What are the spaces in which quantum fields belong and how does that affect the hermitian conjugate of $\partial_{\mu}$?

Question

In this post, I claimed in my answer that $\partial_{\mu}^{\dagger}=\partial_{\mu}$. The reason I claimed that is because $D^{\dagger}_{\mu}=\partial_{\mu}-iqA_{\mu}$ and I assumed that

$$(\partial_{\mu}+iqA_{\mu})^{\dagger}=\partial_{\mu}^{\dagger}+(iqA_{\mu})^{\dagger}=\partial_{\mu}-iqA_{\mu}.$$

(1) Is this assumption correct?

However, a commenter claimed that $\partial_{\mu}^{\dagger}=\partial_{\mu}$ is not true, since the Klein-Gordon field belongs to $L^2(\mathbb{R}^{1,3})$, and therefore

$$\langle f | \partial_\mu g \rangle=\langle f |\partial_\mu | g \rangle =-\langle \partial_\mu f | g \rangle=- \langle f |\partial_\mu^\dagger | g \rangle,$$

where in the second equality we did integration by parts.

(2) Is the commenter is right? According to this equation, does that mean that $\partial_{\mu}=-\partial_{\mu}^{\dagger}$? Unfortunately, I'm not familiar at all with the concept of $L^p$ spaces and therefore I can't judge this on my own.

Additionally, I found this post, in which the answer claims that it does not make sense to take the hermitian conjugate of $\partial_{\mu}$.

(3)If that's the case, how do you show that $D^{\dagger}_{\mu}=\partial_{\mu}-iqA_{\mu}$?

But then again, in this post, the first answer by Javier states that

The vector space (that is, spinor space) being considered here is $\mathbb{C}^4$

and therefore the hermitian adjoint does not affect $\partial_{\mu}$

(4) This seems to be in contradiction with the previous post. What am I missing? Does it make sense to take the hermitian adjoint of $\partial_{\mu}$ or not?

This is all too confusing!

(5) I would like to know which fields belong to which space (like $L^2$ or $\mathbb{C}^2$) and what the consequences of that are. Some that come to mind would be: a real scalar field, a complex scalar field, the Klein-Gordon field, the Dirac field, the vector field $A_{\mu}$ etc. I own a couple of QFT books and I've never seen this pure mathematical aspect.

Answer 1

First, let's consider the derivative $\partial$ as an operator on the Hilbert space $\mathcal{H} = L^2(\mathbb{R}^n)$ of square-integrable functions on $\mathbb{R}^n$ (the space typically encountered in quantum mechanics). Because not all $L^2$-functions are differentiable, we restrict $\partial$ to the (dense) subspace $C_c^\infty(\mathbb{R}^n) \subset L^2(\mathbb{R}^n)$ of smooth compactly supported functions. A simple integration by parts argument leads to the following identity for all $f,g \in C_c^\infty(\mathbb{R}^n)$: $$\langle f, \partial g \rangle = \int_{\mathbb{R}^n} \overline{f(x)}\partial g(x) \mathrm{d} x = -\int_{\mathbb{R}^n} \overline{\partial f(x)} g(x) \mathrm{d} x = - \langle \partial f, g \rangle. $$ The boundary terms appearing after integrating by parts vanish because $f,g$ are compactly supported. In this sense, $\partial^\dagger f = -\partial f$ for all $f \in C_c^\infty(\mathbb{R}^n)$.

Remark: The momentum operator $P_0=-\mathrm{i} \partial$ is symmetric on $C_c^\infty(\mathbb{R}^n)$, and has a self-adjoint extension $P$ (i.e. $P^\dagger = P$) that is defined on the Sobolev space $H^1(\mathbb{R}^n) \subset L^2(\mathbb{R}^n)$ (basically the space of all $L^2$-functions for which the first derivative exists in a weak sense).

A quantum field $\phi(x)$ is, according to the Wightman axioms, an operator-valued distribution. This means that for every test-function $f$ (usually a Schwartz space function), $$\phi(f) := \int_{\mathbb{R}^n} f(x) \phi(x) \ \mathrm{d} x$$ defines an operator on $\mathcal{H}$. The operator-valued distribution $\partial \phi$ is defined as the map that sends a test function $f$ to $\partial\phi(f) = -\phi(\partial f)$. Thus, $\partial \phi$ is not the composition of two operators (the derivative and the quantum field), but another operator-valued distribution.

Lecture notes that discuss some aspects of the mathematical framework of QFT: https://wdybalski.faculty.wmi.amu.edu.pl/Notes-QFT41.pdf