This is called the Fabri-Picasso theorem. Their argument requires both the vacuum and the charge $Q$ to be translationally invariant: $P |0\rangle = 0$, and $[P, Q] = 0$.
The argument goes as follows: Since the charge originates from a symmetry, then according to Noether's theorem:
$$Q = \int d^3x j_0(x)$$
The expectation Consider the correlation function of the charge with itself:
$$\begin{matrix} \langle 0| QQ |0\rangle& = \int d^3x \langle0|j_0(x) Q|0\rangle \\ & =\int d^3x \langle0|e^{iPx} j_0(0) e^{-iPx} Q |0\rangle \\ & =\int d^3x\langle0| e^{iPx} j_0(0) e^{-iPx} Q e^{iPx} e^{-iPx}|0\rangle\\ & =\int d^3x \langle0| j_0(0) Q |0\rangle \end{matrix}$$\begin{align} \langle 0| QQ |0\rangle& = \int d^3x \langle0|j_0(x) Q|0\rangle \\ & =\int d^3x \langle0|e^{iPx} j_0(0) e^{-iPx} Q |0\rangle \\ & =\int d^3x\langle0| e^{iPx} j_0(0) e^{-iPx} Q e^{iPx} e^{-iPx}|0\rangle\\ & =\int d^3x \langle0| j_0(0) Q |0\rangle \end{align} The integrand in the r.h.s. does not depend on the position, therefore its value is proportional to the total space volume, thus $$||Q|0\rangle||^2 = \infty$$ Thus the operator $Q$ does not exist in the Hilbert space unless $Q|0\rangle = 0$.
However, the commutators of $Q$ with the fields for example exist bercuasebecause by Noetheer'sNoether's theorem, they generate the symmetry, in other words the right hand sides of:
$$[Q, \phi] = \delta \phi$$
exist, since they are the symmetry transformed fields.
