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Igor Khavkine
science forum Guru
Joined: 01 May 2005
Posts: 607
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 Posted: Fri May 20, 2005 8:54 am Post subject:
Normal ordering and VEV subtraction
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I'm looking at Wald's book on QFT in curved spacetime. In section 4.6 he
talks about the definition of the stress-energy tensor. It is defined by
a product of two field operators at the same spacetime point. Since the
field operators are actually operator valued distributions, this product
is expected to be divergent (pp.86-87).
A general prescription for regulating this divergence is normal
ordering. For example the product of field operators phi(x)phi(x) can be
expanded in the elemntary mode operators a and a* and manually commuting
all the terms of the form aa* to a*a, neglecting any c-number byproducts
of the commutation. This procedure also renders the vacuum expectation
value of this product zero, <0|phi(x)phi(x)|0> = 0. Neglect technical
difficulties with the choice of the vacuum state. I'm interesting in how
this works for flat backgrounds and even non-relativistic theories.
This suggests another way of regulating the expected divergence, simply
subtracting the vacuum expectation value. Although the product
phi(x)phi(x) and its expectation value <phi(x)phi(x)> with respecto to
any chosen state are ill defined, the product phi(x)phi(y) and its
expectation value <phi(x)phi(y)> is a well defined bidistribution on
test functions in spacetime. So subtraction of the vacuum expectation
value can be defined through the point splitting prescription
lim_{y->x} <phi(x)phi(y)> - <0|phi(x)phi(y)|0>.
In a few simple cases, the normal ordering and the point splitting
prescriptin can be applied explicitly and the results agreee. But this
is still somewhat mysterious to me. Wald claims that "for physically
reasonable states in the standard Fock space, the singular behavior of
the bidistribution <phi(x)phi(y)> is the same as for
<0|phi(x)phi(y)|0>" (paraphrasing slightly).
Now my question: why is Wald's statement true? Why is the singular
behavior of the VEV and expectation value with respect to some other
state the same? Perhaps the reason has more to do with properties of
distributions, bidistributions and test functions than with physical
properties of the theory. I've asked a similar question a while ago, but
in a different context. But I couldn't make sense of the answers to my
satisfaction.
Thanks in advance.
Igor |
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Guest
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| Posted: Fri May 20, 2005 9:20 pm Post subject:
Re: Normal ordering and VEV subtraction & simpler definition of normal ordering
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Igor Khavkine wrote:
| Quote: | I'm looking at Wald's book on QFT in curved spacetime.
[which works within the algebraic approach]. |
| Quote: | In a few simple cases, the normal ordering and the point splitting
prescriptin can be applied explicitly and the results agreee. But this
is still somewhat mysterious to me. Wald claims that "for physically
reasonable states in the standard Fock space, the singular behavior of
the bidistribution <phi(x)phi(y)> is the same as for
0|phi(x)phi(y)|0>" (paraphrasing slightly).
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Reasonable assumption is that the state space is
{ A|0>: A from a class of well-behaved operators }
Then for any state A|0>
<phi(x)phi(y)> = <0| A* phi(x) phi(y) A |0>
and the A's should have no effect on the singularity behavior.
A simpler prescription which subsumes Wald and everyone else -- but
which apparently eluded Wald and others -- is to just define
N(phi(x)phi(y)) = unique (up to constant) operator Z
such that Ad(Z) = Ad(phi(x)phi(y)).
Note that even though phi(x)phi(y) may be singular, it's *adjoint* is
generally not. Various authors try to state this point by pointing out
that the infinity is "only" a c-number, meaning that it isn't even
there under the Ad(). If ths spectrum of phi(x)phi(y) is bounded, the
constant is fixed by convention of setting the lower bound to 0.
[Note: Ad(A) is the adjoint of A, defined as [A, ()]).
In an irreducible representation, the only operator A with Ad(A) = 0 is
a c-number. Therefore, the above prescription does indeed uniquely
specify the operator ... assuming that a solution exists.
The motivation for the foregoing is simply that nearly the only place
(at least in QFT) the stress-energy tensor is used is in the context of
the Lie bracket and the (generalized) Heisenberg equations:
[P_a, O] = i h-bar dO/dx^a.
