I'm reading chapter 4 of the book by Green, Schwarz and Witten. They consider an action $$ S = -\frac{1}{2\pi} \int d^2 \sigma \left( \partial_\alpha X^\mu \partial^\alpha X_\mu - i \bar \psi^\mu \rho^\alpha \partial_\alpha \psi_\mu \right), \tag{4.1.2}, $$ where $\psi^\mu$ are Majorana spinors, $$ \rho^0 = \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}, \qquad \rho^1 = \begin{pmatrix} 0 & i\\ i & 0 \end{pmatrix},\tag{4.1.3} $$ $$ \bar \psi = \psi^\dagger \rho^0. $$
It is claimed that this action is invariant under the following infinitesimal transformations \begin{align} \delta X^\mu &= \bar \varepsilon \psi^\mu,\\ \delta \psi^\mu &= -i \rho^\alpha \partial_\alpha X^\mu \varepsilon, \tag{4.1.8} \end{align} where $\varepsilon$ is a constant (doesn't depending on worldsheet coordinates) anticommuting Majorana spinor.
I can't prove it. Can you show me where I'm wrong? $$ \delta \left( \partial_\alpha X^\mu \partial^\alpha X_\mu \right) = 2 \partial_\alpha X^\mu \partial^\alpha \bar \psi^\mu \varepsilon $$ (I used $\bar \chi \psi = \bar \psi \chi$ identity).
\begin{multline} \delta \left( -i \bar \psi^\mu \rho^\alpha \partial_\alpha \psi_\mu \right) = -i \overline{\left(-i \rho^\alpha \partial_\alpha X^\mu \varepsilon\right)} \rho^\beta \partial_\beta \psi_\mu -i \bar \psi^\mu \rho^\alpha \partial_\alpha \left( -i \rho^\beta \partial_\beta X_\mu \varepsilon \right)\\ = - \overline{\rho^\beta \partial_\beta \psi_\mu} \rho^\alpha \partial_\alpha X^\mu \varepsilon - \bar \psi^\mu \rho^\alpha \partial_\alpha \rho^\beta \partial_\beta X_\mu \varepsilon. \end{multline}
Note that \begin{multline} \overline{\rho^\beta \partial_\beta \psi_\mu} = \partial_\beta \psi_\mu^\dagger (\rho^\beta)^\dagger \rho^0 \equiv \partial_0 \psi_\mu^\dagger (\rho^0)^\dagger \rho^0 + \partial_1 \psi_\mu^\dagger (\rho^1)^\dagger \rho^0\\ = \partial_0 \psi_\mu^\dagger \rho^0 \rho^0 - \partial_1 \psi_\mu^\dagger \rho^1 \rho^0 = \partial_0 \psi_\mu^\dagger \rho^0 \rho^0 + \partial_1 \psi_\mu^\dagger \rho^0 \rho^1 \equiv \partial_\beta \bar \psi_\mu \rho^\beta. \end{multline}
So \begin{multline} \delta \left( -i \bar \psi^\mu \rho^\alpha \partial_\alpha \psi_\mu \right) = - \partial_\beta \bar \psi_\mu \rho^\beta \rho^\alpha \partial_\alpha X^\mu \varepsilon - \bar \psi^\mu \rho^\alpha \partial_\alpha \rho^\beta \partial_\beta X_\mu \varepsilon\\ \equiv - \partial_\alpha \bar \psi_\mu \rho^\alpha \rho^\beta \partial_\beta X^\mu \varepsilon - \bar \psi^\mu \rho^\alpha \partial_\alpha \rho^\beta \partial_\beta X_\mu \varepsilon. \end{multline}
How the variation can vanish? I don't see any chance. I'll remind that the symmetry is global, so we even can't integrate by parts.
