In "Quantum Field Theory" by Peskin and Schroeder, I couldn't understand the where they investigate the commutation (as opposed to the anticommutation) relation for the Dirac field (pg. 53): $$ \begin{align} \psi(x) &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}}e^{ix\cdot p} \sum_{s=1,2} \left(a^s_pu^s(p)+b^s_{-p}v^s(-p)\right) \tag{3.87} \\[5px] \psi^\dagger(y) &= \int \frac{d^3q}{(2\pi)^3} \frac{1}{\sqrt{2E_q}}e^{-iy\cdot q} \sum_{r=1,2}\left(a^{r\dagger}_q u^{r\dagger}(q) +b^{r\dagger}_{-q}v^{r\dagger}(-q)\right). \tag{2} \end{align} $$

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

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?

  • 2
    $\begingroup$ Have you tried putting spinor indices everywhere? That may help you solve you confusion about moving spinors around. $\endgroup$ Sep 19, 2019 at 10:25
  • $\begingroup$ This is confusing to me also because $u^s(p) u^{\dagger r}(q)$ seems to be an outer product and $u^{\dagger r}(q) u^s(p)$ seems to be an inner product. $\endgroup$
    – Mike Flynn
    Nov 27 at 9:58

2 Answers 2


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.


In fact, we're considering the commutator relation of $\Psi_a(x)$, which is just one component of $\Psi(x)$. In this understanding, Commutator relation read: $$ [\Psi_a(x),\Psi_b(y)] = \delta^4(x-y) \delta_{ab} $$

  • $\begingroup$ The components of u and v are just numbers, hence commute. $\endgroup$
    – user377071
    Sep 8 at 5:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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