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
    2 days ago

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

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