Re: Two-slit experiment
- From: "Ron Baker, Pluralitas!" <stoshu@xxxxxxxxxxxxxxxx>
- Date: Wed, 09 Aug 2006 04:30:20 GMT
"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote in message
news:Pine.LNX.4.50.0608081301090.23120-100000@xxxxxxxxxxxx
On Tue, 8 Aug 2006, Ron Baker, Pluralitas! wrote:
"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:
On Sun, 6 Aug 2006, Ron Baker, Pluralitas! wrote:
There is nothing quantized there. delta_E is continuously variable.
Yes, delta_E = h delta_f is continuously variable. The point is that
E=hf
is removed from the monochromatic field of frequency f,
Why? Why do you say that?
Experimental and theoretical support for the idea.
Maybe I'm wrong. Maybe you can explain it to
me. I think there are lurkers following this who would
find this educational too.
For example: photoelectric effect,
Seems to me that it is the work function of the metal (wfm)
that is quantizing the energy level and not an intrinsic
characteristic of the EM field that is quantized.
Certainly if the hf of the photon is equal to the work
function then the photon must give up the whole
hf to liberate the electron. But I don't see anything
in the results of the experiment that precludes a photon
with Ep0 = hf0 > wfm from giving wfm to the electron and
being left with Ep1 = hf0 - wfm (giving f1 = f0 - wfm/h).
(Note that I am not saying that all the photons would
only give up wfm of their energy. They could give anywhere
between wfm and hf0.)
Compton scattering, atomic line
This is precisely the experiment that makes it appear that
a photon can give up part of its hf. A photon goes in
with hf0 and a photon comes out with hf1 where the energy
imparted to the electron is hf0 - hf1. Your contention is
that the photon in and out are different photons.. that the
first is completely absorbed. That seems arbitrary and unnecessary.
spectra, the blackbody spectrum, coincidence rates for low intensity
frequency downconversion, energy of gamma rays following matter-antimatter
annihilation, photoionisation spectra, probability of two-photon
processes, two-step photoionisation, and photomultiplier tube spectral
response, for starters.
Every single one of these involves matter (what else is an EM field going
to exchange energy with?). Many of them involve an initial continuum
energy state and/or a final continuum energy state as far as the matter is
concerned.
To be pertinent the allowed energy level of the matter portion
would have to not be quantize both before and after the interaction.
Would you like to pick one and show how the view that
a photon can give up part of its hf contradicts
experimental observation? Compton scattering seems
like the natural choice.
E = hf for monochromatic fields is compatible with every single one of the
above. IMO opinion, that the radiation field itself is quantised is the
simplest explanation, else one needs to explain why so many different
systems behave in exactly the same way.
[moved from further down due to listing of Compton scattering above]
Ever hear of Compton scattering?
Yes. And it supports E=hf very well. E_in = h f_in, E_out = h f_out.
It doesn't support it. It doesn't disallow it, but it doesn't
support it. It supports E_in = h (f_in - f_out) just
as well. Do you have a counter argument?
Conservation of momentum. If the electron just absorbs a "partial photon"
with zero energy emitted at f_out (this is what you meant?),
No. And I see my statement was not adequately clear.
That 'E_in' above was E into (or transfered to) the electron.
Let's see if this is more clear.
Define Ee0 as energy of the electron before interaction.
Ee1 is after the interaction. Similarly Ep0: energy of photon
before, Ep1: energy of photon after. f0: frequency of photon
before. f1: frequency of photon after.
Planck's law:
Ep0 = h f0 (1)
Ep1 = h f1 (2)
Conservation of energy:
Ee1 - Ee0 = Ep0 - Ep1
From (1) and (2)Ee1 - Ee0 = Ep0 - Ep1 = h (f0 - f1) (3)
Call
delta_E = Ee1 - Ee0 (4)
delta_E is the energy transfered to the electron
from the photon.
From (3) and (4)delta_E = h (f0 - f1) (5)
Call
f0 - f1 = delta_f (6)
From (5) and (6)delta_E = h delta_f
From Compton scattering:delta_E is continuosly variable and not
quatized. 0 <= delta_E <= hf0
then momentum
is not conserved.
The correspondence principle. In classical electrodynamics, a free
electron can't just absorb energy from a monochromatic field without
re-radiating.
Hmm. But is that monochromatic field the field of a
population? Is there a time relation or constraint?
Experimental observation. A. H. Compton, Phys Rev 21, 483 (1923).
I found this link to the actual text.
http://rodin.hep.iastate.edu/jc/321-03/024.PDF
I've already gone over my time budget for this activity
today. I'll at least scan before my next response, that is
assuming you haven't given up on me. ;)
You seem to be saying that a photon can never lose
just part of its energy.
... without changing frequency.
Hedging?
You may or may not want to call the photon
with the changed frequency the "same" photon.
Seems to me your point hinges on it not being
the "same" 'photon'.
