How to Compute charge ##Q## of a particular state in free Dirac field

[Pouramat]

Homework Statement

For a free Dirac field, how can I explicitly compute the charge $Q$ of the state $a_{p1}^{r \dagger}a_{p2}^{s \dagger} b_{p3}^{t \dagger}|0>$.

Relevant Equations

The operator $Q$ is :
\begin{equation}
Q = \int \frac{d^3 p}{(2 \pi)^3} \Sigma_s \bigg(a_p^{s \dagger}a_p^s-b_p^ {s \dagger} b_p^s \bigg)
\end{equation}

suppose I should evaluate $$Qa_{p1}^{r \dagger}a_{p2}^{s \dagger} b_{p3}^{t \dagger}$$ I get lost in the commutator relation. Any help?

 

[vanhees71]

Hint: You need, of course anti-commutator relations, because the Dirac field must be quantized as fermions. The goal is to bring annihilation operators to the right, so that it's acting on the vacuum state, giving 0. Note that for arbitrary operators
$$[\hat{A},\hat{B} \hat{C}]=\{\hat{A},\hat{B} \} \hat{C}-\hat{B} \{\hat{A},\hat{C} \}.$$
Further you have
$$\{\hat{a}_p^{s},\hat{a}_{p'}^{s' \dagger} \}=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}') \delta_{ss'}$$
and similar for the b's. All other anticommutators vanish.

It's also intuitively clear, what the charge of this state is, since obviously any a-particle carries a charge of +1 and any b-particle (the anti-particle of the a-particle) carries charge -1. But it's a good exercise to verify this by the explicit calculation.