Re: QFT question

FrediFizzx wrote: 
"Bjoern Feuerbacher" <bjoern.feuerbacher@xxxxxxxxxxxxxxxxxxxxx> wrote in
message news:d7p7a9$c1$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| FrediFizzx wrote:
| > "Bjoern Feuerbacher" <bjoern.feuerbacher@xxxxxxxxxxxxxxxxxxxxx>
wrote in
| > message news:d7mgun$aj6$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| > | FrediFizzx wrote:
|
| [snip]
|
|
| > | Au contraire, that is precisely the point here: that in states of
the
| > | electromagnetic field with a given number of photons (including
the
| > | ground state!) the electromagnetic field strength is not zero,
| > | although the expectation values are zero.
| >
| > We are talking about *no* photons.
|
| Err, if you didn't notice: *no* photons is just a special case of
| "a state with a given number of photons". So the above is right on the
| point.

You are talking double-talk again. 

In no way.


Measure the EM field strength and tell me what it is. 
> Bet you get zero EM field strength. ;-)

I don't know how to measure that.

However, since it is *you* here whose predictions disagree with theory, it is *your* obligation to measure the field and report your predictions...




No photons = zero EM field strength. 

Wrong. Learn some QED.



| > Give me a reference where someone
| > has measured this non-zero EM field strength when n = 0.
|
| Ever heard of the Casimir effect? That's not a direct measurement, but
| closely related to that.

Sure. However, not everyone is buying the QV explanation.

Who is not buying it? Please give some names - of people who are qualified to judge that.


The Casimer
effect is still somewhat speculation as I am sure it has also been
explained via Van der Waals forces.

If you are so sure, it should be no problem for you to provide a reference.


| Do you deny that QED *predicts* a non-zero field for n=0?

Your question makes no sense.  QED predicts an expectation value of zero
for the E(r, t) and B(r, t) EM fields for n = 0.


Indeed. So what??? I AM NOT TALKING ABOUT THE EXPECTATION VALUE!!!

In Quantum Mechanics for point particles, one has a wave function psi(x) which gives the probability amplitude density to find a particle at x. In QED, one has a "wave functional" Psi[E,B] (more commonly written as Psi[A]) which gives the probability amplitude density to find an electromagnetic field E,B (or A, respectively).
And that wave functional is *not zero* for the ground state; in contrast, it has a Gaussian shape (just like the ground state of the
Harmonic Oscillator in QM!).

For the Harmonic Oscillator, the expectation value of the position is zero in the ground state (and in fact, in all excited states with definite energy). Do you conclude also from that that the particle is always resting at zero, or what?



We are talking about the E and B fields aren't we? 

Yes.


So how do we find your "non-zero" EM fields? 

Don't know how to measure it.


You are the one making the claim so show us.

You are the one making a claim which contradicts the predictions of QED, a well-established theory. So show us.


How do we find and measure these fluctuations? 

Don't know.


[snip]


| You essentially still ignore that argument. Do you admit that the
| fact that the expectation value of E^2 is not zero indicates that
| there still is an electromagnetic field, or not?

The result that E^2 is not zero is still speculation. 

It is a prediction by a well-tested theory. That's not speculation.


It leads to the
result that the EM vacuum state has infinite energy.  Contrary to what
is observed.


Ever heard of renormalization? Cut-offs and the like?


I highly suspect that the EM spectral energy density of
the quantum "vacuum" is how much energy there could be in the "vacuum"
per mode per a certain volume. Not how much energy the vacuum really
has as far as EM fields go.



Feel free to support that suspection somehow.



Now the energy density of the "vacuum" with
respect to fermionic vacua might be a different story, IMHO. Volovik
has the bosonic vacuum and fermionic vacuum canceling each other out. 

Don't know about that. Reference?


[snip]


| > Is it really still an EM field?
|
| There is a non-zero probability to measure a non-zero electric and/or
| magnetic field. What else do you want to call that than an
| electromagnetic field?

How could the probability be non-zero to measure non-zero EM fields of
the quantum "vacuum" if the expectation values are zero?

Think about: How can the probability be non-zero to measure a particle in a Harmonic Oscillator at a non-zero position although the expectation value of its position is zero?


I would like to see that trick. Show your math please. 

psi(x) ~ exp(-x^2)

(~ means proportional)

<x> = 0, but nevertheless there is a non-zero probability to find the
particle at other positions than zero.

It's essentially the same for the electromagnetic field in its ground state. (hint: in the example above, we also have <x^2> != zero, again in analogy to the electromagnetic field).


| Have you ever heard of the "field representation" of the states of the
| electromagnetic field? Of wave "functionals"?

I have heard of them. Your point regarding them is what? 

That you apparently don't know it.

Do you deny that for the ground state of the electromagnetic field,
the wave functional is essentially a Gaussian?


| > Try to measure it. Or show me a
| > reference where someone else measured it.
|
| See above.

See what above? Casimir effect? Not good enough. 

It's good enough as long as you haven't presented an alternative
explanation. "I think that it could also be explained by van-der-Waals
forces" does not count.


[snip]


| > | > 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.
| > |
| > | Do you *really* want to claim that in the vacuum state, the
| > | electromagnetic field is exactly zero? You *do* know that the
| > | electromagnetic field is described in QED as an infinite set of
| > | coupled harmonic oscillators, and that harmonic oscillators in
| > | quantum
| > | theory have a non-zero oscillation amplitude even in the ground
| > | state, don't you?
| >
| > Whoa!  Where do you get this "coupled" from?  That is my line. ;-)
| > Reference for that, please.
|
| For the "coupled"? That should be in all books discussing the
| quantization of the electromagnetic field. Try e.g. Ryder.

Milonni claims the bosonic QHO's of the quantum "vacuum" are un-coupled
on page 45 and I am sure that they are.  Don't have Ryder.  Please give
a specific reference or withdraw your claim.

Sorry, I wasn't very clear. If the oscillators are coupled or not depends on the basis states one uses, IIRC. If you use eigenstates of momentum and energym they are indeed uncoupled.


[snip]


Bye,
Bjoern
.

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