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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).
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I have found an explanation. At the end of Section 22.2, working with Euclidean path integrals, Weinberg shows that $$-\frac{1}{32\pi^2} \int d^4x \epsilon^E_{ijkl}F_{\alpha ij} F_{\beta kl} \operatorname{tr} (t_\alpha t_\beta) = n_+ - n_-$$ where $t_\alpha$ are gauge group generators and $n_\pm$ is the number of zero modes of the gauge covariant Dirac operator $i\gamma^\mu D_\mu$ with positive respectively negative chirality. This is the Atiyah-Singer index theorem at work according to Weinberg. (I have heard of this theorem before but I am not familiar with it.)

As a consequence of this, in the non-trivial sectors where the $\theta$ term is non-zero, there exists at least one zero mode. A massless field enters the path integral as $\overline\psi \gamma^\mu D_\mu \psi$ and it can be integrated out $$\mathcal Z =\int D[A_\mu, \ldots] \det (i\gamma^\mu D_\mu) \exp( S_\text{Yang-Mills} [A_\mu] + S_1[\ldots])$$ where the dots and $S_1$ stand for other fields and their actions, and the determinant of the operator and not its inverse appears, since $\psi$ is Grassmann. But if $\gamma^\mu D_\mu$ has a zero mode in every non-trivial sector... the determinant is 0 in all the non-trivial sectors. This is effectively the same as discarding the $\theta$-term!

(Weinberg's proof uses that $i\gamma^\mu D_\mu$ and $\gamma_5$ anti-commute. Since this is not the case for the bilinear $i\gamma^\mu D_\mu + m$ appearing for a massive field, we don't prove "too much" and rule out $P$ violations when all fermions are massive.)

I found the zero-mode argument in Chapter 94 of Srednicki's book. Srednicki argues that for physical reasons $i\gamma^\mu D_\mu$ must have a zero mode; in Weinberg's book one finds the mathematical proof that this is the case.

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