In general terms, the concept of adjoint-equivalence allows you to
consider a larger class of operators which are all equal to the
original class of bounded, well-behaved operators, modulo the
equivalence. For these operators, normal ordering defined via
adjoint-equivalence then specifies ... up to a constant ... a unique
bounded operator which behaves the same as the original operator in all
contexts involving the commutators. |
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Arnold Neumaier
science forum Guru
Joined: 24 Mar 2005
Posts: 379
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| Posted: Sat May 21, 2005 8:43 am Post subject:
Re: Normal ordering and VEV subtraction
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Igor Khavkine wrote:
| Quote: | Wald claims that "for physically
reasonable states in the standard Fock space, the singular behavior of
the bidistribution <phi(x)phi(y)> is the same as for
0|phi(x)phi(y)|0>" (paraphrasing slightly).
Now my question: why is Wald's statement true?
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To me this seems to be a working assumption, not a theorem.
It delineates 'physically reasonable states' from the others;
so in a sense, it defines what is 'physically reasonable'.
Arnold Neumaier |
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Alexey Popov
science forum beginner
Joined: 09 May 2005
Posts: 14
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| Posted: Sat May 21, 2005 4:48 pm Post subject:
Re: Normal ordering and VEV subtraction
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| Quote: | of the commutation. This procedure also renders the vacuum expectation
value of this product zero, <0|phi(x)phi(x)|0> = 0. Neglect technical
difficulties with the choice of the vacuum state. I'm interesting in how
this works for flat backgrounds and even non-relativistic theories.
|
In same cases, procedure of normal ordering can give anomaly.
For example, axial anomaly emerge when we ordering axial
current.
--
--- Home Page http://ok.novgorod.net/ap --- |
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Igor Khavkine
science forum Guru
Joined: 01 May 2005
Posts: 607
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| Posted: Mon May 23, 2005 8:58 pm Post subject:
Re: Normal ordering and VEV subtraction
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On 2005-05-21, Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
| Quote: | Igor Khavkine wrote:
Wald claims that "for physically
reasonable states in the standard Fock space, the singular behavior of
the bidistribution <phi(x)phi(y)> is the same as for
0|phi(x)phi(y)|0>" (paraphrasing slightly).
Now my question: why is Wald's statement true?
To me this seems to be a working assumption, not a theorem.
It delineates 'physically reasonable states' from the others;
so in a sense, it defines what is 'physically reasonable'.
|
This still leaves me somewhat empty handed. I now have a condition that
I can use to define "physically reasonable". But I still don't know what
that means concretely.
Assuming that I can write phi(x) in terms of creation annihilation
operators a_k, a*_k, each of which creates a smooth localized field
mode. In other words, that's what <0|phi(x)a*_k|0> looks like for each
k. I can build up a Fock space using the |0> and a*_k's. If I know the
singular behavior of <0|phi(x)phi(y)|0>, then what can I say about the
singular behavoir of <psi|phi(x)phi(y)|psi>, where
|psi> = a*_k1 a*_k2 ...|0>? Presumably knowing this I'll be able to
select the states that are in fact physically reasonable.
Thanks.
Igor |
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Alexey Popov
science forum beginner
Joined: 09 May 2005
Posts: 14
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| Posted: Tue May 24, 2005 6:26 pm Post subject:
Re: Normal ordering and VEV subtraction
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| Quote: | If I know the
singular behavior of <0|phi(x)phi(y)|0>, then what can I say about the
singular behavoir of <psi|phi(x)phi(y)|psi>, where
|psi> = a*_k1 a*_k2 ...|0>?
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This seems like Operator Product Expansion in QFT where product
of two operators can be expressed as sum of complete set of
normal ordered products. And singular behavoir accumulated only in
coefficient functions.
--
--- Home Page http://ok.novgorod.net/ap --- |
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Arnold Neumaier
science forum Guru
Joined: 24 Mar 2005
Posts: 379
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| Posted: Thu May 26, 2005 7:03 pm Post subject:
Re: Normal ordering and VEV subtraction
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Igor Khavkine wrote:
| Quote: | On 2005-05-21, Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
Igor Khavkine wrote:
Wald claims that "for physically
reasonable states in the standard Fock space, the singular behavior of
the bidistribution <phi(x)phi(y)> is the same as for
0|phi(x)phi(y)|0>" (paraphrasing slightly).
