# 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?

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}