Notes on BRST I: Representation Theory and Quantum Mechanics

This is the first posting of a planned series that will discuss the BRST method for handling gauge symmetries and related mathematical topics. I’ve been writing a more formal paper about this, but given the substantial amount of not-well-known background material involved, it seems like a good idea to first put together a few expository accounts of some of these topics. And what better place for this than a blog?

Many readers who are used to my usual attempts to be newsworthy and entertaining, often by scandal-mongering or stirring up trouble of one kind or another, may be very disappointed in these posts. They’re quite technical, hard to follow, and of low-to-negative entertainment value. You probably would do best to skip them and wait for more of the usual fare, which should continue to appear from time to time.

Quantum Mechanics and Representation Theory

A quantum mechanical physical system is given by the following mathematical structure:

  • A Hilbert space \mathcal H, the “space of states”. A state of the physical system is determined by a vector |\psi\rangle\in \mathcal H, with unit norm (i.e. ||\psi||^2=\langle \psi|\psi\rangle=1).
  • An algebra \mathcal O that acts on \mathcal H. To each physical observable corresponds a self-adjoint operator O\in\mathcal O. Eigenvectors in \mathcal H of this operator correspond to states where the observable has a well-defined value, which is the eigenvalue.
  • If a physical system has a symmetry group G, there is a unitary representation (\Pi, \mathcal H) of G on \mathcal   H. This means that for each g\in G we get a unitary operator \Pi(g) satisfying

    \begin{displaymath}\Pi(g_3)=\Pi(g_2)\Pi(g_1)\ \text{if}\ g_3=g_1g_2\end{displaymath}

    i.e. the map \Pi from group elements to unitary operators is a homomorphism. The \Pi(g) act on \mathcal O by taking an operator O to its conjugate \Pi(g)O(\Pi(g))^{-1}.

    When G is a Lie group with Lie algebra (\mathfrak g, [\cdot,\cdot]), differentiating \Pi gives a unitary representation (\pi, \mathcal H) of \mathfrak g on \mathcal H. This means that for each X\in \mathfrak g we get a skew-Hermitian operator \pi(X) on \mathcal H, satisfying

    \pi(X_3)=[\pi(X_1),\pi(X_2)]\ \text{if}\ X_3=[X_1,X_2]

    i.e. the map \pi taking Lie algebra elements X (with the Lie bracket in \mathfrak g) to skew-Hermitian operators (with commutator of operators) is a homomorphism. On \mathcal O, \mathfrak g acts by the differential of the conjugation action of G, this action is just that of taking the commutator with \pi(X).

    The Lie bracket is not associative, but to any Lie algebra \mathfrak g, one can construct an associative algebra U(\mathfrak g) called the universal enveloping algebra for \mathfrak g. If one identifies X\in \mathfrak g with left-invariant vector fields on G, which are first-order differential operators on functions on G, then U(\mathfrak g) is the algebra of left-invariant differential operators on G of all orders, with product the composition of differential operators. A Lie algebra representation is precisely a module over U(\mathfrak g), i.e. a vector space with an action of U(\mathfrak g).

    So, the state space \mathcal H of a quantum system with symmetry group G carries not only a unitary representation of G, but also a unitary representation of \mathfrak g, or equivalently, an action of the algebra U(\mathfrak g). X\in \mathfrak g acts by the operator \pi(X). In this way a representation \pi gives a sub-algebra of the algebra \mathcal O of observables. Most of the important observables that show up in practice come from a symmetry in this way. An interesting philosophical question is whether the quantum system that governs the real world is purely determined by symmetry, i.e. such that ALL its observables come from symmetries in this manner.

    Some Examples

    Much of the structure of common quantum mechanical systems is governed by the fact that they carry space-time symmetries. In our 3-space, 1-time dimensional world, these include:

  • Translations in space: G=\mathbf R^3, \mathfrak g =\mathbf R^3, Lie Bracket is trivial.
    For each basis element e_j\in \mathfrak g one gets a momentum operator \pi(e_j)=iP_j
  • Translations in time: G=\mathbf R, \mathfrak g =\mathbf R. If e_0 is a basis of \mathfrak g, i\pi(e_0)=H, the Hamiltonian operator. The fact that this operator generates time-translations is just Schrodinger’s equation.
  • Rotations in 3-space: G=SO(3), or its double cover G=Spin(3)=SU(2), \mathfrak g = \mathbf R^3, with bracket given by the vector product. For each basis element e_j\in \mathfrak g one gets an angular momentum operator \pi(e_j)=iJ_j. These operators do not commute, so cannot be simultaneously diagonalized.
  • Another example is the symmetry of phase transformations of the state space \mathcal H. Here G=U(1), \mathfrak g=R, and one gets an operator Q_e that can be normalized to have integral eigenvalues.

