What Are the States in Quantum Field Theory?

[DrDu]

Dear A Neumaier,
doesn't Haag's theorem prohibit to represent a wavefunction in the Fock space, as you did, for any non-trivial interacting (relativistic) qft?

 

[A. Neumaier]

Dear A Neumaier,
doesn't Haag's theorem prohibit to represent a wavefunction in the Fock space, as you did, for any non-trivial interacting (relativistic) qft?

Indeed, I was representing the free theory, as seemed appropriate in the context of the post by JustinLevy. This is enough for perturbation theory. A nonperturbative discussion of the interacting case is complex, but the principles are the same.

Haag's theorem seems to forbid only that the interacting Fock space is the same as that of the noninteracting theory, while Haag-Ruelle theory seems to be compatible with having an interacting Fock space of a different origin.

 

[schieghoven]

By diagonalizing momenta, one gets the traditional Fock representation in terms of eigenstates $$|p_1,...,p_N>$$, where N=0,1,2,... and each p_k is in R^3, and the wave functions are the coefficients $$\psi_N(p_1,...,p_N)$$ in
$$|\psi> = \sum_N \int dp^N \psi_N(p_1,...,p_N) |p_1,...,p_N>,$$

The N-particle wave function $$\psi_N(p_1,...,p_N)$$ describes an amplitude for N particles with momenta p_1, ... p_N. Could we not simply define a position representation $$\psi_N(x_1,...,x_N)$$ by taking the Fourier transform with respect to each p_j? (Square integrability of either guarantees well-definedness of the Fourier transform.)

From this point of view, the existence of position/momentum representations is purely a consequence of Fourier theory; the two representations are merely two different notations for the same space of states. The question of what dynamics to define on that space is a distinct question (whose answer is not yet clear to me... :-).

 

[A. Neumaier]

The N-particle wave function $$\psi_N(p_1,...,p_N)$$ describes an amplitude for N particles with momenta p_1, ... p_N. Could we not simply define a position representation $$\psi_N(x_1,...,x_N)$$ by taking the Fourier transform with respect to each p_j? (Square integrability of either guarantees well-definedness of the Fourier transform.)

Note that the momenta in relativistic QFT are 4-momenta constrained to a mass shell.
However, parameterizing it by spatial momenta and Fourier-transforming it works indeed for free scalar fields (and is in any textbook on QFT). But for photons, psi must satisfy the transversality condition p_k dot psi =0, which cannot be translated by Fourier transform.
Indeed, a well-known result of Newton and Wigner says that a position operator exists
iff particles are either massive (with arbitrary spin) or massless with spin <1.

For interactive fields, the situation is quite different, since Haag's theorem rules out the Fock representation.

 

[Prathyush]

what is the Haag theorem is saying? Wikipedia is unclear on this.

 

[schieghoven]

But for photons, psi must satisfy the transversality condition p_k dot psi =0, which cannot be translated by Fourier transform.

Each of the p_k are in R^3, and take psi in the temporal gauge (0-component vanishes). Given the Fourier transform is well-defined for all square-integrable functions $\psi_N(p_1,...,p_N)$, doesn't the transversality condition merely restrict physical states for the photon to a proper subspace? The Fourier transform is still well-defined.

Indeed, a well-known result of Newton and Wigner says that a position operator exists
iff particles are either massive (with arbitrary spin) or massless with spin <1.

Once we have a space representation $\psi_1(x_1)$ in the one-particle space, where x_1 in R^3, can we not simply define the space-components of the position operator as usual (multiplication by x_1)? This definition lifts to the overlying Fock space.

For interactive fields, the situation is quite different, since Haag's theorem rules out the Fock representation.

Haag's theorem proves only that the Fock space is not the unique representation of the canonical commutation relations. Suppose we turn the argument around by a priori specifying the space of states as the Fock space over a certain space of functions. Then the CCRs arise as a property of Fock space and the vacuum is unique by construction: there is no problem. I feel that this point of view is to some extent borne out in Weinberg (QTF I pp174-175) taking special effort to present things 'in reverse order' (his words), beginning with the a priori specification of state space and deriving CCRs from that.

