a very small energy range (e.g.
+-10^-14) can be absorbed. The QED explanation that the photon
tries out all possible paths in the whole universe and therefore
recognizes that it can be absorbed by a Mössbauer absorber is
certainly less simple and obvious than the explanation based on
actions-at-a-distance:

If a photon is near enough to an atom with the corresponding
low-energy state, then the whole photon energy and momentum
can instantanously be transferred to the atom.

So the statement "a wave model cannot explain the photoelectic
effect" is not generally valid.

Also because of a lot of further evidence of actions-at-a-
distance, there is absolutely no reason why we should prefer
an explanation based on an axiomatic (theology-like) system
to this simple, concrete and fundamentally physical one.

Wolfgang Gottfried G.

### Paul Lutus

Apr 7, 2000, 3:00:00 AM4/7/00
to
> The whole problem of the photon concept (in the same way as
> many many other problems of Maxwell's theory and of modern
> physics) is still a result of the naive materialistic premise
> that instantanous actions at a distance are impossible.

This is not a presumption of modern physics. What is a presumption is that
no information may he harvested from that action. It is about information,
not events.

The EPR "paradox" shows a kind of instantaneous action at a distance. It is
observable. But you cannot exploit it to extract information.

--

Paul Lutus
www.arachnoid.com

z@z <z...@z.lol.li> wrote in message news:8cl11r$7if$1...@pollux.ip-plus.net...

### Science Hobbyist

Apr 7, 2000, 3:00:00 AM4/7/00
to
In article <8ckj53$76u$1...@miranda.gmrc.gecm.com>,

richard...@gecm.com wrote:
> In article <8ciuap$m0e$1...@nnrp1.deja.com>, Science Hobbyist
(bbe...@microscan.com) wrote:
> > >
> > > > I still must ask my original question. If an atom
> > > > behaves like a dipole antenna,
> > >
> > > It doesn't really. The atom emits individual photons; the dipole
> > > emits vast clouds of coherent photons which behave like waves.
>
> > Individual photons don't behave like waves?
>
> Oh, please. Read what I wrote. Individual atoms don't behave like
> macroscopic dipoles.

Sorry, I thought I WAS reading what you wrote. I thought
you said that atoms are not like macroscopic dipole
antennas BECAUSE macroscopic dipoles emit something wave-
like, but that single atoms do not.

Let's start again. Unless you want to give up! :)

When a single atom emits "a photon", that photon is
associated with a wavefunction which very much RESEMBLES
a classical EM wave. Agreed? We can manipulate the
wavefunction with mirrors, create interference with
beamsplitters, chop it into hunks with high-speed
shutters, etc. OK? The wavefunction often *ACTS* like
EM waves, but there are differences. Objections?

> > I was under the impression that even
> > individual photons have wave-nature, and therefore whenever
> > an atom emits a single photon, that photon has wave-nature.
>
> First define "wave-nature". If you merel mean that the probability
> distribution function for the particle has the properties
> associated with a classical wave, fine.

Yes, that's exactly what I mean. Atoms emit something
wavelike, and so do dipole antennas. Since the electrons
in a macroscopic antenna produce coherent emissions,
the macroscopic antenna acts like a giant atom which
"emits" just one wavefunction, but emits multiple photons
associated with that wavefunction. Antennas emit waves,
and so do atoms. If atoms emit "probability distributions"
rather than EM waves, then the same must be true of
macroscopic dipole antennas. Or is my reasoning screwy?
If so, why? If my original question about "EM waves
emitted by atoms" is wrong, then I want to figure out
the nature of my error.

> > Or from the fields-viewpoint, the atom has emitted a
> > spherewave ripple in the EM field.
>
> If you surround the atom by detectors, that "spherewave ripple"
> only triggers *one* of them. That's not what I would call a sphere.

Of course! But this fact cannot be used to disprove the
existence of the spherewave. That photon certainly isn't
like a flying golf ball which "has" a location while in
flight. If you sample that spherewave at many points
and lead multiple beams through an interferometer, you can
prove that the spherewave "exists", and that it has great
importance for where the photons end up.

Jeeze, it looks like I'm fighting for the idea that
atoms emit SOMETHING WAVELIKE, and you're fighting
against it! I certainly can accept that the
wavefunction has properties in addition to those of
classical EM waves. But if you are certain that atoms
don't emit "wavelike stuff", then I don't understand.

### Tom Roberts

Apr 7, 2000, 3:00:00 AM4/7/00
to
"z@z" wrote:
> Modern physicists often forget, that the interpretation of
> photons as point-like or even virtual entities un-explains all
> that what is explained by classical concepts such as concrete
> electromagnetic fields.

Not true, because photons ARE NOT "point-like". Neither the concept
"particle" nor the concept "wave" applies to photons in the usual
way. Ditto for all other quantum objects (e.g. electrons).

> If we insist on an intuitive explanation
> of photons,

There's your problem right there. Quantum mechanics is not
"intuitive" at all. The quantum world simply does not act as does
our everyday world, and that is where your "intuition" comes from.
Moreover, it is silly to expect that the quantum world would act
as does our everyday world.... IOW it is inappropriate to expect
"an intuitive explanation of photons", much less insist on one.

This is basically a generalization of the Copernican Principle:
just as the earth is not in any special location in the cosmos,
so too the length- and time-scales we humans use are not of any
special importance in the universe.

> The frequency of a photon is fully determined by its energy.

Yes, but neither is sharp. That is, for any given photon (which
is itself a misnomer as photons cannot be distinguished, so I mean
this as the set of photons in a given physical situation), neither
frequency nor energy have any specific value, and all one can do is
say they have a distribution of values. In some physical situations
this distribution can be quite wide, and in others quite narrow.

Note the intermixing of singular and plural there. In general
if one can discuss the energy or momentum of photons, one
cannot count them. English simply does not have words or
grammatical constructs which properly relate to the properties
of quantum objects. English, of course, developed in our
everyday world -- that's why mathematics is the language of
physics, and my translations here into English do not do
full justice to the theory....

> The
> assumption that an atom behaves like a dipole antenna during
> emission and absorption seems untenable to me.

I'll agree with that (:-)). A better assumption is that the atom
acts like a quantum object, and the amplitude for it to emit or
absorb a photon depends upon the overlap of their wavefunctions
(loosely speaking).

expired here?), but I'll take a stab at the question in the Subject.
The length of a "single photon" is determined by how sharply its
energy is constrained by the physical situation in which it appears.
The sharper the energy the longer the "wavetrain". But this is really
a simplistic view of what photons are, and there are subtleties. See
the "Photons, Schmotons" thread(s) in sci.physics.research for more
spinless generalization of photons).

