Bosons and Fermions in a rigorous QFT

[kof9595995]

I'm wondering, is there still a sharp distinction between Bosons and Fermions in a rigorous QFT, if exsits?
My question is motivated by the following, consider one of the equations of motion of QED:
\partial_\nu F^{\nu \mu} = e \bar{\psi} \gamma^\mu \psi
In our familiar perturbative QED (Here I'm not 100% sure if I use the word "perturbative" correctly. I simply mean fields are quantized as free fields, and we introduce an interaction built from free fields operators, like an iteration method), LHS is made of Bosonic operators and RHS is made of Fermionic operators, and since the Bosonic sector and Fermionic sector are independent in the total Fock space, perturbative QED fails to satisfy this equation of motion.
I suppose if a rigourous QED exists, this equation of motion should be satisfied, but this in turn means the fermion operator and bosonic operator must act on a Hilbert space they share together, then is there still a sharp distinction between Bosons and Fermions?

 

[DrDu]

One of the greater successes of axiomatic QFT was the proof that at least in some quantum field theories bosonic and fermionic behaviour results quite naturally for localized charges.
See e.g. the book by R. Haag, Local quantum physics.

 

[tom.stoer]

The above mentioned field equation cannot be quantized directly b/c it has to be gauge fixed. In A°=0 at least for the time-indep. constraint (the Gauss law) this equaton is implemented on the physical Hilbert space. This does not require any "free field approach".

 

[strangerep]

[...] LHS is made of Bosonic operators and RHS is made of Fermionic operators, and since the Bosonic sector and Fermionic sector are independent in the total Fock space, perturbative QED fails to satisfy this equation of motion.

The product of two spin-1/2 operators is bosonic (being a superposition of spin-0 and spin-1 in general). (You might want to review Clebsch-Gordan coefficients and associated angular momentum decomposition theory in ordinary QM if you're not already familiar with it.)

[...] then is there still a sharp distinction between Bosons and Fermions?

Yes, in the sense that there's a superselection rule between them. But when you start forming products, things get more complicated.

 

[tom.stoer]

Perhaps it makes sense to consider a simple qm example; the problem for the two-dim. harmonic oscillator as a toy model would be

H_i = \frac{1}{2}p_i^2 + \frac{1}{2}x_i^2 = a_i^\dagger a_i + \frac{1}{2}

H = H_1 + H_2

Now we can easily solve equations like

(H - N)|N\rangle = 0

for some eigenvalues N, but structurally the equation "bosonic operator = fermionic operator" would be something like

(H - N) = 0

and this is obviously not allowed as an operator equation b/c

(H - N) = 0 \;\Rightarrow\; (H - N)|m,n\rangle = 0 \;\forall m,n \;\Rightarrow\; (m+n+1-N) = 0\;\forall m,n

Gauge fixing introduces some additional structures like resolving "unphysical bosonic operators" in terms of fermionic operators via Gauss law, but I still don't see how this is sufficient to resolve the problems for the remaining operator equations. Neverthelesse there should be some solution e.g. for QE´D and QCD where these problems have been treated w/o using perturbation theory.

 

[DrDu]

I suppose if a rigourous QED exists, this equation of motion should be satisfied, but this in turn means the fermion operator and bosonic operator must act on a Hilbert space they share together, then is there still a sharp distinction between Bosons and Fermions?

Is there anybody who believes that a rigorous QED exists?

 

[tom.stoer]

I don't believe in a rigorous QED ...

... but this problem seems to be trivial and there must be a solution in "standard textbook QED with canonical quantization using Fockspace".

I think the subtlety is the regularization of the operator product on the r.h.s. which requires e.g. point splitting with gauge-field insertion (in order not to destroy gauge invariance); this would introduce a gauge field dependency on the r.h.s. whereas the fermionic contribution on the l.h.s. comes from the solution of the Gauss law constraint i.e. A° expressed in terms of fermionic charge density.

That means that quantization, gauge fixing and regularization translates the equation

"bosonic operator = fermionic operator"

into something like

"bosonic + fermionic operator = fermionic + bosonic operator"

which has a chance to hold as operator equation in terms of standard Fock space creation and annihilation operators.

