1. Apr 25, 2016

### andrew s 1905

I have been trying to improve my understanding of QFT via reading this forum, Zee's QFT in a Nutshell and Lancaster and Blundell's QFT for the Gifted Amateur. I thought I was making some progress but recent post have left me confused so any pointers would be welcome.

As I currently understand it the vacuum is the lowest energy i.e. ground state and does not contain any excitation's.

Virtual particles are just that, an aid to calculation etc. and don't exist.

What I am puzzled about are:

do vacuum fluctuation exist and if so in what sense, and

what is the correct explaination of the Casimir force?

In Zee the Casimir forces is explained in terms of vacuum energy and the effect of the plates on the allowed modes between the plates but I have read here about it being due to van de Waals forces and nothing to do with vacuum energy. If the later is correct what is wrong with Zee's analysis?

I did study physics in the 1970's and have a PhD in it but much has changed since then so be gentle with me.

Thanks Andrew

2. Apr 25, 2016

### Demystifier

This is a somewhat controversial topic because there is no answer on which everyone would agree. I also think that the correct explanation of the Casimir force is the van de Waals force and not the vacuum energy. See also the famous paper by Jaffe
http://lanl.arxiv.org/abs/hep-th/0503158

3. Apr 25, 2016

### andrew s 1905

Thanks that was a very interesting paper. I can see why I was confused.I could understand the papers proposal without having to understand all the equations. The key issues seems to be the assumptions on boundary conditions i.e. ideal v real materials with a finite fine structure constant.

Is there a similar paper or papers that go into the current understanding of the "question of whether zero point fluctuations of the vacuum are or are not “real”". I have tried Google but I have no way to "rate" the large collection of hits many of which describe virtual particle popping in and out of existence.

Regards Andrew

4. Apr 25, 2016

### Staff: Mentor

5. Apr 25, 2016

### Demystifier

This is much less controversial, at least among experts. I think it is safe to say that fluctuations are real, while virtual particles are not.

6. Apr 25, 2016

### Staff: Mentor

That's what the linked paper says

Thanks
Bill

7. Apr 25, 2016

### andrew s 1905

Thank you both the latter paper is somewhat more difficult! I get the broad idea and am happy to accept your views however, I will need to do more study as the following
"
The expectation
sigma^2 = <(q-q_0)^2> (where q_0=<q>)
of a quantum harmonic oscillator in the ground state is nonzero,
because the ground state is not an eigenstate of q.

Does this mean that the ground state of a harmonic oscillator
fluctuates? No. It just means that one cannot assign it a definite
position -- that measurements of the position fluctuate for a particle
in the ground state.

According to the Heisenberg uncertainty relation, the variance
simply tells one something about the theoretical limit accuracy in
trying to measure the oscillator position rather than about any
fluctuation of the oscillator itself. It is the measurement results
that fluctuate, not the underlying object."

Seems to be saying that the state of a quantum harmonic oscillator is well defined (i.e. not fluctuating) but we can't know what it is. If it can't be measured how do we know it is not fluctuating. Clearly more to learn or unlearn!

Thanks Andrew

8. Apr 25, 2016

### Staff: Mentor

That's a misconception.

But he exact meaning of the uncertainty relations requires a new thread.

Thanks
Bil

9. May 16, 2016

### Demystifier

If I had any doubts when I wrote the post above, now I don't. The argument in
http://lanl.arxiv.org/abs/1605.04143
convinced me that, at the fundamental level, the Casimir force is not created by vacuum energy.

10. May 16, 2016

### A. Neumaier

The ground state of a harmonic oscillator, to which these remarks apply, is a well-defined, known state. What we cannot know for sure is whether a particular harmonic oscillator in Nature is exactly in its ground state - it will never be, except at zero temperature, which is unattainable.

11. May 16, 2016

### atyy

In the Hamiltonian picture, the ground state is just the state of lowest energy, and the state itself doesn't fluctuate. However, due to Born's rule, depending on what one measures, the result is still random. In that sense there are quantum fluctuations. Although the Hamiltonian picture is more fundamental, one can use the path integral picture, and one will see that there is a sum over all paths - the multiple paths are the fluctuations.

