Wick rotation from Minkowski Dirac theory to Euclidean Dirac theory: $\gamma^{0} = -i\gamma^{4}$

Question

I am reading Path Integrals and Quantum Anomalies by Kazuo Fujikawa and Hiroshi Suzuki. In chapter 4.2 they calculate the self-energy of photon for QED and say that the actual calculation is performed in Euclidean theory. The recipe they gave to change from Minkowski to Euclidean theory is \begin{align} & \text{time: } x^{0} \to -ix^{4} \\ & \text{contravariant vector component: } V^{0} \to -iV^{4} \\ & \text{covariant vector component: } A_{0} \to iA_{4}\\ & \text{metric: } (+1,-1,-1,-1) \to (-1,-1,-1,-1) \\ & \text{gamma matrix: } \gamma^{0} = -i\gamma^{4}. \end{align} Note that different from the first four transformation which I use "$\to$", in the last one I use "$=$", since we really need the numerical value of $\gamma^{4}$, given the knowledge of $\gamma^{0}$.

I understand that this transformation allows us to compute the integral in Minkowski space in a convergent manner and still obtain the same answer since the mathematical meaning of changing from Minkowski to Euclidean theory is merely a change of integration axis that preserves the answer, provided that we do not hit singularities when we change the integration axis (correct me if I am wrong though).

I am totally fine with the first four transformations I list above. However, I have difficulties to understand the last one: $\gamma^{0} \to -i\gamma^{4}$. Since $\gamma^{0}$ by itself is really just a numerical matrix, it is hard to imagine why it should change in this way.

The following are some of my thoughts. In fact, it is not the $\gamma^{0}$ that change, it is the \begin{align} \bar{\psi} \gamma^{\mu} A_{\mu} \psi \end{align} that we want to preserve upon the change from Minkowski to Euclidean theory, where $A_{\mu}$ is some covariant vector component. We are assuming that in the Lagrangian, whenever $\gamma^{\mu}$ appears, it will always be contracted with some covariant vector component $A_{\mu}$, in order to make the Lagrangian a scalar. And therefore, we may roughly regard \begin{align} \bar{\psi} \gamma^{\mu} \psi \end{align} as some contravariant vector component $V^{\mu}$ (actually indeed $\bar{\psi} \gamma^{\mu} \psi$ transforms as a vector field) and then by requiring $V^{\mu}A_{\mu}$ is preserved we certainly need to change \begin{align} \bar{\psi} \gamma^{0} \psi \to -i\bar{\psi} \gamma^{4} \psi \end{align} which, together with the change of metric and $A_{0} \to iA_{4}$, will preserve the $\bar{\psi} \gamma^{\mu} A_{\mu} \psi$ before and after the transformation.

In short, we should properly say that \begin{align} \bar{\psi} \gamma^{0} \psi \to -i\bar{\psi} \gamma^{4} \psi \end{align} which in literature they will just simply write \begin{align} \gamma^{0} = -i \gamma^{4}. \end{align}

Is my understanding correct? If not is there any suggested reference? It seems like in this book the authors didn't elaborate more on this. Thanks!

Answer 1

One transformation that they didn't write down explicitly is $$\partial_0 \rightarrow i\partial_4.$$ This is just implied by the Wick rotation $x^0\rightarrow -i x^4$, of course. The reason I point this out is that $\gamma^\mu\partial_\mu$ appears in the Minkowski action so after Wick rotation we have $$i\gamma^0 \partial_4 + \gamma^i\partial_i,$$ so they simply define this Euclidean time gamma matrix so as to absorb this factor of $i$, so that it is treated on equal footing to the spatial ones.

This definition means that the Clifford algebra relation $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ is still obeyed using the new Euclidean signature metric, and thus that $\gamma^\mu$ itself transforms like a vector under $SO(4)$ (the Euclidean equivalent to the Lorentz group).

Similarly redefining $A_0 = iA_4$ cancels the new $-i$ from $\gamma^4$ as you point out, but it also would be useful even in a gauged scalar theory, since $A_{0}$ gets mixed with time derivatives through gauge transformations.