This is of course no rigorous proof but is indispensable already for standard textbook QED.

 

[kof9595995]

One of the greater successes of axiomatic QFT was the proof that at least in some quantum field theories bosonic and fermionic behaviour results quite naturally for localized charges.
See e.g. the book by R. Haag, Local quantum physics.

That sounds quite a long way to go, is there a layman explanation of it?

 

[kof9595995]

I don't believe in a rigorous QED ...

Then what's the underlying theory that makes perturbative QED plausible?

 

[kof9595995]

Neverthelesse there should be some solution e.g. for QE´D and QCD where these problems have been treated w/o using perturbation theory.

Can you give a concrete example? Better a QED example since I'm not familiar with QCD.

 

[tom.stoer]

My example is described in #7

We must not mix two issues
- perturbative QED
- Fock space with creation and annihilation operators

The latter does not imply perturbation theory

 

[kof9595995]

My example is described in #7

We must not mix two issues
- perturbative QED
- Fock space with creation and annihilation operators

The latter does not imply perturbation theory

Could you elaborate? I have always thought a Fock space only makes sense for perturbation theory, since I would imagine an interaction should destroy the simple direct product structure of different sectors.

 

[DrDu]

That sounds quite a long way to go, is there a layman explanation of it?

It was formulated about 1969 but there are still people working on it. Its an extremely mathematical approach.
See the article by Roberts:
http://kolxo3.tiera.ru/M_Mathematics/MA_Algebra/MAg_Algebraic%20geometry/Connes%20A.,%20Cuntz%20J.,%20et%20al.%20Noncommutative%20geometry%20(CIME%20lectures,%202000,%20LNM1831,%20Springer%202004)(359s)_MAg_.pdf#page=274

 

[Physics Monkey]

A useful perspective on this question is provided by lattice gauge theory. For example, Gauss' law, which is the 0th component of the equation you wrote, is satisfied as an identity on the physical hilbert space. In other words, in the true hilbert space electric field lines can only end where charges are located. Nevertheless, there are many ways to distinguish bosons and fermions. In this model the fermions carry charge while the gauge bosons do not. There are composite operators made of fermions that are bosonic and carry charge, but there is still a fermion number that remains sensible.

Of course, this is not to say that there is no blurring of the lines. Bosonization in one dimension is a procedure for exchanging bosons and fermions (and is even relatively rigorous). In higher dimensions one can have solitons built from gauge and bosonic matter degrees of freedom that can carry weird charges and even be fermions.

 

[tom.stoer]

Could you elaborate? I have always thought a Fock space only makes sense for perturbation theory, since I would imagine an interaction should destroy the simple direct product structure of different sectors.

Of course going beyond perturbation theory is still difficult, but neither impossible nor excluded in principle

Think about a Hamiltonian

H = \frac{p^2}{2} + V(x)

You can rewrite this as

H = \frac{p^2}{2} + \frac{x^2}{2} + \left[V(x) - \frac{x^2}{2}\right] = \frac{p^2}{2} + \frac{x^2}{2} + \tilde{v}(x) = a^\dagger a + \frac{1}{2} + \tilde{u}

Now it's up to you to solve this problem exactly or to use perturbation theory in u.

I think it's possible to rewrite a QFT in terms of creation and annihilation operators acting on Fock spaces, but I don't see why a perturbative treatment is mandatory.

 

[DrDu]

Of course going beyond perturbation theory is still difficult, but neither impossible nor excluded in principle

Think about a Hamiltonian ...

I don't consider this problem as too representative for the situation in QFT. The main problem in field theory is that the perturbed and free fields live in different Hilbert spaces -- a problem usually absent in ordinary QM.
Superconductivity can be taken as a toy model on how to solve this problem:
The BCS Hamiltonian can be diagonalized introducing new field operators by the Bogoliubov Valatin transformation. The new field operators are also fermionic but describe the interacting particles.
They cannot be obtained perturbationally from the free electron gas.

 

[tom.stoer]

The main problem in field theory is that the perturbed and free fields live in different Hilbert spaces ...
... Superconductivity can be taken as a toy model on how to solve this problem:
The BCS Hamiltonian can be diagonalized introducing new field operators by the Bogoliubov Valatin transformation. The new field operators are also fermionic but describe the interacting particles.
They cannot be obtained perturbationally from the free electron gas.

