So, as in the title, I am looking for a proof that the superstring theory action (in Minkowski spacetime with conformal gauge): $$S=\int d^2\sigma \eta_{AB} (\eta^{ab} \partial_aX^A\partial_bX^B-i\overline \Psi^A\rho^a\partial_a\Psi^B)$$ is invariant under the transformation $$\begin{equation*} \begin{split} \delta X^A &= \bar\epsilon_{\dot \alpha}\Psi_\beta ^A\sigma^{\dot\alpha\beta}\\ \delta \Psi_\alpha^A &= -i\epsilon_\alpha \rho^a\partial_a X^A \end{split} \end{equation*}$$

My attempt:

Taking the differential of $S$, we have $$\delta S \sim 2\eta^{ab}\partial_a\delta X^A\partial_b X^B -i\delta\overline \Psi^A\rho^a\partial_a\Psi^B-i\overline \Psi^A\rho^a\partial_a\delta\Psi^B$$

But if I substitute the transformations, it does not give zero. Do I have to use the Euler-Lagrange equations in fact?

After actually substituting I found $$\delta S \sim 2\eta^{ab}\partial_a \bar\epsilon_{\dot \alpha}\Psi_\beta ^A\sigma^{\dot\alpha\beta}\partial_b X^B -\bar\epsilon_{\dot\alpha} \rho^b\partial_b X^A\rho^a\partial_a\Psi^{B\dot\alpha}+\overline \Psi^{A} _{\dot\alpha} \rho^b\partial_b\epsilon^{\dot\alpha} \rho^a\partial_a X^B$$ $$= 2\eta^{ab}\partial_a \bar\epsilon_{\dot \alpha}\Psi_\beta^A\sigma^{\dot\alpha\beta}\partial_b X^B -\eta^{ab}\bar\epsilon_{\dot\alpha} \partial_b X^A\partial_a\Psi^{B}_\beta\sigma^{\dot\alpha\beta} +\eta^{ab}\overline \Psi^{A} _{\dot\alpha}\partial_b\epsilon_\beta \partial_a X^B\sigma^{\dot\alpha\beta}$$

where I have used $\rho^a\rho^b \partial_a\partial_b= \frac 12\{\rho^a,\rho^b\}\partial_a\partial_b = \eta^{ab}\partial_a\partial_b$. Integration by parts will change the last plus to a minus. I am left with the problem of showing $$\overline\epsilon_{\dot\alpha}\Psi_\beta\sigma^{\dot\alpha\beta} = +\overline\Psi_{\dot\alpha}\epsilon_\beta\sigma^{\dot\alpha\beta}$$ and not the negative $$\overline\epsilon_{\dot\alpha}\Psi_\beta\sigma^{\dot\alpha\beta} = -\overline\Psi_{\dot\alpha}\epsilon_\beta\sigma^{\dot\alpha\beta}$$

  • $\begingroup$ Probably this: $(\bar\epsilon_{\dot\alpha} \Psi_\beta\sigma^{\dot\alpha\beta})^\dagger = \bar\Psi_{\dot\beta} \epsilon_\alpha\sigma^{\dot\beta\alpha} $ $\endgroup$
    – wilsonw
    Mar 11, 2020 at 12:51

1 Answer 1


In order for you to understand why in the final relation you need for your computation, the sign is a plus sign and not a minus, I suggest that you have a look at Table $3.2$, on page 59 in the Supergravity book by Freedman and Van Proeyen.

The aforementioned table is listing the spinors according to dimensions and the antisymmetric spinor bilinears. Looking at the last column of the table you see that the zeroth-order spinor bilinears (the ones that are of the form $\bar{\epsilon}_{1} \epsilon_2$ that you are interested in) are only in 7,8, and 9 dimensions. This means that in the two-dimensional world-sheet a spinor bilinear with a zero-rank $\gamma$-structure is symmetric.

The above argument is along the general lines of Fierzing relations, and you can find this explicitly and very neatly written in the book I mentioned (if you want more details, you want to have a look at Chapter 3 in general).

A final comment: I was studying the Becker, Becker, Schwarz book and I found it easier to show that $\delta S = 0$ using the light-cone coordinates. I have the computations explicitly somewhere. If you are interested I can add this here as well.


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