Bosons and Fermions in a rigorous QFT

[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" :tongue2: 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?

 

[tom.stoer]

... Haag's theorem is the most basic

I have to get the Jaffe paper b/c Haags theorem says something different.

It says that we have to deal with (infinitly many) unitarily inequivalent Hilbert spaces for the free and the interacting theory (theories). Afaik it does not say that there does not exist a Fock space representation for any Hilbert space representation of an interacting theory.

Ogf course one can conclude that if a Fock space representation of an interacting theory does exist, it must be unitarily inequivalent to the free Fock space.

In practice one can avoid the problems of Haags theorem by enclosing the system in a large but finite 3-torus (which is then no longer a rigorous approach)

 

[Physics Monkey]

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" :tongue2: 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.

This is absolutely correct. There are even numerical methods, like stochastic series expansion, that statistically sample thousands of terms of the perturbation series in order to actually find the properties of a relatively large interacting system perturbatively. For example, in any finite system the operator $\exp{(-\beta H)}$ converges, and we can estimate how many terms you need by asking when $(\beta ||H||)^n/n! \sim 1$. If the system contains N spins with typical energy J then this criterion gives $n \sim \beta J N$ which clearly grows with system size.

Another perspective is provided by timescales. Any finite quantum system is quasiperiodic, but the recurrence time may be very long. Typically one must wait for a time of order the smallest spacing between energy eigenvalues. If you have our spin system above, say with Heisenberg interaction, then the largest energy is roughly NJ and the smallest is -NJ but there are 2^N states, hence the typical spacing, in the middle of the spectrum, should be something like $\delta 2 N J /2^N$. To see a recurrence if we have a generic state we must wait for a time of roughly $1/\delta$ which is exponential in system size. Similarly, to see the magnetization flip in a large but finite system, we might have to wait a very long time.

 

[A. Neumaier]

I have to get the Jaffe paper b/c Haags theorem says something different.

It says that we have to deal with (infinitly many) unitarily inequivalent Hilbert spaces for the free and the interacting theory (theories). Afaik it does not say that there does not exist a Fock space representation for any Hilbert space representation of an interacting theory.

It says you need one of the reps that are unitarily inequivalent to the (unique) Fock representation.

In practice one can avoid the problems of Haags theorem by enclosing the system in a large but finite 3-torus (which is then no longer a rigorous approach)

But on a torus, you don't have anymore an S-matrix!

 

[A. Neumaier]

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?

There is no Fock space with only a finite number of particles. The particle number operator in a Fock space has the nonnegative integers as its spectrum - which means that there are states with any number of particles.

 

[A. Neumaier]

Is there anybody who believes that a rigorous QED exists?

I do. There are also quite a number of mathematical physicists who study rigorous versions
of theories closer and closer to true QED. I think it is just a matter of time before someone will find the right limit that allows one to make QED rigorously.

 

[DarMM]

I have to get the Jaffe paper b/c Haags theorem says something different.

It says that we have to deal with (infinitly many) unitarily inequivalent Hilbert spaces for the free and the interacting theory (theories). Afaik it does not say that there does not exist a Fock space representation for any Hilbert space representation of an interacting theory.

Yes, Haag's theorem states that the free and interacting theories are unitarily inequivalent. However generalised free fields "span" the space of Fock representations and hence if the interacting field isn't equivalent to them, it can't have a Fock space structure on its Hilbert Space.

 

[DarMM]

Is there anybody who believes that a rigorous QED exists?

I think there is certainly a greater likelyhood of it existing than $$\phi^{4}$$ for example. There is still some unusual results which suggest it may exist, such as the chiral fixed point discussed by Luscher.

 

[tom.stoer]

... if the interacting field isn't equivalent to them, it can't have a Fock space structure on its Hilbert Space.

Hm, why? What is the (unique) Fock representation? How is it defined?

I understand perfectly that the interacting theory does not live on the Fock space of the free theory. But why is an 'interacting Fock space' not possible?

 

[DarMM]

Hm, why? What is the (unique) Fock representation? How is it defined?

I'm not sure what the (unique) Fock representation is as A. Neumaier mentioned it, perhaps he can explain.

I understand perfectly that the interacting theory does not live on the Fock space of the free theory. But why is an 'interacting Fock space' not possible?

Since generalised free-fields span the space of all Fock representations and the interacting theory is unitarily inequivalent to all free-theories, it is then unitarily inqueivalent to all Fock spaces.

 

[A. Neumaier]

Hm, why? What is the (unique) Fock representation? How is it defined?

I understand perfectly that the interacting theory does not live on the Fock space of the free theory. But why is an 'interacting Fock space' not possible?

The one and only Fock representation of a scalar particle of mass m is the standard representation given in each textbook, the direct sum of all symmetrized tensor product of the one-particle space.

What should an interacting Fock space be?