Quantcast
  • 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.

News

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

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  About the central charge of 4D extended supersymmetry algebra

+ 2 like - 0 dislike
1519 views

The 4D SUSY algebra can be written as

$$\{ Q_{\alpha}^{A} , Q_{\beta}^{B \dagger} \} = 2 m \delta^{AB} \delta_{\alpha \beta} + 2 i Z^{AB} \Gamma^0_{\alpha \beta}, \tag{B.2.37} $$

in a particular reference frame. One can find this formula in the Appendix B, page 448 of Polchinski's String Theory vol.II.

I am confused with the $'i'$ before the central charge. If we do a Hermitian conjugate on both side:

$$\{ Q_{\alpha}^{A \dagger} , Q_{\beta}^{B} \} = 2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{AB} (\Gamma^0_{\alpha \beta})^* $$

and then exchange $(A,\alpha)$ with $(B,\beta)$, the LHS is invariant. But the RHS is

$$2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{BA} (\Gamma^0_{ \beta \alpha})^* = 2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{BA} (\Gamma^0)^{\dagger}_{ \alpha \beta}.$$

Since $Z_{AB}$ is anti-symmetric and $(\Gamma^0)^{\dagger} = -\Gamma^0$, It seems that we have the wrong sign before the central charge term:

$$2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{AB} (\Gamma^0)_{ \alpha \beta}.$$

I think I made a mistake but I can not figure out where is it.


This post imported from StackExchange Physics at 2018-04-18 20:42 (UTC), posted by SE-user JQ Skywalker

asked Apr 10, 2018 in Theoretical Physics by JQ Skywalker (10 points) [ revision history ]
edited Apr 18, 2018 by Dilaton
Is $Z_{AB}^\dagger=+Z_{AB}$ or $Z_{AB}^\dagger=-Z_{AB}$?

This post imported from StackExchange Physics at 2018-04-18 20:42 (UTC), posted by SE-user AccidentalFourierTransform
$Z_{AB}$ is real and anti-symmetric, therefore $Z_{AB}^{\dagger} = - Z_{AB}$. But I don't think the dagger on the both side will involve the indices $A$ and $\alpha$.

This post imported from StackExchange Physics at 2018-04-18 20:42 (UTC), posted by SE-user JQ Skywalker

1 Answer

+ 2 like - 0 dislike

It is important to remember that operator order gets reversed under Hermitian conjugation: $$(ST)^{\dagger}~=~T^{\dagger}S^{\dagger}.$$ Therefore a Hermitian conjugation on the LHS of eq. (B.2.37) effectively exchanges indices $(A,\alpha)\leftrightarrow (B,\beta)$. The same should happen on the RHS. This is implemented by choosing the central charges $Z_{AB}$ to be anti-Hermitian and the gamma matrix $\Gamma^0_{\alpha\beta}$ to be Hermitian.

This post imported from StackExchange Physics at 2018-04-18 20:42 (UTC), posted by SE-user Qmechanic
answered Apr 10, 2018 by Qmechanic (3,120 points) [ no revision ]
I know that $Z_{AB}$ is anti-hermitian but I think the $\Gamma^0$ is also chosen to be anti-hermitian throughout his book, you can check that in (B.1.7a) where $\Gamma^0 = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] \otimes \left[ \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right] $ in 4D.

This post imported from StackExchange Physics at 2018-04-18 20:42 (UTC), posted by SE-user JQ Skywalker

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:
p$\hbar$ysicsOve$\varnothing$flow
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
...