Weinberg [1] has put forth an elegant argument (summarized in [2, Section 3.3] and at the bottom of the question), which suggests that the broken symmetry phase of any $U(1)$ gauge theory will be accompanied by infinite conductivity. I have a question regarding a key step in which he identifies the voltage as the functional derivative of the Hamiltonian with respect to charge density; i.e. $V(x) := \frac{\delta H}{\delta \rho}$.
I’m trying to accumulate evidence for the plausibility for Weinberg’s use of the definition $V(x) := \frac{\delta H}{\delta \rho}$. More precisely, I’d like to show, given two points $x_1,x_2 \in \mathbb{R}^d$, that $V(x_1) - V(x_2)$ is the potential difference that an experimenter would measure.
It’s possible to motivate Weinberg’s definition based on highschool electrostatics. Recall that the energy for a system of static charges is $H = \frac{1}{2}\int\int \rho(x)\rho(x’)/|x-x’|$ in $d=3$ spatial dimensions. Functionally differentiating this expression once with respect to $\rho$ indeed gives the well-known expression for the potential $V(x) = \int \rho(x’)/|x-x’|$ so LHS = RHS. However, it’s non-obvious that this is the right definition with magnetic fields and after coupling to Goldstones.
In order to see the difficulty, consider a minimal Lagrangian density that describes the symmetry broken phase (the Stueckelberg Lagrangian): \begin{align} \mathcal{L} & = - \frac{1}{4} F_{\mu\nu} F^{\mu\nu} - \frac{1}{2}m^2 (A_\mu-\partial_\mu G) (A^\mu-\partial^\mu G) \end{align} In passing to the Hamiltonian formulation, we first need to compute the canonical momenta (now using $\Pi$ to denote the momentum conjugate to the Goldstone): \begin{align} \pi^0 & = 0 \\ \pi^i & = F^{i0} \\ & = \dot{A}^i - \partial_i A_0 \\ \Pi & = m^2(\dot{G} - A_0) \end{align} from which we readily obtain \begin{align} H & = \int \mathrm{d}^d x \left[\Pi \dot{G} + \pi^i \dot{A}_i - \mathcal{L} \right] \\ & = \int \mathrm{d}^d x \left[ \frac{1}{2m^2}\Pi^2 + \frac{1}{2} \pi_i \pi^i + \frac{1}{4} F_{ij} F^{ij} + \frac{1}{2}m^2(A_i-\partial_i G)^2 + (\Pi + \pi^i \partial_i) A_0 \right] \\ & = \int \mathrm{d}^d x \left[ \frac{1}{2m^2}\Pi^2 + \frac{1}{2} \pi_i \pi^i + \frac{1}{4} F_{ij} F^{ij} + \frac{1}{2}m^2(A_i-\partial_i G)^2 + (\Pi - \partial_i\pi^i ) A_0 \right] \end{align} where I integrated by parts. In particular, it’s not clear how functional derivatives with respect to $\Pi$ are related to voltage.
Is it possible to come up with a more convincing argument?
Weinberg’s infinite conductivity argument: Since the effective Lagrangian depends on the Goldstone field $G$ only via the combination $A_\mu - \partial_\mu G$, it follows that $G$ is canonically conjugate to the charge density $\rho$. Thus, in the Hamiltonian formulation, it satisfies an equation of motion $\dot{G} \propto \frac{\delta H}{\delta \rho} = V(x)$. Thus for time-independent field configurations corresponding to steady current, the potential difference vanishes.
[1] https://academic.oup.com/ptps/article/doi/10.1143/PTPS.86.43/1885987