 

[A. Neumaier]

Each of the p_k are in R^3, and take psi in the temporal gauge (0-component vanishes). Given the Fourier transform is well-defined for all square-integrable functions $\psi_N(p_1,...,p_N)$, doesn't the transversality condition merely restrict physical states for the photon to a proper subspace? The Fourier transform is still well-defined.
Indeed, but this only shows thatmy description was sloppy. The correct photon Hilbert space is a quotient space of 4-component wave functions modulo addition of a multiple of 4-momentum. This causes a gauge ambiguity of the position representation. But gauges are unobservable, whence these position representations are unphysical.

Haag's theorem proves only that the Fock space is not the unique representation of the canonical commutation relations.

No. This was known long before Haag; probably even to von Neumann in 1932.

Haag's theorem says that Fock space does not support an interacting Poincare invariant field theory.

Then the CCRs arise as a property of Fock space and the vacuum is unique by construction: there is no problem. I feel that this point of view is to some extent borne out in Weinberg (QTF I pp174-175) taking special effort to present things 'in reverse order' (his words), beginning with the a priori specification of state space and deriving CCRs from that.

But he derives only the free frields in that way. For interactions he resorts to the usual handwaving arguments that ignore on the Hilbert space level not only questions of renormalization but also all issues involving the large volume limit. It is the latter that gives rise to the Non-Fock representation (and to other phenomena like infrared divergences).

 

[schieghoven]

No. This was known long before Haag; probably even to von Neumann in 1932.

Haag's theorem says that Fock space does not support an interacting Poincare invariant field theory.

Ah, then it's likely I didn't actually understand Haag's theorem. I will try to look into it again. Could you comment on whether Haag's theorem is consistent with the work of Glimm and Jaffe in 1+1 and 2+1 dimensions? These authors explicitly construct an interacting relativistic field theory defined on Fock space.

 

[A. Neumaier]

Ah, then it's likely I didn't actually understand Haag's theorem. I will try to look into it again. Could you comment on whether Haag's theorem is consistent with the work of Glimm and Jaffe in 1+1 and 2+1 dimensions?

Yes it is. Look at the paper by Glimm and Jaffe in

Acta Mathematica 125, 1970, 203-267
http://www.springerlink.com/content/t044kq0072712664/

where the (very readable, nontechnical) introduction spells out the relation for the 1+1D case.

These authors explicitly construct an interacting relativistic field theory defined on Fock space.

No. The cited introduction tells exactly the opposite, and clarifies what actually happens.
These authors explicitly construct an interacting relativistic field theory that is only locally Fock, in a sense they make precise.

 

[DrFaustus]

I really wish people and books would stop "teaching" and "explaining" QFT by ONLY talking about Fock space. Not even for free theories it is the be all and end all of QFT (just think about thermal states... they do *not* live in Fock space), let alone for interacting theories where you're thrown out of it as soon as you start *thinking* interactions without even mentioning them, and you don't even realize it.

 

[schieghoven]

I really wish people and books would stop "teaching" and "explaining" QFT by ONLY talking about Fock space. Not even for free theories it is the be all and end all of QFT (just think about thermal states... they do *not* live in Fock space), let alone for interacting theories where you're thrown out of it as soon as you start *thinking* interactions without even mentioning them, and you don't even realize it.

I disagree on all points:

• quantum thermal states live in Fock space just as much as a classical microstate lives in classical configuration space.
• I'm still getting to grips with the finer points of Glimm and Jaffe (thanks A. Neumaier), but without doubt, Fock space is still the starting point for this work, and it is definitely concerned with interacting theory. There aren't any better alternatives to a mathematical formulation of field theory... that's why the Clay mathematics people asked Jaffe to co-write the Problem description.

 

[A. Neumaier]

I'm still getting to grips with the finer points of Glimm and Jaffe (thanks A. Neumaier), but without doubt, Fock space is still the starting point for this work, and it is definitely concerned with interacting theory.

The final Hilbert space is sort of an inductive Limit of a sequence of local Fock spaces.
But the limit has as lmuch to do with a Fock space as the limit of the sequence of rational numbers (1+1/n)^n has to do with a rational number.

Of course, all complex objects are constructed out of simpler ones, but they usually don't remain simple because limiting operations don't preserve all properties of the simpler situation.

Similarly, thermal states are not Fock states because they involve a thermodynamic limit, which moves things from locally Fock to non-Fock, just as in the case of Phi^4_2 theory.

