Re: QFT question

"Bjoern Feuerbacher" <bjoern.feuerbacher@xxxxxxxxxxxxxxxxxxxxx> wrote in
message news:d7k05n$l5s$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| FrediFizzx wrote:
| > "Bjoern Feuerbacher" <bjoern.feuerbacher@xxxxxxxxxxxxxxxxxxxxx>
wrote in
| > message news:d7f6rf$c4n$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > | FrediFizzx wrote:
| > | > "Bjoern Feuerbacher" <bjoern.feuerbacher@xxxxxxxxxxxxxxxxxxxxx>
| > wrote in
| > | > message news:d77h73$p3c$2@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > | > | FrediFizzx wrote:
| > |
| > | [snip]
| > |
| > |
| > | > | Photons are excitations of the EM field in the sense that the
more
| > | > | photons there are, the higher the energy contained in the EM
field
| > is.
| > | >
| > | > What happens when the number of photons is zero?
| > |
| > | Then you still have a non-zero EM field: zero-point fluctuations.
| >
| > Sorry, the expectation value of the E and the B fields are zero. I
| > quote from Milonni's "The Quantum Vacuum: An Intro. to QED" page
41.
| >
| > "In the vacuum state, and in all stationary states |n>, the
expectation
| > values of the electric and magnetic fields vanish:
| >
| > <E(r, t)> = <B(r, t)> = 0 (2.38)
| >
| > since <n|a|n> = 0." Where E, B and r are vectors.
|
| Err, I didn't talk about the expectation values.
|
| Please notice also that for every *excited* stationary state (i.e. for
| every state with a definite photon number!), these expectation values
| vanish. If you want to have a state for which the expectation values
| do *not* vanish, try coherent states.

So what? That is not the point here.

| > However, the expectation value of the square of the electric field,
| > <E^2(r, t)> is non-zero.
|
| Indeed. Don't you think that that's a clear indication that there
still
| is an electromagnetic field, although the expectation values of E and
| B themselves are zero?
|
|
|
| > | > Photons are
| > | > excitations of the "vacuum state EM field" is a more proper way
of
| > | > saying it.
| > |
| > | An excitation of an excited state is still an excitation, so I
don't
| > | see a big need to stress that they have to be an excitation of the
| > | vacuum state.
| >
| > I hope you realize this looks like double-talk. ;-)
|
| Maybe it looks like that to you. Not to me.
|
|
|
| > | > The way you are saying it, it sounds like photons could be
| > | > excitations of photons.
| > |
| > | I don't think that what I said sounds like that.
| >
| > You even say it again above. "An excitation of an excited state..."
|
| Don't you see the difference between "excitation" and "excited state"?
|
|
| > When n = 0 there are no photons and no excitation.
|
| Indeed. So what???

That is the whole point of this exercise. Photons are excitations of
the quantum vacuum and not of an EM field. If you want to claim the
quantum vacuum has an EM field that photons are excitations of, then
measure it and tell me what its value is. Bet you get zero.

FrediFizzx

http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps

.

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