    This last example also comes in a local version, where we make independent phase transformations at different points in space-time. This is an example of a “gauge symmetry”, and the question of how it gets represented on the space of states is what will lead us into the BRST story. Next posting in the series will be about gauge symmetry, then on to BRST.

    If you want some idea of where this is headed, you can take a look at slides from a colloquium talk I gave recently at the Dartmouth math department. They’re very sketchy, the postings in this series should add some detail.

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    63 Responses to Notes on BRST I: Representation Theory and Quantum Mechanics

    1. Tom says:

      Thanks for this BRST post, and those to come! I’m going to enjoy reading some technical stuff. I know I haven’t seen BRST symmetry treated in a really mathematical way before.

    2. anonymous says:

      Please do not use the > character for the ket when you typeset using TeX!
      It is for the mathematical less-than symbol. As such, the spaces around > are automatically widened, which makes your kets super-ugly.
      See p.170 of TeXBook about \mathrel, \mathop etc.
      So please either use at least |\psi\mathclose>. I prefer |\psi\rangle, though.

    3. estraven says:

      Great post, I’m looking forward to the rest fo the series.

      @anonymous at 10.03pm: My .tex files include the lines

      \def\>{\rangle}
      \def\<{\langle}

      because I like good typesetting but am also very lazy.

    4. TSM says:

      I have to ask; which program did you use to make your talk slides? If it is powerpoint can you point me to the template used?

      TIA

    5. anon. says:

      One interesting question which is hard to understand from the standard textbook treatments and limited reading of the literature: does the BRST action for gauge-fixed Yang-Mills theory break down nonperturbatively, due to Gribov copies? Claims in both directions seem to exist, but it seems it would take a great deal of effort and/or reading to settle the matter. (Also: are there Gribov copies for gauge-fixing diffeo and Weyl invariance? I don’t think I’ve seen this mentioned in treatments of BRST quantization of the string.)

    6. anonymous says:

      That’s a LaTeX class called beamer. See http://latex-beamer.sourceforge.net/.
      PowerPoint and Keynote are for losers.

    7. Kurt says:

      Fantastic, I am looking forward to reading about BRST. There is something that has always bugged me about symmetry in QM: Surely we do not want to say that every time independent unitary operator is a symmetry, because then every quantum system with the same dimension Hilbert space would have the same symmetry group (i.e. U(N), where N is the dimension of the Hilbert space). There must be some input about the dynamics (i.e. Hamiltonian) of the system. Usually we say something like: a symmetry is any time independent unitary operator that commutes with the Hamiltonian. This guarantees that the symmetry preserves the Schrodinger equation, but rules out things like the boosts (either Galilean or relativistic), which are represented by unitary operators that depend explicitly on time, and which we would like to have as symmetries. Thus, we might want to say that a symmetry is allowed to carry explicit time dependence, so long as it preserves the Schrodinger equation. But this brings us back to the original problem of all quantum systems (of a given dimensional Hilbert space) having the same symmetry group: given any unitary operator S, we can give it time dependence in such a way that it preserves the Schrodinger equation (just take S(t)=U(t)SU(t)^{-1}, where U(t) is the time evolution operator). So where is the middle ground; what is a definition of symmetry, (using only the notion of Hilbert space and hamiltonian, no fair talking about classical concepts such as action, Noether theorem etc., or extra structure such as the S-matrix) that allows the boosts but does not allow an arbitrary unitary operator?

      KH

    8. N. Nakanishi says:

      The problem of Gribov copies arises only in the path-integral approach. No such trouble is encountered in the BRS-formulated operator formalism.

    9. Peter Woit says:

      Tom and anonymous,

      Thanks for the tex advice.

      TSM,

      The slides were made using the “beamer” latex class.

      anon.,

      One reason I’m interested in BRST is precisely that I don’t think the problem of how to handle gauge invariance is fully understood in theories like the Standard Model. I’ll write more about this later, and the Gribov problem is part of the story. I don’t believe there is any agreement about what happens to it outside of perturbation theory. What I’m writing about is mostly the operator formalism, and this does not have the same problems that the path integral has (as Nakanishi mentions). It may have others…

      In diffeo and Weyl invariance, you don’t get the same topological phenomenon that forces the Gribov problem, so I don’t think you see it in the quantization of the string.

      Kurt,

      Interesting question. I guess when I’m talking about a “symmetry”, this can come from any group action on the classical mechanical system being quantized, whether or not it commutes with the Hamiltonian. These give operators on the state space, which may or may not commute with the Hamiltonian operator. Well-known examples include the algebra of at most quadratic operators in p and q (Heisenberg + symplectic algebra)

    10. A. says:

      Dear Peter and anon,

      Interesting stuff. With regard to Gribov copies, they are not just some mathematical problem with overcounting in the path integral — they have physical implications. It’s been known for a long time that they play a key role in confinement, so it would be really surprising if the operator picture didn’t feel their effects at all.