Note that contrary to classical expectations the detector can
influence the sharpness of the energy of a photon, and therefore
its "length". A well-built single-mode laser can have a coherence
length of several meters, which is essentially the "length" of
its photons, and an incandescent lamp can have coherence lengths
comparable to the wavelengths of the emitted light (< microns).
Note that coherence length includes the properties of both emitter
and detector; for a single-mode laser the emitter is so much
narrower than the detector that the laser essentially determines
the properties of the photons, but for an incandescent lamp this
is reversed and different detectors can measure different coherence
lengths.

So the "length" of an atomically-emitted wavetrain is a rather
nebulous concept. It really depends upon both the properties of
the atomic transition involved (i.e. its width or lifetime) and
also upon _HOW_ it is observed (bandwidth of the detector).

Just in case someone did not get this from my discussion I'll say
it explicitly: I have glossed over many subtleties and details
here. The quantum world is full of such pitfalls, and even the
language we use gets in the way....

Tom Roberts tjro...@lucent.com

### Roy McCammon

Apr 7, 2000, 3:00:00 AM4/7/00
to
Tom Roberts wrote:

> There's your problem right there. Quantum mechanics is not
> "intuitive" at all. The quantum world simply does not act as does
> our everyday world, and that is where your "intuition" comes from.
> Moreover, it is silly to expect that the quantum world would act
> as does our everyday world....

It would be like expecting ants to act like ant beds.

### Tom Roberts

Apr 7, 2000, 3:00:00 AM4/7/00
to
Roy McCammon wrote:

> Tom Roberts wrote:
> > Moreover, it is silly to expect that the quantum world would act
> > as does our everyday world....
> It would be like expecting ants to act like ant beds.

Right. that reminds me of an extended metaphor on this (and many
other things), where your "ant bed" is replaced by "ant colony":

One of the characters in the book is an ant colony who communicates
via patterns of ant motion. This is not at all the central theme of
the book, which is how self-reference is important and exhibited in
the works of the three geniuses named in the title. An excellent read!

Tom Roberts tjro...@lucent.com

### mikegi

Apr 7, 2000, 3:00:00 AM4/7/00
to
> >What causes an unstable nucleus to decay? You don't know, do you?
>
> I didn't restrict the decay to nuclei.
> ...
> Nuclei aren't a very good way to view fundamental decay processes

Then why did you use it as an example? Regardless, do you know why ANY of
asking for the positions of the various decay products before, during, and
after the event.

> >questions (and many more) it seems premature to throw away a rational
> >explanation.
>

> Why is a probabilitic theory not rational? In particular, I don't believe
> it's possible to have a concept like entropy or information without any
> other type of process. Quite the contrary, it is the point of view
> requiring a cause that is irrational. A causal rather than probabilistic
> theory requires action at distance instantaneously.

I guess your viewpoint boils down to, "shit happens". No need to continue

Mike

### Science Hobbyist

Apr 8, 2000, 3:00:00 AM4/8/00
to
In article <38ED0E0E...@mmm.com>,
rbmcc...@mmm.com (Roy McCammon) wrote:

> Science Hobbyist wrote:
> >
> > In article <8ci0ac$c5p$2...@miranda.gmrc.gecm.com>,
> > richard...@gecm.com wrote:
> >
> > > > I still must ask my original question. If an atom
> > > > behaves like a dipole antenna,
> > >
> > > It doesn't really. The atom emits individual photons; the dipole
> > > emits vast clouds of coherent photons which behave like waves.
> >
> > Individual photons don't behave like waves? So if I send

> > my photons through a double-slit experiment INDIVIDUALLY,
> > the fringe pattern which builds up will be different than
> > when I send them through in vast coherent clouds?
>
> The interference pattern is the same. The aggregate effect
> of many photons whether applied one at a time or many at a
> time is well modeled by an EM wave. That doesn't mean that
> the photon is an EM wave or behaves like an EM wave.

Of course. What I'm trying and trying to say is that
SOMETHING WAVELIKE is associated with each emission of
light by an atom, so when an atom emits even a single
photon, there is something wavelike going on there.
It only becomes particle-like when a measurement is
performed.

And the same is true of a macroscopic dipole antenna.
The antenna "emits" something wavelike, even though the
antenna emits many photons while the atom emits just one.

If it's inaccurate to call this wavelike emission by the
name "EM wave", then I want to know what to call it.
If someone tells me that it doesn't exist, then I
can point to simple interference experiments which
demonstrate that it certainly does exist even for
single-photon emissions. When an atom emits a photon,
something wavelike occurs at the same time, and it's not
as if the atom is just spitting out a bullet which
"has" a trajectory.

I don't know, maybe others here imagine that when an
atom emits a photon, that photon does have a distinct
trajectory, but we don't know what it is? If so,
then that's a "hidden variables" interpretation of the
uncertainty principle, and it is false. QM does not
work that way. When a photon is on the fly, it does
not "have position." It takes many paths at once. Or
is there an error elsewhere in my understanding?

It
> could be a hoard of little gremlins that amuse themselves
> by shoving electrons around in a way that makes us think
> there are EM waves.

Certainly. And that hoard of gremlins does the same thing
for macroscopic dipole antennas that they do for single
atoms, the only difference being the number of photons
involved and the shape of the "wavelike entity." It is
not correct to believe that macroscopic antennas emit EM
waves while solitary atoms do not.

>
> > This is news to me. I was under the impression that even

> > individual photons have wave-nature, and therefore whenever
> > an atom emits a single photon, that photon has wave-nature.
>

> It has wave nature, but not EM wave nature.

Yep. What's confusing me is this: what is the difference
between an EM wave and the wavelike probability distributions
of QM physics? The behavior of EM waves must be a subset of
the behavior of wavefunctions, no? When a cellphone antenna
emits "radio waves", what's really happening is that photons
and wavefunctions are imitating "radio waves." What
can wavefunctions do which EM waves do not? How do they
differ besides the fact that photons are involved?

> You can use that model in some circumstance, but in
> others it will let you down.

They certainly let you down when EM quanta are involved.
But aside from the quantized emission and absorbtion, how
are wavefunctions different than EM waves? Or is the
presence of "photons" the only difference?

