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.

  • 1
    $\begingroup$ The commutator tells you that $Q$ is a generator of infinitesimal phase transformations of the field $\phi$. The corresponding finite transformation is $e^{i\alpha Q}\phi e^{-i\alpha Q}=e^{i\alpha}\phi$. $\endgroup$ Dec 16, 2019 at 15:33
  • $\begingroup$ "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. $\endgroup$
    – user4552
    Dec 29, 2019 at 15:21

2 Answers 2


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 $q^a$.

  • $\begingroup$ Thank you so much! $\endgroup$
    – Bernoulli
    Dec 17, 2019 at 15:35

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