      A.

    11. anon. says:

      A. is right: there’s a large literature about Gribov copies and confinement, with which I have only a superficial familiarity, but which seems very interesting. It’s been observed, for instance, that the center vortex configurations that have been primary candidates for explaining confinement since at least the work of ‘t Hooft in the late 70s lie on the Gribov horizon in certain gauges (hep-lat/0401003).

      Fundamentally, Gribov copies are telling us that the Yang-Mills configuration space has a complicated geometry and topology, and I don’t see how that could go away in the operator formalism. (Are there references?)

      I could imagine, though, that the ghosts somehow encode the gauge algebra in a way that allows the BRST action to get all of this right anyway. (To throw out another reference that I haven’t properly digest, Reinhardt in hep-th/9602047 showed that in maximal abelian gauge the path integral with an explicit integration over the group with the correct Haar measure is equivalent to the path integral with a Faddeev-Popov determinant.)

    12. FNesti says:

      Dear Peter,
      thabks for this good idea, it will be interesting and useful to read about BRST and Dirac. I’d like to ask whether you plan to mention the Dirac-Hodge operator (d+delta) and its properties. In my present understanding, this is the same of a standard dirac operator in flat space, while on a generic manifold it gives a different coupling of ‘spinors’ to gravity. I believe this would be a really interesting point to clarify, from your cross-discipline experience, from both the mathematical and physical point of view.

      Cheers,
      Fabrizio

    13. Peter Woit says:

      Fabrizio,

      The point you raise is not one that comes up only tangentially in the BRST story, but it is an important one if you try and understand spinors.

      Basically, the operator you mention operates on differential forms, not spinors, and at each point the differential forms take values not in the spinor space S, but in SxS*. In flat space you get just dim (S*) copies of what you want (this is related to why Kogut-Susskind gives you multiple copies of what you want on the lattice). In non-flat space there’s a non-trivial covariant derivative for the spinors, so you get not just multiple copies, but something different.

    14. Peter Orland says:

      The problem of Gribov copies does not go away in the Hamiltonian formalism. If complete gauge fixing (such as axial or Coulomb gauge) is performed, Gauss’ law needs to be solved.

      As an example I have some familiarity with, if one solves Gauss’ law on a Hamiltonian lattice, to get an axial gauge condition (say $A_{3}(x)=0$ or $U_{3}(x)=1$ on most of space, it is a messy problem to define the conjugate electric field variable ($E_{3}$, or $l_{3}$ on the lattice) everywhere, if the boundary condition is periodic. With open boundary conditions in every direction (so space is a rectangular prism) the situation is much better, and it is possible to further fix the gauge, completely removing all gauge freedom. It is not a trivial task to do this, however.

      Anyway, in the above example, the Gribov problem cannot be ignored, though it can be solved.

    15. Peter Orland says:

      P.S. In the BRST operator formalism, you need to construct an exterior derivative operator from constraints. You need to prove that the square of this operator is zero. I guarantee that there is a problem with this on the Hamiltonian lattice. That is because the degrees of freedom of links of the lattice lie in the group manifold, not a nice linear space.

    16. Chris Oakley says:

      Just to say that I approve of this kind of post. Not being a mathematician, I expect to be left behind soon, although this has not happened yet.

      It is funny that with the existence of half-integral spin, SU(2) and not SO(3) is the symmetry of nature (equivalently SL(2,C) and not O(3,1)). It suggests that Minkowski space is not the fundamental thing, and yet theories that take advantage of this (I am thinking in particular of twistors here, where the fundamental objects live in SU(2,2), the double cover of C(3,1)), although elegant, never seem to help in describing the real world.

    17. FNesti says:

      Thanks Peter,

      yes I also find this quadruplication of ‘spinors’ (e.g. in flat 4D) important, although it is completely unknown to physicists.

      What I was (and is) curious about are the precise properties under possible symmetries (loocal frame rotations but not only?) of these copies. Probably one will have to go through the work of Benn, Tucker et al…

      (Indeed there was a burst of activity on Kahler spinors around the eighties, while now people are much less imaginative (or should one say realists?)).

    18. Aaron Bergman says:

      It is funny that with the existence of half-integral spin, SU(2) and not SO(3) is the symmetry of nature (equivalently SL(2,C) and not O(3,1)).

      The point is that physics deals with projective representations. The half-integral spin reps or SU(2) are perfectly good projective representations of SO(3). You can prove that for a (compact?) f.d. Lie group, any projective rep is an honest rep of the universal cover. This isn’t true in general, however, and that can show up in certain situations.

    19. N. Nakanishi says:

      Concerning the definition of “symmetry”

      The question raised by Kurt can be answered in the following way.
      Suppose that the operator algebra of fields has been found. If a transformation U leaves it invariant, U is a symmetry.
      In the canonical formalism, since the operator algebra is determinded by field equations and equal-time commutation relations, one can check the invariance without constructing the operator solution explicitly.