 

[DrFaustus]

schieghoven -> Seems like you'll have to study Glimm&Jaffe a bit more before being able to meaningfully disagree on QFT discussions :) Don't take it personally or be offended, it's just so big and technical as a subject that, for *physicists*, it's hardly worth the effort of studying properly.

For the record, I'll go even further than just claiming that thermal states don't live in Fock space. The set of all Fock space states is a set of measure zero in the set of *all* QFT states :) This is even true if one only considers Fock space states in addition to thermal states and the reason is simply that Fock space states are countable whereas thermal states are not. There's one thermal state for each value of the temperature...

A small correction on Fock space and interactions. It's true that interacting theories can live in Fock space: Phi_2^4 on the *circle*, i.e. on a spacetime of compact spatial support, *does* indeed live in Fock space. Not completely sure about it, but I think Yukawa_2 on the circle lives on Fock space as well and maybe gauge theories too, I don't remember. But as soon as you want to remove the spatial cutoff you thrown out of it. Incidentally, you'll notice that Glimm&Jaffe always talk about Fock space when considering spatially cutoff theories (together with other cutoffs, if needed) and it is only in 2D that the spatially cutoff interacting theory lives on Fock space.

 

[A. Neumaier]

The set of all Fock space states is a set of measure zero in the set of *all* QFT states :) This is even true if one only considers Fock space states in addition to thermal states and the reason is simply that Fock space states are countable whereas thermal states are not. There's one thermal state for each value of the temperature...

This is not a good argument. Al;ready the state space of a single qubit has uncountably many states. Adding a dimension dependence does not alter the cardinality.

The real situation is complicated since there are many different Fock spaces that can be considered. Haag's theorem only says that the Fock space created from the free vacuum by the free creation operators does not support an interacting theory.

On the other hand, Haag-Ruelle theory suggests that (at least in nice cases) there is a different Fock space formed by the physical vacuum (i.e., that of the interacting theory) by applying creation operators for all asymptotic particle states that describes correctly the interacting theory. However, in terms of that Fock space, the Hamiltonian looks very differently.

 

[cgoakley]

"Haag's theorem only says that the Fock space created from the free vacuum by the free creation operators does not support an interacting theory."

Incorrect. Haag's theorem says that the Fock space created from the free vacuum by the free creation operators is not unitarily equivalent to an interacting Fock space at the same time, i.e. no operator U(t) exists to map the states and operators of one on to the other.

An interacting theory can nonetheless be fabricated from free creation and annihilation operators. See

1. Relativistisch invariante Störungstheorie des Diracschen Elektrons, by E.C.G. Stueckelberg, Annalen der Physik, vol. 21 (1934).

[Reviewed here: http://arxiv.org/abs/physics/9903023]

2. Mass- and charge-renormalizations in quantum electrodynamics without use of the interaction representation, Arkiv för Fysik, bd. 2, #19, p.187 (1950), by Gunnar Källén.

3. Formal integration of the equations of quantum theory in the Heisenberg representation, Arkiv för Fysik, bd. 2, #37, p.37 (1950), by Gunnar Källén.

4. On quantum field theories, Matematisk-fysiske Meddelelser, 29, #12 (1955), by Rudolf Haag

5. Quantum Electrodynamics, by Gunnar Källén, pub. by George, Allen and Unwin (1972), pp.79-85

 

[A. Neumaier]

"Haag's theorem only says that the Fock space created from the free vacuum by the free creation operators does not support an interacting theory."

Incorrect. Haag's theorem says that the Fock space created from the free vacuum by the free creation operators is not unitarily equivalent to an interacting Fock space at the same time, i.e. no operator U(t) exists to map the states and operators of one on to the other.

My statement (which you quoted) and yours are not in conflict.

An interacting theory can nonetheless be fabricated from free creation and annihilation operators.

But not in a mathematically rigorous way in a space that contains both the free and the interacting c/a operators, connected by a homotopy in the coupling constant. This is the importance of Haag's theorem.

In 1+1 and 1+2 dimensions, where one can construct the theories rigorously, one indeed finds that the spaces supporting the free and the interacting theory have only 0 in common.

 

[cgoakley]

My statement (which you quoted) and yours are not in conflict.

