# Confusion about Dirac field operators

I have a little confusion about Dirac field operator. Field operator can be written as $$\psi(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s(a^s_pu^s(p)e^{-ip\cdot x}+b^{s\dagger}_pv^s(p)e^{ip\cdot x}).$$ So I thought that Hermitian conjugate would be $$\psi^{\dagger}(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s(u^{s\dagger}(p)a^{s\dagger}_pe^{ip\cdot x}+v^{s\dagger}(p)b^{s}_pe^{-ip\cdot x}),$$ but the book says $$\psi^{\dagger}(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s(a^{s\dagger}_pu^{s\dagger}(p)e^{ip\cdot x}+b^{s}_pv^{s\dagger}(p)e^{-ip\cdot x}).$$ How can I resolve this?

• They are the same expression. What is the problem? Jul 8, 2021 at 15:28

The order of $$a_p^{s\dagger}$$ and $$u^{s\dagger}$$ doesn't matter because $$a_p^{s\dagger}$$ doesn't act on $$u^{s\dagger}$$. $$u^{s\dagger}$$ is a finite dimensional vector, whereas $$a_p^{s\dagger}$$ is an operator that acts on the infinite dimensional space of field states.
• Thanks for the answer. Then the commutator will be $$[\psi,\psi^{\dagger}]=\int\frac{d^3pd^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_pE_q}}e^{i(p\cdot x-q\cdot y)}\sum_{r,s}(a^r_pu^r(p)a^{s\dagger}_pu^{s\dagger}(p)-a^{s\dagger}_pu^{s\dagger}(p)a^r_pu^r(p)+\cdot\cdot\cdot)$$ How does it become $$\int\frac{d^3pd^3q}{(2\pi)^6}\frac{1}{\sqrt{4E_pE_q}}e^{i(p\cdot x-q\cdot y)}\sum_{r,s}([a^r_p,a^{s\dagger}_p]u^r(p)u^{s\dagger}(p)+\cdot\cdot\cdot)?$$ Jul 8, 2021 at 15:58