Re: Two-slit experiment
- From: Oz <Oz@xxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 26 Jul 2006 20:31:42 +0000 (UTC)
Timo A. Nieminen <timo@xxxxxxxxxxxxxxxxx> writes
On Fri, 21 Jul 2006, Oz wrote:
Timo A. Nieminen <timo@xxxxxxxxxxxxxxxxx> writes"they
The claim regarding electrons having different sizes depending on whether
are undisturbed in an orbital" reads to me as an identification of size with
spread of wavefunction.
Its a very excellent description of size.
We are talking size in meters here, not energy.
You are saying that the "spread of a wavefunction" = "size of particle"?
Of course, its certainly the area where it interacts which is the best
description of 'size' I can think of. However I would distinguish
between the wavefunction, which can vary depending on our knowledge, and
the actual EM wave. MBR emitting atom as an example see below.
The fact that this can sometimes be absolutely HUGE (as in electrons in
metals/graphite and other things) is interesting but not counter-
evidence. As a wave the size will clearly depend on any boxes it gets
put in or other restrictions.
frequenciesThis being independent of wavelength (but see below),
low frequency isn't necessary. In fact, it's best to use the highest
ofavailable, so that the energy required for detection is only a small fraction
the total energy.
Er, yes, but such an object has a very short wavelength so is inherently
more localised (or 'smaller').
No! Not at all! What does wavelength have to do with localisation? It is
relevant to _minimum possible_ localisation, but a monochromatic plane
wave mode is _completely_ unlocalised, in both space and time,
independently of the wavelength.
See above, I agree. Why do you see that as a problem?
Put a gamma source in the middle of some, preferably many,
detectors. Each gamma photon can go in any direction, the radiation field of
each emission is spherically symmetric (well, perhaps a dipole field, but
spherically symmetric averaged over many).
I'm not actually completely in agreement with this statement.
On average this is so, and on average (that is, over many counts) you
get the right answer.
But the uncertainty of momentum may not be spherical because you could
measure the recoil of the source (if the gamma is energetic enough),
which would destroy the spherical symmetry.
So don't measure the recoil. No problem!
Something (unless you go to extreme lengths) will interact with the
particle sooner or later, and this particle will react with others etc
etc. This means that for (taking the age-old example) a MBR emitting
atom its momentum will be fixed well before the 'photon' interacts.
largerI guess that you're wondering about size of photons as it might depend on
wavelength. The above means that it's hard to answer experimentally. I think
that illuminating a group of atoms, all within a wavelength, and only one of
them absorbs and re-emits, is conclusive - the "size" of a photon is no
isthan an atom. Compton can be interpreted as saying that the size of a photon
the size of an electron, ie zero AFAWCT.
Absolutely not, as diffraction will confirm.
No, diffraction just tells you that quantum "particles" are not classical
billiard balls. The wavefunction can have finite (ie non-zero) extent
(and, technically, extends to infinity in all directions, although the
amplitude of the wavefunction may be close enough to zero so you can
pretend it's bounded in space), but that doesn't mean that the "particle"
it prescribes the probability distribution of is as large as the
wavefunction.
Indeed, see above. However I still claim that the MINIMUM (see what I
actually said) size is of the order of a wavelength.
Diffraction tells you about wavefunctions. Diffraction of EM waves tells
you about EM waves, not about whether energy is exchanged between EM waves
and matter in quanta.
Of course. However if you look elsewhere someone with more knowledge
than me has described a typical detector more as I see it but in proper
language.
===From: nightlight <nightlight@xxxxxxxxxxxxxx>
Newsgroups: sci.physics.research
Subject: Re: Two-slit experiment
Approved: (sci.physics.research)
Message-ID: <1151472082.880906.31520@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
The facts are that the response of an array of photodetectors is
statistically indistingushable from droplets statistics of an array of
dripping faucets. Specifically, the best (the most narrow distribution
of triggers) that you will ever get here is a Poissonian distribution
(as long as QED vacuum is there and its vacuum fluctuations).
There isn't anything 'single-photon-like' about it that can
distinguish it from a thresholded EM field measurements (with
photo-electron current measurements results reduced via AD
conversion by the 'pulse analayzer & discriminator' electronics
to a single bit precision, 0/1 values, based on the detector
thresholds) of a classical field.
=======end nightlight
If all the energy goes to one atom, and isn't divided among many,
Of course, where is the problem here?
the
quantum is one atom in size, at most, regardless of how many atoms wide
the wave is.
No, you are confusing the detector and the wave.
To understand more clearly imagine the atome emitting such a huge long
wave, then play this backwards. Not a problem.
Remember that an aerial does NOT have to be a wavelength long to
receive. In fact its normal for low frequencies to put appropriate L & C
on a short aerial so as to tune it to the required low frequency/long
wavelength. Similarly you can see an absorbing (or emitting) atom as
tuned to its transition frequency, with the masses of the
electron/nucleus and associated electromagnetic field interaction tuning
it to the very low (in wavelength terms) transition frequency.
Just as a properly tuned small aerial can transmit a very long
wavelength signal, so a very small atom is perfectly capable of being
tuned to a very long wavelength em wave.
Clearly most transitions take very many cycles of emission before the
transition is completed.
No. The energy difference between the upper and lower levels is hf. To go
through many cycles would mean pumping out and sucking back that energy
over and over.
This isn't so in non-linear systems.
See parametric amplifier.
Quantum mechanics is typically rather non-linear.
In any case consider the reverse of an atom emitting a photon.
Yet the energy is launched out at c, and you can suck it
back. Yes, for a short dipole, you put energy into the near field, and get
it back. Yes, the wavefunction of the electron oscillates (and see my
recent post in sci.physics on a content-related thread for more on this,
and some nifty references supplied by others). However, classically, you
never have "the transition is completed".
Well, of course you can have equivalent systems, see expert example
above. You cannot, however, expect to model a pure quantum example non-
quantum-mechanically. For the non-relativistic H-atom emission, though,
I believe it can be done using schroedinger only.
Never. This is a quantum
concept. Classically, you have a damped oscillator,
Why damped? Damping energy is going where?
which never decays to
zero, but a photon is emitted, and maybe detected, and the atom is
demonstrably in the lower state.
Of course. Why do you see that as a problem? Atoms are, as I have said
repeatedly, quantum devices, they have energy levels. Thats where
quantum mechanics works well. The fact that the detectors and emitters
are quantum mechanical does not mean the EM wave is, although its hugely
convenient to consider it to be so.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
.
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