# Dimension of the parameter of SUSY transformation

I'm trying to gain some intuition on the dimension on the parameter of the supersymmetric transformation.

Assume I'm creating SUSY from scratch. I am making a wrong choice and defining my supersymmetry transformation $\Delta$ as: $$\Delta\psi = \xi \phi \quad,\qquad$$ where the anticommuting parameter $\xi$ has the dimension: $$[\xi] = [\text{length}]^{-1/2} = [\text{mass}]^{1/2} \quad.$$ Here $\psi$ and $\phi$ are the fermion and boson, correspondingly. I would also define the following transformation of the dummy field $F = m \phi$: $$\Delta F = \xi \psi\quad.$$ Then I would have: $$\Delta (m \psi \bar{\psi}) = \ldots+m\dfrac{1}{m^2}\xi^2\phi^2\quad.$$ Through the $F$-term, I would also get $$\Delta (m^2 \phi^2) =\ldots+ m^2\dfrac{1}{m^2}\xi^2\psi^2\quad.$$ Both lines are fine from dimensional point of view. Not sure how to define the other transformations in my alt-SUSY so far.

Obviously, the construction built on the first two equations in this question should quickly fail. What it the easiest way to see this?

• Your transformation cannot be a symmetry of a Lagrangian containing dynamical bosonic and fermionic fields. A boson kinetic term contains two spacetime derivatives, while a fermion kinetic term contains one, and so without derivatives in your transformation law the changes in these two terms under the transformation cannot cancel. (I may have misunderstood your transformation however, as I am confused as to why you have not specified the transformation of $\phi$ explicitly.) – diracula Oct 14 '17 at 22:03