• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,743 comments
1,470 users with positive rep
818 active unimported users
More ...

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

+ 1 like - 0 dislike

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
& \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}.
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 
\bar{\psi} \gamma^{\mu} A_{\mu} \psi
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 
\bar{\psi} \gamma^{\mu} \psi
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 
\bar{\psi} \gamma^{0} \psi \to -i\bar{\psi} \gamma^{4} \psi
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 
\bar{\psi} \gamma^{0} \psi \to -i\bar{\psi} \gamma^{4} \psi
which in literature they will just simply write 
\gamma^{0} = -i \gamma^{4}.

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!

asked Mar 2, 2021 in Theoretical Physics by ocf001497 (15 points) [ no revision ]

1 Answer

+ 0 like - 0 dislike

Yes, you are right. In fact, it is just a definition of one matrix via another. Your arrows "$\to$" are also such definitions :-)

answered Mar 3, 2021 by Vladimir Kalitvianski (102 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights