# 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.

• The commutator tells you that $Q$ is a generator of infinitesimal phase transformations of the field $\phi$. The corresponding finite transformation is $e^{i\alpha Q}\phi e^{-i\alpha Q}=e^{i\alpha}\phi$. Dec 16 '19 at 15:33
• "The question is..." makes it sound like this is homework. If so, then please follow our homework policy: physics.meta.stackexchange.com/questions/714/… This means adding the homework-and-exercises tag, citing the source of the question, making an effort at solution, and having a conceptual aspect to your question. Even if this was not homework, you could tell us what you tried.
– user4552
Dec 29 '19 at 15:21

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 $$1$$.

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 $$q^a$$.

• Thank you so much! Dec 17 '19 at 15:35