Chiral symmetry of the Euclidean action for fermions

Question

In the literature, such as QFT Volume-II by Weinberg, p.368, the chiral anomaly is derived using Euclidean path integral. To formulate the question, let's start with the Minkowski space with signature $(+,-,-,-)$. The relevant action in the Minkowski space is given as $$S_M=\int d^4x \bar{\psi}i\gamma^\mu D_\mu\psi,$$ where $\bar{\psi}=\psi^\dagger\gamma^0$. The minkowski action is invariant under the chiral transformation $\delta\psi=i\alpha \gamma^5\psi $ and $\delta\bar{\psi}=i\bar{\psi}\alpha \gamma^5$. Here the transformation of $\bar{\psi}$ is implied by that of $\psi$ since $\gamma_0$ and $\gamma_5$ anticommutes. We now go to Euclidean coordinates ($x^\mu_E$) defined by $x^0_E=x^0, x^k_E=ix^k$ and accordingly the gamma matrices $\gamma^0_E=\gamma^0, \gamma^k_E=i\gamma^k$. In the Euclidean space, the Lorenz symmetry becomes $SO(4)$ symmetry. The $SO(4)$ invariant action should be of the form $$S_E=i\int d^4x\psi^\dagger i\gamma^\mu_E D^E_\mu\psi.$$ Notice that the $SO(4)$ symmetry forces us to use $\psi^\dagger$ instead of $\bar{\psi}$ in the action. It is easy to see that the action $S_E$ does not have the chiral symmetry $\delta\psi=i\gamma^5 \alpha \psi$ and $\delta \psi^\dagger=-i\gamma^5\alpha$, because of the absence of $\gamma^0$. Where am I going wrong?

Answer 1

I've struggled with some similar issues recently. The problem is that usually the Euclidean continuation of fermions is not considered with great care. Anyway, the whole point that you should keep in mind is that in the Path Integral $\psi$ and $\psi^\dagger$ are independent integration variables (if you recall the steps to properly define the path integral for fermions, going through an isomorphism with a space of Grassmann variables, this is quite clear). This means that $\psi^\dagger$ is really just a name, but it is not the hermitian of $\psi$. Keeping this in mind, everything else makes more sense.

In Euclidean space you are forced to give up the relation between $\psi$ and $\psi^\dagger$. To avoid confusion you can also change names: $\psi$ and $\chi^\dagger$ (instead of $\psi^\dagger$). The action in the Euclidean is not hermitian, but this is not a problem, since we require unitary just for a theory with real time. If you want to read something about Euclidean continuation of fermions a simple article is arXiv:hep-th/9608174 (On Euclidean spinors and Wick rotations). The first ones to properly define it, from a canonical point of view, are Osterwalder and Schrader in Phys. Rev. Lett. 29 (1972). They show that, even from a canonical point of view, it is necessary to "double" the fermionic degrees of freedom to go to Euclidean (with double I mean to consider $\psi$ and $\psi^\dagger$ independent from each other and it is in quotation because it is not really a doubling, since, as said at the beginning, in the path integral they are just independent integration variables!).

To give an answer to your question: since $\psi$ and $\psi^\dagger$ are independent from each other, it follows that you cannot derive the transformation law of $\psi^\dagger = \chi^\dagger$ from that of $\psi$. For the chiral symmetry case, we just define $\delta \psi = i \alpha \gamma^5 \psi$ and $\delta \psi^\dagger = \psi^\dagger i \alpha \gamma^5$. You see that the Euclidean action is invariant under this transformation (notice that it is a bit weird your rotation of the 3 spatial coordinates, it is more common to rotate just $x^0$).

** I add two remarks that I think could be useful:

1. The Euclidean action is still $SO(d+1)$ invariant. $\psi$ and $\psi^\dagger$ are independent from each other, but $\psi^\dagger$ still transforms as a hermitian spinor under a $SO(d+1)$ rotation (this is why I choose to use $\psi^\dagger =\chi^\dagger$). This ensures the correct invariance of the action.
2. You can see that the Euclidean action is invariant also under the axial transformation \begin{equation} \psi \rightarrow e^{a\gamma^5}\psi, \qquad \psi^\dagger \rightarrow \psi^\dagger e^{a\gamma^5}, \end{equation} with $a \in \mathbb{R}$. In this case $\psi$ and $\psi^\dagger$ transform effectively as one the hermitian of the other and one can think that is more natural, but this is really just a choice for what I said: $\psi$ and $\psi^\dagger$ are not the hermitian of each other, that's it. Also this chiral transformation is not compact, but it is more like a scale transformation, and (crucially perhaps) it is not a unitary transformation.