Re: Two-slit experiment
- From: Timo Nieminen <timo@xxxxxxxxxxxxxxxxx>
- Date: Wed, 9 Aug 2006 17:14:42 +1000
On Wed, 9 Aug 2006, Ron Baker, Pluralitas! wrote:
"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:
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.)
OK, your main point here seems to be a matter of semantics. I'll with it
further below. I'll cut where I think you're making the same point to
avoid excess length; feel free to restore if you meant a different point.
Millikan's paper, Phys Rev 7, 355-388 (1916) is very worthwhile reading on
the photoelectric effect in general.
[big cut, I think I see your point now]
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'.
No, that's not the point, that's just a matter of semantics. Basically, to
say E = hf only makes sense for a monochromatic field, where there is an f
to talk about.
Quantisation of the radiation field is, in essence, that energy is added
to or removed from a _monochromatic_ field in chunks of hf. If added, we
say we add a photon, if removed, we remove a photon. To ask what a photon
actually _is_ is somewhat of a dubious question, given that "adding a
photon" and "removing a photon" are the things that are actually defined.
Given that no radiation field is actually monochromatic, is the above
definition of a photon useless? No more so than the concept of
monochromatic fields in classical EM. Often, the field is close enough to
monochromatic, so we just pretend that it is - works for both the quantum
and classical cases.
Given a polychromatic field with a discrete spectrum, then you can just
write E_total = sum a_1 E_1 + a_2 E_2 + a_3 E_3 ...
where E_1,2,... are monochromatic fields, with frequencies f_1, f_2, etc.
It's entirely reasonable to talk about the energy in field E_1, energy in
E_2 etc - again, something done classically. Quantisation simply means
that the energy in field E_1 is exchanged with matter or matter+other
spectral components in chunks of h f_1.
QED-wise, I don't think the question of whether or not it's the same
photon if you take h f_1 from field E_1 and add h f_2 to field E_2 really
makes sense.
A little more complicated with a continuous spectrum, since then you have
integrals over spectral mode densities, rather than sums over mode
amplitudes, but still, the correspondence between quantum and classical is
very, very close.
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?
Sure. But why those two frequencies in particular? OK, this is explainable
for bound-bound transitions in an atom. Bound-free and free-free
transitions, and transitions that have no resonant frequency - why should
they also behave in this same manner?
That quantisation is a property of the field seems to me to be far simpler
than trying to ascribe it purely to matter.
The transport of _all_ the energy of a single emission by an atom to a
single absorbing atom when this is classically impossible then has to be
either explained as a property of the field, or a property of the atom
interacting with a purely classical field. The absorbing atom could be
light-years away. To have the atom do this would be truly remarkable, far
more so IMO than it being a result of quantisation of the field (still a
remarkable result).
[cut more]
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.
Why not? If you have a single excited atom as the source of the field, and
you don't know _exactly_ how the environment perturbs the energies of the
upper and lower state, or you don't know which lower state the atom will
de-excite to, or you don't know the motion of the atom _exactly_, or the
upper state has a finite lifetime, you'll get a polychromatic photon.
Diffraction grating + CCD can give you the wavelength. If the main thing
is which lower state the atom goes to, the probabilities are something
that might well be known from previous measurements.
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
The photon doesn't have components, the wavefunction has components of
different energies. Total probability = 1 if you have a single photon.
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.
If you want to say that the "new" photon of frequency f_2 is the same
individual photon as the "old" photon of frequency f_1, then sure, it can
give up part of its energy. But only by changing frequency, with the
energy change = h delta_f.
The key point is: if this is because this is how EM radiation fields work,
then the field is quantised. If this is because all matter that the fields
interact with work that way, then you'd see the same thing, but the
explanation is, IMO, far more complex and demanding of the suspension of
disbelief.
And finally, just to return to "is it the same photon?". Frequency
downconversion: one photon in, two identical photons out. Which one is the
same individual photon as the original? (It's just the splitting amoeba
identity problem again!)
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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