Yes they are. You are making the additional assumption that if U(t) does not exist then free and interacting theories cannot live in the same Fock space. If you look at Källén's work you will see that interacting creation/annihilation operators are formed as sums of tensor products of free c/a operators. These cannot be unitarily transformed back to their non-interacting counterparts. Yet using Stückelberg's covariant perturbation theory (my ref. 1 above), scattering amplitudes can still be calculated. Mathematical rigor should not be an issue as we have not left free field Fock space.

 

[A. Neumaier]

Yes they are. You are making the additional assumption that if U(t) does not exist then free and interacting theories cannot live in the same Fock space. If you look at Källén's work you will see that interacting creation/annihilation operators are formed as sums of tensor products of free c/a operators. These cannot be unitarily transformed back to their non-interacting counterparts. Yet using Stückelberg's covariant perturbation theory (my ref. 1 above), scattering amplitudes can still be calculated. Mathematical rigor should not be an issue as we have not left free field Fock space.

Mathematical rigor is essential for claiming that one can evade Haag's theorem, since the latter is a rigorous theorem; so non-rigorous arguments do not prove that one can escape his conclusions.

On less rigorous levels, physicists have always used Fock space, closing the eyes to the problems exposed by Haag's theorem. This is possible since all physical representations seem to be locally Fock, so that Fock space techniques together with a non-rigorous use of Bogoliubov transformation (which typically leave Fock space) are sufficient for approximate arguments.

 

[cgoakley]

Mathematical rigor is essential for claiming that one can evade Haag's theorem, since the latter is a rigorous theorem; so non-rigorous arguments do not prove that one can escape his conclusions.

Where was I trying to evade Haag's theorem?

As I said, the interacting c/a operators in Stückelberg's covariant perturbation theory cannot be unitarily transformed to the non-interacting ones, even though they live in the same Fock space. This is consistent with Haag's theorem.

A matrix element for a real process in ordinary QM will normally involve this calculation: $$\langle f\vert \exp(i(H_0+V)t) \vert i\rangle$$. The fact that this tends not to be covariant is closely connected with Haag's theorem; only the final results for cross-sections, etc. tend to be covariant. In Stückelberg's covariant perturbation theory, though, although the final results are the same, the intermediate expressions will be different as covariance is maintained throughout.

 

[meopemuk]

As far as I know, the proof of Haag's theorem uses the assumption that *interacting* fields have specific transformation laws under the *interacting* representation of the Lorentz group. Namely, the transformation is assumed to be of the form

$$\Psi_i(x) \to \sum_j D_{ij}(\Lambda^{-1}) \Psi_j(\Lambda x)$$

Can somebody explain me the reason for this formula? In particular, it is interesting why there are no traces of interaction there?

Thanks.
Eugene.

 

[cgoakley]

Eugene,

I do not understand your question. As you point out, what you have written down is just a definition.

 

[meopemuk]

Eugene,

I do not understand your question. As you point out, what you have written down is just a definition.

Why do you think this definition makes physical sense?

Don't you find it suspicious that we use *interacting* representation of the Poincare group to transform *interacting* fields and then assume that the transformation formula does not depend on interaction at all? This is either a remarkable coincidence or simply a wrong assumption (definition).

There is a paper

H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties" Prog. Theor. Phys. 98 (1966), 934

where it is shown that the above field transformation formula does not hold in a model interacting theory.

Eugene.

 

[A. Neumaier]

Where was I trying to evade Haag's theorem?

As I said, the interacting c/a operators in Stückelberg's covariant perturbation theory cannot be unitarily transformed to the non-interacting ones, even though they live in the same Fock space.

Since the construction cannot be made rigorous, it is irrelevant for the purposes of constructive quantum theory (which is under discussion in this thread).

There are other reasons why Fock space cannot be the answer for real QFT, even in its approximate version as used by most physicists: Fock space cannot accommodate any of the nonperturbative stuff that is being discussed (e.g., solitons and instantons) .

 

[A. Neumaier]

As far as I know, the proof of Haag's theorem uses the assumption that *interacting* fields have specific transformation laws under the *interacting* representation of the Lorentz group. Namely, the transformation is assumed to be of the form

$$\Psi_i(x) \to \sum_j D_{ij}(\Lambda^{-1}) \Psi_j(\Lambda x)$$

Can somebody explain me the reason for this formula? In particular, it is interesting why there are no traces of interaction there?

