Quantum field theory, interpretation of commutation relation

Question

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.

Answer 1

Q is a charge for the boson field $\phi$, which, e.g., annihilates a $\pi^-$ or creates a $\pi^+$. Its eigenvalue q for a state is that state's charge: the number of $\pi^+$s minus the number of $\pi^-$s.

Thus, by virtue of the commutation relation $[Q,\phi] = \phi$, the quantum field $\phi$ raises the charge of a state by 1, since $$ Q|\alpha\rangle= q|\alpha\rangle \implies Q(\phi|\alpha\rangle)= \phi (Q+1) |\alpha\rangle =(q+1) ( \phi|\alpha\rangle). $$

Conjugately, $[Q,\phi^\dagger] = -\phi^\dagger$, so $\phi^\dagger$ has the opposite charge: it kills a $\pi^+$ or creates a $\pi^-$.