# Quantum field theory, interpretation of commutation relation

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 $1$.