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  About states, observables and the wave functional interpretation in QFT with gauge fields

+ 4 like - 0 dislike

First of all, I'm a mathematician, so forgive me for my possible trivial mistakes and poor knowledge of physics.

In a QFT, we just start with a field (scalar, vectorial, sponsorial, gauge etc), so I would like to know what are the observables and the states in this context.

In QFT, the general approach would be by using the Fock space (for the free field case, since I don't really know if this would be true for the interacting one) and getting down, by using the particles associated to the operators $a$ and $a^{\dagger}$, to QM particles (I don't really know if this is true, because the number of particles is not constant and depends on the observer) or by using the wave functional interpretation (a functional on the space of field configurations satisfying Schrödinger equation), though I've heard that this functional is not Lorentz covariant (by the way, any proof?). However, according to this article (http://core.ac.uk/download/pdf/11921990.pdf) the wave functional interpretation is equivalent to the Fock space, so, in any case, this interpretation is not physically reasonable.

In AQFT, in contrast, the operators are already given (so we already have the observables). Furthermore, if the Lorentzian manifold is globally hyperbolic, a Cauchy hyper surface would be a possible interpretation for a state.

In other aspect, are the quantized fields of a given QFT really observables in the sense that they measure?

Now, adding gauge fields, everything will be grupoid valued and observables would be defined on quotients by the gauge group. In this context, I haven't really seen anything written about states and I have no idea on how the Fock space would be. The naive approach would be to consider the wave functional interpretation with domain in a grupoid.

Furthermore, if we restrict ourselves to TQFT, CFT or other specific class of field theories, would all this problem be solved?

Thanks in advance.

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user user40276

asked Apr 17, 2015 in Theoretical Physics by user40276 (140 points) [ revision history ]
edited May 1, 2015 by Dilaton
An historical remark: the pdf reference you cite seems to be quite out of date concerning the references...the interpretations of QFT provided there and the related discussions/problems are known since the end of the fifties of the last century ;-)

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user yuggib

2 Answers

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The algebraic approach gives the better idea of what the states and observables of a quantum theory are, and this holds in infinite dimensional systems as well.

In the modern mathematical terminology, observables of quantum mechanics are the elements of a topological $*$-algebra, and states are objects of its topological dual that are positive and have norm one. The most usual case is to take the $*$-algebra to be a $C^*$ or $W^*$ (von Neumann) algebra; however with such choice unbounded operators are not, strictly speaking, observables (but they can be "affiliated" to the algebra if their spectral projections are in the algebra). The advantage of this abstract approach is that, by the GNS construction, one can immediately associate an Hilbert space to the given $*$-algebra (and a particular state), where the elements of the algebra act as linear operators, and the given state as the average w.r.t. a specific Hilbert space vector.

In usual physical terms, only self-adjoint operators are considered to be observables, for an observable should have real spectrum (and could be associated to a strongly continuous group of unitary operators). The quantum field is, usually, considered to be an observable in a QFT (it is self-adjoint but unbounded, so often it would be affiliated to the $W^*$ algebra generated by its family of exponentials, the Weyl operators); and it is perfectly possible, theoretically, to measure its average value on states (to do it really in experiments, that is all another problem).

Quantum field theories are almost always represented in Fock spaces. However, since the Heisenberg group associated with an infinite dimensional symplectic space is not locally compact, the Stone-von Neumann theorem does not hold and there are infinitely many irreducible inequivalent representations of the Weyl relations, the Fock space being only one of them. To complicate things more, the Haag's theorem states that, roughly speaking, the free and interacting Fock representations are unitarily inequivalent (but that is a problem mostly for scattering theory, not at a fundamental level).

The "wave functional interpretation" (never heard this terminology) is just the functorial nature of the second quantization procedure that can associate to each Hilbert space the corresponding Fock space. This is due to Segal and you may also consult Nelson. The idea is that to each Hilbert space $\mathscr{H}$ one can associate a Gaussian probability space $(\Omega,\mu)$ such that the Fock space $\Gamma(\mathscr{H})$ is unitarily equivalent to $L^2(\Omega,\mu)$, and the map between $\mathscr{H}$ and $\Gamma(\mathscr{H})$($L^2(\Omega,\mu)$) is a functor in the category of Hilbert spaces with self-adjoint and unitary maps as morphisms. The $L^2(\Omega,\mu)$ point of view becomes very natural if one is interested to study QFTs by means of the stochastic integral approach (Feynman-Kac formulas) in euclidean time.

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user yuggib
answered Apr 17, 2015 by yuggib (360 points) [ no revision ]
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Sorry, but what do you mean by the Fock representation. There is no sympletic space at the beginning of the construction, so ,given a QFT, how can you associate a Fock representation?

