# Fabri-Picasso theorem: Why can we assume that the charge commutes with momentum?

To prove the Fabri-Picasso theorem, we assume that the charge $$Q$$ is translationally invariant, i.e. it commutes with the momentum operator:

$$[Q,P^\mu]=0$$

Why can we assume this?

$$Q = \int \!\!\text{d}^3x\;j^0(x^\mu)$$
it will no longer depend on $$x^i$$ after the integration. Therefore $$[Q,P^i]=0$$. As for $$\mu=0$$:
\begin{align} [Q,P^0]f &= QP^0f-P^0Qf\\ &= \text{i}Q\frac{\partial f}{\partial t} - \text{i}\frac{\partial}{\partial t} (Qf)\\ &= \text{i}Q\frac{\partial f}{\partial t} - \text{i}\frac{\partial Q}{\partial t} f - \text{i}Q\frac{\partial f}{\partial t}\\ &= - \text{i}\frac{\partial Q}{\partial t} f \\ &= -\text{i}\int\!\!\text{d}^3x\;\frac{\partial j^0(x)}{\partial t}\;\cdot f\qquad(\text{use }\partial_\mu j^\mu=0)\\ &= -\text{i}\int\!\!\text{d}^3x\;\vec\nabla\vec j\;\cdot f\\ &= -\text{i}\int _{\partial V} \text{d}^2\vec A\; \vec j\;\cdot f\\ &\propto j(x) \text{ at the boundary} \to 0 \end{align}