I never said that.

What I am saying is that one may start with Fock space operators and then use some non-perturbative techniques. Bogoljubov transformation, bosonization, ... are examples. That is more than just a solution, it's a kind of formal re-definition of the theory.

There is no need to use perturbation theory only b/c of Fock space states, neither before nor after Bogoljubov transformation.

 

[kof9595995]

Of course going beyond perturbation theory is still difficult, but neither impossible nor excluded in principle

Think about a Hamiltonian

H = \frac{p^2}{2} + V(x)

You can rewrite this as

H = \frac{p^2}{2} + \frac{x^2}{2} + \left[V(x) - \frac{x^2}{2}\right] = \frac{p^2}{2} + \frac{x^2}{2} + \tilde{v}(x) = a^\dagger a + \frac{1}{2} + \tilde{u}

Now it's up to you to solve this problem exactly or to use perturbation theory in u.

I think it's possible to rewrite a QFT in terms of creation and annihilation operators acting on Fock spaces, but I don't see why a perturbative treatment is mandatory.

But here if you creat the Hilbert space using creation operators, it might not be the same with the true Hilbert space, then the best you can get is a perturbation theory.

 

[kof9595995]

Is there anybody who believes that a rigorous QED exists?

I'd like the ask the question again, if a rigorous QED doesn't exist, what would be underlying theory of perturbative QED?

 

[tom.stoer]

But here if you creat the Hilbert space using creation operators, it might not be the same with the true Hilbert space, then the best you can get is a perturbation theory.

At least in QM tis is not true.

Which states do I miss?

I can construct nearly arbitrary operators from the creation and annihilation operators; look at the method of coherent states, for example.

And please not that up to now we haven't defined the Hilbert space, neither for x and p, nor for the creation and annihilation operators! So if we need to change the entire Hilbert space for some reason we are not forced to do this by introducing creation and annihilation which are nothing else but linear combinations of x and p. There is physics behind it so far.

 

[kof9595995]

At least in QM tis is not true.

Which states do I miss?

I can construct nearly arbitrary operators from the creation and annihilation operators; look at the method of coherent states, for example.
And please not that up to now we haven't defined the Hilbert space, neither for x and p, nor for the creation and annihilation operators! So if we need to change the entire Hilbert space for some reason we are not forced to do this by introducing creation and annihilation which are nothing else but linear combinations of x and p. There is physics behind it so far.

I kinda get what you mean, but I think in QFT the situation is quite different, and I'm more curious about QFT senario.
In QM, before we define any operator, we set the full Hilbert space as all square-integrable functions, and we define how x and p (so are a and a+) act on this full Hilbert space, then according to the specific Hamiltonian we have, we can construct the subspace using a and a+, though the construction may not be as simple as harmonic oscillator case. I guess this is what you mean.
However in an interacting QFT, the true structure of the full Hilbert space is not clear yet, let alone defining how a and a+ act on the space. The normal treatment is to use the Hilbert space from free theory (a Fock space) and then do things perturbatively. And my point is, if the interaction actually destroys any existence of direct-product structure of the true Hilbert space, then whatever kind of Fock space you use will be just an approximation at best, and if this is the case, I think Fock space does necessarily imply a perturbative approach.

 

[Physics Monkey]

I'd like the ask the question again, if a rigorous QED doesn't exist, what would be underlying theory of perturbative QED?

Lattice gauge theory or an asymptotically free gauge theory in a partial higgs phase can provide a UV completion of low energy QED.

 

[atyy]

Googling "Grassmann Osterwalder-Schrader" produced these papers which may be relevant.

Jaffe, Constructive Quantum Field Theory
"The Euclidean methods also apply to theories with fermions, at least for examples with interactions that are quadratic in the fermions. This is the case for free and for “Yukawa type” interactions, used extensively in physics."

Benfatto, Falco, Mastropietro, Functional Integral Construction of the Thirring model
"Proposed by Thirring half a century ago, the Thirring model is a Quantum Field Theory of a spinor field in a two dimensional space-time, with a self interaction ..."

