Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 48 / 3 / c / Solution

Denote the momentum in the printed mode expansion by ${\tilde{\pi }}^{\nu }$, as it equals $-{\dot{A}}^{\nu }$ for the boundary-subtracted density discussed in part (b). Take a real complete basis of polarization vectors. Its completeness relation with one index raised isHere raising or lowering the polarization labels uses the same diagonal matrix $\eta$, so this is just the completeness relation in the PDF. Write $\mathbf{r}=\mathbf{x}-\mathbf{y}$. In the mixed field-momentum commutator, the $a{a}^{†}$ term gives the product of the minus sign in $\tilde{\pi }$'s creation part and the minus sign in the oscillator commutator. The $a^{†}a$ term gives the same sign. The two mode normalization factors multiply to $1/2$, and the momentum delta function removes one integral. ConsequentlyFor the field-field commutator, the polarization sum gives insteadbecause the integrand is odd under $\mathbf{p}\mapsto -\mathbf{p}$. Similarly,These are the requested canonical commutation relations, derived by photon oscillator completeness and canonical brackets.

They also hold for the literal momenta of part (b). The shifts areSince the fields commute at equal times, their mixed brackets with the momenta do not change. The only potentially new momentum bracket isAll other momentum brackets vanish separately. Thus both canonical conventions have $[A_{\mu },{\pi }^{\nu }]=i{\delta }_{\mu }{}^{\nu }{\delta }^3$ and vanishing coordinate-coordinate and momentum-momentum brackets, even though the momentum operators differ by spatial derivatives. This resolves the density/expansion mismatch without changing the printed oscillator algebra.