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

Use the spatial transverse-longitudinal decompositionUnder the U(1) gauge symmetry $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }ϵ$,The assumption ${\nabla }^2ϵ\ne 0$ makes the longitudinal split unambiguous. The scalaris therefore gauge-invariant.

The electric components of the electromagnetic field tensor arewhile $F_{ij}$ depends only on the transverse vector. The scalar-vector cross term integrates to zero because ${\partial }_i{A}_i^V=0$, so the scalar action isVarying the nondynamical scalar gives the constraint equation in field theoryFor every nonzero Fourier momentum this fixes ${\Phi }_A=0$, confirming that a source-free massless vector has no propagating scalar polarization.

Applying $ϵ=b_i(t)x^i$ to $A_{\mu }=0$ givesIts field strength vanishes, so it is a large gauge transformation of the background. To arise as the zero-momentum limit of a physical transverse perturbation, $b_i(t)$ must obey the zero-momentum vector equation of motion. In the variables used in the question this iswith a constant growing solution and a decaying solution proportional to ${\int }^{t}d{t}^{\prime }/a^3(t^{\prime })$. This condition makes the large gauge mode an adiabatic mode that can be continued to small nonzero momentum.

For constant $b_i$, the transformation is $\delta A_i^V=b_i$. Its Noether charge is thereforeup to the Fourier-sign convention. Expand a transverse polarization asOnly the creation term survives on the vacuum, while $A_i^V(\mathbf{q})|0⟩=f_q^{\ast }{a}_i^{†}(-\mathbf{q})|0⟩$. Hence

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.