Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 312 / 2 / e / Solution

Hermitian conjugation of part d gives the corresponding charge insertion on the bra. The Ward identity $i⟨[Q,\mathcal{O}]⟩=⟨{\delta }_b\mathcal{O}⟩$ therefore becomesFor a neutral operator, ${\delta }_b\mathcal{O}=0$, so the two soft limits, multiplied by their respective wavefunctional coefficients, are equal. If $\mathcal{O}$ is charged, the right side is nonzero. For a product of fields of charges $e_a$ at positions ${\mathbf{x}}_a$, it is proportional to ${\sum }_a{e}_a{b}_i{x}_a^i⟨\mathcal{O}⟩$; in momentum space this becomes the corresponding momentum derivative. This is a soft-vector Ward-Takahashi identity.