### Science Hobbyist

Apr 8, 2000, 3:00:00 AM4/8/00
to
In article <38EE2723...@lucent.com>,

Tom Roberts <tjro...@lucent.com> wrote:
> "z@z" wrote:
> > Modern physicists often forget, that the interpretation of
> > photons as point-like or even virtual entities un-explains all
> > that what is explained by classical concepts such as concrete
> > electromagnetic fields.
>
> Not true, because photons ARE NOT "point-like".

Possible terminology problem here. I've always thought
that "photon" was the pointlike particle, while "EM field"
or "wavefunction" or even "light" was the wavelike entity.
It's a similar situation with electric charge: "electron"
is the pointlike entity, while "charge" is the
uniformly distributed "stuff." If your definition is
different than mine, we might have misunderstandings.

> frequency nor energy have any specific value, and all one can do is
> say they have a distribution of values. In some physical situations
> this distribution can be quite wide, and in others quite narrow.

And if we create a light beam with narrow frequency
distribution, but then put it through a high speed
chopper, we can alter the frequency distribution of
"the photons." ( Heh. cut them up into smaller ones!)

> > The
> > assumption that an atom behaves like a dipole antenna during
> > emission and absorption seems untenable to me.
>
> I'll agree with that (:-)). A better assumption is that the atom
> acts like a quantum object, and the amplitude for it to emit or
> absorb a photon depends upon the overlap of their wavefunctions
> (loosely speaking).

I'll agree, but I'll also point out that this also applies to
a macroscopic dipole antenna in this sense: as we make a
metal antenna smaller and smaller, the wave nature of its
emissions does not suddenly change. (Or does it?)
Instead, quantization of emission and absorbtion becomes
more and more of an issue. But do other issues arise
as well?

> Note that coherence length includes the properties of both emitter
> and detector; for a single-mode laser the emitter is so much
> narrower than the detector that the laser essentially determines
> the properties of the photons, but for an incandescent lamp this
> is reversed and different detectors can measure different coherence
> lengths.
>
> So the "length" of an atomically-emitted wavetrain is a rather
> nebulous concept. It really depends upon both the properties of
> the atomic transition involved (i.e. its width or lifetime) and
> also upon _HOW_ it is observed (bandwidth of the detector).

to my question really is that the "length of the wave" is
not a correct concept, and it's incorrect to insist that the
wave "is" less than half a cycle long, but its also incorrect
to insist that it "is" many cycles long.

Things are starting to make a bit more sense.

### Bilge

Apr 8, 2000, 3:00:00 AM4/8/00
to
>
>What causes an unstable nucleus to decay? You don't know, do you? What are

I didn't restrict the decay to nuclei. Anything, but I'm excluding gravity
here since no one can say a whole heck of a lot about it in this regard.
Things decay because it's energetically advantageous. Things are hindered
when energy barriers hinder they accessibility of lower energy states.
Nuclei aren't a very good way to view fundamental decay processes, because
it involves having to distinguish between interactions. The fact that the
W,Z are heavy makes weak decays appear weaker than strong decays. The
standard model states that the existence of different couplings is not
due to having 3 intrinsic couplings and at higher temperatures there is
only one. At unification temperature, all processes have the same
interaction strength and particles become massless, so "decay process"
doesn't have a meaning that differs from "scattering process".

So, for what you see as decay processes around you, why is a cause
required? The only "cause" is due to the conservation laws that
prohibit the change of state (energy, angular momentum). Those
may be violated for a time given by the uncertainty relation.
That time is the mean decay time. What your question is asking,
is what "causes" the uncertainty relation?

>the initial conditions of the unstable nucleus and how does the process of
>generating it affect them? What external stimuli is the nucleus exposed to
>while you wait for it to decay? Without knowing the answers to these

The point is, that the decay is spontaneous. Nuclei which don't decay
don't for the sole reason they don't have sufficient energy to do so.
That prevents the neutron from decaying in the nucleus. Soon as you pull
the neutron out, it lives about 900 sec on average. The only reason it
requires 900 seconds is kinematic factors. The W is a massive particle,
so beta decay appears to occur slowly. The effect of external stimuuli
is only to raise or lower the barrier to decay, not "cause" it. In that
sense, the binding energy of the nucleus is just such an external factor.

>questions (and many more) it seems premature to throw away a rational
>explanation.
>
Why is a probabilitic theory not rational? In particular, I don't believe
it's possible to have a concept like entropy or information without any
other type of process. Quite the contrary, it is the point of view
requiring a cause that is irrational. A causal rather than probabilistic
theory requires action at distance instantaneously.

>BTW, any group will interact. The question is whether the interaction is

>large enough to affect the outcome.
>

Quantum mechanics tells you what that scale is. That's practically
the single deefining feature that separates it from classical
mechanics. Classical mechanics is quantum mechanics in the limit
where h->0. If that limit isn't valid, you can no longer make
measurements that have a negligibile effect on what you measure.
All your statement is saying is that you've affected the \Delta t
that is the characteristic time of the interaction by affecting
\Delta E. That doesn't imply that there's an underlyin process being
influenced by the interaction. It also means any light is fundamentally
a quantum process and classical E&M has to be a limit of quantum
mechanics. The people that want to create a basis for quantum
phenomena from classical theories and retain classical ideas
are the ones that have it backwards.