 

[strangerep]

In QM, before we define any operator, we set the full Hilbert space as all square-integrable functions, and we define how x and p (so are a and a+) act on this full Hilbert space,

In modern QM, that's become kinda backwards. We start with an algebra of observable quantities which characterize the (class of) system being modeled, and then (try to) construct a unitary representation (i.e., construct a Hilbert space) in which those quantities are represented as operators on the Hilbert space. For many interesting case (like x,p) above, it turns out that a rigged Hilbert space is more convenient than an ordinary Hilbert space. See the early chapters of Ballentine for a more detailed exposition of all this.

However in an interacting QFT, the true structure of the full Hilbert space is not clear yet, let alone defining how a and a+ act on the space. The normal treatment is to use the Hilbert space from free theory (a Fock space) and then do things perturbatively. And my point is, if the interaction actually destroys any existence of direct-product structure of the true Hilbert space, then whatever kind of Fock space you use will be just an approximation at best, and if this is the case, I think Fock space does necessarily imply a perturbative approach.

That's essentially what renormalization tries to fix. At each order of perturbation, we adjust the space we're working with, either by adjusting the Hamiltonian, or applying a Bogoliubov-like transformation to the basic c/a operators (cf. Shirokov's approach: http://arxiv.org/abs/nucl-th/0102037[/url]).

BTW, the presence of interaction doesn't necessarily "destroy" direct-product structure. It simply means that the Hamiltonian can now mix stuff from different sectors.

 

[atyy]

That's essentially what renormalization tries to fix. At each order of perturbation, we adjust the space we're working with, either by adjusting the Hamiltonian, or applying a Bogoliubov-like transformation to the basic c/a operators (cf. Shirokov's approach: http://arxiv.org/abs/nucl-th/0102037[/url]).[/QUOTE]

How does this tie in with the view that renormalization, say in QED, just preserves the important low energy terms, and that at high energies non-renormalizable terms or even new degrees of freedom should enter?

 

[kof9595995]

BTW, the presence of interaction doesn't necessarily "destroy" direct-product structure. It simply means that the Hamiltonian can now mix stuff from different sectors.

But even the perturbative Hamiltonian can "mix stuff from different sectors". Let me elaborate my original point: my starting point is that in a mathematically rigorous QED, the equation of motion in my original post must be satisfied. However the perturbative approach definitely fails to satisfy it, because LHS and RHS are defined on independent sectors, so for example, if I sandwich (i.e. taking rayleigh quotient) both sides using photon states, then LHS is mostly nonzero but RHS is always 0. So to satisfy the equation, the true Hilbert space must somehow take Bosonic and Fermionic sector as one, but I can't proceed with the reasoning because I haven't acquired enough maths.

 

[kof9595995]

Lattice gauge theory or an asymptotically free gauge theory in a partial higgs phase can provide a UV completion of low energy QED.

But as far as I know, an constant issue of Lattice theory is about how to get to the continuum limit, and this isn't very well-established yet. And what I'm trying to argue is, QED is probably some limit of underlying bigger theory, but within this limit we should have a mathematically rigorous framework. Just like classical mechanics and quantum mechanics, the former is a limit of the latter, but still classical mechanics is a mathematically rigorous theory.

 

[atyy]

That's essentially what renormalization tries to fix. At each order of perturbation, we adjust the space we're working with, either by adjusting the Hamiltonian, or applying a Bogoliubov-like transformation to the basic c/a operators (cf. Shirokov's approach: http://arxiv.org/abs/nucl-th/0102037[/url]).[/QUOTE]

How does this tie in with the view that renormalization, say in QED, just preserves the important low energy terms, and that at high energies non-renormalizable terms or even new degrees of freedom should enter?

I took a look at Haag's "Local Quantum Physics". He describes both views of renormalization, but doesn't give any link between them. The view you describe is related to the "Algebriac QFT" formalism, while the latter Wilsonian view is related to the "Constructive QFT" formalism, which typically constructs a statistical field theory, and checks that it satisfies the Osterwalder-Schrader conditions in order to make it a Minkowski spacetime QFT.