      Concerning the problem of Gribov copies

      In the BRS-formulated covariant operator formalism of gauge theories, there arises no problem concerning the topology of the
      gauge fixing. The BRS generator can be proved to be nilpotent. The problem of topology is encountered in the path-integral formalism, because it is based on classical quantities. One should note that there is no justification of the non-perturbative path-integral approach; indeed, in that framework, one cannot discuss the unitarity of the physical S-matrix.

    20. anonymous says:

      @Prof. Nakanishi
      I think you’re the man who introduced the Nakanishi-Lautrup field in the quantization of the Maxwell field, is it correct? If so, it is a great honor to have you here!

    21. N. Nakanishi says:

      anonymous:

      Yes. Thank you for your courtesy.

    22. Kurt says:

      Re: Nakanishi-

      I wonder, however, if such a definition can be made without additional structure beyond a Hilbert space and a Hamiltonian. If we are dealing with a generic quantum system, not necessarily a field theory and not necessarily possessing an algebra of field operators, is there still a good notion of symmetry?

      KH

    23. This is very cool; thanks Peter! So nice to have a mathematician doing the talking after half a term of a physicist teaching us “quantum mechanics” (i.e. linear algebra) here at Caltech :-|.

    24. Peter Orland says:

      It is not correct that Hamiltonian BRST quantization solves the Gribov problem. If the Hamiltonian gauge theory is ultraviolet-regularized in a gauge-invariant way (and the only way to do this, except in some special cases, is on a lattice), there is no nilpotent BRST operator.

      From what I understand, Hamiltonian BRST is fine for dealing with theories perturbatively, but that’s it.

    25. Coin says:

      Peter, thanks, this is actually really helpful. I have a couple of questions that I hope aren’t too stupid.

      When G is a Lie group with Lie algebra, (frac-g, [•,•]), differentiating ∏ gives a unitary representation (π, H) of g on H.

      I’m a little confused as to why this is– I.E. why it is exactly that differentiating ∏ happens to produce this thing that you want, or exactly what it means to “differentiate” this ∏ object.

      Examining the wikipedia article for lie algebras, they mention: “Given a Lie group, a Lie algebra can be associated to it… by endowing the tangent space to the identity with the differential of the adjoint map.” In linked articles they describe the “adjoint map” as something that sounds extremely similar to the ∏ map you define and which they use to construct an “adjoint representation” of a lie algebra; and they define “differential” in this context as a particular operation they also call a “pushforward”. Is this in fact what is happening here? I.E. is it the case that ∏ is the same thing as the adjoint map of your symmetry group G, and the reason why differentiating ∏ gives a representation of frac-g is because this is just the canonical way of getting a lie algebra’s “adjoint representation” given the associated lie group?

      (By the way, a minor thing– you say “a unitary operator ∏(g) satisfying [conditions], i.e. the map P from group elements to unitary operators is a homomorphism”. Is the “P” here a typo for ∏?)

      An interesting philosophical question is whether the quantum system that governs the real world is purely determined by symmetry, i.e. such that ALL its observables come from symmetries in this manner.

      Are there any known examples of quantum observables that are not specifically known to come from symmetries?

      Thanks!

    26. Kasper Olsen says:

      Hi Peter,

      In mentioning different groups of symmetries, you need to say what the group action is; just to be more precise..

    27. Chris Oakley says:

      Aaron,

      I am not quite sure what you are saying here … a 2π rotation leaves Minkowski space invariant, but does not leave a spinor invariant, yet a spinor is a necessary part of physics as we understand it. It suggests that requiring a unitary rep of the isometry group of Minkowski space is not quite the right thing to do, and that there is something more fundamental lurking beneath.

    28. Marco Frasca says:

      Dear All,

      About Gribov copies and Yang-Mills there is a quite interesting situation both from theoretical and lattice point of views. There are two ways people managed confinement and these were Gribov-Zwanzinger and Kubo-Ojima scenarios. Both scenarios, that imply that Gribov copies are relevant in the infrared, require the gluon propagator going to zero at lower momenta and the ghost propagator going to infinity in the same limit faster than the free particle one. Functional methods devised in the ’90 seemed to support such a view.

      Of course, several other people turned out to computers to see if these views are true. These computations, done on lattices having huge volumes (till (27fm)^4!), completely disproved above results showing as the gluon propagator goes to non-null finite value at lower momenta and the ghost propagator is the one of a free particle. Above confinement scenarios and functional methods that heavily rely on Gribov copies are blatantly wrong. The main conclusion to be derived from this is that Gribov copies do not play any role for Yang-Mills theory, not even in the infrared limit. Indeed, the concept of Gribov copies appears useless to our understanding of Yang-Mills theory.

