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

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!

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.