What Are the States in Quantum Field Theory?

[genneth]

All this is definitely true in condensed matter physics. However, I don't think that the same ideas are valid for fundamental relativistic quantum fields. There is no physical "lattice" or "medium" whose "displacements" can be interpreted as particle quantum fields and whose "excitations" can be interpreted as particles.

Eugene.

I don't think the Count was trying to say that things are exactly the same, more just giving a concrete example of how the formalism works. After all, it's Hilbert spaces all the way down...

 

[meopemuk]

But you can pretend that there is such a lattice and then work in the scaling limit where the effects of any such lattice are invisible. It then doesn't matter whether or not in reality such a lattice exists.

If it doesn't matter, then I would prefer to think that this lattice doesn't exist. It is always good to reduce the number of unprovable assumptions to a minimum.

QFT is only an effective theory, like thermodynamics was before we knew that it is based on statistical physics (you can do thermodynamics without knowing that a gas consists of molecules).

But today we *do know* that gas consists of molecules and we *do not know* if there is "the lattice". So, it is well established that thermodynamics is an effective theory. The claim that relativistic QFT is also an effective theory does not have any experimental support. My opinion is that this claim is just another attempt to "sweep problems under the rug".

Eugene.

 

[Count Iblis]

Don't you think the renormalizability of the Standard Model is indirect evidence for it to be an effective theory? Similarly to why you have a renormalizable phi^4 theory for the Ising model in the scaling limit: Because all the irrelevant operators flow to zero in the scaling limit.

 

[meopemuk]

Don't you think the renormalizability of the Standard Model is indirect evidence for it to be an effective theory?

"indirect evidence" is not the same as "proof".

Eugene.

 

[genneth]

If it doesn't matter, then I would prefer to think that this lattice doesn't exist. It is always good to reduce the number of unprovable assumptions to a minimum.

But today we *do know* that gas consists of molecules and we *do not know* if there is "the lattice". So, it is well established that thermodynamics is an effective theory. The claim that relativistic QFT is also an effective theory does not have any experimental support. My opinion is that this claim is just another attempt to "sweep problems under the rug".

Eugene.

I'd say reason works the other way. To assume that you have a "true" theory, without any upper cut off, is wildly optimistic. The question of where that cut off is, and what is beyond, are good questions.

 

[meopemuk]

ITo assume that you have a "true" theory, without any upper cut off, is wildly optimistic.

I am optimistic because I know a theory which

1. does not have ultraviolet cutoff
2. does not have divergences in S-matrix calculations
3. produces the same scattering amplitudes as the traditional renormalized QFT

E. V. Stefanovich, "Quantum field theory without infinities", Ann. Phys. (NY), 292 (2001), 139.

E. V. Stefanovich, "Relativistic quantum dynamics", http://www.arxiv.org/abs/physics/0504062

Eugene.

 

[xepma]

So you're basically slowly but surely hijacking this thread to promote your own book?

 

[DrDu]

meopemuk, in some sense for an infinite system, the wavefunction never exists, but one has to define states as functionals on the space of operators. So maybe the difference between relativistic and non-relativistic qft is not so fundamental after all.

 

[meopemuk]

meopemuk, in some sense for an infinite system, the wavefunction never exists, but one has to define states as functionals on the space of operators. So maybe the difference between relativistic and non-relativistic qft is not so fundamental after all.

I do not agree with the usual characterization of QFT as a theory studying systems with infinite number of degrees of freedom. I think it is more appropriate to call them "systems with uncertain number of degrees of freedom". The number of degrees of freedom is proportional to the number of particles in the system, and this number can change in the process of time evolution. For example, within QFT one can describe a state with only two particles. There is no problem in describing such a state by a 2-particle wave function. The only "problem" is that in the course of time evolution the number of particles in the system can change, so the form of the wave function becomes more complicated and the number of degrees of freedom increases.

Eugene.

 

[A. Neumaier]

In the Fock basis, there would be a function of N positions for the N particle basis, right? And then there'd be a sum over all N.
Could someone please show the steps connecting the state in the "field eigenbasis" to the state in the "Fock basis" a bit more explicitly? Not just the math of "here is an equation", but the procedure ... ie. how we derive the connection between the two.
Can someone show what the "wavefunctional" equation would be for QED? I still don't understand how you are deriving these things since there are fields in the Lagrangian, but no wavefunctional.

First consider the single-particle case, i.e., ordinary quantum mechanics.
Working in the position representation just means working with the wave function whose values are the coefficients in the eigenbasis |x> of the commuting position operators x_1, x_2, x_3, short x:

|\psi> = \int dx \psi(x) |x>, where \psi(x)=<x|psi>

Working in the momentum representation just means working with the wave function whose values are the coefficients in the eigenbasis |p> of the commuting momentum operators p_1, p_2, p_3, short p:

|\psi> = \int dp \psi(p) |p>, where \psi(p)=<p|psi>

To translate between the two, one needs to know how to find the eigenstates
of p in the x-representation, and the eigenstates of x in the p-representation.
This is given by the Fourier transform.

Now consider a field theory. To get a representation one needs to pick a maximal commuting family of operators and their eigenstates. Depending on the choice, one gets different but isomorphic representations. 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>,

Note that any 1-particle operator A lifts to the resulting Fock space by means of
\hat A = \int dp a^*(p) A a(p) with the corresponding c/a operators,
giving in particular for the total momentum

\hat p |p_1,...,p_N>=(p_1+...+p_N) |p_1,...,p_N>,

showing that we have indeed eigenstates. This is the appropriate representation for
scattering experiments, where the input configurations are prepared in momentum eigenstates.

By diagonalizing instead Hermitian field operators at time t=0 (which commute because of the canonical commutation relations), one gets the functional Schroedinger representation in terms of eigenstates |\phi> (one state for each possible classical field configuration \phi at time t=0), and the wave functions \psi(\phi) are coefficients in a corresponding functional integral over all fields \phi.
(To get references, go to scholar.google.com and enter the key words functional schroedinger.) This is the appropriate representation when the field was prepared at time t=0.

Again conversion from one to the other representation requires the solution of the corresponding eigenproblems, but I haven't seen anyone do this explicitly.

 

[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.