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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,847 answers , 20,601 comments
1,470 users with positive rep
501 active unimported users
More ...

Why is the mass dimension of anticommuting coordinates $[Mass]^{-1/2}$

+ 3 like - 0 dislike
141 views

I am reading a review ( http://arxiv.org/abs/hep-ph/9709356 ) about supersymmetry. On page 29 I have read that the mass dimension of the Grassmann anticommuting coordinates is $-1/2$. Why this? Why don't they have the same mass dimensions as the bosonic coordinates?

asked May 13, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]
edited May 13, 2015 by Arnold Neumaier

You can work out the dimension of any field from the fact that the free action must be dimensionless (in units where $\hbar=1$). 

1 Answer

+ 5 like - 0 dislike

By definition supersymmetry transformations square to spacetime translations. In a superspace formalism the supersymmetry operator is constructed from the vector field $\partial_\theta$ with respect to the odd coordinates $\theta$. As this operator has to square to the vector field $\partial_x$ with respect to the even coordinates $x$, which is of dimension $1$, the vector field with respect to the odd coordinate has to be of dimension $1/2$ and so the odd coordinate as to be of dimension $-1/2$.

Equivalently, a typical superfield is of the form

$\phi + \theta \psi +...$

where $\phi$ is a scalar and $\psi$ a spinor. In $d$ spacetime dimensions, a scalar is of dimension $(d-2)/2$, a spinor is of dimension $(d-1)/2$ and so $\theta$ has to be of dimension $-1/2$.

answered May 14, 2015 by 40227 (4,660 points) [ revision history ]
edited May 15, 2015 by 40227

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$\hbar$ysicsO$\varnothing$erflow
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
...