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user user40276
Furthermore, the Fock space depends on the observer (different observers will result in different Fock spaces).

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user user40276
The algebra of observables you (almost always) give in a QFT contains the so-called CCR (CAR) algebra, that is the algebra of Weyl operators $\{W(f), f\in H\}$, where $H$ is a real Hilbert space, satisfying the relation $W(f)W(g)=e^{-i\sigma(f,g)/2}W(f+g)$ for any $f,g\in H$, with $\sigma(\cdot,\cdot)$ a nondegenerate antisymmetric bilinear form. The one-particle symplectic structure is therefore $\Sigma(H)=(H,\sigma)$, and the Fock representation associated is then natural. However this is only of the infinitely many possible irreducible representations of the CCR algebra.

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user yuggib
Sorry, but, again, I can't see what do you mean by the Hamiltonian in the Fock representation.

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user user40276
Let us continue this discussion in chat.

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user yuggib
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The interacting Fock space cannot be rigorously constructed in most interesting QFTs; however you may take a look to the second book of Bratteli-Robinson to get an idea (applied on a different context) of the Haag's theorem and the inequivalent vacuum/ground-state representations associated to different QFTs. Also the book by Derezinski and Gerard gives some detail (in the end) on quantization of interacting theories. Finally, you may also try to take a direct look at the original works by Haag himself.

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user yuggib
Concerning the wave functional, the Hamiltonian in that case would be, roughly speaking, the same as in the Fock representation but with the field replaced by the multiplication by the gaussian functional, and the momentum replaced by the derivative w.r.t. to the aforementioned functional. In general the Hamiltonian has to be a self-adjoint operator on the $L^2(\Omega,\mu)$ space. Anyways I am not completely familiar with this type of description, so take these informations with benefit of the doubt ;-)

This post imported from StackExchange Physics at 2015-05-01 12:37 (UTC), posted by SE-user yuggib
+ 3 like - 0 dislike

From the rigorous point of view, the observable vacuum sector of a relativistic quantum field theory (QFT) on flat Minkowski space is defined by the Wightman axioms. (There are also variations of these in terms of nets of local algebras, but the Wightman axioms are considered most basic; they are also the criterion to be met for a solution of the Clay Millenium problem to construct a QFT for Yang-Mills. There you can also see how the vacuum sector of a gauge theory fits in conceptually. The unsolved conceptual problems that you allude to concern the charged sectors only.)

Given the Wightman axioms, the observables (in the sense of potentially measurable operators) are the smeared fields obtained by integrating the distribution-operator valued fields with an arbitrary Schwartz test function, their products, the linear combinations of these, and their weak limits, as far as they exist.

The state vectors are the products $\psi=A|0\rangle$ where $A$ is an observable and $|0\rangle$ is the vacuum state. (Of course, many different $A$ produce the same $\psi$; e.g., for a free QFT, one can change $A$ by adding any operator of the form $Ba(f)$ where $a(f)$ is a smeared annihilation operator, without changing the state.)

The dynamics is dependent on the choice of a time direction along positive multiples of a timelike vector $v$, and is given by $\psi(t):=A(t)|0\rangle$, where $A(t)$ is obtained from $A$ by replacing all arguments $x$ of field operators in the expression defining $A$ by $x-tv$.  The latter operation is an algebra automorphism believed to be always inner, i.e., induced by conjugation with a strongly continuous 1-parameter group generated by a $v$-dependent Hamiltonian $H$ with $H|0\rangle=0$. Assuming this, the Schroedinger equation holds.

To get a more concrete view of the Hilbert space and the dynamics one must either consider exactly solvable QFTs (of which nontrivial examples currently are known only in spacetime dimensions $<4$, and indeed, in 2-dimensional conformal field theory one can give a much more specific picture.), or sacrifice rigor and consider renormalized perturbation theory. In 4 dimensions, the latter builds the Hilbert space as a formal deformation of a Fock space and the fields as formal power series in $\hbar$ or a renormalized coupling constant, although to get physical results one hopes that these formal power series can be evaluated numerically by appropriate trickery. In case of QED this works exceedingly well, but less so in other QFTs.

Alternatively, one discretizes the QFT on a finite lattice, and reduces the problem in this way to one of ordinary quantum mechanics, hoping that for a fine enough and large enough lattice, the results close to the continuum results.

One can also use the functional Schroedinger representation, though this is not mathematically well-defined. Note that contrary to the false claim unlike the functional field equation discussed in the article cited by the OP (which is a philosophical, not a physics paper), the functional Schroedinger equation is in general not equivalent to the Fock representation. In particular, unlike the Fock representation, the functional Schroedinger equation is able to explain many nonperturbative features of interacting QFT. See the discussion of Jackiw's work.

