Quantum field theory, interpretation of commutation relation Let $\phi$ be the quantum field
$$
\phi(x) = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}} \Big[ b_\mathbf{p}e^{-ip\cdot x} + c_\mathbf{p}^\dagger e^{ip\cdot x} \Big]
$$
with commutation relations
$$
[b_\mathbf{p}, b_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
$$
[c_\mathbf{p}, c_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
all other commutators zero. Let $Q$ be the charge operator
$$
Q = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \Big[c_{\mathbf{p}}^\dagger c_{\mathbf{p}} - b_{\mathbf{p}}^\dagger b_{\mathbf{p}} \Big].
$$
We calculate the commutator $[Q,\phi] = \phi$. The question is what is an interpretation of this commutation relation? We know that $Q$ is the number of antiparticles minus the number of particles.
 A: One interpretation is as follows: $[Q,\phi(\vec{x})] = \phi(\vec{x})$ means that $Q\phi(\vec{x}) = \phi(\vec{x})(Q + 1)$. Thus, if $\vert q \rangle$ is a charge eigenstate with eigenvalue $q$ (i.e. $Q\vert q \rangle = q\vert q \rangle$) then
$$ Q \phi(\vec{x}) \vert q \rangle = \phi(\vec{x}) (Q + 1) \vert q \rangle = (q + 1) \phi(\vec{x}) \vert q \rangle, $$
which means that $\phi(\vec{x}) \vert q \rangle$ is a charge eigenstate with eigenvalue $q + 1$. Thus, acting with $\phi(\vec{x})$ increases the charge by $1$.
In fact, a common interpretation of the operator $\phi(\vec{x})$ is that it creates a particle at position $\vec{x}$ (being a kind of Fourier transform of the creation operator $c^\dagger_\vec{p}$, which creates a particle with momentum $\vec{p}$). So the commutation relation says that if you add a particle, the total charge will increase by $1$.
A: The commutation relation $[Q,\phi] = n \phi$ tells you that the field $\phi$ has charge (or eigenvalue) $n$ under $Q$ in our case $n=1$. An easy way to see this is to note
$$\left| \phi \right> := \phi \left| 0 \right> \implies Q \left| \phi \right> = Q \phi  \left| 0 \right> = [Q,\phi]  \left| 0 \right> = n \phi  \left| 0 \right> = n \left| \phi \right>  $$
where going from $Q\phi$ to commutator we used the fact that $Q \left| 0 \right> =0$. 
In general an equation of the form $[Q^a,\phi] = q^a \phi$ for charge operators $Q^a$, which commute amongst each other, defines the charge of $\phi$ to be $q^a$.
