Apply parity to the mixed anticommutator. The transformed left-hand side is
$$
\left\{\eta_P\sigma^0_{\alpha\dot\gamma}\bar Q_{\dot\gamma},
-\eta_P^*\bar\sigma^0_{\dot\beta\delta}Q_\delta\right\}
=-2\sigma^0_{\alpha\dot\gamma}
\sigma^\mu_{\delta\dot\gamma}
\bar\sigma^0_{\dot\beta\delta}P_\mu,
$$
because $|\eta_P|=1$. The Pauli-matrix identity represented by this product leaves the temporal matrix unchanged and reverses the three spatial matrices, so it equals
$$
2\sigma^\mu_{\alpha\dot\beta}(P_0,-\mathbf P)_\mu.
$$
But a <parity> transformation acts on momentum as $P^0\mapsto P^0$ and $P^i\mapsto-P^i$. This is exactly the parity transform of
$$
2\sigma^\mu_{\alpha\dot\beta}P_\mu,
$$
so $\{Q_\alpha,\bar Q_{\dot\beta}\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu$ is consistent with the stated transformation law. The arbitrary intrinsic phase cancels.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 307 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-307/1/c/solution