      About BRST symmetry and Yang-Mills theory one has the following conclusion recently drawn in

      http://arxiv.org/abs/0810.1987 .

      This symmetry is violated by the lattice solution that implies that the gluon acquires a mass. Indeed, lattice computations done so far show us the very existence of a mass gap and this mass gap has the effect to break BRST symmetry. This is rather a screening mass than a true pole in the propagator and this means that the true particles of Yang-Mills theory in the infrared are no more gluons but gluon bound states making the above mass a constituent mass (see also my blog

      http://marcofrasca.wordpress.com/2008/10/14/a-new-point-of-view/ .

      This is our current understanding about confinement and Yang-Mills theory. Some theoretical results have also been known that are in agreement with it (e.g. see my paper to appear in PLB

      http://arxiv.org/abs/0709.2042

      and refs therein).

      This situation is quite interesting and shows clearly the dynamics of the understanding in physical science at work.

      Marco

    29. N. Nakanishi says:

      Re: Kurt

      Symmetry should be defined at the operator level, because symmetry may be broken spontaneously at the level of representation. Any quantum system is described by an operator algebra and its representation in terms of state vectors. Symmetry is the invaraince property of operator algebra. Hamiltonian is nothing more than a particular symmetry generator for time translation.

    30. Peter Orland says:

      Marco,

      It simply isn’t clear what the relevance of Gribov copies is to confinement and the mass gap. We won’t know until confinement and the gap are understood.

      No calculation or theorem indicates confinement at weak bare coupling (strong bare coupling has been understood since the mid-70’s). Ruling out a proposed scenario invoking certain concepts does not mean those concepts are not relevant. We simply don’t know whether they are relevant. None of the scenarios proposed to explain confinement have been successful (including some I have worked on). That doesn’t mean that the ideas have no merit, just as it doesn’t mean that they do.

      The best that theorists can do on tough problems, is to study what they find interesting, follow hunches, prove or disprove hypotheses, etc. But we also have to be intellectually honest; we have to be prepared to abandon our most cherished ideas when they don’t work.

      But anyway, I don’t think that Peter W. is after confinement per se, just a better understanding of gauge theories and BRST.

    31. Marco Frasca says:

      Peter,

      This is the work of an entire community and I was thinking that some of these results were leaking out to a lot of people outside. By the way you are speaking it seems like all this people’s work went out useless. Do not you believe on lattice results for Yang-Mills propagators? It seems to me that the a clear picture is emerging by the work of this people and its acceptance may be just a matter of time. Meanwhile, I am trying to let their (and mine) work widely known. But, of course, your considerations seem simply out of track.

      Marco

    32. woit says:

      Coin,

      Thanks for pointing out the typo.

      Many quantum observables don’t come in any known way from a symmetry. For example, in a gauge theory, take the field-strength operator (or some gauge invariant quantity constructed out of it).

      The way I stated the relation between representations of the group and the Lie algebra is standard, if a bit abstract for most physicists. The point is that a very good way to think of a representation of a group is as a map of groups

      \Pi: G\rightarrow GL(n,\mathbf C)

      that is a homomorphism of groups. For a unitary representation, the map is to a unitary group U(n)..

      Such a homomorphism is a differentiable map, its differential is a map from the tangent bundle of G to the tangent bundle of U(n). Evaluating this at the identity of the group gives a homomorphism of Lie algebras, from the Lie algebra of G (which can be identified with the tangent space at the identity) to the Lie algebra of U(n) (which can be identified with both the tangent space to U(n) at the identity, and the space of n by n skew-hermitian matrices).

      The above applies to any representation, What you’re quoting from Wikipedia is how one constructs a very specific representation, the adjoint represention of a group on its Lie algebra. One way to do this is to look at the conjugation map of the group to itself

      g_0\in G \rightarrow gg_0g^{-1}\in G

      and take the differential of that. For each element g of the group you get a linear map of the tangent space at the identity to itself, this is the adjoint representation of the group on its own Lie algebra.

    33. Peter Woit says:

      Marco,

      Please, don’t try and turn this into a discussion of your favorite ideas about confinement. Peter Orland is right that the question of confinement is not one that I’m trying to address here, it involves a whole host of very different issues.

      My point of view the significance of BRST is that a better understanding of its mathematical context can lead to new ways of using it to understand physics. I don’t have any particular reason to think that this will help with the understanding of confinement. More likely would be some new insight into electroweak symmetry breaking, where the issue of how gauge symmetry gets handled is more directly relevant.

    34. Peter Woit says:

      Nakanishi,

      Many thanks for your comments. The point that the right way to think about symmetries of a quantum system is as the group of automorphisms of the algebra of operators was especially helpful to me.

    35. Marco Frasca says:

      Peter,

      Sorry for my improper posting but some people turned your very fine argument into something about Yang-Mills and Gribov copies and for this there exist a lot of lattice computations. I just aimed to answer to this.

