Re: Critique of the "photon" theory of electromagnetic radiation
- From: Ralph Hartley <hartley@xxxxxxxxxxxxxxxx>
- Date: Wed, 13 Apr 2005 05:23:32 +0000 (UTC)
Robin Whittle wrote:
> Short version:
>
> Interference between two microwave transmitters cannot be explained
> with the "photon" theory of electromagnetic radiation (emr). This
> theory cannot be a true representation of Nature, and so should not
> be used to evaluate observations or other theories.
Short answer:
You have some misunderstandings of the theory. Most of them are in the
form of general statements of the sort that are often used to describe
the theory in non-technical language, but are not adequate for an in
depth look.
I will answer in detail only the points you would learn the most from.
> 1 - The photon begins when quantum of energy is lost, somewhere, in
> one specific location - such as an electron dropping an energy level
> in an atom. (Actually, I think conventional photon theory may allow
> for probability distributions for when and where the photon began.
It also allows for a quantum superposition. A superposition is *almost*
exactly like a probability distribution. The difference is that they
have "amplitudes", which are just like probabilities except that they are
complex numbers instead of positive reals. You can get probabilities
from them by taking their squared magnitude, so in a sense a
superposition is a square root of a probability distribution.
> 7 - Photons do not interfere with each other - but the wavefunction
> of an individual wavefunction does interfere with itself.
> Consequently, the probabilities of where photons will deliver their
> energy in any experiment are not affected by the number of photons in
> the system.
This oversimplification seems to give you the most trouble.
Photons are indistinguishable from one another in a fairly absolute
sense. One can reasonably talk about two photons, but not about photon A
and photon B.
Several of your problems below disappear if you stop trying to
distinguish between different photons.
Photons are not necessarily independent. There are situations in which
knowing where one photon was detected tells you a lot about where
another will be.
> See also the talking-to he [Lamb] gave the delegates to the 1995 US
> Army workshop on quantum cryptography and computing:
>
> http://www.aro.ncren.net/phys/proceed.htm
I suggest you take his first point to heart:
Lamb:
> Anyone wanting to discuss a quantum mechanical problem had better
> understand and learn to apply quantum mechanics to that problem.
Robin Whittle wrote:
> But if "photons" are considered to have a superposition of all
> polarisation states except those at 90 degrees to their official
> state, then I think the "photon" concept can be made to do pretty
> much whatever people want. I think this adds no explanatory or
> predictive power to the classical view of emr polarisation - and is
> far less elegant.
As a counter example I would suggest the Aspect experiment or it's
variants, but I don't think you are ready for them yet.
[Snipped description of two slit experiment performed with two microwave
transmitters]
> According to my understanding of "photons", this experiment does not
> allow a "photon" to interfere with itself, since a photon can only
> arise from one transmit antenna or the other. Yet we observe perfect
> interference.
Indeed we do. But the indistinguishability of photons means we can't
discriminate between photons from one source and those from the other.
According to quantum mechanics, photons from different sources *can*
interfere.
Even when there is only one photon, two possible sources of that photon
can interfere. (I'm leaving out some fine print)
> We can detect near visible infra-red quanta - with photographic film
> and silicon photodiodes - but this involves frequencies where we need
> lasers, rather than electronic oscillators, as transmitters.
Interference between separate lasers has been done.
> Photon's can't be "massless". Saying they are "massless at rest" is
> pointless, since they cannot be at rest.
Would you rather people said "photons have no rest mass" instead?
It makes things simpler to think of photons (and other particles that
can never be at rest) as having a rest mass of zero. It never causes any
problems (except confused questions). The equations that describe
particles with well defined rest masses also work for particles that
travel at the speed of light, but only if you use 0 for the rest mass.
I any case, when we say photons are "massless", we *mean* rest mass. We
all know that a box full of photons (e.g. with perfect mirrors on the
inside) will weigh slightly more than an empty box.
> I understand that the traditional view of photons is that each one
> has a specific wavelength / frequency - and therefore quantum of
> energy. Are there any exceptions to this view?
This view is actually a very special case, and can never be perfectly
realized in practice.
Only an ideal laser (which must run for all eternity) can produce
photons of an exact wavelength.
> Do some theories involving photons allow for a probability
> distribution for the photon's wavelength?
There is really only *one* such theory, and it does. Not only
probability distributions, but quantum superpositions.
Ralph Hartley
.
- References:
- Critique of the "photon" theory of electromagnetic radiation
- From: Robin Whittle
- Critique of the "photon" theory of electromagnetic radiation
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