Under <spatial reflection>, a <scalar field> obeys $P\phi(\mathbf x)P=\phi(-\mathbf x)$, and hence each <spatial derivative> changes sign. The interaction contains three such derivatives, so changing the integration variable from $\mathbf x$ to $-\mathbf x$ gives
$$
PH_{\rm int}P=-H_{\rm int}.
$$
Thus this is a <parity-odd scalar interaction>.

Put $O(\mathbf k_a)=\prod_{a=1}^n\phi_a(\mathbf k_a)$ and $A(\mathbf k_a)=\langle H_{\rm int}O(\mathbf k_a)\rangle$. <Hermitian conjugation> and commutativity of equal-time scalar fields give
$$
\langle O(\mathbf k_a)H_{\rm int}\rangle
=A(-\mathbf k_a)^*.
$$
The <parity-invariant vacuum> and the odd parity of $H_{\rm int}$ imply $A(-\mathbf k_a)=-A(\mathbf k_a)$. Therefore
$$
\boxed{\langle[H_{\rm int},O]\rangle
=A-A(-\mathbf k_a)^*=A+A^*=2\operatorname{Re}A}.
$$
There is no conflict with the usual $2i\operatorname{Im}A$ rule for two <Hermitian operators>: a momentum-space product at fixed $\mathbf k_a$ is generally not itself Hermitian, since its adjoint carries momenta $-\mathbf k_a$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 312 / 1 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-312/1/i/solution