Past exam of the theoretical physics major option course of the University of Oxford / 2022 / 5 / f / Solution

Write $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2{\omega }_p]$, $p^{\mu }=({\omega }_p,\mathbf{p})$ and ${\omega }_p=\sqrt{{\mathbf{p}}^2+m^2}$. The Fourier expansion in the source is compatible with the oscillator canonical commutation relationsFor the free real scalar field, $\pi =\dot{\varphi }$ andSubstitution of the Fourier expansion produces spatial Dirac delta functions. Terms with two annihilation operators or two creation operators cancel: in $P_0$ their coefficient is $-{\omega }_p^2+{\mathbf{p}}^2+m^2=0$, and in $P_i$ their integral is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The two mixed terms combine, with normal ordering removing the vacuum constant, to giveIn particular, $P_i$ contains the lower component $p_i=-{\mathbf{p}}_i$ in this Minkowski metric convention.

To check commutativity without manipulating a square of a Dirac delta function, define $N_f=\int d{\Pi }_{p}f(p)a_p^{†}{a}_p$. The oscillator algebra gives $[N_f,a_q]=-f(q)a_q$ and $[N_f,a_q^{†}]=f(q)a_q^{†}$. Thus $[N_f,a_q^{†}{a}_q]=0$ and $[N_f,N_g]=0$. Taking $f=p_{\mu }$, $g=p_{\nu }$ provesThese identities hold on the usual dense domain of finite-particle states, or can first be proved in a finite box and then continued to the continuum.