12
$\begingroup$

From Matthew D. Shwartz Quantum Field Theory textbook, he writes:

"Most of the time, a symmetry of a classical theory is also a symmetry of the quantum theory based on the same Lagrangian. When it is not, the symmetry is said to be anomalous."

A priori, how is this even possible? If a symmetry is present in a classical Lagrangian, and the quantum theory is based on the same Lagrangian, and the Lagrangian completely specifies the dynamics, then wouldn't the symmetry being present in the Lagrangian necessitate the symmetry is in the quantum theory?

$\endgroup$
1
  • 2
    $\begingroup$ The most important things have been said in the answers below, but if I can add a small remark: think of QFT as a tool to build theories on demand. You input an algebraic framework and a set of symmetries, and after that you're "on rails" to build a lagrangian, equations of motion, and so on. However, there's no guarantee that these inputs can lead to a consistent construct. Symmetry breaking, anomalies, non-renormalisable theories... The list of things that can go wrong is long. $\endgroup$
    – Miyase
    May 10 at 23:32

3 Answers 3

11
$\begingroup$

It's true as the others have said that the path integral measure may not be invariant.

However I don't really like that quote as a general description of anomalies, since it sounds like they rely on a Lagrangian or a classical limit, and there are theories without any classical counterpart for which we can still discuss their anomalies.

There are even symmetries which you can't see in the action, which don't act on the fields, but only on solitonic objects like magnetic monopoles.

Meanwhile, some anomalies, such as those involving the theta angle, are perfectly apparent at the level of the classical action and one doesn't need to think about the measure.

In modern times we realize that 't Hooft anomalies are a part of the very definition of the symmetry in a quantum system. We understand them by the behavior of topological defects associated with the symmetries, or by special contact terms in the Ward identities.

$\endgroup$
2
  • $\begingroup$ Hi Ryan, when you say there are anomalies that appear at the level of the classical action, do you mean when we stick the action in the partition function, or can we really see the anomaly in the classical theory alone? I would be interested in any references you could provide for some of the statements you've made. I have never really been able to get my head about the standard explication of 't Hooft anomalies. I (vaguely) understand they are obstructions to coupling to background gauge fields, but I feel like I don't understand the practical consequences of this for a theory... $\endgroup$
    – d_b
    May 13 at 15:24
  • 1
    $\begingroup$ @d_b I was thinking of the examples in these papers arxiv.org/abs/1703.00501 arxiv.org/abs/1705.04786 . Eg. there could be some symmetry which when you gauge it increases the periodicity of the theta angle, by changing the global structure of the gauge group, like SU(2) to SO(3). It will have an anomaly with time reversal which can pin \theta = \pi in the SU(2) theory. But then SO(3) has a 4pi periodic angle and TRS is "broken". The action sort of knows something non-perturbative in this case, right, since the \theta angle is only gonna contribute to some instanton sectors. $\endgroup$ May 13 at 16:01
9
$\begingroup$

This is such a short answer I'm tempted to just leave it as a comment, but besides specifying the Lagrangian you need to also specify your regularization, and the regularization is what introduces the anomaly. You can think of this regularization as specifying the measure of the path integral, and the Fujikawa perspective on anomalies is that the Lagrangian is invariant under the symmetry but the path integral measure is not.

$\endgroup$
8
$\begingroup$

The reason for this is that in the classical theory the Lagrangian fully specifies the dynamics of the system, however, in the Quantum system this is not true. Rather the Quantum theory is given by (in the path integral formulation) by the partition function $Z = \int D F e^{iS[F]}$ (with appropriate boundary condition. Now the idea is that even though you know that the action is invariant under a transformation the path integral measure $D F$ does not necessarily have to be. This can essentially break your symmetry fro the quantum theory.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.