Re: QFT question

FrediFizzx wrote: 
"Bjoern Feuerbacher" <bjoern.feuerbacher@xxxxxxxxxxxxxxxxxxxxx> wrote in
message news:d8bnta$a5s$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
| FrediFizzx wrote:



[snip]


| I didn't see you define it. You merely gave a value for it. And the
| value you gave for <E_o> was the square root of <E^2>. Since <E> is
| zero, this implies that E_o is essentially the standard deviation
| of the probability distribution for E.

Sorry, my fault.  I was taking expectation value to mean something
somewhat different than this.  I think you are using this definition;

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/expect.html

Milonni isn't quite using it like that. Or if he is, then it is hidden.

Well, then how is he using it? "expectation value" in math and physics
essentially always means that one is averaging a quantity which can take on different values according to a probability distribution, with
exactly that distribution. In Dirac's language of QM, the expectation
value of an observable O in state |psi> is
<O> = <psi|O|psi>
If one goes over to wave functions, this becomes
<O> = int psi^*(x) O(x) psi(x) dx,
where the integral runs over the whole space on which psi is defined.
psi^*(x) psi(x) is the probability distribution.


[snip]


OK.  Let me refine what I mean.  When I make the replacement with
1/volume, then the expression is no longer an expectation value by the
definition at the link above.


Well, right - it is then simply something like an approximation to the
real expectation value.


IOW, <E^2(r, t)> goes to E^2 

What is E supposed to mean here, exactly? You do know that in QED, E
is an *operator*, don't you?


and is a scalar; 

I never disputed that. <E^2> is also a scalar. So what's your point?


we are just dealing with the scalar values at that point. 

If you write E^2, you are dealing with an *operator* at that point.


[snip]


| > I suppose that is somewhat the wrong question.  There is no absolute
| > experimental *fact* that <x^2> and <p^2> are non-zero for the EM
| > vacuum state field.
|
| I think the experiments done on coherent states in harmonic oscillator
| potentials lend at least strong support to that, although they don't
| show that directly.

Got any references? 

Not directly at hand, sorry. Ever heard of "Schroedinger cat states"?
IIRC, there was an article on that in Nature about 3 years ago.


[snip]


| > I suspect that
| > the EM zero-point energy also vanishes similarly
|
| Err, where should the negative energy for the cancelling come from?

See below. 

From some fermionic fields? Could work in some versions of MSSM, AFAIK.
But not in the SM.


[snip]


| Don't think so. I still think that the Casimir effect is evidence for
| the fluctuations. You have claimed that this could also be explained
| otherwise, but haven't shown support for that claim so far.

Study the following document "Of Some Theoretical Significance:
Implications of Casimir Effects" carefully.

http://arxiv.org/abs/quant-ph/0105002 

Thanks, I'll look at it.


[snip]


| > Sure.  But no one has really figured out why the "vacuum" should
| > have this tremendous EM energy
|
| Due to the uncertainty principle.

Don't hack messages in the middle of a sentence.  It is bad enough that
you can't write a coherent paragraph in reply to a paragraph.  Learn
some proper netiquette.  Did you have any writing classes at the "frame
school" you went to?  Last warning.  Answer a paragraph with a coherent
paragraph.

Hint: none of the actual physicists in this group ever complained about my writing style. Only some of the cranks.

And yes, I did have some writing classes. No teacher ever said that I write incoherent paragraphs.


| > contrary to GR.
|
| In what way is that contrary to GR?

Sheesh! 

Care to tell me?



| > Maybe Volovik is right and
| > the fermionic vacuum energy is canceling out this bosonic vacuum
| > energy?
|
| Could be, I don't know his work. But AFAIK, this doesn't work in the
| Standard Model. It could work in the MSSM.

What is MSSM?

Minimally supersymmetric extension of the Standard Model. Never heard that abbreviation? It is very common, IIRC even listed in Wikipedia.


It does work in the Standard Model as currently
interpreted.  The Dirac fermionic vacuum has negative energy.  See
Milonni "The Quantum Vacuum" sect. 10.6 "Dirac Vacuum in Field Theory".

If you are talking about Dirac's idea of a "sea" of fermions with negative
energy: I always thought that's an outdated idea, and since we have QED, vacuum energies are only positive. I'll try to get a look at Milonni's book.


| > | > The fermionic
| > | > quantum "vacuum" from Dirac theory is actually much more
| > | > interesting, IMHO.
| > |
| > | Indeed, but that was not the topic here.
| >
| > It's related.  See above.  Let's take an electron and positron to
| > illustrate a point.  They both have the same rest mass energy as a
| > positive magnitude.  However, there is some kind of difference
| > energy-wise.
|
| Why do you think so?

[Replacing what you *hacked out*;]

"Like one energy is "left-handed" and the other energy is
"right-handed".  IOW, another feature that distinguishes them as an
electron or positron besides their positive and negative charge.  I
believe it is this "feature" that is missing."


I don't see in any way how that answers my question above why you think
so, or clarifies what you said, sorry.



Or another explanation is that we are "riding" on a sea of tremendous
positive energy and there is a mean value that looks like zero energy to
us macroscopically.


What's the difference between a "sea" of positive energy and the QFT
concept of vacuum energy?



[snip]

Bye,
Bjoern
.

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