0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Have you tried putting spinor indices everywhere? That may help you solve you confusion about moving spinors around. $\endgroup$ – Oбжорoв Sep 19 '19 at 10:25
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.