 

[strangerep]

I took a look at Haag's "Local Quantum Physics". He describes both views of renormalization, but doesn't give any link between them. The view you describe is related to the "Algebriac QFT" formalism, while the latter Wilsonian view is related to the "Constructive QFT" formalism, which typically constructs a statistical field theory, and checks that it satisfies the Osterwalder-Schrader conditions in order to make it a Minkowski spacetime QFT.

Which sections of Haag? There's some stuff in the early sections, but also in the later ch VIII.
(I should probably refresh my memory of Haag a bit more thoroughly, else I risk talking out of my rear... :-)

 

[strangerep]

[...] Let me elaborate my original point: my starting point is that in a mathematically rigorous QED, the equation of motion in my original post must be satisfied. However the perturbative approach definitely fails to satisfy it, because LHS and RHS are defined on independent sectors, so for example, if I sandwich (i.e. taking rayleigh quotient) both sides using photon states, then LHS is mostly nonzero but RHS is always 0. So to satisfy the equation, the true Hilbert space must somehow take Bosonic and Fermionic sector as one, [...]

This is an example of how the space of states in the free theory does not satisfactorily span the space of physical states in the full physical (interacting) theory.

The field operators in the exact form of Gauss' law correspond to the physical fields, not the free fields. Some approaches to QFT try to construct physical field operators perturbatively in terms of (increasingly-complicated) products of the free field operators. Among other things, this procedure must ensure that the new field operators still correspond to suitable Poincare unirreps with the physically correct spin, etc.

Do you recall the earlier thread where I talked about dressing the asymptotic electron states using coherent photon operators? After such dressing has been applied, the commutation relations are a bit different. E.g., the commutator between an electron operator and the electric field operator is no longer zero. Instead, it gives the usual Coulomb field of a charged electron. (This is all at low momenta, since the main task there was to deal with IR divergences.) A similar construction (of Dirac) also shows how to banish some parts of the unphysical EM gauge freedom.

At least one of the rigorous QFT results that I know proceeds via a related process of dressing transformations applied to the basic operators:

J. Glimm,
"Boson Fields with the :Φ^4: Interaction in Three Dimensions",
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1103840981

Warning: very few people can safely read that on an empty stomach...

 

[atyy]

Which sections of Haag? There's some stuff in the early sections, but also in the later ch VIII.
(I should probably refresh my memory of Haag a bit more thoroughly, else I risk talking out of my rear... :-)

In the second edition, the first sort of renormalization is in section 2.4, while the second type is in the section starting p323 "Algebraic Approach versus Euclidean Quantum Field Theory". I think that in his language "renormalization" always means the first type, because when discussing the second type he says something like it does away with renormalization (can't remember the exact words, I read it in the library, and am getting the section references from a search on Amazon).

 

[tom.stoer]

The field operators in the exact form of Gauss' law correspond to the physical fields, not the free fields.

...

Do you recall the earlier thread where I talked about dressing the asymptotic electron states using coherent photon operators? After such dressing has been applied, the commutation relations are a bit different. ...

This is exactly what I had in mind in #3

The above mentioned field equation cannot be quantized directly b/c it has to be gauge fixed. In A°=0 at least for the time-indep. constraint (the Gauss law) this equaton is implemented on the physical Hilbert space. This does not require any "free field approach".

It is interesting that in the very end the physical fields can be decomposed using physical creation and annihilation operators spanning a physical Fock space. Now I understand where the confusion comes from; people seem to confuse Fock space with free particle Fock space plus perturbation theory. This (implicit) assumption is wrong of course.

A Fock space is nothing else but a direct sum of tensor products of copies of a single-particle Hilbert space H.

But it is not necessary that the Hilbert space H is spanned by free fields; any single particle Hilbert space with the correct creation and annihilation operator algebra is sufficient. So if it's possible to construct a suitable transformation from free to 'physical' or 'dressed' fields and a physical Hilbert space then the latter one can be decomposed into physical Fock states. This has been done in QCD in orer to study confinement in the canonical formulation (the problem with the construction of the physical Hilbert space is of course always the same: complete gauge fixing, taming Gribov ambiguities etc.; anyway - these problems do by no means spoil the Fock space approach using physical fields)

 

[Physics Monkey]

But as far as I know, an constant issue of Lattice theory is about how to get to the continuum limit, and this isn't very well-established yet. And what I'm trying to argue is, QED is probably some limit of underlying bigger theory, but within this limit we should have a mathematically rigorous framework. Just like classical mechanics and quantum mechanics, the former is a limit of the latter, but still classical mechanics is a mathematically rigorous theory.