### jddescr...@my-deja.com

Apr 8, 2000, 3:00:00 AM4/8/00
to
In article <38EE2723...@lucent.com>,
Tom Roberts <tjro...@lucent.com> wrote:
> "z@z" wrote:
> > Modern physicists often forget, that the interpretation of
> > photons as point-like or even virtual entities un-explains all
> > that what is explained by classical concepts such as concrete
> > electromagnetic fields.
>
> Not true, because photons ARE NOT "point-like". Neither the concept
> "particle" nor the concept "wave" applies to photons in the usual
> way. Ditto for all other quantum objects (e.g. electrons).
>
> > If we insist on an intuitive explanation
> > of photons,
>
> There's your problem right there. Quantum mechanics is not
> "intuitive" at all. The quantum world simply does not act as does
> our everyday world, and that is where your "intuition" comes from.
> Moreover, it is silly to expect that the quantum world would act
> as does our everyday world.... IOW it is inappropriate to expect
> "an intuitive explanation of photons", much less insist on one.
>
> This is basically a generalization of the Copernican Principle:
> just as the earth is not in any special location in the cosmos,
> so too the length- and time-scales we humans use are not of any
> special importance in the universe.
>
> > The frequency of a photon is fully determined by its energy.
>
> Yes, but neither is sharp. That is, for any given photon (which
> is itself a misnomer as photons cannot be distinguished, so I mean
> this as the set of photons in a given physical situation), neither
> frequency nor energy have any specific value, and all one can do is
> say they have a distribution of values. In some physical situations
> this distribution can be quite wide, and in others quite narrow.
>
> Note the intermixing of singular and plural there. In general
> if one can discuss the energy or momentum of photons, one
> cannot count them. English simply does not have words or
> grammatical constructs which properly relate to the properties
> of quantum objects. English, of course, developed in our
> everyday world -- that's why mathematics is the language of
> physics, and my translations here into English do not do
> full justice to the theory....
>
> > The
> > assumption that an atom behaves like a dipole antenna during
> > emission and absorption seems untenable to me.
>
> I'll agree with that (:-)). A better assumption is that the atom
> acts like a quantum object, and the amplitude for it to emit or
> absorb a photon depends upon the overlap of their wavefunctions
> (loosely speaking).
>
> I have not seen any previous messages of this thread (already
> expired here?), but I'll take a stab at the question in the Subject.
> The length of a "single photon" is determined by how sharply its
> energy is constrained by the physical situation in which it appears.
> The sharper the energy the longer the "wavetrain". But this is really
> a simplistic view of what photons are, and there are subtleties. See
> the "Photons, Schmotons" thread(s) in sci.physics.research for more
> details, and also any threads there about "Fotons" (which are a
> spinless generalization of photons).
>
> Note that contrary to classical expectations the detector can
> influence the sharpness of the energy of a photon, and therefore
> its "length". A well-built single-mode laser can have a coherence
> length of several meters, which is essentially the "length" of
> its photons, and an incandescent lamp can have coherence lengths
> comparable to the wavelengths of the emitted light (< microns).
> Note that coherence length includes the properties of both emitter
> and detector; for a single-mode laser the emitter is so much
> narrower than the detector that the laser essentially determines
> the properties of the photons, but for an incandescent lamp this
> is reversed and different detectors can measure different coherence
> lengths.
>
> So the "length" of an atomically-emitted wavetrain is a rather
> nebulous concept. It really depends upon both the properties of
> the atomic transition involved (i.e. its width or lifetime) and
> also upon _HOW_ it is observed (bandwidth of the detector).
>
> Just in case someone did not get this from my discussion I'll say
> it explicitly: I have glossed over many subtleties and details
> here. The quantum world is full of such pitfalls, and even the
> language we use gets in the way....
>
> Tom Roberts tjro...@lucent.com
>

----------------------------------------------------------------------

I think if you invested as much good effort into trying to understand
science as you do in alibing for the deliberate deceptions of the KMS
[King's Men of Science] phony, balony stories about why they should be
richly "funded" even though they don't 1) know what they are talking
about and 2) know where their magic game starts and stops then you
would make some important truth telling progress. Dennis McCarthy has
given you many historical examples of similar huge misunderstandings
and inabilities to explain by material cause and effect in the past.
Why do you think these current magic, mysteries in quantum mechanics
and relativity and such will not be resolved in similar ways?

Good seeing. JD

----------------------------------------------------------------------

### Bilge

Apr 8, 2000, 3:00:00 AM4/8/00
to
>
>Then why did you use it as an example? Regardless, do you know why ANY of

So that I had something familiar. I didn't intend to start a discussion
that required going back to the big bang just to define the terminology.

>asking for the positions of the various decay products before, during, and
>after the event.
>

You get positions and momenta of the final states EXACTLY from their
relationship given by distribution you get from the decay energy.
That decay energy is the same energy that determines the lifetime.

>I guess your viewpoint boils down to, "shit happens". No need to continue

At least I can make verifiable predictions with what you call
a "shit happens" theory. That's preferable to a theory for
which the only know result is no one can predict anything unless
someone finds out conclusively why "shit happens". You asked.
I'm sorry if you don't like the answer. I can't make you like it
any more than I can predict when an electron will go from the
2p-1s level in an atom. At least the atom lends itself to a
probabilistic answer from a rational premise.

If you think there is more, why am I explaining something that agrees
with observation while you cant even postulate something that
describes a single observation so that it isn't ruled out by
other observations?

### Bilge

Apr 8, 2000, 3:00:00 AM4/8/00
to
Science Hobbyist said some stuff about

> Of course. What I'm trying and trying to say is that

> SOMETHING WAVELIKE is associated with each emission of
> light by an atom, so when an atom emits even a single
> photon, there is something wavelike going on there.
> It only becomes particle-like when a measurement is
> performed.

How do you make an experiment show your "SOMETHING" is
"wave-like"? If you cant, what exactly are you arguing?

> If it's inaccurate to call this wavelike emission by the
> name "EM wave", then I want to know what to call it.

It's inacurrate to use what you seem to think ang E&M wave
is.

> demonstrate that it certainly does exist even for
> single-photon emissions. When an atom emits a photon,

So. If I show you a screen on which one photon has
left a bright point, you can tell whether or not it
went through a slit? Can you also determine how many
slits? If I looked a a hundred pictures, each with one
bright point corresponding to a single photon, I'd have
a hard enough time telling them apart, nevermind what
kind of experiment produced each photagraph.

> something wavelike occurs at the same time, and it's not
> as if the atom is just spitting out a bullet which
> "has" a trajectory.
>

More like an atom probably spitting out a bullet along a trajectory.
And probably along another. And probably along another. But definitely
spitting out a bullet, probably at time t_0. And definitely spitting
out a bullet, probably at time t_1. And Defintely spitting out a
bullet at a time that is definitely between the time I started waiting
and the rest of eternity. By a wise choice of apparatus, one
can choose "probably" well enough to make a lightbulb.

> trajectory, but we don't know what it is? If so,
> then that's a "hidden variables" interpretation of the

No it isn't. A hidden variable implies you could predict
the result if you knew the hidden variable.

> uncertainty principle, and it is false. QM does not
> work that way. When a photon is on the fly, it does
> not "have position." It takes many paths at once. Or
> is there an error elsewhere in my understanding?

At least the statement "QM does not work this way" is correct.
The other errors may be found by removing that statement.

> Certainly. And that hoard of gremlins does the same thing
> for macroscopic dipole antennas that they do for single
> atoms, the only difference being the number of photons

well-known, common experiments.

>
> Yep. What's confusing me is this: what is the difference
> between an EM wave and the wavelike probability distributions

In your case, the interpretation. Do you know what "overconstrained"
means with respect to imposing too many boundary conditions? If,
not, you are an example. The boundary conditions you seem to
think define a logical situation don't permit a solution.
Hence, you will never get an answer, much less the one you
want. On the plus side, that means your answer is as good as
any, as you've suspected. Just don't expect us to go along.

