Let
$$
N=\int\frac{d^3q}{(2\pi)^3}a_{\mathbf q}^\dagger a_{\mathbf q},
\qquad
K=\int\frac{d^3q}{(2\pi)^3}a_{\mathbf q}^\dagger a_{-\mathbf q}.
$$
The canonical commutation relations give $[N,a_{\mathbf k}]=-a_{\mathbf k}$ and $[K,a_{\mathbf k}]=-a_{-\mathbf k}$. Hence exponentiating the adjoint action gives
$$
\mathcal P_1a_{\mathbf k}\mathcal P_1^{-1}
=e^{i\pi/2}a_{\mathbf k}=ia_{\mathbf k}.
$$
Writing $R a_{\mathbf k}=a_{-\mathbf k}$, one has $R^2=1$ and $\operatorname{ad}_K=-R$ on annihilation operators. Therefore
$$
\mathcal P_2a_{\mathbf k}\mathcal P_2^{-1}
=e^{-i\gamma_p\pi R/2}a_{\mathbf k}
=-i\gamma_p a_{-\mathbf k}.
$$

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/c/solution