In Zee's quantum field theory in a nutshell, 2nd edition, pg 551 he has the charge of a Dirac field written as
$Q=\int {d^3p \over (2\pi)^3(E_p/m)} \sum_s \{b^\dagger(p,s)b(p,s)-d^\dagger(p,s)d(p,s)\}$
He then goes onto write
We find $[Q,\psi(0)]=-\psi(0)$, thus showing that $b$ and $d^\dagger$ must carry the same charge.
I would very much appreciate if someone could explain to me how one deduces this. He writes the Dirac field as $\psi(x)=\int\frac{d^3p}{(2\pi)^{3/2} (E_p/m)^{1/2}}\sum_s [b(p,s)u(p,s)e^{-ipx}+d^\dagger(p,s)v(p,s)e^{ipx}]$. So doesn't this imply that $[Q,\psi(0)]\left|0\right>=Q\psi(0)\left|0\right>=-\psi(0)\left|0\right>$ and if so this seems to be independent of $b(p,s)$ so I don't understand why he writes
We find $[Q,\psi(0)]=-\psi(0)$, thus showing that $b$ and $d^\dagger$ must carry the same charge.