> of QM physics? The behavior of EM waves must be a subset of
> the behavior of wavefunctions, no? When a cellphone antenna
> emits "radio waves", what's really happening is that photons

It's not hard to explain, but what's the point of just providing
you something else to confuse you?

>
> They certainly let you down when EM quanta are involved.

No. It let's you down. Some people know that equivalent physical
descriptions doesnt imply equally illuminating and tractable
results for any situation.

> But aside from the quantized emission and absorbtion, how
> are wavefunctions different than EM waves? Or is the
> presence of "photons" the only difference?

Well, the E&M wave carries energy. The wavefunction tells you
the probablity that you'll find the energy at some point in
space. If you don't like this, argue heisenberg via a seance.
He died without leaving "why" in the margin of goldstein.

### Bilge

Apr 8, 2000, 3:00:00 AM4/8/00
to
Science Hobbyist said some stuff about

> not a correct concept, and it's incorrect to insist that the

> wave "is" less than half a cycle long, but its also incorrect
> to insist that it "is" many cycles long.

No. It's incoreect to say anything in terms of cycles. Cycles implies
repitition. repition rates cycles. Do you see any repition involved
with a single photon? Is there a 12*pi long sinewave that lands in
the spectrometer? Times may imply energy, but it doesn't imply cycles.
The energy in multiple cycles of something, in fact the number cycles
x the energy in a single "cycle". So, a 3 "cycle" 1 MHz photon
of energy 2*pi*1x10^6 ev-nm/c has 3 times that much energy, right?
Doesn't that sound a bit off-the wall?

Just once, before continuing to try and pry the answer you want
by being deliberately obtuse, so you can (deceptively) recast it
later to support (to the surprise of your "supporter") your
opinion of what an electromagnetic wave is and why the rest of
the world is wrong, why don't you take a pen and paper and draw
what you're asking. multiply a few numbers and see if the question
makes sense. What does it mean to have something with a period of
1 sec, that's 3 seconds long, emitted every 30 seconds? What does
it mean if it's emitted every 1 usec? (besides the second case
being impossible, strictly speaking).

recruiting people from another newsgroup in the hope you can
take advantage of some person that hasn't seen the weeks
of posts indicating the futility of the idea. If you hit enough
newsgroups, I suppose someone will unwittingly give you a
phrase you can twist. If you dont want to make the effort to

### Ken H. Seto

Apr 8, 2000, 3:00:00 AM4/8/00
to
On Sat, 08 Apr 2000 01:41:21 GMT, Science Hobbyist
<bbe...@microscan.com> wrote:

>In article <38EE2723...@lucent.com>,
> Tom Roberts <tjro...@lucent.com> wrote:
>> "z@z" wrote:

>> frequency nor energy have any specific value, and all one can do is
>> say they have a distribution of values. In some physical situations
>> this distribution can be quite wide, and in others quite narrow.
>

> And if we create a light beam with narrow frequency
> distribution, but then put it through a high speed
> chopper, we can alter the frequency distribution of
> "the photons." ( Heh. cut them up into smaller ones!)

That's exactly what Aspect did in his experiment to determine the
correlation between two photons separated by a distance. He switch
the photon from the horizontal polarizer to the vertical polarizer at
a rate of 200 million times per second. The switching operation cut
the photons to pieces and thus gave a much higher correlation than
expected. This higher rate of correlation is interpreted as FTL action
at a distance.

Ken Seto

### David Evens

Apr 9, 2000, 3:00:00 AM4/9/00
to
In article <38ef3dad$0$25...@news.voyager.net>, ken...@erinet.com says...

Obviously he didn't conclude this, since he obviously didn't get that result.
All experiments of this type do is detect coupling effects between states of
particles that were created or forced into coupled states. Unless the
particles are put into different states, then they will remain in the coupled
states, even if they move on seperate paths.

### David Green

Apr 9, 2000, 3:00:00 AM4/9/00
to

Bill Beatty wrote:

> > Or from the fields-viewpoint, the atom has emitted a
> > spherewave ripple in the EM field.
>

Richard Herring replied:

> If you surround the atom by detectors, that "spherewave ripple"
> only triggers *one* of them. That's not what I would call a sphere.
>

Yes, this would be proof that light consisted of photons,
if it really happened this way. Of course, this experiment
has never been done.

Martin Green

### Roy McCammon

Apr 9, 2000, 3:00:00 AM4/9/00
to
Science Hobbyist wrote:
>
> In article <38ED0E0E...@mmm.com>,
> rbmcc...@mmm.com (Roy McCammon) wrote:

> Of course. What I'm trying and trying to say is that
> SOMETHING WAVELIKE is associated with each emission of
> light by an atom, so when an atom emits even a single
> photon, there is something wavelike going on there.
> It only becomes particle-like when a measurement is
> performed.
>

> And the same is true of a macroscopic dipole antenna.
> The antenna "emits" something wavelike, even though the
> antenna emits many photons while the atom emits just one.
>

> If it's inaccurate to call this wavelike emission by the
> name "EM wave", then I want to know what to call it.

The effect that the macroscopic antenna has on other
conductors
is well modeled by classical maxwell em theory. In that
sense
it is accurate to say it emits an em wave. Likewise, that
effect
is well modeled by QED and it is likewise accurate to say
the
antenna emits photons. Single atoms, in general, are not
well
modeled by classical em waves so it is not accurate to sat
that
they emit em waves. They are well modeled by QED so it is
accurate to say they emit QED photons.

> I don't know, maybe others here imagine that when an
> atom emits a photon, that photon does have a distinct

> trajectory, but we don't know what it is? If so,
> then that's a "hidden variables" interpretation of the

> uncertainty principle, and it is false. QM does not
> work that way. When a photon is on the fly, it does
> not "have position." It takes many paths at once. Or
> is there an error elsewhere in my understanding?

You appear to believe that the photon has an existence
between the moment of emission and the moment of absorption.
There is, of course, no evidence of that. QED is rather
silent on that. The equations are consistent with
interpretations such as "it is as if the photon goes
everywhere at all possible speeds both greater than and
less than the speed of light and interferes with itself".
That doesn't sound very wavelike, does it (except for the
interfering part).

> It is
> not correct to believe that macroscopic antennas emit EM
> waves while solitary atoms do not.

But it is correct to believe that macroscopic antennas
are well modeled by classical maxwell em waves while
solitary atoms are not.

