# Elementary question about global supersymmetry of a worldsheet [closed]

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.

## closed as off-topic by Kyle Kanos, Ryan Unger, yuggib, ACuriousMind♦, Kyle OmanMay 29 '15 at 16:01

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• Welcome to Physics! Please see this Meta post for "check my work" problems, as they are generally considered off-topic here. We prefer our questions to be about physics concepts, rather than 'how did I do this wrong' (and similar questions). – Kyle Kanos May 28 '15 at 17:42
• @Kyle, okay... Really, my question is not very conceptual and too specific. But I hope there's a chance for it not to be a completely irrelvant, as, probably, I miss something important. Something not reducing to sign error or something. – vanger May 28 '15 at 18:02

## 1 Answer

Hints:

1. The Majorana spinor is real. For instance $\bar{\psi}=\psi^T\rho^0$ without complex conjugation.

2. The SUSY transformation $\delta{\cal L}$ of the Lagrangian density ${\cal L}$ does not have to vanish. It is enough if it is a total divergence. See the notion of quasi-symmetry, cf. e.g. this and this Phys.SE posts.

• Thank you! 1. I used $\bar \chi \psi = \bar \psi \chi$, which is true for real spinors only even for complex ones. That game me the sign error in the first term in the variation of the fermionic part. 2. The book confused me with "the action is invariant under transformations". I was sure it meant to be "cmpletely invariant", not "modulo boundary terms". – vanger May 29 '15 at 0:43