Charge operator action in QFT

Question

I have a quick question regarding the action of the charge operator in quantum field theory. I am reading Schwartz's book on QFT, and he proves the two following commutation relations $$[Q,\psi]=-\psi\qquad [Q,\psi^\dagger]=\psi^\dagger,$$ where $Q=\int d^3x\;\psi^\dagger\psi$, and $\psi$ is the normal quantized fermionic field. He then concludes that this implies that $Q$ counts the number of particle minus antiparticles. I do not see how this is true. Any help is much appreciated!

Answer 1

A quick answer. You should think of $\psi^\dagger$ as the creation operator for a particle, and conversely you can think of $\psi$ as the creation operator of an anti-particle.

Thus since the commutation relation has different sign for $[Q,\psi]$ and $[Q,\psi^\dagger]$, $Q$ will count the particles as positive charge, and anti-particles as negative charge.

The counting would work like this: Assume you have state $|\psi\rangle=\psi^\dagger|0\rangle$ where $|0\rangle$ is the vacuum (i.e. no particles).

The action of $Q$ on this state is

$$ Q|\psi\rangle=Q\psi^\dagger|0\rangle=(\psi^\dagger+\psi^\dagger Q)|0\rangle=\psi^\dagger|0\rangle=|\psi\rangle $$
where we assumes that $Q$ annihilates the vacuums state. Similarly for an antiparticle state $|\bar\psi\rangle=\psi|0\rangle$ we have

$$ Q|\bar\psi\rangle=Q\psi|0\rangle=(-\psi+\psi Q)|0\rangle=-\psi|0\rangle=-|\bar\psi\rangle $$

So we see that $Q$ has positive eigenvalues for particles and negative eigenvalues for anti-particles. Thus $Q$ measures the total charge $Q=N_\psi - N_{\bar\psi}$.

Answer 2

Another quick answer. Hope it be somehow helpful. If one looks at the Fourier expansion of $\psi$ in terms of momenta, they would find an expression for $Q$ in terms of annihalation operators, but with the minus sign between them. (And of course an infinity.) So, it can be deduced that $Q$ actually counts the difference between number of particles and anti-particles.