Showing that electron and positrons have the same absolute charge

Question

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.

Answer 1

$$[Q,\psi(0)]=-\psi(0)\implies [Q,d^\dagger]=-d^\dagger $$ If the charge of a general state is $q$:

$$Q|\Psi\rangle=q|\Psi\rangle $$ Then the charge of a state with one more $d$-quantum is: $$Qd^\dagger|\Psi\rangle=(q-1)|\Psi\rangle$$ Then the $d$-quantum has negative charge.

Analogously for $b$.

Answer 2

From the basic anticommutators such as $[b,b^\dagger]\sim \delta$ – according to your normalizations etc. – and similarly for the $d$'s, it follows that the commutator of the first, $b^\dagger b$ term, with any $b$ is $-b$ (the same one). The second term doesn't contribute. This $[Q,b]=-b$ is interpreted as "$b$ carries charge $-1$". Similarly, the other term in $Q$, schematically of the form $d^\dagger d$, has the commutator with $d^\dagger$ that is $[Q,d^\dagger]=-d^\dagger$. A sign flip from the added dagger cancels against a sign flip from the minus sign in front of the second term in $Q$.

The same coefficient on the right hand side of $[Q,b]$ and $[Q,d^\dagger]$ means that the particles created/annihilated by $b$ and $d$ and their conjugates have the same charge, up to the sign.

One may also directly extra this self-evident conclusion from $[Q,\psi(0)]=-\psi(0)$. It's because $\psi(0)$ can be written as a linear combination (sum of integral) of operators $b$ and those proportional to $d^\dagger$. Because $[Q,\psi(0)]$ is proportional to $\psi(0)$, $\psi(0)$ is an "eigenstate", it carries a well-defined charge given by the coefficient on the right hand side of the commutator, $-1$. Because $\psi(0)$ is composed of two independent pieces, both of them must be eigenstates with the same eigenvalue, too.

There are lots of things to prove those statements. My point is that this is a totally trivial thing and you should be able to answer the question yourself.