> > It has wave nature, but not EM wave nature.
>

> Yep. What's confusing me is this: what is the difference
> between an EM wave and the wavelike probability distributions

> of QM physics?

Good question, but I want to check some references

> The behavior of EM waves must be a subset of
> the behavior of wavefunctions, no?

A waves and wave functions are nothing more than a
solutions of a wave equation. In the sense that
"wave equations" have something in common, then so
do all waves.

> When a cellphone antenna
> emits "radio waves", what's really happening is that photons

> and wavefunctions are imitating "radio waves."

I'd say "QED photons" are imitating EM waves.

> What
> can wavefunctions do which EM waves do not? How do they
> differ besides the fact that photons are involved?

QED photon theory can be used to accurately model
atomic interactions.

> > You can use that model in some circumstance, but in
> > others it will let you down.
>

> They certainly let you down when EM quanta are involved.

> But aside from the quantized emission and absorbtion, how
> are wavefunctions different than EM waves? Or is the
> presence of "photons" the only difference?

QED photons belong to a wave function that is different
from the EM wave function.

### Maarten Hoogerland

Apr 10, 2000, 3:00:00 AM4/10/00
to

What has been done is measure the recoil of the atom from emitting that
photon. And yes, the atom does recoil in a particular direction. That
should be equivalent. If you do the experiment many times, you recover
the spherical probability distribution.

Cheers,

Maarten Hoogerland

### Maarten Hoogerland

Apr 10, 2000, 3:00:00 AM4/10/00
to

What is a QED photon?

Not much.

>
> > The behavior of EM waves must be a subset of
> > the behavior of wavefunctions, no?
>
> A waves and wave functions are nothing more than a
> solutions of a wave equation. In the sense that
> "wave equations" have something in common, then so
> do all waves.
>
> > When a cellphone antenna
> > emits "radio waves", what's really happening is that photons
> > and wavefunctions are imitating "radio waves."
>
> I'd say "QED photons" are imitating EM waves.
>
> > What
> > can wavefunctions do which EM waves do not? How do they
> > differ besides the fact that photons are involved?
>
> QED photon theory can be used to accurately model
> atomic interactions.
>
> > > You can use that model in some circumstance, but in
> > > others it will let you down.
> >
> > They certainly let you down when EM quanta are involved.
> > But aside from the quantized emission and absorbtion, how
> > are wavefunctions different than EM waves? Or is the
> > presence of "photons" the only difference?
>
> QED photons belong to a wave function that is different
> from the EM wave function.

??? What the heck does that mean?

we usually confine our universe to a box. Then, we say that the field on
the sides of the box has to be zero, or some variation of that. That
means that the radiation can only exist for certain modes, with an
integer number of waves fitting in the box. Most of the time we let the
size of the box go to infinity at some point, but there is still a
countable number of modes. Each of those modes is treated as a QM
harmonic oscillator, with equally spaced energy levels. Now the zeroth
energy level still has some excitation in it, and fluctuations can give
rise to a photon in this mode. This means that even the vacuum field (no
light at all) has fluctuations. This correction solves most problems
that we have with Maxwell's equations. In equation form, this is much
easier to explain, but I did my best.

Cheers,

Maarten

### Bilge

Apr 10, 2000, 3:00:00 AM4/10/00
to
Maarten Hoogerland said some stuff about

>rise to a photon in this mode. This means that even the vacuum field (no
>light at all) has fluctuations. This correction solves most problems
>that we have with Maxwell's equations. In equation form, this is much
>easier to explain, but I did my best.
>

As an additoional point to anyone that thinks qm is somehow less
"physical" or by its nature (rather than just personal experience)
than classical E and B fields, the fields are only observed through
intensities, which are identical to the mod squared of the wf. Which
is to say that the people who have it backwards are those that treat
fields as being something which has no abstraction and which has a
nice obvious classical picture while treating the observale prob-
abilities of an intensity at some point in space as being unphysical.
I'll save the next question for someone with a firm handle of what
the "real" field looks like, since it's a good bet little oscillating
arrowrs whizzing by is only an abstraction....

### Mark Samokhvalov

Apr 10, 2000, 3:00:00 AM4/10/00
to

Tom Roberts пишет в сообщении <38EE2723...@lucent.com> ...

>"z@z" wrote:
>> Modern physicists often forget, that the interpretation of
>> photons as point-like or even virtual entities un-explains all
>> that what is explained by classical concepts such as concrete
>> electromagnetic fields.
>
>Not true, because photons ARE NOT "point-like". Neither the concept
>"particle" nor the concept "wave" applies to photons in the usual
>way. Ditto for all other quantum objects (e.g. electrons).
>
>
>> If we insist on an intuitive explanation
>> of photons,
>
>There's your problem right there. Quantum mechanics is not
>"intuitive" at all. The quantum world simply does not act as does
>our everyday world, and that is where your "intuition" comes from.
>Moreover, it is silly to expect that the quantum world would act
>as does our everyday world.... IOW it is inappropriate to expect
>"an intuitive explanation of photons", much less insist on one.
>
>This is basically a generalization of the Copernican Principle:
>just as the earth is not in any special location in the cosmos,
>so too the length- and time-scales we humans use are not of any
>special importance in the universe.
>
>
>> The frequency of a photon is fully determined by its energy.
>
>Yes, but neither is sharp. That is, for any given photon (which
>is itself a misnomer as photons cannot be distinguished, so I mean
>this as the set of photons in a given physical situation), neither
>frequency nor energy have any specific value, and all one can do is
>say they have a distribution of values. In some physical situations
>this distribution can be quite wide, and in others quite narrow.
>

