# Elementary question about global supersymmetry of a worldsheet

+ 1 like - 0 dislike
84 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
retagged May 30, 2015

+ 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 (2,860 points)
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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.