 Solution

The number operator $N$ is self-adjoint. Alsoafter $\mathbf{q}\mapsto -\mathbf{q}$. Thus both exponents are anti-Hermitian and ${\mathcal{P}}_1,{\mathcal{P}}_2$ are unitary. Their product obeysand likewise for creation operators. Both generators annihilate the vacuum, so the product leaves it invariant. Substitution in the free-field mode expansion yieldsHence $\mathcal{P}={\mathcal{P}}_1{\mathcal{P}}_2$ implements parity with intrinsic parity ${\gamma }_p$.