Re: photon's wavefunction
- From: Arnold Neumaier <Arnold.Neumaier@xxxxxxxxxxxx>
- Date: Thu, 8 Feb 2007 15:03:26 +0000 (UTC)
kvblake schrieb:
In fact I have seen a formula for <x/ in a book by the russian
Prohorov (2003) ---- <0/A(x) where one easy calculates for
wavefunction of the photon exp(-ikx)eps(k).
Yes; this corresponds to my description with the Fourier transform undone.
I am bothered that the function is a complex function. This
is necessity for de Broglie function but how can one calculate E and M
which have to be real.
Read about the complex analytic signal in
L. Mandel and E. Wolf,
Optical Coherence and Quantum Optics,
Cambridge University Press, 1995.
to see, how real E and M are described naturally in terms of complex
quantities. The complex version is very natural for the discussion of
causality issues.
Arnold Neumaier schrieb:A single photon has the same degrees of freedom as a classical
electromagnetic field; its shape is characterized by an arbitrary
4-potential A(x)
Do you mean by this that the photon generates(or is equal to,
identical to) A(x)
No. Just that there are as many distinct photons as there are distinct
(gauge orbits of) A(x).
and that two photons may have different A(x)??
Yes.
Or if one puts it another way that any EM field may in fact can be one
photon?
No. The electromagnetic field is the expectation of the EM field operator, whereas the wave functions define the ensemble for which
the expectations are taken.
In an analogy to quantum mechanics, x, A (the photon amplitude),
E(x) (the field operator) correspond to
k, x, and p_k. Thus the coordinate index k is inflated to the spacetime
position x, the argument of the wave function is inflated to a solution
of the free Maxwell equations, the momentum operator is inflated to
a field operator, and the integral over x becomes a functional integral
over photon amplitudes.
As the observed components of the mean momentum, say, are
<p_k> = integral dx psi(x)^* p_k psi(x),
so the observed values of the electromagnetic field are
<E(x)> = integral dA psi(A)^* E(x) psi(A).
Here psi(A) is the most general state vector in Fock space; for a
single photon, psi depends linearly on A,
psi(A) = integral d\p^3/p_0 A(\p)|\p> = |A>.
I'm especially impressed by the words 'its shape is
characterized by an arbitrary 4-potential A(x)' which I understand as
above described.
If you have a coherent state whose single photon part has A(x) as its
wave function then it generates a classical coherent electromagnetic field with 4-potential A(x). If you have superpositons or mixtures of
states whose whose single photon part have different A(x), you get
instead some kind of average field.
In fact I thought that if you has a box full of EM radiation
(arbitrary A(x)) and if you let the photons to leave one by one at the
end you will stay with a specific A(x) which will be the photons
wavefunction.
Each photon is the whole ray - a mode of the electromagnetic field, and not an object that leaves the source one by one...
Yes, this is somewhat counterintuitive, but that's what it is.
Not everything in quantum mechanics has an easy intuition.
Arnold Neumaier schrieb:
In momentum space, single photon states have the form
psi = integral d\p^3/p_0 A(\p)|\p>,
I haven't also found a monentum representation of the wavefunction of
the photon in any textbook. I would appreciate if you can write where
you found it. I can't understand why it is integrated in dp - psi will
then not depend on p
This is how things always look in second quantization. Even a harmonic
oscillator. The wave function psi(x) or psi(p) in standard (first quantized) quantum mechanics becomes the state vector
psi = integral dx psi(x) |x> or integral dp psi(p) |p>
in Fock space; the wave function at x or p turns into the coefficient
of |x> or |p>.
Arnold Neumaier schrieb:4=2E Thus a general photon is a superposition of monochromatic waves of
arbitrary frequencies and directions, though for experiments one
usually uses nearly monochromatic photons bundled into narrow beams.
This here is very very suprising for me. I always thought that a photon
is a particle and wave of 1 direction and 1 frequency.
This may be the basis of your difficulties with the concept of a photon.
Only a monochromatic beam has a fixed frequency and direction.
As far as I
know a decay of an atom emits one photon of one frequency.
But already arbitrary direction - since its shape is a radial wave and
not a plane wave. Written in terms of plane waves, it is a
superposition.
How can we then
regard a photon to be superposition of monochromatic waves of
arbitrary frequencies and directions?
It just means that some photons are harder to produce than others.
Atomic decays only produce special kinds of photons.
Arnold Neumaier schrieb:
>A(x) can be regarded as the photon's wave function in
momentum space. Since photons are not localizable there is no photon's
wave function in coordinate space. One could regard the 4-potential
A(x) as coordinate space wave function, but because of its gauge
dependence, this is not really useful.
Sorry but I can't also agree with that that photons are not
localizable. For localization of course one can use the photoelectric
effect,
Of course they are approximately localizable, otherwise we couldn't do experiments with them. But unlike electrons, they cannot be localized
arbitrarily well. This is because they lack the longitudinal part of
a general wave function.
You can read about all this in the standard reference for quantum optics,
L. Mandel and E. Wolf,
Optical Coherence and Quantum Optics,
Cambridge University Press, 1995.
Arnold Neumaier
.
- Follow-Ups:
- Re: photon's wavefunction
- From: kvblake
- Re: photon's wavefunction
- References:
- Re: photon's wavefunction
- From: Arnold Neumaier
- Re: photon's wavefunction
- From: kvblake
- Re: photon's wavefunction
- Prev by Date: Re: Coherence length and bandwidth of photons
- Next by Date: Re: Actions, symmetries, and gauge theories
- Previous by thread: Re: photon's wavefunction
- Next by thread: Re: photon's wavefunction
- Index(es):
Relevant Pages
|
