Quantum Anomalies and Quantum Symmetries

Question

In Quantum Field Theories (QFT) there is a well known phenomenon of anomalies, where a classical symmetry is broken in the quantum theory due to a so called anomaly. This symmetry breaking can be understood in the path intergral formulation to be a result of the non-invariance of the functional measure under the classical symmetry transformation. Although the action itself is invariant, the functional measure might not be, and therefore, if that is the case the path integral wouldn't be invariant either. For further reading you can look in wikipedia.

My question is: is it possible that the action wouldn't be invariant, and the measure neither, but the path integral would? That is to say, the lack of invarince of the action and the measure would cancel each other, to form an invariant path integral, in such a way that would give have a symmetry in the quantum theory, but not in the classical one.

I would guess that it isn't possible, but is there a proof?

Answer 1

Comments to the question (v2):

1. Traditionally, the classical action $S$ sits in the Boltzmann factor $\exp\left[\frac{i}{\hbar} S\right]$ behind an inverse power of $\hbar$ in the path integral, while the path integral measure is independent of $\hbar$. In the conventional way of counting, we say that the Jacobian $J$ from the path integral measure is a one-loop effect proportional to $\hbar$, while the variation of the classical action $S$ is tree-level, i.e. independent of $\hbar$. Anyway, the upshot is, that in a usual setting, the two variations carry different $\hbar$-orders, and cannot cancel.

2. However, in principle one may introduce a quantum action $$\tag{A} W(\hbar)~=~S+ \sum_{n=1}^{\infty}\hbar^n M_n~=~ S+\hbar M_1+{\cal O}(\hbar^2)$$ with quantum terms. Then the Boltzmann factor becomes $$\tag{B} \exp\left[\frac{i}{\hbar} W\right]~=~\exp\left[\frac{i}{\hbar} S\right]e^{iM_1}(1+{\cal O}(\hbar)),$$ so that a cancellation may formally take place between the $M_1$-action factor and the path integral measure.

3. It seems appropriate to mention that such cancellation is the main idea behind the quantum master equation (QME) $$ \frac{1}{2}(W,W)~=~i\hbar\Delta W\tag{QME}$$ in the Batalin-Vilkovisky formalism. The lhs. and rhs. of the above QME are associated with the action and Jacobian, respectively, leading to a (generalized) BRST symmetry of the path integral.

4. Nevertheless, in practice in a local QFT, the BV operator (aka. the odd Laplacian) $\Delta$ is singular object. The QME is typically only satisfied if both sides of the QME are zero separately, i.e. the action and the measure parts cancel separately in practical applications.

References:

1. I.A. Batalin & G.A. Vilkovisky, Gauge Algebra and Quantization, Phys. Lett. B 102 (1981) 27–31.

2. W. Troost, P. van Nieuwenhuizen & A. Van Proeyen, Anomalies and the Batalin-Vilkovisky lagrangian formalism, Nucl. Phys. B333 (1990) 727.

3. nLab.

Answer 2

There is a simple proof that the cancellation is impossible (at least unless you are willing to add to the classical a term proportional to $\hbar$), I am reformulating an answer by @Qmechanic in a simpler language:

The anomaly, or the measure variation which is the typical source of the anomaly, contributes a term which is independent of $\hbar$, while any variation of the action comes with a $\frac{1}{\hbar}$ coefficient. Therefore, they just cannot cancel each other.