The details of the statements You make here could well be discussed, using
the following example.
An oscillator generates continuous 10 Mhz signal, which through a chopper
and a potentiometer is fed to an antenna, say, of 100 Ohm radiative
resistance. The minimum energy produced by a metallic potentiometer would be
about 1.10^(-38) J, corresponding to an electron transition in the metal's
conduction band. Actuate the chopper to produce 1 ms pulses (10^4 cycles).
The energy of the respective photon would be calulated as hnu = 6.6 x
10(^-27)J. This would require the antenna to radiate the power of about
10(^-23) W, which it would do with a voltage of ~ 3 x 10(^-11) V applied to
it. Now, therre's a set of questions for You to answer:
a) Could smaller voltages be applied?
b) If yes, would the antenna stop radiating?
c) If not, what would it radiate?
The next thing would be to try to receive the signals. The mean energy
received some distance away from the source would be smaller than the
radiated. There would be noise problems, and they could be solved, using a
set-up consiting of a receiver with a band width sufficient to pass the
pulse, amplifier and synchronous detector with its phase set by a signal
coming from the oscillator via an electric connection and signal stacking
integrator- I, myself, was able to measure 50 KHz signals of 10(^-11) V
across a 300 Ohm resistor at room temperature. And some new questions:
a) D'You expect the signals to be received?
b) If yes, will their repetion rate be the same as emitted, or less?
I, myself, haven't the slightest doubt, that sub-quantum signals can be
radiated just as easily, as regular photons. And by now, there's adequate
proof of it, even with light. Just to mention a series of experiments made
in the seventies by independent French teams with ultra-weak light said to
consist of individual photons. This light was divided by a semi-transparent
mirror, and the two beams were directed by additional mirrors on to a
photomultiplier. It was established, that neither beam by itself produced a
response, whereas from the photon theory one would expect a two-fold
decrease in the response rate as compared to that from the undivided beam.
With both beams ariving at the multiplier, a regular path-dependent
intrerference pattern was obtained, which means that both beams displayed
typical e/m wave behaviour (transmission, deflection, diffraction,
interference), except that each was unable by itself to cause optical
transitions. All sorts of weird explanations were advanced to account for
the effect, including such idiotic ones as a probability wave propagating
ahead of photons with v>c. Apparantly, no one had the pluck to voice the
only rational explanation: that passing through the semi-transparent mirror
produced waves of the same frequency, but with an energy E<hnu
(sub-photons). This would not violate QM, because interaction of the wave
with the mirror's electrons doesn't require portions of ~1 eV - it'll go
ahead with portions ~ 10(^19) times less. Other experiments with chopping
light have also been mentioned.
This means, that Planck's law is valid for what it was originally intended -
obey it.
This indivisible photon idol is another case of thoughtless face-saving
(for a particular interested group) superstion quite frequent not just in
modern physics, not just in modern science, but throughout modern life.

### Richard Herring

Apr 10, 2000, 3:00:00 AM4/10/00
to
In article <8clplt$p23$1...@nnrp1.deja.com>, Science Hobbyist (bbe...@microscan.com) wrote:
> In article <8ckj53$76u$1...@miranda.gmrc.gecm.com>,
> richard...@gecm.com wrote:
> > In article <8ciuap$m0e$1...@nnrp1.deja.com>, Science Hobbyist
> (bbe...@microscan.com) wrote:

[snip the details; the real point is this:]

> Jeeze, it looks like I'm fighting for the idea that
> atoms emit SOMETHING WAVELIKE, and you're fighting
> against it! I certainly can accept that the
> wavefunction has properties in addition to those of
> classical EM waves. But if you are certain that atoms
> don't emit "wavelike stuff", then I don't understand.

I'm trying to talk only about those aspects of the theory
which have measurable consequences. The only events in the life
of the photon that can be measured are its creation and destruction.
The probabilities of those events are described by a wavefunction;
the "life" of the photon is not.

That may seem unsatisfactory, and you may want to construct
some more concrete description of what happens in between.

But once you do that, you're constructing not a theory
but an interpretation. It may or may not be valid, but it can't
be proved or falsified by the measured facts.

--
Richard Herring | <richard...@gecm.com>

### David Green

Apr 10, 2000, 3:00:00 AM4/10/00
to

Richard Herring wrote:
> >
> > > If you surround the atom by detectors, that "spherewave ripple"
> > > only triggers *one* of them. That's not what I would call a sphere.
> > >

Martin Green replied:

> >
> > Yes, this would be proof that light consisted of photons,
> > if it really happened this way. Of course, this experiment
> > has never been done.
> >
> >

Marten Hoogerland replied:

>
> What has been done is measure the recoil of the atom from emitting that
> photon. And yes, the atom does recoil in a particular direction. That
> should be equivalent. If you do the experiment many times, you recover
> the spherical probability distribution.
>

This experiment is easier said than done. I don't mean to call anyone a
liar, but measuring the recoil of a single atom sounds practically
impossible.

Of course, to prove the event was really a photon emission and not
just an ordinary e-m wave, the experiment would have to be carried
out in the complete absence of any external light sources. Because
when the transition is driven, or stimulated, by an external source,
there will be an atomic recoil even in the case of a classical wave.

There are all kinds of "thought-experiments" that "prove" the
existence of photons, but very few real ones.

Martin Green

### Bilge

Apr 10, 2000, 3:00:00 AM4/10/00
to
Mark Samokhvalov said some stuff about

>An oscillator generates continuous 10 Mhz signal, which through a chopper
>and a potentiometer is fed to an antenna, say, of 100 Ohm radiative
>resistance. The minimum energy produced by a metallic potentiometer would be
>about 1.10^(-38) J, corresponding to an electron transition in the metal's
>conduction band. Actuate the chopper to produce 1 ms pulses (10^4 cycles).
>The energy of the respective photon would be calulated as hnu = 6.6 x
>10(^-27)J. This would require the antenna to radiate the power of about
>10(^-23) W, which it would do with a voltage of ~ 3 x 10(^-11) V applied to
>it. Now, therre's a set of questions for You to answer:

I dont see the intent here. How about separating this into two
parts. First, just assume the that practical considerations
aren't an issue and phrase it in terms of what this is designed
to show. Once what it's intended to show is clearer, then the
praactical considerations can be thrown in. However I see little
point in checking what the ability of the latest lock-in's or
whatever are until I know what the exact restrictions will be.
That requires knowing what I'm trying to show, not simply some
circumstances.

>c) If not, what would it radiate?

If you can accelerate a charge, you can get it to radiate. Can you
accelerate a charge? I don't know. Let's check. You need an energy

E = h\nu = 2*pi*[6.57 * 10^-16 eV-sec]*50000

= 2.7*10^-10 eV

So, it would appear you are underfunded at 10^-11 volts. But that isn't
quite correct. Your applied potential difference acts to accelerate
an electron and the radiation occurs at a collision. The normal situation
is that you are interested in the average so that on average, you need
to supply the entire difference over the thermal energy of the conduction
electron. The thermal energy is very large by comparison and despite
being random, your antenna sits in a heat bath, so there is no need
to leave an energy deficit deficit. So, you really should only decrease
the efficiency by the factor that the solid angle for scatters of the
energy needed is decreased by the deficit. It just means that you'll
have to wait longer for the emission because fewer scatters will produce
the correct energy vs the background. I would expect there to be
an energy tail as well.

>integrator- I, myself, was able to measure 50 KHz signals of 10(^-11) V
>across a 300 Ohm resistor at room temperature. And some new questions:
>a) D'You expect the signals to be received?

