# Problema with Grassmann Derivative: two solutions for the same function

I'm confused as to why mathematica is not giving me the correct answer. I'm using the grassmannOps package and defining $\theta$, $\bar\theta$ as grassmannians and $X_{\alpha\,\dot\alpha}$ as a bosonic term.

I define

${\mathcal D}[x,\{\alpha,k\}]=\frac{\partial x}{\partial\theta^{\alpha\,k}}+4{\rm i} \epsilon_{\beta\,\dot\beta}\epsilon_{\alpha,\beta}\bar\theta^{\dot\mu}_k **\frac{\partial x}{\partial X_{\beta\,\dot\beta}}$

(the $\alpha$ index is a SU(2) Lorentz, and the $i$ index is a $SU(2)_R$, I work with 2 supersymetries in 4 dimensions)

Where is the problem?, I define

$\mathcal D\mathcal D[\{x,y\},\{i,j\}]=\epsilon^{\alpha\,\beta}\mathcal D[x,\{\alpha,i\}]**\mathcal D[y,\{\beta,j\}]$

where the partial derivative is the GD function that comes with gasssmannops.

If I take

$\mathcal D\mathcal D[\{X^2,X^2\},\{i,j\}]$

I obtain something different than

$\epsilon^{\alpha\,\beta}\mathcal D[X^2,\{\alpha,i\}]**\mathcal D[X^2,\{\beta,j\}]$

but they are the same function!!!

Anyone knows why mathematica does not compute this properly?

• Pretty much impossible to guess at this without the actual Mathematica code. – Daniel Lichtblau Dec 21 '15 at 19:30
• Sorry for not showing the code. I haven't realized that grassmannOps.m is not longer on the web... Anyway, I found the problem: when defining $\mathcal D\mathcal D[\{x,y\},\{i,j\}]=\epsilon^{\alpha\,\beta}\mathcal D[x,\{\alpha,i\}]**\mathcal D[y,\{\beta,j\}]$ GD do something funny with the term inside the bracket {x,y}. To solve the issue I have to define my derivative square as $\mathcal D\mathcal D[x,y,\{i,j\}]=\epsilon^{\alpha\,\beta}\mathcal D[x,\{\alpha,i\}]**\mathcal D[y,\{\beta,j\}]$ And that's it. – CGH Dec 21 '15 at 20:09