• 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.


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

202 submissions , 160 unreviewed
4,973 questions , 2,138 unanswered
5,333 answers , 22,595 comments
1,470 users with positive rep
812 active unimported users
More ...

  Hamiltonian in SUSY (SUSY algebra)

+ 2 like - 0 dislike

I was reading the book Supersymmetry, Theory, Experiment and Cosmology by P. Binétruy, and on page 25 the author goes from

$$ 1)[Q_r,Q_t]_+ \gamma^{0}_{ts}=2\gamma^{\mu}_{rs}P_{\mu} $$ $$ 2)\mbox{Contracting with $ \gamma^{0}_{sr}$, } \sum_{r,t}[Q_r,Q_r]_{+}(\gamma^{02})_{tr} = 2Tr(\gamma^{0}\gamma^{\mu})P_{\mu} $$ $$ 3) \sum_{r}Q^{2}_{r}=4P^{0}$$ $$ 4) H=\frac{1}{4}\sum_{r}Q^{2}_{r} $$

I don't really understand how to go from stage 1 to 2 or 3 to 4, can anyone help?

This post imported from StackExchange Physics at 2014-04-15 16:43 (UCT), posted by SE-user user21119
asked Mar 1, 2013 in Theoretical Physics by user21119 (10 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

The step from 1 to 2 is: multiply by $\gamma^0$ and trace over the result.

2->3: $Tr((\gamma^0)^2)=Tr(1)=1$ on the left hand side and $2Tr(\gamma^0\gamma^\mu)P_\mu=8\eta^{0\mu}P_\mu=8P_0=8P^0$ on the right hand side. Moreover the anticommutator gives $2Q_r^2$ so you can just strip off a factor of 2 on both sides.

3->4: $P^0$ is the total energy by definition, as the right hand side is in fact mulitplied by an unit operator (as $Q$ are operators) one has $P^0 1 = H$ giving the final result.

For some magic involving the gamma matrices see the wikipedia article http://en.wikipedia.org/wiki/Gamma_matrices

This post imported from StackExchange Physics at 2014-04-15 16:44 (UCT), posted by SE-user A friendly helper
answered Mar 1, 2013 by A friendly helper (320 points) [ no revision ]
Am i correct in thinking that $[Q_r,Q_t]_+ = 0$ if r doesn't equal t?

This post imported from StackExchange Physics at 2014-04-15 16:44 (UCT), posted by SE-user user21119

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).
To avoid this verification in future, please log in or register.

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

Your rights