If transmitted, I expect them to be received. Since I think they
are transmitted, the real question is finding them in the noise.

>b) If yes, will their repetion rate be the same as emitted, or less?

I have to say, that depends:

+--+ +--+ +--+ +--
| | | | | | |
--+ +----+ +----+ +----+

Which is the repetition and which is the signal? Based
soley on the statistical considerations above, the lower
energy (freq) component will be reduced further.

>transitions. All sorts of weird explanations were advanced to account for
>the effect, including such idiotic ones as a probability wave propagating
>ahead of photons with v>c. Apparantly, no one had the pluck to voice the
>only rational explanation: that passing through the semi-transparent mirror

I'll have to read the original to see what was actuall done in
detail. Have a reference?

>produced waves of the same frequency, but with an energy E<hnu

No. What it would appear to be is the energy spread caused by the
diffraction being seen at the repition rate. It sounds like what
is usually called quantum beats, but I would need to see all of
the details to make any statement about what it is or isn't.

>This means, that Planck's law is valid for what it was originally intended -
>obey it.

I don't right off hand know how you can state that. What is the
difference between E = hw and E = h'w'? where h' is arbitrary
in inelastic scattering? In principle, by looking at the angular
cutoff in the scattering, but your antenna scenario has further
difficulty from thermal motion and I havent seen the experiments
to which you refer.

>This indivisible photon idol is another case of thoughtless face-saving
>(for a particular interested group) superstion quite frequent not just in
>modern physics, not just in modern science, but throughout modern life.

I think perhaps this declaration is premature.

### Bilge

Apr 10, 2000, 3:00:00 AM4/10/00
to
David Green said some stuff about

>
>There are all kinds of "thought-experiments" that "prove" the
>existence of photons, but very few real ones.
>

I personally prefer the approach that something took place
and various different measurements have produced a set of
self-consisten attributes that let me choose the description
that gives me the right answer for the least effort. One
thing about long calculations -- each step over and above a
shorter, equivalent, simpler one is an opportunity to
make a mistake you didn't have with a shorter calculation.

### Tom Roberts

Apr 10, 2000, 3:00:00 AM4/10/00
to
Science Hobbyist wrote:
> I've always thought
> that "photon" was the pointlike particle,

That's an overly-simplistic interpretation of the way photons appear
in a perturbative approximation to QED, and it's not the whole story.
In such an approximation one must integrate each vertex of each
Feynman diagram over all posible spacetime positions (equivalently
over all possible 4-momenta), and it's rather difficult to think of
such an integral as "pointlike"!!! Especially when applied to _both_
ends of the "photon"! Yes the _vertex_ is evaluated at a single point
within the integral, but that is not really interpretable by itself.

IOW: a "pointlike" particle ought to have a definite position in
space, but both ends of a photon are integrated over all spacetime.
This is what is known as a "sum over histories" inb= is the path
integral approach to QED.

> It's a similar situation with electric charge: "electron"
> is the pointlike entity, while "charge" is the
> uniformly distributed "stuff."

I disagree. I would say that "electron" is the particle and "charge"
is the property of a particle which couples to the E&M interaction.
Nobody has ever observed any sort of "uniformly distributed" charge,
it is only observed as attributes of particles. But "pointlike"
applied to electrons and other particles has the same difficulties
as for photons above. Note, however, that electron scattering at
high energies indicates that the effective radius of an eletron is
smaller than current resolutions (well under 1 Fermi).

> And if we create a light beam with narrow frequency
> distribution, but then put it through a high speed
> chopper, we can alter the frequency distribution of
> "the photons." ( Heh. cut them up into smaller ones!)

Sure. You still have difficulty counting them....

> > So the "length" of an atomically-emitted wavetrain is a rather
> > nebulous concept. It really depends upon both the properties of
> > the atomic transition involved (i.e. its width or lifetime) and
> > also upon _HOW_ it is observed (bandwidth of the detector).

> AH! This is something I didn't know about. So the answer
> to my question really is that the "length of the wave" is

> not a correct concept, and it's incorrect to insist that the
> wave "is" less than half a cycle long, but its also incorrect
> to insist that it "is" many cycles long.

As I said, it depends upon the entire physical situation in which
the photons appear. Some situations contain photons with an effective
length of many cycles, and other situations may only have photons
with effective lengths less than one cycle long.

Note that a light ray can have a wave which is very long. The
problems arise when one attempts to ascribe a length to a photon.

Tom Roberts tjro...@lucent.com

### Peter Somlo

Apr 10, 2000, 3:00:00 AM4/10/00
to
r...@gmrc.gecm.com (Richard Herring) wrote:

>I'm trying to talk only about those aspects of the theory
>which have measurable consequences. The only events in the life
>of the photon that can be measured are its creation and destruction.
>The probabilities of those events are described by a wavefunction;
>the "life" of the photon is not.
>
>That may seem unsatisfactory, and you may want to construct
>some more concrete description of what happens in between.
>
>But once you do that, you're constructing not a theory
>but an interpretation. It may or may not be valid, but it can't
>be proved or falsified by the measured facts.
>
>--
>Richard Herring | <richard...@gecm.com>

------------------------------------------------------
Hello Richard,
With reference to the above, I have often contemplated the fact that
according to relativity theory, for anything that moves at the speed
of light, time stops. From a photon's point of view, it gets launched,
and it gets absorbed - with no time in between.
Peter
_____________________________________________________________________
Dr.Peter I Somlo FIEEE | M1: "Every coin has 3 sides - at least"
Microwave Consultant | email: so...@ieee.org
tel/fax: 61-2-9451-2478| ICQ: 1032408
Mobile(AU):041-926-3168| <http://www.zeta.org.au/~somlo/default.htm>

### Maarten Hoogerland

Apr 11, 2000, 3:00:00 AM4/11/00
to
> There are all kinds of "thought-experiments" that "prove" the
> existence of photons, but very few real ones.
>
> Martin Green

There is a large recent range of experiments that is classified as "atom
optics". Do a web search on that and see what you find.
A real proof for the existence of photons is that the light field can
have sub-Poissonian statistics, i.e. the variation in measured photon
number is less than Poisson statistics. If you look at the counting
statistics of a classical, coherent laser field, which is as quiet as it
classically gets, it has Poissonian photon statistics, i.e. the spread
in the detected number goes as the square root of the number. Light
fields that are less noisy than that can be generated and observed,
another recent line of research of "quantum optics". Try a web search
and see what you find.

Cheers,

Maarten