Now my question: why is Wald's statement true?
To me this seems to be a working assumption, not a theorem.
It delineates 'physically reasonable states' from the others;
so in a sense, it defines what is 'physically reasonable'.
This still leaves me somewhat empty handed. I now have a condition that
I can use to define "physically reasonable". But I still don't know what
that means concretely.
Assuming that I can write phi(x) in terms of creation annihilation
operators a_k, a*_k, each of which creates a smooth localized field
mode. In other words, that's what <0|phi(x)a*_k|0> looks like for each
k. I can build up a Fock space using the |0> and a*_k's. If I know the
singular behavior of <0|phi(x)phi(y)|0>, then what can I say about the
singular behavoir of <psi|phi(x)phi(y)|psi>, where
|psi> = a*_k1 a*_k2 ...|0>? Presumably knowing this I'll be able to
select the states that are in fact physically reasonable.
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Wald delineates a class of 2-point functions called Hadamard states
that have locally the same kind of singular behavior as the flat
free 2-point functions. This class of states is also natural from
several other points of view, though I cannot give details off-hand
since this is slightly outside my field of knowledge.
Associated to each Hadamard state is a Gaussian state |0>
of the quantum field which is constructed from the 2-point function
via Wick's theorem. This state is often called a 'vacuum state',
though this is not quite appropriate, unless you allow the vacuum
to carry gravitational and electromagnetic fields. A more appropriate
name would be a 'coherent state' since it is the generalization of
coherent states in the Fock spaces considered in optics.
Each Gaussian state produces a Hilbert space of wave functions
consisting of linear combinations of the a*_k1 a*_k2 ...|0>,
weighted by sufficiently smooth functions of the k's to render
their norm finite.
All states in this Hilbert space are also physically reasonable,
but they do not have the same basic (vacuum-like)
status as the Hadamard states since they are no longer Gaussian,
and hence are harder to work with.
But you can evaluate <psi|phi(x)phi(y)|psi> in such a state by
expanding everything in terms of vacuum expectations of expressions
in a's and a^*'s and applying Wick's theorem. Their leading singular
behavior is probably the same as for the Gaussian state itself,
though I haven't tried to check this.
Arnold Neumaier |
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Igor Khavkine
science forum Guru
Joined: 01 May 2005
Posts: 607
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| Posted: Thu May 26, 2005 7:06 pm Post subject:
Re: Normal ordering and VEV subtraction
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On 2005-05-24, Alexey Popov <avp@novgorod.net> wrote:
| Quote: |
If I know the
singular behavior of <0|phi(x)phi(y)|0>, then what can I say about the
singular behavoir of <psi|phi(x)phi(y)|psi>, where
|psi> = a*_k1 a*_k2 ...|0>?
This seems like Operator Product Expansion in QFT where product
of two operators can be expressed as sum of complete set of
normal ordered products. And singular behavoir accumulated only in
coefficient functions.
|
Ok, this makes sense, but again only shifts the problem to another
place. Where can I find a proof/derivation of the operator product
expansion?
I just looked in Weinberg's QFT vol.2. He has a chapter on OPEs, but I'm
finding the discussion a bit dense. Moreover, his derivation of the OPE
is done in the path integral picture, and I'm not exactly sure how to
translate it to the Fock space picture. Clarifications or references
appreciated.
Thanks.
Igor |
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Alexey Popov
science forum beginner
Joined: 09 May 2005
Posts: 14
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| Posted: Sat May 28, 2005 9:14 am Post subject:
Re: Normal ordering and VEV subtraction
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| Quote: | Ok, this makes sense, but again only shifts the problem to another
place. Where can I find a proof/derivation of the operator product
expansion?
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Unfortunaly, OPE not have rigorous proof, except in pertrubative
expantion. In the framework of axiomatic field theory, OPE can not
be obtained from Wightman axioms. But OPE often arise in same
toy (and not only toy) models.
I think that your problem not easy than problem of OPE proof
in general case.
| Quote: | I just looked in Weinberg's QFT vol.2. He has a chapter on OPEs, but I'm
finding the discussion a bit dense.
