Insert the <mode expansion of a free field> into the classical expression for $\mathbf P$. The $aa$ and $a^\dagger a^\dagger$ terms cancel after $\mathbf p\mapsto-\mathbf p$, while the mixed terms give, after the stated <normal ordering>,
$$
\boxed{\mathbf P=\int\frac{d^3p}{(2\pi)^3}\,
\mathbf p\,a_{\mathbf p}^\dagger a_{\mathbf p}.}
$$
Thus the momentum operator counts each occupied mode with weight $\mathbf p$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 1 / ii / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-301/1/ii/a/solution