I would disagree about the continuum limit. We understand very well that the lattice theory flows to the continuum theory plus irrelevant operators. The essential properties of the low energy theory are there. For example, the heat capacity in lattice qed at low temperatures in the deconfined phase is proportional to T^3 just as you would get for free photons. Similarly, the long distance decay of gauge invariant correlation functions is exactly what you would expect for free photons. Lattice theories of qcd are also quite advanced, including fairly good numbers for hadron masses, although treating fermions dynamically is always troublesome because of the sign problem.

I also think it is interesting to note that classical electromagnetism is also not a rigorous theory, at least when thinking about point charges. Indeed, classical fluid dynamics and general relativity are also not known to be free of singularities.

 

[strangerep]

It is interesting that in the very end the physical fields can be decomposed using physical creation and annihilation operators spanning a physical Fock space.

...except that they might not satisfy exactly the same CCRs/CARs as in the free case.
DarMM once mentioned something about this, but we never got to hear the full story.

people seem to confuse Fock space with free particle Fock space plus perturbation theory. This (implicit) assumption is wrong of course.

A Fock space is nothing else but a direct sum of tensor products of copies of a single-particle Hilbert space H.

Yes, yes, and yes.

But it is not necessary that the Hilbert space H is spanned by free fields; any single particle Hilbert space with the correct creation and annihilation operator algebra is sufficient.

It's still not clear to me what the "correct" c/a operator algebra should be. E.g., physical electron operators probably shouldn't commute with physical photon operators like they do in the free theory. Not sure about this, though. I recall a theorem (in Barut?) about how a very large class of algebras can be expressed in terms of operators from a Heisenberg algebra.

 

[tom.stoer]

I remember the 'dressing' and the 'gauge fixing by unitary transformations'; both approaches seem to be similar b/c they partially 'solve' some field equations.

The first approach (dressing) changes the operator algebra; it can be solved explicitly in some 1+1 dim. field theories like the Schwinger model (here one formally solves the Dirac equation by exponentiation using a gauge field string with a path ordered product).

The second approach does not change the operator algebra; the Gauss law is solved but b/c a unitary trf. is used, all operator algebras remain unchanged. This may be spoiled by regularization which requires gauge-invariant point splitting (I can only remember the two-dim. case).

In the first case the interaction is "hidden" in the dressed fields; they create the physical Coulomb interaction, but the interaction term itself looks trivial algebraically. In the second case the interaction terms are constructed explicitly and in principle they can be expressed using physical Fock space operators.

In the second case the (A°=0 & Coulomb gauge) Hamiltonian contains one piece which shows directly the color-electric Coulomb potential:

H_C = g^2 \int d^3x \int d^3y \,\text{tr}\,J^{-1}(x)\,\rho(x)\,(-D\partial)^{-1}\,(-\partial^2)\,(-D\partial)^{-1}\,J(y)\,\rho(y)

with D = ∂ + gA, A being the gauge-fixed gluon field, J being the Fadeev-Popov determinant J = det(-D∂), ρ = ρ[q] + ρ[A] being the total color charge with quark and gluon contribution (w/o J, D and ρ[A] in H_C the usual Coulomb gauge interaction in QED is recovered)

 

[atyy]

@strangerep, BTW, another interesting comment in Haag was that since the Hilbert spaces for each representation of the CCRs are different, presumably the selection of the representation depends on dynamics. He then says that the advantage of the Lagrangian approach is that it makes it easy to choose the dynamics based on symmetries, and then construct the appropriate Hilbert space after that. (Again, I don't have the page reference, but it should be in one of the two sections I mentioned above.)

Also, it's really interesting to me that BCS has this "rigourous treatment" - I'd always taken it to be unrigourous since it's modelling a condensed matter phenomenon where one can definitely take a lattice cut-off so that Haag's theorem won't apply.

