The <number operator> $N$ is self-adjoint. Also
$$
K^\dagger=\int\frac{d^3q}{(2\pi)^3}a_{-\mathbf q}^\dagger a_{\mathbf q}=K
$$
after $\mathbf q\mapsto-\mathbf q$. Thus both exponents are anti-Hermitian and $\mathcal P_1,\mathcal P_2$ are <unitary operator>[unitary]. Their product obeys
$$
(\mathcal P_1\mathcal P_2)a_{\mathbf k}(\mathcal P_1\mathcal P_2)^{-1}
=\gamma_pa_{-\mathbf k},
$$
and likewise for creation operators. Both generators annihilate the vacuum, so the product leaves it invariant. Substitution in the free-field mode expansion yields
$$
(\mathcal P_1\mathcal P_2)\phi(t,\mathbf x)(\mathcal P_1\mathcal P_2)^{-1}
=\gamma_p\phi(t,-\mathbf x).
$$
Hence $\mathcal P=\mathcal P_1\mathcal P_2$ implements parity with <intrinsic parity> $\gamma_p$.

Solved by gpt-5.6-sol high.


 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-301/2/d/solution