Spinor notation excercise with grassman numbers

[Onamor]

Spinor notation exercise with grassman numbers

I'm checking a term when squaring a vector superfield in Wess-Zumino gauge, but its really just an excercise in index/spinor notation:

I need to square the term $\left(\theta^{\alpha}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\right)V_{\mu}$

where $\theta^{\alpha}$ is a Weyl spinor of grassman numbers, $\bar{\theta}^{\dot{\beta}}$ is a spinor in conjugate rep, $\sigma^{\mu}$ is a Pauli 4-vector (ie $\left(1,\sigma^{i}\right)$) and $V_{\mu}$ is just a 4-vector.

The answer is $\frac{1}{2}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)V^{\mu}V_{\mu}$ where $\left(\theta\theta\right)\equiv\theta^{\alpha}\theta_{\alpha}$ and $\left(\bar{\theta}\bar{\theta}\right)\equiv\theta_{\dot{\beta}}\theta^{\dot{\beta}}$.

So, in detail, to square the term I want to multiply it by a similar term with upper and lower indices switched:
$\left(\theta^{\alpha}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\bar{\theta}^{\dot{\beta}}\right)\left(\theta_{\delta}\left(\sigma_{\nu}\right)^{\delta\dot{\gamma}}\bar{\theta}_{\dot{\gamma}}\right)V^{\nu}V_{\mu}$.

First of all I switch the $\bar{\theta}^{\dot{\beta}}\theta_{\delta}$ for free as they commute(?)

Then since $\theta^{\alpha}\theta_{delta}=\theta^{\alpha}\theta^{\lambda}\epsilon_{\lambda\delta}=-\frac{1}{2}\left(\theta\theta\right)\epsilon^{\alpha\lambda}\epsilon^{\lambda\delta}=-\frac{1}{2}\left(\theta\theta\right)\delta^{\alpha}_{\delta}$,
and similarly $\bar{\theta}^{\dot{\beta}}\bar{\theta}_{\dot{\gamma}}=\frac{1}{2}\left(\bar{\theta}\bar{\theta}\right)\delta^{\dot{\beta}}_{\dot{\gamma}}$, I have

$-\frac{1}{4}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)\delta^{\alpha}_{\delta}\delta^{\dot{\beta}}_{\dot{\gamma}}\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\delta\dot{\gamma}}V^{\nu}V_{\mu}$.

Or $-\frac{1}{4}\left(\theta\theta\right)\left(\bar{\theta}\bar{\theta}\right)\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\alpha\dot{\beta}}V^{\nu}V_{\mu}$.

So, does $\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\alpha\dot{\beta}}=-2\delta^{\mu}_{\nu}$?

Its possible that a minus sign comes from the commutation of the two thetas in the first step, and perhaps I should have used the conjugate pauli 4-vector $\left(-1,\sigma^{i}\right)$ to square the term?

 

[Nate810]

Yes, it is correct that \left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}\left(\sigma_{\nu}\right)^{\alpha\dot{\beta}}=-2\delta^{\mu}_{\nu}. The minus sign from the commutation of the two thetas does not play a role, since it cancels out with the minus sign in the definition of \left(\theta\theta\right) and \left(\bar{\theta}\bar{\theta}\right). And no, you do not need to use the conjugate Pauli 4-vector to square the term.