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

Under spatial reflection, a scalar field obeys $P\varphi (\mathbf{x})P=\varphi (-\mathbf{x})$, and hence each spatial derivative changes sign. The interaction contains three such derivatives, so changing the integration variable from $\mathbf{x}$ to $-\mathbf{x}$ givesThus this is a parity-odd scalar interaction.

Put $O({\mathbf{k}}_a)={\prod }_{a=1}^n{\varphi }_a({\mathbf{k}}_a)$ and $A({\mathbf{k}}_a)=⟨H_{\mathrm{int}}O({\mathbf{k}}_a)⟩$. Hermitian conjugation and commutativity of equal-time scalar fields giveThe parity-invariant vacuum and the odd parity of $H_{\mathrm{int}}$ imply $A(-{\mathbf{k}}_a)=-A({\mathbf{k}}_a)$. ThereforeThere is no conflict with the usual $2i\mathrm{Im}A$ rule for two Hermitian operators: a momentum-space product at fixed ${\mathbf{k}}_a$ is generally not itself Hermitian, since its adjoint carries momenta $-{\mathbf{k}}_a$.