Bosons and Fermions in a rigorous QFT

[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

 

[DarMM]

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

I'm not sure what you mean. For \phi^{4}_{2} for example the theory has a well-defined Hamiltonian on a non-Fock Hilbert Space. So there is no problem with the Hamiltonian or the canonical commutation relations, the theory just happens to live on a different space. (There's especially no problem with the Hamiltonian being unbounded, even the simple harmonic oscillator in QM has an unbounded Hamiltonian, in fact even the free particle does). You can also explicitly prove for this theory that only the in/out fields can live on a Fock space.

 

[atyy]

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.

Whereas with finite number of particles, the Hilbert spaces spanned by the different Fock spaces are the same? Or do they get more and more orthogonal with increasing numbers of particles?

Actually, I don't even know something so basic as whether the single-particle harmonic oscillator energy eigenfunctions span the space of wavefunctions for any potential.

 

[Physics Monkey]

Whereas with finite number of particles, the Hilbert spaces spanned by the different Fock spaces are the same? Or do they get more and more orthogonal with increasing numbers of particles?

The issue is whether one can reach one state from another by acting with a finite number of creation or annihilation operators. In any finite volume fermion system on a lattice, any state can be reached from any other state with a finite number of creation and annihilation operators, but the number needed grows with system size. So if one first takes the thermodynamic limit, then systems at different chemical potentials or temperatures have states that differ by an infinite number of creation and annihilation operators (they have infinitely different total energy).

 

[DrDu]

Whereas with finite number of particles, the Hilbert spaces spanned by the different Fock spaces are the same? Or do they get more and more orthogonal with increasing numbers of particles?

Yes, a finite number of particles in an infinite volume corresponds to density 0 and, in the free particle case, to mu=0. For a finite volume, all states with different particle number lie in the same Hilbert space. So to say the chemical potential is a new classical variable which enumerates inequivalent Hilbert spaces in the thermodynamic limit.

 

[atyy]

The issue is whether one can reach one state from another by acting with a finite number of creation or annihilation operators. In any finite volume fermion system on a lattice, any state can be reached from any other state with a finite number of creation and annihilation operators, but the number needed grows with system size. So if one first takes the thermodynamic limit, then systems at different chemical potentials or temperatures have states that differ by an infinite number of creation and annihilation operators (they have infinitely different total energy).

Yes, a finite number of particles in an infinite volume corresponds to density 0 and, in the free particle case, to mu=0. For a finite volume, all states with different particle number lie in the same Hilbert space. So to say the chemical potential is a new classical variable which enumerates inequivalent Hilbert spaces in the thermodynamic limit.

What's a good way to think about this in terms of modelling phenomena? Is it something interesting associated with, say, phase transitions requiring a thermodynamic limit? Or is it a mathematical curiosity due to a convenient approximation for what is in reality is a large but finite number of particles?

 

[DrDu]

What's a good way to think about this in terms of modelling phenomena? Is it something interesting associated with, say, phase transitions requiring a thermodynamic limit? Or is it a mathematical curiosity due to a convenient approximation for what is in reality is a large but finite number of particles?

That's a question that has been discussed a lot. Certainly sharp phase transitions only occur in the thermodynamic limit. These new features which arise as a consequence of some asymptotic idealization are called "emergent". Some people argue that almost all statements we can make about nature in fact refer to emergent qualities. There are some nice articles by Hans Primas, e.g.
http://books.google.de/books?hl=de&...4pV4l-ciXDkmQ7dw#v=onepage&q=Primas-H&f=false

It is also important in the calculation of any broken symmetry phase -- ferromagnetic, superconducting or the like -- that e.g. using Greens function techniques -- they cannot be obtained adiabatically from a non interacting ground state but the parameter which distinguishes the different inequivalent representations -- called anomalous Greens functions -- has to be taken into account ad hoc.

 

[atyy]

That's a question that has been discussed a lot. Certainly sharp phase transitions only occur in the thermodynamic limit. These new features which arise as a consequence of some asymptotic idealization are called "emergent". Some people argue that almost all statements we can make about nature in fact refer to emergent qualities. There are some nice articles by Hans Primas, e.g.
http://books.google.de/books?hl=de&...4pV4l-ciXDkmQ7dw#v=onepage&q=Primas-H&f=false

It is also important in the calculation of any broken symmetry phase -- ferromagnetic, superconducting or the like -- that e.g. using Greens function techniques -- they cannot be obtained adiabatically from a non interacting ground state but the parameter which distinguishes the different inequivalent representations -- called anomalous Greens functions -- has to be taken into account ad hoc.

All that seems pretty physical and interesting. It also seems to clarify the "foundational" question since the real system is finite, so that one could in principle solve it perturbatively from the non-interacting case, then take the thermodynamic limit. It's just that that's too hard, and it's "easier" to take the thermodynamic limit and guess the "non-perturbative" ground state. The non-perturbative nature of the limit then explains why "qualitative" differences appear to exist for large but finite systems.

 

[tom.stoer]

Regarding

\phi^{4}_{2} ... [one] can also explicitly prove that only the in/out fields can live on a Fock space.

That sounds interesting.

Just to make sure that I understand you correctly: you are saying that not only does the fully interacting theory not live on the free-particle Fock space, but that not even a different "interacting" Fock space basis can be constructed in principle?

I still can't believe that w/o a rigorous proof.

Looking at a Bogoliubov transformation for example, the free and the interacting theory live on different Hilbert spaces, but both can be given a Fock space structure with a precise mapping.

 

[DarMM]

That sounds interesting.

Just to make sure that I understand you correctly: you are saying that not only does the fully interacting theory not live on the free-particle Fock space, but that not even a different "interacting" Fock space basis can be constructed in principle?

Yes, only the in/out fields live on a Fock space. If you want a proof there are several, Haag's theorem is the most basic, but in the paper:

James Glimm "Boson Fields with the :ϕ4: Interaction in Three Dimensions", Comm. Math. Phys. 10, 1-47.

You can see an explicit construction of the non-Fock Hilbert space.

 

[kof9595995]

@DarMM
If a Fock structure doesn't exist, is there still the concept of "number of particles"? And are there still Bosonic and Fermionic sectors, if not, how do we make sense of of Fermions and Bosons?
I don't have enough math background to understand too technical stuff like James Glimm's work, so could you explain my above questions in a not-too-technical way?