      So, I turn back into my little corner.

      Marco

    36. Peter Orland says:

      Hi Peter,

      I had a minor question concerning your formalism. Are you concerned about unbounded operators and non-normalizable states in all of this? The reason I ask is that you are defining states the mathematician’s way as Hilbert space vectors, which square-normalizable. Depending on the manifold you start with, you might need to deal with a spectrum defined more generally than eigenvalues on such states, e.g. by singularities of resolvents.

      If you work on a bounded manifold (such as a Lie group), there is no need to worry about this issue since the entire spectrum consists of eigenvalues of Hilbert vectors. But I assume you eventually want to consider field theory in the thermodynamic limit, which means the spectrum contains numbers which are not eigenvalues (or in the physicist’s language, not eigenvalues of normalizable states).

      I suspect you’ve already dealt with these functional-analytic considerations, or they are just not important. I was just wondering…

    37. Serifo says:

      Hi professor Peter,

      I suppose the considered Hilbert space is a separable Hilbert space, right ? Or it doesn`t really matter ?!

      I also suppose you are assuming the Hilbert space as a complex vector space ( right ? ), now since the field of complex numbers is an extension of the field of real numbers, the same Hilbert space can be viewed as a real vector space . Does it help when we view the space as real Hilbert space ?

      Sincerely

    38. N. Nakanishi says:

      Peter Woit:

      There is no nilpotent selfadjoint operator defined on a Hilbert space. If you discuss the BRS generator in your next presentation, I expect that you will introduce an indefinite-metric Hilbert space. Is that so?

    39. Hendrik says:

      Dear Peter, I agree that it is high time that there is some serious discussion on quantum constraints;- there is a wide spectrum of different methods, not all of them equivalent. A few reactions to specific points:

      1) The initial framework which you propose, of operators on a Hilbert space, binds you to a specific representation. It is better to start with an abstract algebra (e.g. a C*-algebra) as your setting, and then to select the convenient representations. For instance, as Peter Orland pointed out, in some representations you cannot solve your constraints due to continuous spectrum problems. There is a discussion of these points here (Erratum here) In any case, a constraint algorithm constructs a different representation for the observable subalgebra, than the starting representation.

      2) There is not one BRST-formalism, but several, of which the two major approaches are the BFV-approach (as in the book by Hennaux and Teitelboim) and the Kugo-Ojima approach. Whilst these are superficially similar, they are in fact inequivalent, i.e. produce different results for the same system (I have a PhD student whose thesis contains substantiation of this). Moreover, the BFV-approach is a general constraint algorithm (so does not need a gauge group action to be applied), whereas the KO-approach is specifically written for gauge theories, so we don’t know how to apply it to general constraints.
      It is not difficult to show that the BFV-approach produces different results than the usual Dirac constraint method (see Sect 7 of this old scanned preprint -11MB, sorry). On the other hand, for Quantum Electromagnetism (the only gauge QFT which we can handle explicitly in the C*-algebraic framework), the KO-approach to BRST produces the correct results.

      3) A common problem of the BFV-BRST method is that it selects multiple copies of the physical state space (this has been noted all over the literature, e.g. p50 of Van Holten, and I also found it in my old preprint above). This should not be confused with the Gribov ambiguity problem, which as I understand it, is a classical and global result, stating that there is no continuous section of the gauge orbit space of the connections (as the set of orbits of the connections is a principal fibre bundle, I think this just means that this bundle is nontrivial). Since the operator theory with which BRST works is local, infinitesimal (not global) and “forgets” the smooth structures of the classical objects, I cannot see how it can detect a Gribov problem. Of course, it is a different matter for the path integral approach, which starts from the classical theory. Having said that, I know that McMullan did some work on BRST and Gribov ambiguity, using Mackey quantization and analogies (as the groups are infinite dimensional, Mackey theory does not apply directly).

      4) Serifo’s question is an interesting one;- if you want to avoid continuous spectrum problems, you may need to use representations which are discontinuous w.r.t. the exponentiated commutation relations (“nonregular representations”) and these must be on nonseparable Hilbert spaces. However, the final constrained Hilbert space must always be separable.

    40. Peter Woit says:

      Thanks to all for the interesting comments. The questions about BRST that I’m interested in are independent of the analysis issues people raise, they occur even for toy models with finite dimensional state spaces.

      The state spaces in question are complex vector spaces, QM requires this. Similarly in representation theory I’ll be discussing complex representations. Representations on real vector spaces don’t give the examples and structures I’m interested in.

      The question Nakanishi raises is an important one: having a self-adjoint BRST operator with square zero requires an indefinite metric. The Dirac cohomology idea I’ve been working with evades this, since the analog of the BRST operator has a square that is not zero, but a central element in the operator algebra. So, it acts as zero on the algebra of observables, but is not zero on states. This is one of the main reasons I think there may be something new here.

