How to show that $\sigma^\mu_{\alpha \dot{\alpha}} \bar\sigma_\mu^{\dot{\beta} \beta} = 2\delta_\alpha ^\beta \delta_\dot{\alpha}^\dot\beta$?

Question

Consider the usual definitions $\sigma^\mu = (1, \sigma^i)$ and $\bar\sigma^\mu = (1, -\sigma^i)$, is it possible to show

$$\sigma^\mu_{\alpha \dot{\alpha}} \bar\sigma_\mu^{\dot{\beta} \beta} = 2\delta_\alpha ^\beta \delta_\dot{\alpha}^\dot\beta$$

without resorting to going through each index manually?