# Identity Involving Grassmann Variables and Pauli Matrices

I am trying to prove the following identity: $$\theta\sigma^{\mu}\bar{\theta}\theta\sigma^{\nu}\bar{\theta}=\frac{1}{2}g^{\mu\nu}\theta\theta\bar{\theta}\bar{\theta}$$ Where $$\theta$$ and $$\bar{\theta}$$ are anticommuting grassmann variables, and $$g^{\mu\nu}$$ is the metric.

I have written the above in component notation;

$$\theta^{\alpha}\sigma_{\alpha\dot{\alpha}}^{\mu}\bar{\theta^{\dot{\alpha}}}\theta^{\beta}\sigma_{\beta\dot{\beta}}^{\nu}\bar{\theta^{\dot{\beta}}}$$

Also, given the identity: $$\sigma^{\mu}_{\alpha\dot{\alpha}}\sigma_{\mu\beta\dot{\beta}}=2\epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}$$, I have managed to show that; $$\sigma^{\mu}_{\alpha\dot{\alpha}}\sigma_{\beta\dot{\beta}}^{\nu} = \frac{1}{2}\epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}g^{\mu\nu}$$ Which seems like a step in the right direction.

Also, I can't seem to figure out if the barred and unbarred grassman variables commute, i.e. do we have that $$\theta^{\alpha}\bar{\theta^{\dot{\beta}}}=-\bar{\theta^{\dot{\beta}}}\theta^{\alpha}$$?

I have been stuck for quite a while with no useful progress, but will update if I make any. A push in the right direction would be much appreciated.

The anti commuting grassman variables satisfy

$$\{ \theta^\alpha , \bar{\theta}_{\dot{\gamma}}\} = 0.$$

If your concern is about the position of the indices just multiply both sides by $$\epsilon^{\dot{\beta}\dot{\gamma}}$$ and you obtain

$$\{ \theta^\alpha , \bar{\theta}^{\dot{\beta}}\} = 0.$$

Now, as for your main question I'll go ahead and try the proof.

First note that you have a minus sign missing in your stated identity. The correct identity is

$$\sigma^{\mu}_{\alpha\dot{\alpha}}\sigma_{\beta\dot{\beta}}^{\nu} = -\frac{1}{2}\epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}g^{\mu\nu}$$

as stated here in the very colorful fun with spinor indices

Proof: We wish to show that

$$\theta\sigma^{\mu}\bar{\theta}\theta\sigma^{\nu}\bar{\theta}=\frac{1}{2}g^{\mu\nu}\theta\theta\bar{\theta}\bar{\theta}.$$

Writing the LHS in index notation we have that

\begin{align*} &\theta\sigma^{\mu}\bar{\theta}\theta\sigma^{\nu}\bar{\theta} = \theta_\alpha\sigma^{\mu}_{\alpha\dot{\alpha}}\bar{\theta}_{\dot{\alpha}} \theta_\beta\sigma^{\mu}_{\beta\dot{\beta}}\bar{\theta}_{\dot{\beta}}\\ &= \theta_\alpha\bar{\theta}_{\dot{\alpha}} \theta_\beta\bar{\theta}_{\dot{\beta}} \sigma^{\mu}_{\alpha\dot{\alpha}}\sigma^{\mu}_{\beta\dot{\beta}}\\ &= - \frac{1}{2} \theta_\alpha\bar{\theta}_{\dot{\alpha}} \theta_\beta\bar{\theta}_{\dot{\beta}} \epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}g^{\mu\nu} \tag{a}\\ &= \frac{1}{2} (\epsilon_{\alpha\beta}\theta_\alpha\theta_\beta)(\epsilon_{\dot{\alpha}\dot{\beta}}\bar{\theta}_{\dot{\alpha}} \bar{\theta}_{\dot{\beta}}) g^{\mu\nu} \tag{b}\\ &= \frac{1}{2}g^{\mu\nu}\theta\theta\bar{\theta}\bar{\theta} \end{align*}

where to get from (a) to be we have anti-commuted through the innermost grassman variables.

• Thanks, that really helps! It turns out the source I was using had a sign error in that identity! (th.physik.uni-bonn.de/Groups/drees/ident.pdf equation 4A) Commented Feb 27, 2019 at 11:11
• Those pesky minus signs ;) Commented Feb 27, 2019 at 19:01