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

Question

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?

Answer 1

Since the charge is defined as

$$ Q = \int \!\!\text{d}^3x\;j^0(x^\mu) $$

it will no longer depend on $x^i$ after the integration. Therefore $[Q,P^i]=0$. As for $\mu=0$:

$$ \begin{align} [Q,P^0]f &= QP^0f-P^0Qf\\ &= \text{i}Q\frac{\partial f}{\partial t} - \text{i}\frac{\partial}{\partial t} (Qf)\\ &= \text{i}Q\frac{\partial f}{\partial t} - \text{i}\frac{\partial Q}{\partial t} f - \text{i}Q\frac{\partial f}{\partial t}\\ &= - \text{i}\frac{\partial Q}{\partial t} f \\ &= -\text{i}\int\!\!\text{d}^3x\;\frac{\partial j^0(x)}{\partial t}\;\cdot f\qquad(\text{use }\partial_\mu j^\mu=0)\\ &= -\text{i}\int\!\!\text{d}^3x\;\vec\nabla\vec j\;\cdot f\\ &= -\text{i}\int _{\partial V} \text{d}^2\vec A\; \vec j\;\cdot f\\ &\propto j(x) \text{ at the boundary} \to 0 \end{align}$$

Answer 2

The charge of any particle, or any charged object, does not depend on its position, where it is placed. That's what translation invariance means. Indeed, it would be a strange world if for a given object the value of its charge changes as we move around the room.

On a more abstract level, translation invariance has to do with spacetime symmetries, while charge (and its conservation) has to do with the so called internal symmetries. The two act on different spaces, in a sense they do not mix, therefore their generators have to commute.