For nonrelativistic QFTs, the situation is somewhat simpler, as particle number is conserved. In the vacuum representation, the Hilbert space is a proper Fock space, and splits into a direct sum of $N$-particle spaces to which standard quantum mechanics applies. However other representations such as those relevant for equilibrium thermodynamics, some of the problems from the relativistic case recur, since the appropriate Hilbert space is no longer a Fock space.

In curved space, no good system of axioms is known, and one generally uses a Fock space perturbation approach with all its limitations.

answered May 1, 2015 by Arnold Neumaier (15,787 points) [ revision history ]
edited May 2, 2015 by Arnold Neumaier
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1) Well, I would like to see a proof of this, because the following reasoning does not seem to support the assertion: given the Fock representation, the dynamics can be seen as an automorphism of a suitable (von Neumann) algebra $U$ covering the CCR $C^*$ algebra (and induces an automorphism on the dual, i.e. on the suitable quantum states; the covering algebra is included in the algebra of bounded operators in the Fock space). If we now take the algebra $U$ with the given automorphism as the starting abstract dynamical system, it should again be represented in at least one inequivalent non-Fock way (inequivalent w.r.t. the CCR subalgebra; but, admittedly, I am not completely sure about that), and I don't see why the non-Fock representations should violate the Wightman axioms.

3) You inference is unjustified: you can infer that the interacting representation is inequivalent to the free Fock one; but you have admitted yourself (point 2) that there are inequivalent Fock representations, so even the interacting one may be Fock. It is actually the case in a simple interacting situation where the renormalized theory is just a free theory with renormalized mass: it has been rigorously proved by Ginibre and Velo in 1970 that for a quadratic interaction (I know it is an easy one, but nevertheless it respects Wightman axioms and Haag's theorem) in dimension $3+1$ the renormalized theory requires a change of Hilbert space, but the final result is indeed in the Fock representation (and it is a free theory with renormalized mass).

1) I need to find the right references. I think that the existence of a Poincare invariant vacuum state singles out the Fock representation from the many possible ones.

3) Of course, a quadratic Poincare-invariant ''interaction'' changes a free theory into another free theory, so the renormalization can only have the effect of a change of mass. It is not really an interaction. 

@yuggib: I couldn't find an appropriate reference; so I retract my claims about 1) and 3). They reflected my intuition rather than definite knowledge, and after the present discussion I am no longer convinced that my intuition was correct.

Concerning 1), my intuition (but it may be also wrong) is that if there are other non-fock representations of the free dynamics they should satisfy the Wightman axioms as well, for these axioms seem "representation-independent", at least to me.

Concerning 3), I think that even if you cannot say for sure that the interacting representations are non-Fock, this may indeed be the case for many (or maybe most) interesting theories. Looking a little bit around, I found for example that the representation associated to the renormalized $\phi^4_3$ Hamiltonian given by Glimm is non-Fock (link to where I found the assertion: https://projecteuclid.org/euclid.cmp/1103857837, in page 2); however Glimm's hamiltonian has still the volume cutoff, so this seems unrelated to Haag's theorem. I think that the possibilities are many, and it is difficult to know a priori in which type of representation one may end up after renormalization.

I now found a weak reference; Arthur Jaffe, shorthly after minute 04:00 in http://media.scgp.stonybrook.edu/video/video.php?f=20120117_1_qtp.mp4 says that Fock space is not appropriate for interacting fields.

Most recent comments show all comments

@yuggib: "My point was not to argue [...]" - You had said "Well, that this picture is more appropriate than Fock space is your opinion, and you are entitled to it (even if I do not agree)." - expressing both disagreement and downgrading a fact to a mere opinion. 

For a free field, every construction must of course be equivalent to the Fock representation. but Haag's theorem implies rigorously that an interacting QFT is definitely not Fock. 

That the functional Schrödinger representation is not a rigorous representation is nothing bad as long no rigorous interacting relativistic QFT is known. 

"but also two free representations corresponding to different masses are inequivalent." - These are not representations of the same CCR, hence it is meaningless to label them as inequivalent. 

I wouldn't argue if there weren't a significant difference in content.

1) "For a free field, every construction must of course be equivalent to the Fock representation." This is not true, there are infinitely many inequivalent representation of the (free field) CCR.

2) "These are not representations of the same CCR, hence it is meaningless to label them as inequivalent. " The time zero fields and momenta corresponding to different masses are two inequivalent representations of the CCR over the one-particle space $L^2(\mathbb{R}^3)$. Therefore of the same CCR, that fixed the one-particle space is unique up to $*$-isometric isomorphisms.

3) "Haag's theorem implies rigorously that an interacting QFT is definitely not Fock" Haag's theorem says that two representations of the CCR for free and interacting theories must be inequivalent, not non-Fock.

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