# 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 $$1$$.