Re: QFT question

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.


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???


Bye,
Bjoern
.

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