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.

This post imported from StackExchange Physics at 2015-05-30 00:22 (UTC), posted by SE-user vanger