 

[DrDu]

Also, it's really interesting to me that BCS has this "rigourous treatment" - I'd always taken it to be unrigourous since it's modelling a condensed matter phenomenon where one can definitely take a lattice cut-off so that Haag's theorem won't apply.

In solid state physics the appearance of inequivalent representations is due to the system being idealized as infinite. So its not the discreteness of the lattice but the lattice being infinite.

 

[atyy]

In solid state physics the appearance of inequivalent representations is due to the system being idealized as infinite. So its not the discreteness of the lattice but the lattice being infinite.

That's very unintuitive. Naively, I think of the situation as electrons in a potential. The free electrons should be like Fourier modes. Then, the ground state wavefunction when they are put in a potential can't be represented by Fourier decomposition when the system is infinitely large?

 

[kof9595995]

It is interesting that in the very end the physical fields can be decomposed using physical creation and annihilation operators spanning a physical Fock space. Now I understand where the confusion comes from; people seem to confuse Fock space with free particle Fock space plus perturbation theory. This (implicit) assumption is wrong of course.

A Fock space is nothing else but a direct sum of tensor products of copies of a single-particle Hilbert space H.

So what's the justification for the physical Hilbert space being Fock-like? I mean, for a free theory Fock structure is sort of natural, but for an interacting theory I can't convince myself about it.

 

[DrDu]

That's very unintuitive. Naively, I think of the situation as electrons in a potential. The free electrons should be like Fourier modes. Then, the ground state wavefunction when they are put in a potential can't be represented by Fourier decomposition when the system is infinitely large?

When you compare e.g. a normal and a superconducting ground state (or, even simpler, two normal states of different temperature), the occupation of an infinite number of the Fourier modes is different so that the two state have no overlap with each other and no operator localized in a finite region has a matrix element between the two states. Hence the representation of the algebra of observables is distinct.

 

[tom.stoer]

So what's the justification for the physical Hilbert space being Fock-like? I mean, for a free theory Fock structure is sort of natural, but for an interacting theory I can't convince myself about it.

What's your problem?

Suppose you have field operators and their canonical conjugate momenta; you can decompose them (as usual) in k-space and you can introduce the standard linear combination for creation and annihilation operators. These creation and annihilation operators satisfy the usual commutation relations and generate a Fock space; this is a purely 'algebraic' construction, the relevant operators are linear (no operator products are involved) and do not depend on details of the dynamics. For the Fock space construction it's irrelevant whether the field operators are 'free' or 'physical' fields as long as the correct commutation relations are satisfied.

In the above mentioned formulation of (non-perturbatively quantized, fully gauge fixed) QCD the field operators act on a Fock space. The creation and annihilation operators generate physical (gauge fixed) quarks and gluons.

I see what may bother you - the k-space decomposition. Is your problem the plane wave basis for interacting fields?

 

[DarMM]

Suppose you have field operators and their canonical conjugate momenta; you can decompose them (as usual) in k-space and you can introduce the standard linear combination for creation and annihilation operators. These creation and annihilation operators satisfy the usual commutation relations and generate a Fock space; this is a purely 'algebraic' construction, the relevant operators are linear (no operator products are involved) and do not depend on details of the dynamics. For the Fock space construction it's irrelevant whether the field operators are 'free' or 'physical' fields as long as the correct commutation relations are satisfied.

The fields constructed in this manner however will necessarily obey the free Klein-Gordon equation, the only way to prevent this is if there exists no state which all the a_{p} operators annihilate, which eliminates the Fock structure. Hence if you perform this construction for an interacting theory, you're building the in/out fields, not the field that appears in the Hamiltonian.

 

[tom.stoer]

The fields constructed in this manner however will necessarily obey the free Klein-Gordon equation, ...

Why? Canonical commutation relations and all that are much more general than a specific field equation.

Hence if you perform this construction for an interacting theory, you're building the in/out fields, not the field that appears in the Hamiltonian.

I don't understand.

Suppose you have

H_0 = \sum_k \sum_i\omega_k {a^i_k}^\dagger a^i_k

in k-space with additional (internal) index i and the the usual commutation relations; that's trivial.

