Supercharge of supersymmetry transformations of the super-world-sheet coordinates

Question

In the book "String theory and M-theory", the authors mentions on page 114 that "the generators of supersymmetry transformations of the super-world-sheet coordinates, called supercharges, are $$Q_A =\frac{\partial}{\partial \bar{\theta}^A}-(\rho^{\alpha}\theta)_A\partial_{\alpha}.$$ How does the author get it? The action is $$S=-\frac{1}{2\pi}\int d^2\sigma (\partial_{\alpha}X_{\mu}\partial^{\alpha}X^{\mu}+\bar{\psi}^{\mu}\rho^{\alpha}\partial_{\alpha}\psi_{\mu})$$ and the superfield is $$Y^{\mu}(\sigma^{\alpha},\theta)=X^{\mu}(\sigma^{\alpha})+\bar{\theta}\psi^{\mu}(\sigma^{\alpha})+\frac{1}{2}\bar{\theta}\theta B^{\mu}(\sigma^{\alpha})$$ where the super-world-sheet coordinates are given by $(\sigma^{\alpha},\theta_A)$. Shoud we tranform the action by replacing $X^{\mu}$ to $Y^{\mu}$ and calculate the supercurrent and then the supercharge? or is there another way?

My try: Operating on Y with $Q$, we have $$Q_A Y^{\mu}(\sigma^{\alpha},\theta)=\psi^{\mu}(\sigma^{\alpha})+\theta B^{\mu}(\sigma^{\alpha})-(\rho^{\alpha}\theta)\partial_{\alpha}X^{\mu}(\sigma^{\alpha})-(\rho^{\alpha}\theta)\bar{\theta}\partial_{\alpha}\psi^{\mu}(\sigma^{\alpha}).$$

Answer 1

Hint: up to suitable Grassmann signs, etc, & Fierzing, you have, from the above, something like $$\delta Y^\mu= \delta X^\mu +\bar{\theta}\delta \psi^{\mu}+\frac{1}{2}\bar{\theta}\theta ~\delta B^{\mu} ,$$ so that $$ \delta X^\mu = \bar \epsilon \psi^\mu ,\\ \delta \psi^\mu = \partial \!\!/ X^\mu \epsilon+ B^\mu \epsilon ,\\ \delta B^\mu = \bar \epsilon \partial \!\!/ \psi^\mu , $$ but, on shell ($B^\mu=0$), the middle increment simplifies to a standard susy transformation. Inserting these increments into Noether's construction on S, you find the supercurrent/supercharge you'd expect.

This is the superspace analog of a 2d rotation of $$ \vec V= v_x \hat x + v_y \hat y \\ \vec V \cdot \vec V= v_x^2+ v_y^2 $$ by an infinitesimal angle ε, $$ \delta \vec V= v_x ( ε \hat y) + v_y (-ε \hat x) = (-ε v_y)\hat x + ( ε v_x)\hat y \qquad \leadsto \\ \delta v_x= -ε v_y , \qquad \delta v_y= ε v_x . $$