# Perturbative and non perturbative vaccum states.

1. Mar 13, 2012

### dpa

Hi all,

what is the meaning/difference between perturbative and non perturbative vaccum.

2. Mar 13, 2012

### tom.stoer

The perturbative or Fock vacuum |0> is simple described by

an|0> = 0

for all possible quantum numbers n.

The non-perturbative ground state is describe by something like

(H-E°)|Ω> = 0

with minimum E°.

Alternatively one could write something like

<0|H|0> ≥ <Ω|H|Ω>

3. Mar 14, 2012

### dpa

hi,
could you give physical meaning rather than mathematical.
I am not an expert you see.

4. Mar 14, 2012

### lfighter

A non perturbative vacuum may be topologically different from the trivial or ordinary vacuum. That is, one cannot use a topologically trivial transformation(homotopic to identical mapping) to transform it to the trivial vacuum. You can read some references on instanton to get more information.

5. Mar 14, 2012

### tom.stoer

Typically a non-perturbative vacuum is not 'empty'. In QCD you have chiral symmetry breaking with a non-vanishing order parameter indicating a phase transition. The order parameter is the so-called quark condensate $\langle \bar{q}q\rangle \neq 0$. That means that in the phase where chiral symmetry is broken the 'vacuum' is not 'empty' but 'contains quark-antiquark pairs'.

Usually you would assume that $\langle \bar{q}q\rangle = 0$ b/c of normal ordering, but this applies only to the trivial vacuum state.

6. Mar 18, 2012

### Naty1

Hi dpa:
I'm not getting this yet......

"Typically a non-perturbative vacuum is not 'empty'....

is a perturbative vacuum 'empty'.....???? that doesn't sound like this description:...

[...the article provides some interesting background]

http://en.wikipedia.org/wiki/Vacuum_state#Non-vanishing_vacuum_state

and this:

http://en.wikipedia.org/wiki/Zero-point_energy

Last edited: Mar 18, 2012
7. Mar 18, 2012

### tom.stoer

Yes, in a certain sense the perturbative vacuum is 'empty'; it's annihilated by typical field operators, so the result for counting particles in the vacuum is zero (after normal ordering); the examples you give (condensates like BCS, QCD ground state, non-vanishing vev for Higgs, ...) are all examples for non-perturbative vacuum states.

8. Mar 18, 2012

### Naty1

you Are referring to particals, here, not virtual particles...right???

9. Mar 18, 2012

### tom.stoer

yes!

10. Mar 18, 2012

### marcus

Is the number of particles in a region always well-defined?
Say in the case the geometry is curved, or there are different observers?
I've heard people say it's not a well-defined concept.

11. Mar 18, 2012

### atyy

The non-perturbative vacuum is the true ground state of a system.

If the system is strongly interacting, then we may not know how to solve our equations to get the true ground state. If there is a non-interacting system whose ground state we do know, then we may try to write the true ground state approximately as the ground state of the non-interacting system plus some, hopefully small, corrections. (That doesn't always work.)

12. Mar 19, 2012

### martinbn

No, it is not well defined in curved beckground, in the sense that it is coordinate dependent. But i have also heard that even in flat spacetime (and only inertial systems considered) 'particle' is not well defined in a bounded region.

13. Mar 19, 2012

### marcus

I have seen what you heard demonstrated mathematically*. It seems that the idea of "particle" and the number of particles taking part in any given circumstance is highly observer dependent and geometry dependent.

"Particle" seems very far from being a fundamental, background independent, concept. More of a mathematical convenience useful in specific circumstances. Rather than something in nature.

Vacuum also observer dependent.

*Google "rovelli particle" and get http://arxiv.org/abs/gr-qc/0409054 What is a particle?

14. Mar 19, 2012

### Ilmrak

Exactly, even in QFT the number of particles in any bounded region is not an observable.
The particle number operators in 2 bounded disjoint region do not commute for space-like separations (i's not a local function of the fields), they only approximately commute for high separations.

What is an observable is for example the charge contained in any bounded region.

Ilm

15. Mar 19, 2012

### martinbn

Yes, I have seen that too. My comment was that even after fixing the background even in flat spacetime QFT, the concept of particle is difficult. Ilmrak's post explains better, what I've only heard.

16. Mar 19, 2012

### Naty1

Starting with just

I'd be really doubtful on a conceptual basis that a particle would be well defined if we know all our best theories falter at the singularity of a black hole....are there even 'particles' [mass, space] in those extreme conditions......where do they 'go' ....in fact where does space go....that seems a 'singularity' in time....it only takes a single exception like this to hint the rest of GR and QM are most likely approximations....it's perhaps not just particles that may not be so well defined as we take them to be in everyday less extreme conditions.

String theory suggests that it may be the configuration of higher dimensional spaces that influences string [particles] properties....their vibrational patterns and energies for example ....so when spacetime jiggles around or morphs from one region to another it seems plausible that our perception of particles might also change...because they change.

Further, the Unruh effect [regarding vacuum state temperatures] suggests different observers read coincident spacetime vacuum conditions differently....another hint that things globally and locally may be more surprising then we expect.

Both these concepts seem to support Marcus post.

I read Rovelli's Introduction and this caught my attention:

Not surprising......
I put the article on my reading list....

17. Mar 20, 2012

### Naty1

For those who might be interested in a complementary discussion, with some good explanations of the mathematical apparatus involved, check this one from March 2010:

What is a Particle.