How to take hermitian conjugate of an operator containing multiple elements? The annihilation operator in the Dirac field could be written as
$$
a_p^s = \frac{e^{iE_pt}}{\sqrt{2E_p}}u^s(p)^\dagger\int d^3xe^{-ipx}\psi(t,x)
$$
Where
\begin{equation*}
\begin{split}
\psi(t, x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_{s = 1,2}\left[a_p^su^s(p)e^{-ipt}+ b_{-p}^{s\dagger}v^s(-p)e^{ipt}\right]e^{ipx}
\end{split}
\end{equation*}
My question is how do we find $a_p^{s\dagger}$? For operators $A, B, C$, we have $(ABC)^{\dagger} = C^{\dagger}B^{\dagger}A^{\dagger}$. Can we split $a^\dagger$ into 4 parts and use this identity to get
$$
a_p^{s\dagger} = \int d^3x\psi^{\dagger}(t,x)e^{+ipx}u^s(p)\frac{e^{-iE_pt}}{\sqrt{2E_p}}
$$
 A: Your final formula is right, but I'm confused why you used $(ABC)^\dagger = C^\dagger B^\dagger A^\dagger$ since there is just a product of two objects, namely $u^s(p)^\dagger \psi(x)$. Moreover, it is important to justify the formula here, because the dagger on $u^s(p)$ is just the $\mathbb{C}^4$ dagger while on $\psi(x)$ it is both the $\mathbb{C}^4$ dagger and the Hilbert space dagger combined.
In other words, I believe your problem is that an object like $\psi(x)$ is both a $\mathbb{C}^4$ column vector and an operator in a Hilbert space. A simple way to proceed is to observe that in $u^s(p)^\dagger \psi(x)$ the dagger is the $\mathbb{C}^4$ one. Now $u^s(p)^\dagger$ is a row vector whose entries are numbers, while $\psi(x)$ is a column vector whose elements are operators. The matrix product $u^s(p)^\dagger \psi(x)$ is therefore a single operator.
Let us denote $u^s(p) = (u^s_\alpha(p))$ where $\alpha$ is the $\mathbb{C}^4$ index and also $\psi(x) = (\psi_\alpha(x))$. Then we have $$u^s(p)^\dagger \psi(x)=\sum_\alpha u^s_\alpha(p)^\ast \psi_\alpha(x).$$
Now it is easy to take the adjoint of this operator
$$[u^s(p)^\dagger \psi(x)]^\dagger=\sum_\alpha u_\alpha^s(p) \psi_\alpha^\dagger(x)=\psi^\dagger(x)u^s(p).$$
Everything else on the $a_p^s$ formula are numbers, so that the adjoint becomes just the complex conjugate. Indeed rewrite $a_p^s$ by bringing $u^s(p)^\dagger$ into the integral
$$a_p^s =  \frac{e^{iE_pt}}{\sqrt{2E_p}}\int d^3xe^{-ipx}u^s(p)^\dagger\psi(t,x).
$$
Now take the adjoint. It is
$$(a_p^s)^\dagger =  \frac{e^{-iE_pt}}{\sqrt{2E_p}}\int d^3xe^{ipx}[u^s(p)^\dagger\psi(t,x)]^\dagger=\frac{e^{-iE_pt}}{\sqrt{2E_p}}\int d^3xe^{ipx}\psi^\dagger(x)u^s(p),
$$
as you proposed.
