# Quantum symmetries: $S$ or $Z$?

Let $I$ be the action of some QFT (gauge-fixed and including all the necessary counter-terms); $S$ the associated scattering-matrix; and $Z$ the partition function (in the form of, say, a path integral). There are three notions of symmetries that are typically discussed,

1. Action symmetries, that is, transformations of the form $\phi\to\phi'$ that leave $I$ invariant,
2. $S$-matrix symmetries, that is, operators that (super)commute with $S$, and
3. Quantum symmetries, that is, transformations of the form $\phi\to\phi'$ that leave the volume form $\mathrm e^{I[\phi]}\mathrm d\phi$ invariant.

It is a well-known phenomenon that the symmetries of $I$ need not agree with those of neither $S$ nor $Z$ (e.g., anomalies, SSB, etc.). What is not-so-clear is whether $S$ symmetries and $Z$ symmetries are equivalent. What I want to know is whether

For each symmetry of $S$ there is a symmetry of $Z$ and vice-versa

or a counter-example. If the equivalence is actually true, I would like to have a more-or-less precise statement, in the form of a theorem (to the usual level of rigour in physics textbooks).

No cheating please. Counter-examples are only valid for "real-life" QFT's (e.g. in a free theory everything commutes with $S$ but not everything leaves $Z$ invariant; this is not a valid counter-example because it is completely trivial). No TQFT's either. Thanks.

Someone mentioned in the comments the effective action $\Gamma[\phi]$, which is defined as the Legendre transform of $\log Z$. I didn't want to bring this object into the picture, because I wanted to leave it to the discretion of the rest of users whether to mention this object or not. In principle, I don't need answers to analyse the symmetries of this object, but they may if they believe it could be useful to do so. In any case, let me stress that $\Gamma$ is not the same object as $I$.

• If you consider the quantum action $I$ (which seems to be so since you include counter-terms), then, of course, it will contain mostly the same symmetries as $S$ and $Z$. Commented Nov 14, 2017 at 7:59
• In retrospective, I should have called the first kind of symmetries "Lagrangian symmetries". This way, the title could have been "L, S or Z" :-P Commented Mar 6, 2018 at 16:25
• Translations are not Lagrangian symmetries, only symmetries of the action. Commented Mar 6, 2018 at 17:28
• @ArnoldNeumaier It was more of a joke (as in, "LSZ reduction formula"). Admittedly, not a very funny one... Commented Mar 6, 2018 at 17:31
• My point was that the joke is too lame because of the lack of fatual correspondence of the underlying language. Commented Mar 6, 2018 at 17:33

People sometimes talk about on-shell symmetries: symmetries of the equations of motion or the S-matrix, which do not hold off-shell (i.e. at the level of the action, path integral, correlators, etc). Electric-magnetic duality transformations \begin{aligned} F &\rightarrow \cos(\theta)\,F - \sin(\theta)\,\star F \\ \star F &\rightarrow \sin(\theta)\,F + \cos(\theta)\,\star F \end{aligned} are an example of this. These transformations leave the equations of motion and the S-matrix invariant (with a corresponding selection rule), so they would qualify as a symmetry.

However, this symmetry is usually broken by non-perturbative effects (e.g. monopoles). In addition, it is well known that dualities are not symmetries of the full theory but instead a better understood as a change in our description of the physics. For instance duality transformations relate the partition function of the theory at different values of the coupling (see this paper for an example).

• Thanks, this is interesting. I'll have to think about it. For now, can you clarify how the electric-magnetic duality works at the level of the $S$-matrix? Is there some operator that generates this symmetry, $[Q,F]=\theta \star F$, that commutes with $S$? Commented Mar 10, 2018 at 4:28
• Yes, there is a charge operator $Q$ which is given as the integral of a "current" which is the Chern-Simons 3-form. I write quotes around current because it is not gauge invariant. However, it only changes by a total derivative under a gauge transformation so $Q$ is well defined. This operator commutes with $S$ and gives a usual "charge conservation"-type selection rule.
– jpm
Commented Mar 10, 2018 at 4:33
• It's confusing, but I think it's wrong to think of a duality as a redundancy. At special self-dual values of the coupling constants you really do have new symmetries in the theory. They are not automatically gauged. Take $R \to 1/R$ duality of a compact boson in 2d as an example. At the self-dual radius the symmetry is enhanced to $SU(2)$. Commented Mar 10, 2018 at 5:00
• Can you please give a reference for the statement that S-matrix is not sensitive to non-perturbative effects? Commented Mar 10, 2018 at 7:06
• @jpm I'm just commenting because you said "it is well known that dualities are not symmetries of the full theory but instead a better understood as a change in our description of the physics", which is the difference between the self-dual points having an extra symmetry or just a gauge symmetry. Commented Mar 10, 2018 at 19:16

I think that a symmetry in Z implies perturbatively a symmetry in S since the perturbative S-matrix can be computed from the effective action (which is directly related to Z) in a straightforward way that leaves no room for breaking a symmetry.

Whether the converse holds is questionable as it is not even clear whether the S-matrix determines the action.

• Thanks. Re. your first paragraph: I think so too. But the purpose of this post was to have a more precise statement beyond "I think it should work". Re. your second paragraph: indeed, the $S$ matrix does not determine the action, and that makes the question non-trivial. But in principle it is perfectly valid that whatever the action is, if $S$ has a symmetry then so does the action. The actual realisation of the symmetry at the level of the action depends on the action itself, sure. But is it there, or it need not? Commented Mar 6, 2018 at 17:42
• Well, for one direction I gave reasons that can probably be turned into a proof at the usual level of rigor of theoretical physics. But I haven't seen the problem discussed in the literature, and unfortunately don't have enough time to work out the details. - For the other direction, I admit that your question makes sense, but nonuniqueness makes a positive bet very risky. Commented Mar 6, 2018 at 17:48