Now suppose you have an interaction term

H = H_0 + \mathcal{V}[{a^i_k}^\dagger, a^{i^\prime}_{k^\prime}]

How does this destroy the Fockspace structure?

 

[strangerep]

Now suppose you have an interaction term
H = H_0 + \mathcal{V}[{a^i_k}^\dagger, a^{i^\prime}_{k^\prime}]
How does this destroy the Fockspace structure?

The interaction term typically has products consisting only of creation operators, hence can't annihilate the ordinary vacuum.

This problem can't be removed by simple normal ordering, nor by subtracting an infinite constant as is done in the free case.

This issue is the motivation for some less well known perturbation approaches such as Kita, or Shebeko-Shirokov, or Stefanovich.

 

[atyy]

When you compare e.g. a normal and a superconducting ground state (or, even simpler, two normal states of different temperature), the occupation of an infinite number of the Fourier modes is different so that the two state have no overlap with each other and no operator localized in a finite region has a matrix element between the two states. Hence the representation of the algebra of observables is distinct.

So we can still do a Fourier decomposition, but the Fock states built off the two normal states at different temperature are completely orthogonal?

Is it still ok to think of the Fourier states as "free particle states"?

 

[tom.stoer]

The interaction term typically has products consisting only of creation operators, hence can't annihilate the ordinary vacuum.

This is neither a problem nor does it contradict any Fock space property.

As I already said (and you agreed), a Fock space is nothing else but a direct sum of tensor products of copies of a single-particle Hilbert space H. This is still valid. The a's are exactly the single particle operators, they are acting on Fock states.

H is an operator on Fock space; H may not violate the Fock vacuum |0> i.e. there may be a different physical vacuum state |Ω> and so on. All this is not in contradiction with Fock space.

Are you assuming that H|0> = 0? A Fock space does not require any special property of H.

btw.: if you may have a look at the physical interaction term of QCD you will find that it has by no means a product structure; it has non-trivial operators in the denominator.

 

[strangerep]

This is neither a problem nor does it contradict any Fock space property. [...]

Aargh! We're talking at crossed purposes. I was trying to relate it back to what DarMM said in post #42.

The problem is in trying to find a irreducible set of (annihilation)operators a (and their adjoints) and a fiducial state (vacuum) \Omega which is annihilated by all the a, and also by the Hamiltonian, but also such that the entire physical Hilbert space is generated by the creation ops acting on \Omega.

 

[DarMM]

H = H_0 + \mathcal{V}[{a^i_k}^\dagger, a^{i^\prime}_{k^\prime}]

How does this destroy the Fockspace structure?

It destroys it as \mathcal{V} will always map out of the Fock space, so that:
\left(H\phi,H\phi\right) = \infty, for any Fock space state. You need to go to another Hilbert space for it to be well defined.

 

[DrDu]

So we can still do a Fourier decomposition, but the Fock states built off the two normal states at different temperature are completely orthogonal?

Is it still ok to think of the Fourier states as "free particle states"?

With states of non-zero temperature the problem is that we are used to describe them not as pure states but as statistical mixtures. However, alternative descriptions in terms of reducible representations are possible.
Maybe a clearer example is that of non interacting Fermions at zero temperature but one at zero chemical potential and the other one at non-zero potential. It is clear that the two vacuum states have zero overlapp and that action of a finite value of creation or anihilation operators won't change this. However, both vacua can be used to define a Fock space. However, in the space with non-zero mu, the role of creation and anihilation operators has to be inversed for E<mu (which can be seen as a special case of a Bogoliubov Valatin trafo). Nevertheless, also this new operators are Fourier components of the field.

 

[tom.stoer]

It destroys it as \mathcal{V} will always map out of the Fock space, so that:
\left(H\phi,H\phi\right) = \infty, for any Fock space state. You need to go to another Hilbert space for it to be well defined.

This is not a problem of Fock space.

It's a problem of all canonical formulations using Hilbert spaces and not properly regularized (unbounded) Hamiltonians. It can even be a problem in ordinary QM.

I think we are constantly mixing different issues and are running round in circles