Dirac Current Derivations I am currently self-studying Quantum Field Theory and am using the book An Introduction to Quantum Field Theory by Peskin and Schroeder. I am confused about a derivation presented in section 3.5 (called "Quantization of the Dirac Field"). Equation 3.111 derives a "rotation current density" J which splits up into an orbital angular momentum part and a spin momentum part. This derivation I understand. Next, the book wishes to prove that the Dirac Equation deals with particles of spin $1/2$. To do so, the authors consider the case when the particle is at rest; this allows us to ignore the orbital angular momentum term in equation 3.111. From this it follows (again I understand this):
$J_z = \int d^3x \int\frac{d^3p d^3p'}{(2\pi)^6} \frac{1}{\sqrt{2E_p 2E_p'}}e^{-ip'\cdot x}e^{ip\cdot x} 
 \sum_{r,r'}\bigg(a_{p'}^{r'\dagger}u^{r'\dagger}(p') + b_{-p}^{r'}v^{r'\dagger}(-p')\bigg)\frac{\Sigma^3}{2}\bigg(a_{p}^{r}u^{r}(p) + b_{-p}^{r\dagger}v^{r}(-p)\bigg)$
However, the next equation says that
$J_z a_0^{s\dagger}|0\rangle = \frac{1}{2m}\sum_{r}\bigg(u^{r\dagger}(0) \frac{\Sigma^3}{2}u^s(0)\bigg)a_0^{r\dagger}|0\rangle$
I am unsure why this result is true. The book gave the commutator relation $[a_p^{r\dagger}a_p^{r'},a_0^{s\dagger}] = (2\pi)^3 \delta^3(p)a_0^{r\dagger}\delta^{r's}$. Note that all $p$'s represent three momentum here and not four momentum. I tried using this commutator relationship when I expanded out the parenthetical terms in $\sum_{r,r'}$ but the $\Sigma^3$ matrix got in between these operators. Can anyone explain mathematically how the second equation follows from the first?
 A: For the commutator $[J_z, a_{ 0}^{s\dagger}]$ we have non-zero terms only for those that have an $a$ in $J_z^{(s)}$
\begin{align}
[J_z, a_{  0}^{s\dagger}] = &\, \int d^3 x \, \int \frac{d^3p\, d^3q}{(2\pi)^6} \frac{1}{\sqrt{2 E_{  p} 2 E_{  q}}}  \sum_{r, r'}  e^{+i (  p -   q)\cdot   x}
\Big[\Big(a_{  q}^{r \dagger} u^{r\dagger}(  q) +b_{-  q}^{r } v^{r\dagger}(-  q)  \Big)  \left(\frac{1}{2}\Sigma^3 \right) 
 a_{  p}^{r'} u^{r'}(  p) , a_0^{s\dagger} \Big]
\end{align}
If we act with this on the ground state $| 0 \rangle$ then the $b$ terms vanish and so we get
\begin{align}
[J_z, a_{  0}^{s\dagger}] | 0 \rangle= &\, \int d^3 x \, \int \frac{d^3p\, d^3q}{(2\pi)^6} \frac{1}{4\sqrt{ E_{  p}  E_{  q}}}  \sum_{r, r'}  e^{+i (  p -   q)\cdot   x}
u^{r\dagger}(  q)  \Sigma^3   u^{r'}(  p)[a_{  q}^{r \dagger}a_{  p}^{r'}  , a_0^{s\dagger} ]  | 0 \rangle \nonumber\\
=&\,\int d^3 x \, \int \frac{d^3p\, d^3q}{(2\pi)^6} \frac{1}{4\sqrt{ E_{  p}  E_{  q}}}  \sum_{r, r'}  e^{+i (  p -   q)\cdot   x}
 u^{r\dagger}(  q)  \Sigma^3   u^{r'}(  p)a_{  q}^{r \dagger}  (2\pi)^3 \delta^3(  p) \delta^{r's}  | 0 \rangle \nonumber\\
=&\,\int d^3 x \, \int \frac{d^3q}{(2\pi)^3} \frac{1}{4 E_{  q} }  \sum_{r}  e^{-i   q \cdot   x}
  u^{r\dagger}(  q)  \Sigma^3   u^{s}(  0)a_{  q}^{r \dagger}  | 0 \rangle
  \end{align}
Note that the $a$'s carry no Dirac index, so we can move them across the Dirac spinors without penalty. We can now also perform the $x$ integration, which will give a $\delta^3 (  q)$ which we can then do as well. This gives
\begin{align}
[J_z, a_{  0}^{s\dagger}] | 0 \rangle
 =&\, \frac{1}{4 E_{  0} }  \sum_{r} 
  u^{r\dagger}(  0)  \Sigma^3   u^{s}(  0)a_{  0}^{r \dagger}  | 0 \rangle
  \end{align}