[to continue after that Compton interlude ...]
and (E-delta_E) is
added to a monochromatic field of frequency (f-delta_f).
I worded it quite badly, I see. My apologies. Energy is removed/added
to/from the _total_ field in chunks of delta_E = h delta_f, while
removed/added from/to spectral components of the total field in chunks
of
E=hf.
What is this? What is that supposed to mean? Seems
like some strange accounting. The 'spectral components'
exchange energy in quantized amounts but when you
total quantized amounts it is a continuously variable
amount?
Can you reference published accepted theory that describes that?
Would you like to illustrate that with an example?
What's so strange about it? Resonantly illuminate an atom in energy
state of energy E0 at f1, excite it to an energy state of energy
(E0 + h f1 = E0 + E1), get spontaneous emission with de-excitation to an
energy of E3, and you get radiation at a frequency of
f2 = E2/h = (E0 + E1 - E3)/h, with energy E2 at that frequency. Meanwhile,
there is E1 less energy in the radiation field at frequency f1, and the
total energy change in the EM field is a gain of E2 - E1.
It is the atom that is resonant. It is the atom that has
quantized energy levels. The EM field is not. The EM
field will take what the atom gives. The EM field can
only give what the atom will take. N'est ce pas?
Reference? Try 186-188 in:
Claude Cohen-Tannoudji, Jacques Dupont-Roc, and Gilbert Grynberg
Photons and atoms: Introduction to quantum electrodynamics
Wiley, New York, 1989
Note the use of annihilation/creation operators a_i corresponding to
monochromatic modes i of differing frequencies omega_i.
And delta_E = h delta_f is not what you said in objecting
to the OP. Are you withdrawing that objection?
If memory serves, the original objection was to "electron" =
"wavefunction
of electron".
No need to rely on memory. It is recorded.
Nightlight said:
The exchange is quantized only for bound electrons, not for
free electrons. In the case of bound or spatially constrained
electrons, the available frequencies for the electron matter
field are discrete for the same reason that the frequencies
of constrained guitar wires are discrete. The discrete
spectrum is a property of the solutions of those types
of PDEs with those boundary conditions.
To which you replied:
"No, that's a different story. If, in a free-free process, why is the
change in frequency given by E=hf?"
But now later you have said that it is delta_E = h delta_f
and that that is continously variable.
Note well that you quote me as writing the _change_ in frequency is given
by E=hf. How is that different in meaning delta_E = h delta_f?
Perhaps I am wrong but it seems to me you have
effectively said delta_f = f . Now there may be some ambiguity.
If you say that the input photon is always totally absorbed
then delta_f = f for that absorbtion. Now one might define
delta_f as the difference between the f of the input photon
and a subsequently reemitted different photon, but then we would
have two different delta_f's. You haven't made a distinction
between them. Neither have I but you are the one who has said that
the input photon is always totally absorbed.
Note that
in a free-free process, the change in energy (E in the first writing,
delta_E in the latter, but the same thing nonetheless) is continuously
variable - this is a change from an energy state within a continuum of
states to another energy state in that same continuum.
This is just the "discrete" vs "quantised" issue.
(The original original objection was to "electron" = "wavefunction", but
that was in sci.physics.moderated where the thread started.)
OK.
As above. For a monochromatic field, E=hf. For a non-monochromatic
field,
for the total field, delta_E = h delta_f, while for any spectral
component, E=hf, with delta_E = E1 - E2 = h (f1 - f2) = h delta_f.
But you are assuming populations again, right?
It doesn't work for a single photon, does it?
In theory, it works for single photons. The whole point is that it's the
standard description in QED of a single photon being removed from the
field and a single photon being added. Why wouldn't it work for a single
photon?
Experimentally verifying it for a single photon is another matter, which I
assume is the point of:
What are the spectral components of a non-monochromatic
single photon?
Depends on the spectrum. Basically, the probability distribution of energy
or frequency or wavelength measurements.
That is not a single photon.
Not something you're going to
measure if you only have a single photon available.
That's my point. You can't say that an individual
photon has components
If you have a
repeatable source of single photons, then you can measure the spectrum.
Which one loses hf?
You won't know, unless the change in energy in the interaction is known -
you're not going to measure an incoming single photon.
How does one calculate or measure the delta_f?
If you can measure the change in energy of the system the radiation
interacts with, and the energy/frequency/wavelength of the outgoing
photon, you'll have it. Sounds difficult to do. I don't know if this is
possible; it's out of my field so I don't follow the relevant literature.
Not a single-atom single-photon experiment, but laser cooling of a warmish
atomic beam with a narrow-band laser supports this. See Phillips, Rev Mod
Phys 70, 721 (1998). Laser cooling of atoms, in general, supports
delta_E = h delta_f very well.
I may be wrong but it still seems to me that
you can't say that an individual photon cannot give
up just part of its energy.
--
rb
.
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