Suppose that we have a Lagrangian density like $$\mathcal L = -\frac{1}{4} \operatorname{tr} F_{\mu\nu}F^{\mu\nu} + \frac{\theta}{32\pi^2} \operatorname{tr} \big( \epsilon^{\mu\nu\rho\sigma} F_{\mu\nu}F_{\rho\sigma}\big) + \overline{\psi}\gamma^\mu D_\mu \psi$$
where $F_{\mu\nu}$ is the gauge field strength and $D_\mu$ the gauge covariant derivative, and $\psi$ is a fermion field. This Lagrangian is not $P$ conserving because of the $\theta$ term.

However if we redefine the fields $\psi \mapsto \exp(i\alpha \gamma_5)\psi$ we can make $\theta$ go away, by choosing $\alpha = \theta/2$ as per the Fujikawa method (described in [Weinberg], Chapter 22 or [Fujikawa]); this is due the path integral measure also transforming under the redifinition. With this redefinition of fields $\mathcal L$ is manifestly $P$ conserving. But surely I can't get more or less symmetry by redefining fields, so **how should I understand that the $P$ symmetry is not manifest with the original definition of the fields?**

I suspect that the $P$ transformation too transforms the path integral measure, in a way that sends $\theta \mapsto -\theta$, but I do not know how to show this.

- [Weinberg] Weinberg, S. The Quantum Theory of Fields. 2: Modern Applications
(Cambridge, 2005).
- [Fujikawa] Fujikawa, K. Path-Integral Measure for Gauge-Invariant Fermion Theories.
Phys. Rev. Lett. 42, 1195{1198 (18 Apr. 1979).

This post imported from StackExchange Physics at 2014-04-13 14:05 (UCT), posted by SE-user Robin Ekman