I am self-studying QFT and I came to the point of quantizing the Dirac field. The Dirac field expanded in terms of creation/annihilation operators is:
$$\psi(\vec{x})=\sum_{s=1}^{2}\int{\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{p}}}}\left(b_{\vec{p}}^su^s(\vec{p})e^{i\vec{p}\vec{x}}+c_{\vec{p}}^{s\dagger}\upsilon^s(\vec{p})e^{-i\vec{p}\vec{x}}\right)}$$
Then, the notes I am studying suggest that:
$$\psi^\dagger(\vec{x})=\sum_{s=1}^{2}\int{\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{p}}}}\left(b_{\vec{p}}^{s\dagger}u^{s\dagger}(\vec{p})e^{-i\vec{p}\vec{x}}+c_{\vec{p}}^s\upsilon^{s\dagger}(\vec{p})e^{i\vec{p}\vec{x}}\right)}$$
It looks like the creation/annihilation operator commutes with the spinors so that for example
$$(b_{\vec{p}}^{s}u^{s}(\vec{p}))^{\dagger}=u^{s\dagger}(\vec{p})b_{\vec{p}}^{s\dagger}=b_{\vec{p}}^{s\dagger}u^{s\dagger}(\vec{p}).$$
Does this happen because the creation and annihilation operators act on states $|p\rangle$, and not on spinors?
Best Answer
The spinors $u^s(p), v^s(p)$ are not operators, they are just (an array of) numbers and therefore commute with the creation/annihilation operators.