• 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 , 22,734 comments
1,470 users with positive rep
818 active unimported users
More ...

  Annoying sign error in gamma matrix calculation

+ 2 like - 0 dislike

I am trying to show that

\[\bar{\psi}\gamma^{\mu}\psi =\bar{\psi}_L\gamma^{\mu}\psi_L + \bar{\psi}_R\gamma^{\mu}\psi_R\]

For $\mu = 0$, this is using $\gamma^0 \gamma^0 = 1$ rather straightforward.

However, for the case $\mu \neq 0$ I obtain an annoying sign error (using the Weyl representation for the gamma matrices:

\[\bar{\psi}\gamma^1\psi = \left( \begin{array}{cc} \psi_L^*,\psi_R^*\\ \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 0 & -\sigma_1 \\ \sigma_1 & 0 \\ \end{array} \right) \left( \begin{array}{c} \psi_L \\ \psi_R \\ \end{array} \right)\]

which when multiplying out stuff gives me the wrong result

\[\bar{\psi}\gamma^{1}\psi =\bar{\psi}_L\gamma^{1}\psi_L - \bar{\psi}_R\gamma^{1}\psi_R\]

For the other two components, the same would happen. How can I get rid of this annoying minus sign in front of the second term? Or is my way to show this identity not a good one in the first place?

asked Nov 24, 2015 in Theoretical Physics by anonymous [ revision history ]
edited Nov 29, 2015 by Arnold Neumaier

Just a note: I fixed your equations to what I think was your intention, considering that your "wrong result" evaluated to zero, and the "right one" was \(2\bar\psi\gamma^\mu\psi\), which probably wasn't what you intended. Feel free to rollback.

1 Answer

+ 4 like - 0 dislike

Start noticing that the identity operator $\textbf{1}$ can be written as $\textbf{1} = P_L + P_R$, where $\psi_{R,L} = P_{R,L}\psi$, respectively. As such we have:
$$\bar{\psi} = \bar{\psi}(P_L+P_R)$$ (and likewise for $\psi$). From there one can derive

$$\bar{\psi}\gamma^{\mu}\psi = \bar{\psi}(P_L+P_R)\gamma^{\mu}(P_L+P_R)\psi$$ the trick being to make the projection operators come into play. Expanding the above equation one gets: $$\bar{\psi}P_R \gamma^{\mu}P_L\psi + \bar{\psi}P_L \gamma^{\mu}P_R\psi + \bar{\psi}P_L\gamma^{\mu}P_L\psi + \bar{\psi}P_R\gamma^{\mu}P_R\psi. $$

Notice now that $\bar{\psi}P_R = \bar{\psi}_L$ (and viceversa), so that the previous equation becomes $$\bar{\psi}_L\gamma^{\mu}\psi_L + \bar{\psi}_R\gamma^{\mu}\psi_R + \bar{\psi}\left(P_L\gamma^{\mu}P_L + P_R\gamma^{\mu}P_R\right)\psi$$

the last contribution being zero due to the anti-commutation relations $\left\{\gamma^5,\gamma^{\mu}\right\}=0$.

answered Nov 28, 2015 by GennaroTedesco (80 points) [ revision history ]
edited Nov 30, 2015 by GennaroTedesco

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