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

+ 3 like - 0 dislike

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 12, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 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

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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 (5,140 points) [ revision history ]
edited May 15, 2015 by 40227

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