      Hendrik is right that there are multiple versions of “BRST”, and I’ve always found keeping them straight and understanding the relations between them incredibly confusing. What I’ll be writing about is yet a different variant. It’s not one that can be applied to any constrained dynamical system, but is intended to work for gauge theories.

    41. Hendrik says:

      Dear Peter, if your BRST-variant is intended for gauge theories, then Krein representations (indefinite metric) are required for reasons other than BRST (unless you use the exponentiated C*-algebra version with nonregular representations). The reasons come from a whole swag of theorems:- (1) Strocchi, F.: Gauge Problem in Quantum Field Theory, Phys. Rev. 162, 1429-1438 (1967) – there is no representation of the vector potential (i.e. local gauge field) which is covariant w.r.t. the usual action of the Poincare group, having a vacuum and satisfying the Maxwell equations, in which the fields (i.e. local curvature) are nonzero. Also see Comm. Math. Phys. 35, 25 (1974), and other Strocchi papers at this time or earlier. (2) Wightman, A.S., Garding, L.: Arkiv Fysik 28, 129 (1964) a covariant representation of a vector potential which is weakly local, cannot be a normal Hilbert space representation, it must be done w.r.t. an indefinite inner product. (3) Barut and Raczka proved that the only zero mass representation of the Poincare group on tensor-valued functions (e.g. Fock rep) must necessarily have indefinite metric.

      These theorems meant that everyone in Wightman Field Theory, did their gauge fields in the indefinite metric framework.
      The only way in which you can avoid using indefinite metric, is by using the exponentiated (Weyl) fields in a C*-algebra framework. Then you can work with ordinary Hilbert space representations, but they must be necessarily nonregular to avoid the theorems above. Nevertheless, you still get sensible results after constraining. You can read about that approach here. Unfortunately, the only gauge field for which we have an explicit C*-algebra is the Electromagnetic field.

    42. The point that the right way to think about symmetries of a quantum system is as the group of automorphisms of the algebra of operators was especially helpful to me.

      By the way, the essence of this is extracted by the Doplicher-Roberts reconstruction theorem applied to the symmetric monoidal category of local endomorphisms of the net of local quantum observables:

      the theorem proves that just from the net of local algebras alone the global symmetry group of the QFT can be reconstructed, hence in particular its particle content be determined.

      The DR-reconstruction theorem is one of the earliest instances where physicists found in their application a fundamental purely category-theoretic fact.

    43. tomate says:

      I’m sorry for the down-to-earth question in such elevate discussion. Is every symmetry of a quantum system connected to a conserved quantity or charge (as in your first few examples)? If so, does the U(1) global symmetry uniquely imply conservation of probability and vice-versa?

    44. Chris Oakley says:

      Tomate,

      Yes, every symmetry has a conservation law, a result known as Noether’s theorem. This is a classical result, but it applies also to quantum field theory, and is the only thing I can think of in elementary QFT that is named after a woman. U(1) global symmetry, if present, may be connected with charge conservation but has nothing to do with conservation of probability. One of the reasons for requiring the inner product or metric of Hilbert space to be positive definite in the foregoing discussion is that these quantities sometimes have the interpretation of probabilities, and probabilities cannot be negative.

    45. Peter Woit says:

      Tomate,

      The point Kurt was asking about and Nakanishi clarified, is that I’m thinking of a “symmetry” more generally, as any automorphism of the algebra of observables (i.e. any map from this algebra to itself that preserves the algebra structure). Of these “symmetries”, some will commute with the Hamiltonian operator, some won’t. The ones that commute with the Hamiltonian are the ones that correspond to conserved quantities. Ones that don’t commute with the Hamiltonian are sometimes given names like “dynamical symmetry”, or “spectrum generating algebra”. A few seconds of Googling turned up this more detailed explanation:

      http://eom.springer.de/S/s110230.htm

      As Chris explains, conservation of probability comes from the condtion that the Hamiltonian be self-adjoint, generating a unitary time evolution. The U(1) action corresponds to a charge, which may or may not be conserved.

    46. Hendrik says:

      Dear Peter, I think the concept of “symmetry of a quantum system” is not an easy one to sort out:
      In the literature the concept is defined in a number of inequivalent ways:(1) As a continuous map on the projective sphere of a Hilbert space, preserving (moduli of) inner products. The content of Wigner’s theorem is that these are either unitaries or antiunitaries, and in the case of a continuous one-parameter group of them, then they must be unitaries.(2) As automorphisms of the *-algebra A of observables (where A can be either abstract, or concrete operators on a Hilbert space H – if they are unbounded, a dense invariant domain must be specified).(3) As superderivations, e.g. in the case of BRST or supersymmetry.