Moreover, his derivation of the OPE
is done in the path integral picture, and I'm not exactly sure how to
translate it to the Fock space picture.
|
Yes, path integral picture is not rigorous proof. But it give nice
intuitive motivation for OPE in the standart QFT.
--
--- Home Page http://ok.novgorod.net/ap --- |
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Arnold Neumaier
science forum Guru
Joined: 24 Mar 2005
Posts: 379
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| Posted: Mon May 30, 2005 5:18 am Post subject:
Re: Normal ordering and VEV subtraction
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Alexey Popov wrote:
| Quote: | Unfortunaly, OPE not have rigorous proof, except in pertrubative
expantion. In the framework of axiomatic field theory, OPE can not
be obtained from Wightman axioms.
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In 2D field theory, it can, with full rigor.
There is a nice booklet on this:
VG Kac,
Vertex algebras for beginner,
American Mathematical Society, Providence, RI, 1997.
Arnold Neumaier |
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Igor Khavkine
science forum Guru
Joined: 01 May 2005
Posts: 607
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| Posted: Mon May 30, 2005 5:21 am Post subject:
Re: Normal ordering and VEV subtraction
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On 2005-05-28, Alexey Popov <avp@novgorod.net> wrote:
| Quote: | Ok, this makes sense, but again only shifts the problem to another
place. Where can I find a proof/derivation of the operator product
expansion?
Unfortunaly, OPE not have rigorous proof, except in pertrubative
expantion. In the framework of axiomatic field theory, OPE can not
be obtained from Wightman axioms. But OPE often arise in same
toy (and not only toy) models.
I think that your problem not easy than problem of OPE proof
in general case.
|
I would not have expected the level of rigor to be higher than at the
level of perturbation theory. I'd be happy for a reference where such a
derivation is given.
| Quote: | I just looked in Weinberg's QFT vol.2. He has a chapter on OPEs, but I'm
finding the discussion a bit dense.
Moreover, his derivation of the OPE
is done in the path integral picture, and I'm not exactly sure how to
translate it to the Fock space picture.
Yes, path integral picture is not rigorous proof. But it give nice
intuitive motivation for OPE in the standart QFT.
|
While being non-rigorous, since there is a non-rigorous equivalence
between the path integral and the operator pictures. What's not clear to
me is how to translate the non-rigorous motivation of the OPE in a path
integral formalism to a non-rigorous motivation of the OPE in the Fock
space picture. All this non-rigorousness is probably bad for one's
health, but rare indulgences can be tolerated. :-)
Igor |
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Alexey Popov
science forum beginner
Joined: 09 May 2005
Posts: 14
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| Posted: Tue May 31, 2005 4:07 pm Post subject:
Re: Normal ordering and VEV subtraction
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| Quote: | I would not have expected the level of rigor to be higher than at the
level of perturbation theory.
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At the level of pertrubation theory OPE for given operator productcan be derived
by direct computation of Feynman diagrams.
| Quote: | I'd be happy for a reference where such a
derivation is given.
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Classical most cited references are:
1) Nonlagrangian Models Of Current Algebra. K.G. Wilson, Phys.Rev. 179 (1969) 1499
2) K.G. Wilson and W. Zimmermann, Commun.Math.Phys. 24 (1972) 87
But they are so old. I don know more fresh papers.
| Quote: | Yes, path integral picture is not rigorous proof. But it give nice
intuitive motivation for OPE in the standart QFT.
While being non-rigorous, since there is a non-rigorous equivalence
between the path integral and the operator pictures.
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Troubles only in cases with non trivial vacuum. But, for example,in QCD widely used hypothesis of
quark-hardron duality. This can
viewed as manifestation of equivalence of path integral and Fock space
pictures. Of course this only applicable in phenomenological level.
| Quote: | What's not clear to
me is how to translate the non-rigorous motivation of the OPE in a path
integral formalism to a non-rigorous motivation of the OPE in the Fock
space picture.
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In my opinion, Fock space OPE arise simple by proposition that any
dynamic can not make singularity in operator product as in non interacting
case. In non singular case we can simple use Taylor expantion 
Most interesting topic is possible anomalies in operator product. For
example, well known Casimir force is anomaly in normal ordering
procedure.
--
--- Home Page http://ok.novgorod.net/ap --- |
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