Quantum Anomalies and Quantum Symmetries

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.


  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.