This is the meaning of carrying a representation of the Lorentz group preserving the causal commutation relations.

Informally (and in 1+1D and 1+2D without bound states rigorously locally - ignoring large volume questions), the interacting Psi_i(x) is a unitary transform of the free Psi(x), hence satisfies the same transformation rules.

 

[cgoakley]

Since the construction cannot be made rigorous, it is irrelevant for the purposes of constructive quantum theory (which is under discussion in this thread).

I hope so, as this constructive QFT seems not to be able to generate cross-sections in 3+1 dimensions.

We seem not even to be able to agree about the basic rules of logic. Stückelberg's covariant P.T. uses free field theory as its framework. You agree that free theory is rigorous. Yet you say that Stückelberg's methods are not. Why?

As there seems to be little danger of you (or anyone else) actually looking at the references I gave I will once again give up. Maybe I will come back in another five years, though I am not optimistic that constructive/axiomatic/algebraic/whatever field theorists will be calculating cross-sections for real processes even then.

 

[A. Neumaier]

We seem not even to be able to agree about the basic rules of logic. Stückelberg's covariant P.T. uses free field theory as its framework. You agree that free theory is rigorous. Yet you say that Stückelberg's methods are not. Why?

Stueckelberg constructs only the leading terms in a formal power seires. But it is well-known that formal power series never define a function: There are always infinitely many functions whose Taylor expansion agrees with the formal power series.

What is missing to make his methods rigorous is a recipe for picking the right function in a way that is consistent with the action of some self-adjoint Hamiltonian acting on Fock space (or another space).

 

[meopemuk]

the interacting Psi_i(x) is a unitary transform of the free Psi(x), hence satisfies the same transformation rules.

Can you prove this statement or, at least, point me to the reference, where it is proved? It doesn't look obvious to me.

Eugene.

 

[strangerep]

As far as I know, the proof of Haag's theorem uses the assumption that
*interacting* fields have specific transformation laws under the
*interacting* representation of the Lorentz group. Namely, the
transformation is assumed to be of the form

$$\Psi_i(x) \to \sum_j D_{ij}(\Lambda^{-1}) \Psi_j(\Lambda x)$$

Can somebody explain me the reason for this formula?
In particular, it is interesting why there are no traces of
interaction there?

This is the meaning of carrying a representation of the Lorentz group
preserving the causal commutation relations.

I don't see how it's reasonable to insist that an interacting theory
(involving accelerations of particles wrt each other) must preserve
the same causal structure of spacetime as the free theory.
Consider Rindler horizons for mutually accelerated observers...
this implies a very different causal structure compared to that
perceived by inertial observers.

Informally (and in 1+1D and 1+2D without bound states rigorously
locally - ignoring large volume questions), the interacting Psi_i(x) is
a unitary transform of the free Psi(x), hence satisfies the same
transformation rules.

Was there a typo in your last sentence above? (If the free and interacting
fields are related by a unitary transform then the spectra of the two
theories are the same, aren't they?)

 

[A. Neumaier]

I don't see how it's reasonable to insist that an interacting theory
(involving accelerations of particles wrt each other) must preserve
the same causal structure of spacetime as the free theory.
Consider Rindler horizons for mutually accelerated observers...
this implies a very different causal structure compared to that
perceived by inertial observers.

Quantum gravity is beyond my expertise. I am assuming a renormalizable QFT in flat space. This has a well-defined causal structure independent of any interactions.

Informally (and in 1+1D and 1+2D without bound states rigorously locally - ignoring large volume questions), the interacting Psi_i(x) is a unitary transform of the free Psi(x), hence satisfies the same transformation rules.

Was there a typo in your last sentence above?

Not a typo but I was too sloppy. The interaction Psi_i(x) is a limit of a sequence of unitary transform of the free Psi(x), and the conclusion still holds. For all known relativistic QFTs in 1+d dimensions (d=1,2) for which the analysis could be made rigorous, the infinite volume limit changes the representation. The same is expected to hold for
the case d=3 where no rigorous analysis has been completed so far.

(If the free and interacting fields are related by a unitary transform then the spectra of the two theories are the same, aren't they?)