      The mathematical structure of collections of symmetries vary, e.g. they can be groups (discrete, Lie, infinite dimensional) or they can even be quantum groups, which are not groups at all. In the case that they are Lie groups, then of course the infinitesimal generators (selfadjoint operators or *-derivations) encode the symmetry group.

      A symmetry is relative, i.e. it is a symmetry of something, and that something must be specified. An algebra A of operators on a Hilbert (or Krein) space by itself is not a physical system (different quantum systems have isomorphic algebras of observables). Some extra structure must be specified, together with some map that takes you from the labels in your theory to something outside, which can e.g. be measured in experiment. That is why we use suggestive labels like “mass” or “charge” for our mathematical objects. Then a symmetry is a transformation of the system which leaves the defining structures of the system invariant. So for instance, a “dynamical symmetry” is a transformation which leaves some specified one-parameter time evolution group invariant, and a “relativistic symmetry” is something which e.g. respects causality requirements. Even elementary descriptions of symmetry, include such a requirement. I am very uncomfortable with a “radical” interpetation of symmetry as any automorphism of your algebra of observables (though it appears in the literature). This seems to me without physical content.

      The discussion regarding Tomate’s question above, is not quite complete:If you have a continuous one-parameter group of symmetries in the Wigner sense above (they preserve probability), then it is given by a continuous one-parameter group of unitaries t -> U(t), and by Stone’s theorem this must be of the form U(t) = exp(itS) for some selfadjoint operator S. If you assume that you have a time-evolution which commutes with the symmetries, then S is a conserved quantity, i.e. does not change in time. However, if you start from a discrete group or a discontinuous action, then there is no such conserved quantity. This picture becomes much more problematic if you view your symmetries as automorphisms on the algebra. For instance, you need not have that it is unitarily implemented, so even if you have a continuous one-parameter group, there need not be a concrete generating operator. (At the most, you have a *-derivation of the algebra).

    47. Thomas Larsson says:

      It seems to me that this discussion is closely related to the eternal confusion about diffeomorphism symmetry in GR. From one POV, every model has a diffeo symmetry because it can be written in curvilinear coordinates, and hence the concept is empty. By the same token, every (massive or massless) model has scale symmetry; after all, the scaling group is a subgroup of diffeomorphism group, since the dilatation D = x^u d/dx^u is a particular kind of vector field.

      From this viewpoint, the difference between massive and massless is that the representations of the scaling algebra are qualitatively different. In the massless case, the Hilbert space is a direct sum of one-dimensional reps, spanned by a vector |h>:

      D |h> = h |h>.

      In the massive case, we need to look at a family of models at different scales, and the reps act on an infinite-dimensional vector space with basis |n>:

      D |n> = |n+1>.

      Although this viewpoint on massive models is not necessarily wrong, it is not fruitful neither; unlike in the massless case, we don’t learn anything about the model by pondering dilatations.

      More generally, assume that we have a big group G, a subgroup H, and an H rep R acting on our Hilbert space. Then G also acts on the Hilbert space, possibly after we have completed it to include new states. The G rep is necessarily a suprep of the induced rep Ind R. If the G rep is all of Ind R, considering G is fruitless; all information is already encoded in H. OTOH, if the G rep is a proper subrep of Ind R, we learn something new, and this viewpoint is fruitful.

      This suggest one definition of “symmetry”. Let G be the full group of automorphisms on the Hilbert space, and let H be the biggest subgroup such that the rep of G is induced from H. Then H is a “symmetry”, in the sense that it is the largest group that says something nontrivial about the Hilbert space.

    48. tomate says:

      Hendrick,

      > However, if you start from a discrete group or a discontinuous action, then there is no such conserved quantity.

      I know this is a typical statement about discrete symmetries, but couldn’t we put it this way: symmetry is an unobservable degree of freedom (as assumed in Wigner’s theorem) and conservation laws (of continuous quantities like momentum or discrete charges) are their observable realization. A discrete example: inverting spatial axis is unobservable but has a corresponding conserved quantity, parity, which you can infer from experiment (up to conventions) and might be conserved in a process if the theory has such symmetry. I know it sounds a little bit tautologic, but it seems to me like a coherent view. Does this make any sense?

      Peter and Chris,

      thanks for explanation.

    49. Peter Woit says:

      Thomas,

      I’m confused. You’re starting with an H rep, for H a subgroup of G. There’s the induced G rep, but what’s the other one you have in mind? Can you give an example?

    50. Thomas Larsson says:

      1. The above scale symmetry of a massive model, on the group level. G = scaling group, H = trivial group, R = trivial rep.

      2. Give the massive model a conformal symmetry: G = conformal group, H = Poincare group, R = a Poincare rep.

      3. Make a model diff symmetric: G = diffeomorphism group, H = Poincare or conformal, R = a rep.

      The last example is similar in spirit to saying that any model has a diffeomorphism symmetry because it can be written without coordinates, but not exactly the same, I think.

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