Very basic question about quantum field operators For a matrix $A$, the notation $A^\dagger$ implies the transpose of the complex conjugate of $A$ i.e., $A^\dagger=(A^*)^T$. 
What does the symbol $\hat{\phi}^\dagger$ mean for a quantum operator corresponding to a classical field $\phi(x)$? Is it okay to think of $\hat{\phi}(x)$ as an infinite dimensional column vector and $\hat{\phi}^\dagger$ as a row vector with $\hat{\phi}^\dagger=(\hat{\phi}^*)^T$?
However, there are two problems that I can immediately see. 
1. Operators in ordinary quantum mechanics are square matrices while (if my representation is valid) $\hat{\phi},\hat{\phi}^\dagger$ are column and row vectors. 
2. For a complex scalar field $$[\hat{\phi}(t,\textbf{x}),\hat{\phi}^\dagger(t,\textbf{y})]=0\implies \hat{\phi}(t,\textbf{x})\hat{\phi}^\dagger(t,\textbf{y})=\hat{\phi}^\dagger(t,\textbf{y})\hat{\phi}(t,\textbf{x}).$$ If my representation is valid, this equation becomes meaningless because on one side we have number and on the other side we have a matrix.
What is is the correct way to visualize quantum field and interpret the commutation relation? 
 A: Don't think about matrices. Let $\mathcal{H}$ be a Hilbert space, then one linear operator in $\mathcal{H}$ is a function $A : D(A)\subset \mathcal{H}\to \mathcal{H}$ which satisfies
$$A(\alpha v+\beta w)=\alpha A(v)+\beta A(w),\quad\forall\alpha,\beta\in\mathbb{C},v,w\in\mathcal{H}.$$
This isn't a matrix. The adjoint $A^\dagger$ is defined as the unique linear operator in $\mathcal{H}$ satisfying:
$$\langle Av,w\rangle=\langle v,A^\dagger w\rangle,\quad v,w\in \mathcal{H}.$$
It is also not a matrix. When you introduce a basis on $\mathcal{H}$ you can associate a matrix to $A$ and $A^\dagger$ inasmuch as every other operator, but matrices are matrices and operators are operators. It is then a result that the matrices of $A$, $A^\dagger$ are related by
$$[A^\dagger]=([A]^\ast)^T.$$
Now, in QFT field are operator valued distributions. For now think of them as actual functions on spacetime. Thus for any event $x\in M$, a quantum field $\phi$ associates an operator in some Hilbert space $\mathcal{H}$ denoted $\phi(x)$.
Since $\phi(x)$ is an operator in some Hilbert space it has one adjoint $\phi(x)^\dagger$ in the exact same manner as I said above.
By the way, when you use the Fock space representation (which is the standard one in QFT textbooks) you have that by definition
$$\phi(x)=\int \dfrac{d^3k}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_k}}(a_ke^{-ikx}+a_k^\dagger e^{ikx})$$
where $a_k$ and $a_k^\dagger$ are creation and annihilation operators in a certain Fock space (which is a space of states of a system of a variable number of particles).
Interestingly now $a_k$ has the meaning of removing a particle with momentum $k$ from a state, $a_k^\dagger$ has the meaning of creating a particle with momentum $k$ and this is due to the commutation relations they should satisfy in order for the Quantum Field satisfy the Canonical Commutation relations, namely
$$[a_k,a_q]=0,\quad [a_k^\dagger,a_q^\dagger]=0,\quad [a_k,a_q^\dagger]=(2\pi)^3 \delta(k-q).$$
But this is a particular case. Adjoints are adjoints, as defined above in the mathematical setting of Hilbert spaces.
A: That for a matrix the dagger denotes the transpose conjugate is really just a special (namely the finite-dimensional) case of the general definition of the Hermitian adjoint:
For any operator $A$ on a Hilbert space $H$, the adjoint $A^\dagger$ is the operator such that
$$ \langle v, Aw\rangle = \langle A^\dagger v,w\rangle$$
for all $v,w$ in the domain of definition of $A$.
Since for any quantum field $\phi$, $\phi(x)$ is an operator (neglecting the case where we treat the field as an operator-valued distribution rather than a function, replace $\phi(x)$ with $\phi(f)$ for some test function $f$ in that case), there is no problem in applying this definition to a quantum field.
A: The dagger $\dagger$ can have two meanings. 
First, there is the linear algebra meaning. We use $\dagger$ to denote a Hermitian conjugate, that is transposition and complex conjugation. 
Second, there is the quantum field operator meaning. Here, $\dagger$ is used to change a creation operator into an annihilation operator $-$ and vice versa. 

To answer your question: if $\phi(x)$ is a scalar quantum field, then $\phi^\dagger(x)$ has the QFT-meaning. 


Fun fact: the only ambiguity arises when dealing with spinors, 
$$
  \chi = \begin{pmatrix} \chi_1 \\ \chi_2\end{pmatrix},
$$
because what does $\chi^\dagger$ mean? Is it the quantum field operator interpretation, 
$$
  \chi^{\dagger,LA} = \begin{pmatrix} \chi_1^\dagger \\ \chi_2^\dagger \end{pmatrix},
$$
or is the Hermitian conjugate applied in both the operator and the linear algebra sense, as in 
$$
  \chi^{\dagger,QFT+LA} = \begin{pmatrix} \chi_1^\dagger , \chi_2^\dagger \end{pmatrix}?
$$
This depends on the convention chosen by the author. 
