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
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Supersymmetry invariants

+ 2 like - 0 dislike
411 views

On page 158 of Fields, the supersymmetry algebra is represented in terms of the action on supercoordinates as

$$\delta \theta^\alpha = \epsilon^\alpha$$ $$\delta\bar{\theta}^{\dot{\alpha}} = \bar{\epsilon}^{\dot{\alpha}}$$ $$\delta x^{\alpha\dot{\beta}} = \frac{1}{2}i(\epsilon^\alpha \bar{\theta}^{\dot{\beta}} + \bar{\epsilon}^{\dot{\beta}}\theta^\alpha)$$

Further down on the page, the following statement is made:

In the supersymmetric case the infinitesimal invariants under the q's (and therefore p) are

$$d\theta^\alpha, \qquad d\bar{\theta}^{\dot{\alpha}}, \qquad dx^{\alpha\dot{\beta}} + \frac{1}{2}i(d\theta^\alpha)\bar{\theta}^{\dot{\beta}} + \frac{1}{2}i(d\bar{\theta}^{\dot{\beta}})\theta^\alpha$$

I get a sense that these should follow from the definition of the supersymmetry transformations, but what exactly does the quoted statement mean?

Thinking of the SUSY variation $\delta_\epsilon$ as something similar to a BRST variation, is this statement about nilpotency of the algebra in form notation or something?


This post imported from StackExchange Physics at 2015-05-22 21:05 (UTC), posted by SE-user leastaction

asked May 20, 2015 in Theoretical Physics by leastaction (425 points) [ revision history ]
edited Sep 29, 2015 by leastaction

It looks like $d\theta$ is simply the Berezin measure for Grassmann integration. 

Yes, that may as well be the case. I can't seem to edit my original post right now, but there's a distinction between $d\theta$ and $\delta \theta$ of course which I seem to have overlooked in my post. I suppose the claim the author is making is that in superspace, the stated quantities are in fact invariants under the supersymmetry transformations.

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$\hbar$ys$\varnothing$csOverflow
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
...