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Re: Is photon a wave packet ?



Roger Haar wrote:

I am not sure why you believe that a single
photon must be monochromatic. It is true that the
often formalism deals with it that way.

This is a very good question. It cuts to the heart
of the matter. It caused a light to go on for me.
So, here is a explanation that I should have given a
couple of days ago, but I was unable to dredge it
up from my subconscious until now:

The short answer is that the photon is purely a creature
of the formalism. For better or worse, the formalism
of physics contains more than a few such creatures, things
that exist in the formalism but cannot be realized in the
laboratory.

An example that is very often used and very often
misunderstood is the expansion of the identity operator
in terms of projection operators.
1 = sum_over_i P_i in general
= sum_over_i |i> <i| in a given basis
and we can use this in many ways, e.g. to write the Hamiltonian
in the form:
H = sum_over_i |i> E_i <i|
where Ei is the energy of the ith eigenstate.

But this does not mean that we can construct the projection
operator P_i in the laboratory!!!!!!!!! The only things
that can actually be constructed are unitary, and P_i
is not unitary. The existence of a P_i operator would
violate the uncertainty principle, violate the 2nd law
of thermodynamics, and violate Liouville's theorem
(which is three ways of saying the same thing).

Expanding things in the Fourier basis (using pure sine
waves) is a special case of the above. You can't make
a pure sine wave ... yet Fourier analysis remains a
useful formalism.



If there
is just a single frequency associated with a
photon, how long must it be? From the Fourier
analysis I think you get infinite. If there is no
uncertainty in the energy, what is the uncertainty
in when the photon was somewhere. Again infinite.
( These two questions are related.)

All true.

One can experimentally measure the length of a
single photon.

No, according to the way experts use the term "photon",
one cannot.

Use an interferometer, such as a
Michelson, and a light source that emits a single
transition (use filters or gratings as needed),
and that emits at a single photon rate. The
interference pattern falls apart if the difference
of the two optical paths lengths exceeds the
length of a photon because it can no longer
interfere with itself. Thus photons have a finite
length and the photons are not truly
monochromatic.

No, this shows that the wave packet has a finite length.
It has nothing to do with the photon number(s) [i.e.
occupation number of any mode(s)]. In particular, the
falling-apart behavior is independent of intensity,
and the foregoing phrase that specifies "single photon
rate" is unnecessary. It is a red herring. Remove it,
and the true physics becomes clear.

We, physicists, almost always work in systems of
energy eigenstates, but it is not required.

True.

A single isolated, at rest, atom emits a photon with
a energy spread associated with the natural line
width of the transition.

1) That is mixing up a couple of ideas. In particular
it is mixing some wave ideas with some particle ideas.

2) See previous statement about the non-mandatory nature
of energy eigenstates. The atom emits a photon if and
only if it is making a transition between energy eigenstates.

3) If the states have a lifetime, then they are not truly
eigenstates. The only way to describe a state with a
nonzero life (i.e. nonzero energy spread) is as a superposition
of modes. Each of these modes will have its own occupation
level (i.e. photon number).