No, since the Hamiltonians are different. But if the free and interacting fields are related by a unitary transform then the representations of the equal-time CCRs of the two theories are the same, and both would be Fock spaces, which contradicts Haag's theorem.

 

[A. Neumaier]

Can you prove this statement or, at least, point me to the reference, where it is proved? It doesn't look obvious to me.

As just mentioned, the correct, intended statement was ''The interaction Psi_i(x) is a limit of a sequence of unitary transform of the free Psi(x), hence satisfies the same transformation rules.'' This is indeed a nontrivial statement; for the 1+1d case, see, e.g.,
the paper by Glimm and Jaffe in
Acta Mathematica 125, 1970, 203-267
http://www.springerlink.com/content/t044kq0072712664/
and the two papes that preceded their part III.

 

[cgoakley]

Stueckelberg constructs only the leading terms in a formal power series.

The power series is obtained by expanding the QED equations of motion. If one simply defines one's theory such that the interacting fields are the free fields plus these leading terms then everything will be perfectly finite and rigorous, and the simple scattering amplitudes will be correctly reproduced. The cost will just be that the interacting CCRs will not be the expected ones.

But it is well-known that formal power series never define a function: There are always infinitely many functions whose Taylor expansion agrees with the formal power series.

True, but irrelevant.

What is missing to make his methods rigorous is a recipe for picking the right function in a way that is consistent with the action of some self-adjoint Hamiltonian acting on Fock space (or another space).

This is not in my view the issue. The Hamiltonian for the free fields works just as well for the interacting fields in this approach. The thing that does not exist is a "free Hamiltonian" that time-displaces the interacting field as though it was the free field (and if it did, it would violate Haag's theorem).

 

[A. Neumaier]

The power series is obtained by expanding the QED equations of motion. If one simply defines one's theory such that the interacting fields are the free fields plus these leading terms then everything will be perfectly finite and rigorous, and the simple scattering amplitudes will be correctly reproduced. The cost will just be that the interacting CCRs will not be the expected ones.

But then the commutation relations are not causal, and what you get is not Poincare invariant. It is very easy to construct theories that are well-defined and approximate QED one way or another. It is also easy to construct Poincare invariant theories if you make compromises with causality.

But real QED at the same time
(i) is Poincare invariant,
(ii) is causal, i.e., field commutators at space-like related arguments vanish,
(iii) satisfies the cluster decomposition property,
and no amount of tinkering a la Stueckelberg so far has lead to a consistent and nontrivial 4D field theory with these properties.

 

[cgoakley]

But then the commutation relations are not causal,

Spacelike (anti)commutativity is not guaranteed, unless one introduces higher-order terms, certainly. What this has to do with causality is not so clear, though, as defining what one means by this in the quantum world is a lot harder than in the classical world.

and what you get is not Poincare invariant.

This I absolutely do not get. How and why is this not Poincare invariant?

It is very easy to construct theories that are well-defined and approximate QED one way or another. It is also easy to construct Poincare invariant theories if you make compromises with causality.

Examples?

 

[meopemuk]

As just mentioned, the correct, intended statement was ''The interaction Psi_i(x) is a limit of a sequence of unitary transform of the free Psi(x), hence satisfies the same transformation rules.'' This is indeed a nontrivial statement; for the 1+1d case, see, e.g.,
the paper by Glimm and Jaffe in
Acta Mathematica 125, 1970, 203-267
http://www.springerlink.com/content/t044kq0072712664/
and the two papes that preceded their part III.

Thanks. I'll check it out.

Since this nontrivial statement (which was assumed to be self-evident by Haag) was proven only much later, does it mean that Haag's proof is not complete? Shall we call it Haag-Glimm-Jaffe theorem now? Does it mean that this theorem is rigorously valid only in the 1+1d case? The story of Haag's theorem becomes rather confusing.

Eugene.

 

[A. Neumaier]

Since this nontrivial statement (which was assumed to be self-evident by Haag) was proven only much later, does it mean that Haag's proof is not complete? Shall we call it Haag-Glimm-Jaffe theorem now? Does it mean that this theorem is rigorously valid only in the 1+1d case? The story of Haag's theorem becomes rather confusing.

No. Haag's theorem is a theorem that holds in general when the Wightman axioms are satisfied, while the statement under discussion is a significantly stronger statement that can be proved for the specific theories that were explicitly constructed.