What Are the States in Quantum Field Theory?

[meopemuk]

I didn't say anything about instantaneous dressing or infinite numbers of photons at the "next time instant", I said the real time dynamics of the dressing process can be quite interesting. I think it's fair to say that the process is less interesting in QED than QCD where something really dramatic happens.

The dynamics of a single "bare" electron in QED is very curious too. If you formally make the unitary time evolution operator out of QED Hamiltonian (even ignoring the fact that the true Hamiltonian of QED contains divergent counterterms) and apply this operator to one-bare-electron state you'll see that the time evolution is a complicated process of creation and annihilation of photons and electron-positron pairs.

As far as I know nobody has seen these kinds of processes in experiments. My conclusion is that "bare" electrons and the associated Hamiltonian do not provide adequate description of particles seen in nature. In order to study the time evolution one should switch to the "dressed particle" description.

Eugene.

 

[Physics Monkey]

The dynamics of a single "bare" electron in QED is very curious too. If you formally make the unitary time evolution operator out of QED Hamiltonian (even ignoring the fact that the true Hamiltonian of QED contains divergent counterterms) and apply this operator to one-bare-electron state you'll see that the time evolution is a complicated process of creation and annihilation of photons and electron-positron pairs.

I feel like you're not actually reading what I write. I have argued that QED has a cutoff, but your posts continue to ignore that fact and speak about divergences. I have also said that the dressing process occurs with all kinds photons and particle-antiparticle pairs appearing, I only took exception with your statement that it was somehow instanteous.

As far as I know nobody has seen these kinds of processes in experiments. My conclusion is that "bare" electrons and the associated Hamiltonian do not provide adequate description of particles seen in nature. In order to study the time evolution one should switch to the "dressed particle" description.

Eugene.

In my opinion this statement just doesn't make sense. I am not saying that bare electrons are the same as dressed electrons. Sure they have the same quantum numbers, but they are different states. On the other hand, for any finite regulated theory one can prepare states identical to the state of the bare particle and study their time evolution. Just because bare electron states are potentially hard to prepare doesn't mean their evolution is unphysical. Conceptually identical experiments are carried out routinely in atomic physics and condensed matter physics labs. I have also argued that this kind of thinking is experimentally relevant for QCD.

 

[meopemuk]

... one can prepare states identical to the state of the bare particle and study their time evolution. Just because bare electron states are potentially hard to prepare doesn't mean their evolution is unphysical.

There is a contradiction in your position. When scattering amplitudes are calculated in QED (and compared with real observable cross-sections) they are calculated exactly for "bare" particle states. So, QED assumes that electrons available in routine experiments are "bare". This disagrees with your claim that "bare" states are difficult to prepare.

Eugene.

 

[Physics Monkey]

There is a contradiction in your position. When scattering amplitudes are calculated in QED (and compared with real observable cross-sections) they are calculated exactly for "bare" particle states. So, QED assumes that electrons available in routine experiments are "bare". This disagrees with your claim that "bare" states are difficult to prepare.

Eugene.

This is also not true. Even your beloved Weinberg makes this point (p 110), that we must include some interactions in the description of asymptotic states so that these states are the physical particles. At the level of perturbation theory this is related to the amputation of external lines in Feynman diagrams. The issue is particularly evident in qcd where the bare quarks are not at all like good asymptotic states. The question of proper asymptotic states is a dynamical one that qft solves just fine. Note also that these comments have little bearing on how hard or easy it is to prepare a state describing a bare particle.

 

[meopemuk]

The issue is particularly evident in qcd where the bare quarks are not at all like good asymptotic states. The question of proper asymptotic states is a dynamical one that qft solves just fine. Note also that these comments have little bearing on how hard or easy it is to prepare a state describing a bare particle.

I agree that free quarks do not exist. That's why I was talking about QED not about QCD. Let me summarize what we've discussed so far and see if we can agree on that.

1. In QED the state of one electron is described as

a^{\dag}_p |0 \rangle......(1)

Perhaps being naive, I think that this state corresponds to a single free electron that can be easily prepared in the laboratory. This view is supported by the fact that when calculating scattering amplitudes in QED we take matrix elements on states exactly as (1). The results of these calculations are very accurate.

2. It is also true that "bare" states like (1) are not eigenstates of the total QED Hamiltonian. In particular, the time evolution of such states (generated by the total QED Hamiltonian) is rather non-trivial. It involves creation and annihilation of multiple (virtual) photons and electron-positron pairs.

3. My point is that this non-trivial dynamics has never been observed. A single free electron always looks like a single free electron and nothing else. In my opinion, this is a serious contradiction in the QED formalism.

4. In my understanding, you think that the non-trivial dynamics of a single "bare" electron can be observed if we learn how to prepare such states. This implies that states participating in (well-described) scattering experiments are not "bare" states like (1), but some other states. What are they?

Eugene.

 

[DrDu]

Maybe as usual in PF when somebody asks a question related to QFT it is immediately highjacked. I think, most of the problems which have been addressed are highly peculiar to relativistic field theories and not to qft in general. E.g. the basic states of a free non-relativistic electron gas are simply Sslater determinants of Bloch waves. Or, to bring a bosonic example, for a set of Einsteinian phonons, just products of harmonic oscillator functions.

 

[genneth]

I think we should return to the original poster's question. I believe he has not realized that it is possible to talk about probability distributions (or in this case, amplitudes) over field configurations. That is, the object \Psi[\phi] is the corresponding object to \psi(x) in usual QM. Just like x labels a possible configuration of a single particle, \phi labels the possible configuration of a whole field (in some space-like surface). Thus, the states could be written \left|\phi\right\rangle, analogous to \left|x\right\rangle. These would then form a complete basis. In practise, it's hard to construct these things, and instead we use the Fock representation, where the particle nature of things are more apparent.

 

[DrDu]

Exactly, genneth. Maybe an example would be helpfull, e.g. the linear chain of particles sitting in equilibrium at equally spaced positions labelled by index i. Then the deviations from these positions is u_i. The set of all displacements is the field phi and the basis states would be |\phi>=\prod_i |u_i> the |u_i> corresponding to the position eigenstates |x> of genneth, phi itself would be the vector of all the displacements u_i or the function u(r) in the continuum limit. This basis is seldomly used, especially in the continuum limit, as we are usually interested in the solutions where the u_i vary continuously. Hence one uses often an alternative basis of Fourier transformed states |k>=\sum_l |u_l>\exp ilk the field is then characterized by |\tilde{\phi}>=\prod_k |u_k>. The latter basis has the advantage that it remains discrete in the continuum limit. Furthermore, we can get rid easily of the non-smooth configurations by cutting of the product at high momenta k. That's called regularization.

 

[meopemuk]

Maybe an example would be helpfull, e.g. the linear chain of particles sitting in equilibrium at equally spaced positions labelled by index i. Then the deviations from these positions is u_i. The set of all displacements is the field phi and the basis states would be |\phi>=\prod_i |u_i> the |u_i> corresponding to the position eigenstates |x> of genneth, phi itself would be the vector of all the displacements u_i or the function u(r) in the continuum limit.

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.

 

[Count Iblis]

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.

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.

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).

 

[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.

 

[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.