Solution

Apply parity to the mixed anticommutator. The transformed left-hand side is

{η_{P} σ_{α \overset{γ}{˙}}^{0} \overset{ˉ}{Q}_{\overset{γ}{˙}}, - η_{P}^{*} \overset{σ}{ˉ}_{\dot{β} δ}^{0} Q_{δ}} = - 2 σ_{α \overset{γ}{˙}}^{0} σ_{δ \overset{γ}{˙}}^{μ} \overset{σ}{ˉ}_{\dot{β} δ}^{0} P_{μ},

because

∣ η_{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 σ_{α \dot{β}}^{μ} (P_{0}, - P)_{μ} .

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 σ_{α \dot{β}}^{μ} P_{μ},

{Q_{α}, \overset{ˉ}{Q}_{\dot{β}}} = 2 σ_{α \dot{β}}^{μ} P_{μ}

is consistent with the stated transformation law. The arbitrary intrinsic phase cancels.

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