• 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

201 submissions , 159 unreviewed
4,961 questions , 2,127 unanswered
5,328 answers , 22,576 comments
1,470 users with positive rep
809 active unimported users
More ...

  Strange Grassmann double integration

+ 5 like - 0 dislike

I can unterstand why because the integration over Grassman variables has to be translational invariant too, one has

$$ \int d\theta = 0 $$


$$ \int d\theta \theta = 1 $$

but I dont see where the rule for this double integration

$$ \int d^2 \theta \bar{\theta}\theta = -2i $$

comes from.

So can somebody explain to me how this is motivated and/or derived?

asked Apr 14, 2013 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
Is the last integral supposed to read something like $\int d^2\!\theta \, \bar{\theta}\theta$?

This post imported from StackExchange Physics at 2014-03-09 16:25 (UCT), posted by SE-user Olof
@Olof yes, I just corrected the typo thanks.

This post imported from StackExchange Physics at 2014-03-09 16:25 (UCT), posted by SE-user Dilaton

1 Answer

+ 6 like - 0 dislike

As with anything that has to do with supersymmetry the details will be dependent on your exact conventions, but we can obtain the result as follows:

Assume we have two Grassman variables $\theta_1$ and $\theta_2$. By applying your first formula twice we find $$\int d\theta_1 d\theta_2 \, \theta_2 \theta_1 = 1$$ Now combine these into $$\theta = \theta_1 + i\theta_2 \qquad \text{and} \qquad \bar{\theta}=\theta_1-i\theta_2.$$ We then have $$\bar{\theta} \theta = - 2i\theta_2\theta_1$$ and hence $$\int d\theta_1 d\theta_2 \bar{\theta} \theta = - 2i$$ which is exactly your second integral, if we identify the measure $$d^2\theta = d\theta_1 d\theta_2.$$

This post imported from StackExchange Physics at 2014-03-09 16:25 (UCT), posted by SE-user Olof
answered Apr 14, 2013 by Olof (210 points) [ no revision ]
Ah thanks Olof, that is exactly what I needed.

This post imported from StackExchange Physics at 2014-03-09 16:25 (UCT), posted by SE-user Dilaton

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