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I have a basic understanding of the Wightman axioms for QFT. I was reading the about the Mass Gap problem for simple compact gauge groups and was wondering how the gauge group is supposed to be implemented into the framework. The official problem description (pdf) says:

To establish existence of four-dimensional quantum gauge theory with gauge group $G$, one should define a quantum field theory (in the sense of the Wightman Axioms) with local quantum field operators in correspondence with the gauge invariant local polynomials in curvature and its covariant derivatives.

Can someone clarify what this means and what it means in terms of the physics?

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  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/69287/2451 $\endgroup$ – Qmechanic Jan 5 '14 at 22:47
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    $\begingroup$ I could blather about Yang-Mills all day, but it's hard to know what to clarify without knowing what you're finding confusing. $\endgroup$ – user1504 Jan 6 '14 at 1:39
  • $\begingroup$ @user1504 I don't know much about gauge theory other than that classically, a gauge group (varying point-wise) leaves the Lagrangian invariant. I would like to know intuitively how this classical picture is supposed to manifest itself in QM and why the above requirement in my question is the mathematical image of this manifestation. $\endgroup$ – SWV Jan 6 '14 at 13:15
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    $\begingroup$ Nonetheless, I wouldn't mind blathering. $\endgroup$ – SWV Jan 6 '14 at 14:02
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    $\begingroup$ for a fairly precise discussion of what the official problem description means see physicsoverflow.org/21786/… $\endgroup$ – Arnold Neumaier Aug 29 '14 at 15:59
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OK, blathering it is.

The key thing to understand in Yang-Mills theory (whether or not you're trying to be mathematically rigorous) is that gauge symmetry is not a physical symmetry. (Its job is to keep track of the overcounting in a redundant description.)

Classically, this means that the observables -- the numerical quantities which we can measure -- are not functions on the space $\mathcal{F}$ of connections but rather functions on the quotient space $\mathcal{F}/\mathcal{G}$ of connections modulo the action of the group of gauge transformations. So, for example, for any $G$-invariant polynomial $P$ on the Lie algebra $\mathfrak{g}$ of $G$ and any point $x \in \mathbb{R}^4$, there's an observable $\mathcal{O}_{P,x}$, which sends the equivalence class $[A]$ of a connection $A$ to $P(F_A(x))$, where $F_A$ is the curvature of $A$. But there isn't any physical way of measuring the value of a connection $A$ at the point $x$; this quantity is useful for intermediate steps in the computation, but it is not an observable.

In the Wightman framework, one describes a QFT in terms of a Hilbert space which carries a representation of the Poincare group and comes equipped with a collection of local observables. So for Yang-Mills theory, naively, what we'd like is an operator $\hat{\mathcal{O}}_{P,x}$ for every classical observable like $\mathcal{O}_{P,x}$. This doesn't quite work, however. You can't have an observable $\hat{\mathcal{O}}_{P,x}$ which measures the value of something precisely at $x$, for basically the same reason that you can't have a particle whose wave-function is supported precisely at $x$. Things have to get fuzzed out a little bit in quantum theories. So instead of a function $x \mapsto O_{P,x}$, what we get in the quantum theory is an operator-valued distribution, which sends any test function $f$ to the observable $\hat{\mathcal{O}}_{P,f}$ which measures the value of the smeared observable $\int P(F_A(x))f(x)dx$.

You will occasionally find claims (on wikipedia, for example) that the Wightman axioms fail for gauge theories because gauge theories involve states with negative norms. This isn't really correct. Negative norms show up at intermediate stages in some quantization procedures (but not all of them), but they should never be present in the final result, which is the only thing the Wightman axioms are concerned with.

Note that what I've written above isn't quite the usual statement of the Wightman axioms. People usually give them only for scalar field theory, where they can take a shortcut because the basic fields $\phi(x)$ that generate the observables are actually observable themselves. However, all one really needs for Wightman's setup is local observables whose classical analogues provide a complete set of coordinates on the space of classical fields.

Finally, the really interesting subtlety: If you're being careful, the group of gauge transforations is the subgroup of $Map(\mathbb{R}^4,G)$ consisting of $G$-valued functions which take the identity value at $\infty$. This would suggest that the values of invariant differential polynomials in the curvature tensor are close to but not quite a complete set of coordinates on $\mathcal{F}/\mathcal{G}$, since they're also invariant under the gauge transformations which are constant over $\mathbb{R}^4$. So why should these be enough? The answer is confinement. We expect that, in pure Yang-Mills theory, the only physically realizable states are color singlets, transforming trivially under the global $G$.

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  • $\begingroup$ Why just polynomials in $F_A$ are observables? This just looks like the a variant of the Chern class. $\endgroup$ – user40276 Apr 17 '15 at 9:37
  • $\begingroup$ Furthermore, these observables (since they're similar to the Chern class) will not differentiate between two choices of gauge (they just know the topology of the underlying $G$-bundle) $\endgroup$ – user40276 Apr 17 '15 at 9:43
  • $\begingroup$ @user40276: A long overdue response: Observables don't distinguish between gauge choices. That's a critical part of the definition of observable. $\endgroup$ – user1504 Sep 12 '16 at 0:30
  • $\begingroup$ Sorry, I think I misused the word gauge here according to the physical usage and I probably thought you're picking the cohomology class of the polynomial applied to the connection. In the case that you pick the cohomology class, the connection chosen does not matter at all (the values of the polynomial will be cohomologous). But now I see that even if you don't pick the cohomology class, the observables will depend only on the curvature and not on the connection, so probably you're using a topologically trivial space-time, otherwise you would have to add Wilson loops to correct this. $\endgroup$ – user40276 Sep 12 '16 at 11:14
  • $\begingroup$ … , otherwise the observables wouldn't distinguish different connections on the same vector bundle. And Thanks for the response. $\endgroup$ – user40276 Sep 12 '16 at 11:17

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