# Commutation relation for Dirac field

In "Quantum Field Theory" by Peskin and Schroeder, I couldn't understand the commutation relation calculation for Dirac field (pg. 53): \begin{align} \psi(x) &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}}e^{ix\cdot p} \sum_{s=1,2} \left(a^s_pu^s(p)+b^s_{-p}v^s(-p)\right) \tag{1} \\[5px] \psi^\dagger(y) &= \int \frac{d^3q}{(2\pi)^3} \frac{1}{\sqrt{2E_q}}e^{-iy\cdot q} \sum_{r=1,2}\left(a^{r\dagger}_q u^{r\dagger}(q) +b^{r\dagger}_{-q}v^{r\dagger}(-q)\right) \tag{2} \end{align}

They postulate $$[a^s_p,a^{r\dagger}_q]=[b^s_{p},b^{r\dagger}_{q}]=(2\pi)^3\delta^3(p-q)\delta^{sr}$$ and the commutation relation is written as below (eq (3.89)) \left[\psi(x),\psi^\dagger(y)\right] = \int{ \begin{align} & \frac{d^3pd^3q}{(2\pi)^6} \frac{1}{\sqrt{2E_p2E_q}} e^{i(x\cdot p-y\cdot q)} \\ & \sum_{s,r=1,2} \left([a^s_p,a^{r\dagger}_q]u^s(p)u^{\dagger r}(q) +[b^s_{-p},b^{r\dagger}_{-q}]v^s(-p)v^{r\dagger}(-q) \right) \end{align} } \tag{3.89}

However, I can not understand spinor calculation part. For example, the first term of $$\psi(x)\psi^\dagger(y)- \psi^\dagger(y)\psi(x)$$ is $$\sum_{s,r=1,2}\left( a^s_pu^s(p) a^{r\dagger}_qu^{r\dagger}(q) - a^{r\dagger}_qu^{r\dagger}(q) a^s_pu^s(p) \right) \,, \tag{4}$$ and it becomes $$\sum_{s,r=1,2} (a^s_pa^{r\dagger}_q - a^{r\dagger}_qa^s_p)u^s(p) u^{\dagger r}(q) \,. \tag{5}$$

But I think $$u^s(p) u^{\dagger r}(q) \neq u^{\dagger r}(q) u^s(p)$$. Then, how can it reach to the part?

And if $$u^s(p) u^{\dagger r}(q) = u^{\dagger r}(q) u^s(p)$$, why they don't do $$\sum_{s,r=1,2} u^{\dagger r}(q) u^s(p)$$, which gives a scalar quantity?

• Have you tried putting spinor indices everywhere? That may help you solve you confusion about moving spinors around. – Oбжорoв Sep 19 '19 at 10:25

First of all, in general $$(AB)^\dagger = B^\dagger A^\dagger$$, so your $$\psi^\dagger$$ expression is not entirely correct.
Secondly, the spinors $$u^s, v^s, u^{r\dagger}, v^{r\dagger}$$ are not operators but just numbers (columns of numbers technically) hence they commute with the creation and annihilation operators. The creation/annihilation operators act on the vacuum (or any other) state $$|0\rangle$$, not on the spinors.