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Elementary question about global supersymmetry of a worldsheet

+ 1 like - 0 dislike
53 views

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
asked May 28, 2015 in Theoretical Physics by vanger (5 points) [ no revision ]
retagged May 30, 2015

1 Answer

+ 0 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-05-30 00:22 (UTC), posted by SE-user Qmechanic
answered May 28, 2015 by Qmechanic (2,790 points) [ no revision ]
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".

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

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