In Zee's analysis, he always uses the path integral picture. But the whole path integral picture is in a sense not real, since it is just an expedient way to calculate. It is the Hamiltonian picture that is real.

So in Copenhagen, at the top level there is real reality - the measurement apparatus and outcomes.

Then there is the fake reality, which is the quantum state. This fake reality is what I call realer than the path integral picture.

Then there is the fake fake reality of the path integral, which Zee uses.

The virtual particles are what intermediate terms that one gets when one does a perturbative expantive expansion of the path integral - so in that sense they are fake fake fake reality.

12. May 16, 2016

### Demystifier

I like your levels of fake reality.

How about the difference between Heisenberg picture and Schrodinger picture? Would you say that one is more fake than the other?

13. May 16, 2016

### atyy

Only if one is a Bohmian (or MWIer) :)

14. May 16, 2016

### Demystifier

Section I.8 is an exception.

15. May 16, 2016

### vanhees71

What difference do you mean? It's just two ways to shuffle the extreme ways to spread to the time dependence of highly abstract mathematical objects, from the point of view of physics auxilliary variables, to describe the unique time dependence of observable quantities. There is no difference between quantum theory in the Heisenberg and the Schrödinger picture or any other picture of time evolution. There's just only quantum theory!

16. May 16, 2016

### A. Neumaier

The difference is that the Schroedinger picture allows only one time variable, while the Heisenberg picture allows many. The difference shows when studying times correlations.

17. May 16, 2016

### Demystifier

See the context in which the question about the "difference" has been asked. If somehow (as atyy argued) path-integral formalism is "less real" than the Hamiltonian formalism, then it is not meaningless to ask what I asked.

18. May 16, 2016

### rubi

I think it makes no sense to say that any description is more fundamental if it is equivalent. However, one can say that the Heisenberg picture has a wider range of applicability, since it even applies in situations where the Schrödinger picture makes no sense, such as fully constrained systems (such as time reparametrization invariant systems like for example canonical quantum gravity). One can also argue that the Hilbert space viewpoint contains more physically irrelevant arbitrariness than the $C*$ algebraic viewpoint.

19. May 16, 2016

### atyy

But that assumes canonical quantum gravity makes sense.

20. May 16, 2016

### rubi

Well, canonical quantum gravity is just one example. Every time reparametrization invariant system is fully constrained. The Klein-Gordon equation is a constraint equation for example, rather than a Schrödinger time evolution.

21. May 16, 2016

### atyy

https://en.wikipedia.org/wiki/Klein–Gordon_equation This Klein-Gordon equation? Isn't this the theory of a free massive relativistic scalar field?

22. May 16, 2016

### rubi

The free relativistic quantum scalar field is described by a quantum KG equation, but I'm talking about a free relativistic spin 0 particle, which you get when you quantize the Lagrangian $L=-m\sqrt{-\eta_{\mu\nu} \dot x^\mu \dot x^\mu}$. The classical constraint is just the mass-shell condition $p_\mu p^\mu = m^2$ and its quantization yields the KG equation.

23. May 16, 2016

### DarMM

Interestingly in QFT the Schrodinger picture requires additional renormalizations beyond that of the Heisenberg picture.

24. May 16, 2016

### Staff: Mentor

Dirac, who was a famous critic of renormalisation, managed to do the QED calculations in the Heisenberg picture without using it:
https://en.wikipedia.org/wiki/Paul_Dirac
'He found a rather novel way of deriving the anomalous magnetic moment "Schwinger term" and also the Lamb shift, afresh in 1963, using the Heisenberg picture and without using the joining method used by Weisskopf and French, and by the two pioneers of modern QED, Schwinger and Feynman. That was two years before the Tomonaga–Schwinger–Feynman QED was given formal recognition by an award of the Nobel Prize for physics.'

I don't agree with our modern understanding of renormalisation its anything to worry about, but one hell of a theorist Dirac most certainly was - awe inspiring really.

Thanks
Bill

25. May 17, 2016

### atyy

Could you give me a pointer to some place that explains this? I haven't come across this before.