Consider the real Klein-Gordon scalar;
Taking the Hermitian Hamiltonian, and (spatial) momentum operators as
$$\hat H = \int \frac{d^3 \vec p}{\left(2 \pi\right)^32E(\vec p)}E(\vec p)\hat a^\dagger(\vec p)\hat a(\vec p)\tag{1}$$ $$\hat P^i = \int \frac{d^3 \vec p}{\left(2 \pi\right)^32E(\vec p)}p^i\hat a^\dagger(\vec p)\hat a(\vec p)\tag{2}$$ and writing the total 4-momentum operator $\hat P^\mu=\left(\hat H, \hat{P}^i\right)$, show that these commute with the ladder operators as
$$\left[\hat P^\mu,\, \hat a(\vec q)\right]=-q^\mu \hat a(\vec q)\tag{A}$$ and $$\left[\hat P^\mu,\, \hat a^\dagger(\vec q)\right]=q^\mu \hat a^\dagger(\vec q)\tag{B}$$ where $q^\mu=\left(E(\vec q),\,\vec q\right)$
After a long calculation (which I would prefer not to typeset) I can show the first relation, $(\mathrm{A})$. I can also show that $$\left[\hat H,\, \hat a(\vec q)\right]=-E(\vec q)\hat a(\vec q)\tag{3}$$ and $$\left[\hat P^i,\, \hat a(\vec q)\right]=-q^i\hat a(\vec q)\tag{4}$$
Now instead of repeating a similar calculation to show $(\mathrm{B})$ I would like to use $(\mathrm{A})$ instead and use the fact that $$\left[\hat P^\mu,\, \hat a(\vec q)\right]=\left[\hat H +\hat P^i,\, \hat a(\vec q)\right]$$ $$=\left[\hat H,\, \hat a(\vec q)\right]\stackrel{\color{red}{?}}{+}\left[\hat P^i,\, \hat a(\vec q)\right]=-E(\vec q)\hat a(\vec q)-q^i\hat a(\vec q)\stackrel{\color{red}{?}}{=}-q^\mu \hat a(\vec q)\tag{5}$$
Taking the Hermitian adjoint of both sides of $(5)$, $$\left[\hat P^\mu,\, \hat a(\vec q)\right]^\dagger=\left(\left[\hat H,\, \hat a(\vec q)\right]+\left[\hat P^i,\, \hat a(\vec q)\right]\right)^\dagger$$ $$=\left[\hat H,\, \hat a(\vec q)\right]^\dagger+\left[\hat P^i,\, \hat a(\vec q)\right]^\dagger$$ $$=\left(\hat H\hat a(\vec q)-\hat a(\vec q)\hat H\right)^\dagger+\left(\hat P^i\hat a(\vec q)-\hat a(\vec q)\hat P^i\right)^\dagger$$ $$=\left(\hat H\hat a(\vec q)\right)^\dagger-\left(\hat a(\vec q)\hat H\right)^\dagger+\left(\hat P^i\hat a(\vec q)\right)^\dagger-\left(\hat a(\vec q)\hat P^i\right)^\dagger$$ $$=\left(\hat a^\dagger(\vec q)\hat H -\hat H\hat a^\dagger(\vec q)\right)+\left(\hat a^\dagger(\vec q)\hat P^i -\hat P^i\hat a^\dagger(\vec q)\right)$$ $$=\left[\hat a^\dagger(\vec q),\,\hat H\right]+\left[\hat a^\dagger(\vec q),\,\hat P^i\right]=-\left(\left[\hat H,\,\hat a^\dagger(\vec q)\right]+\left[\hat P^i,\,\hat a^\dagger(\vec q)\right]\right)$$ $$=-\left(E(\vec q)\hat a(\vec q)\right)^\dagger-\left({\color{red}{q^i}}\hat a(\vec q)\right)^\dagger=-E(\vec q)\hat a^\dagger(\vec q)-\hat a^\dagger(\vec q){\color{red}{q^i}^\dagger}\ne q^\mu \hat a^\dagger(\vec q)\tag{6}$$
I've written many steps on purpose to show that it is the presence of the 3-momentum vector, $q^i$, colored red that is causing the problem. Typically in QM/QFT the Hermitian adjoint, $\dagger$, acts as an invertible map between the bras and kets as elements of a Hilbert space, which, in Dirac notation would look like for states $|\phi\rangle$, $|\psi\rangle$ $$|\phi\rangle=\hat O|\psi\rangle\overbrace{\iff}^\dagger\langle\phi|=\langle\psi|\hat O^\dagger\tag{7}$$
Here is the problem: I do not know how to interpret the notation of this $q^i$ object, is it a vector or a vector component? I'm confused as physicists sometimes abuse notation and write a 4-vector as $P^\mu=\left(p^0,\,p^i\right)$ or $P^\mu=\left(p^0,\,\vec p\right)$ which I find very annoying.
I don't understand whether to switch the order and take the Hermitian adjoint for the second term in parentheses after the first equality (shown in red) of eqn. $(6)$ or, treat $q^i$ as a component of a vector and not switch the order.
In short, how does one know when to interpret say $x^j$ as a component or a 3-vector and hence when to switch the order when taking the Hermitian adjoint?
Update
Now a comment and an answer both say that the notation $p^i$ is a vector component (the ith component) and not a vector. That said, why am I still seeing statements like the following:
Hence the confusion. Sorry I don't have a source for the extract above.
I've been told in a comment by @Gonçalo that $\hat{P}^\mu\neq\hat H +\hat P^i$, which means that the first line of my eqn. $(5)$ is incorrect. For the second line of eqn. $(5)$ I have put question marks above the parts for which I am unsure. Basically, I don't think I can add commutators like that to get the result of eqn. $(\mathrm{A})$, namely, $$\left[\hat P^\mu,\, \hat a(\vec q)\right]=-q^\mu\hat a(\vec q)$$ but I don't know how else to arrive at the result.
Any help would be most appreciated, thanks.
There is a sign error in the last line of $(6)$.
$q^i$ are real-valued scalar coordinates and $E(q^1, q^2, q^3) = \left((q^1)^2+(q^2)^2+(q^3)^2+m^2\right)^{1/2}$ is also a real-valued scalar.
For any bounded operators $A,B$ on a Hilbert space $$[A,B]^\dagger = (AB-BA)^\dagger=B^\dagger A^\dagger - A^\dagger B^\dagger = [B^\dagger,A^\dagger]=-[A^\dagger,B^\dagger]$$
Therefore $$ \begin{split} \left[\hat H,\, \hat a(\vec q)\right]^\dagger = \left[\hat a^\dagger(\vec q),\,\hat H^\dagger\right] &= \left[\hat a^\dagger(\vec q),\,\hat H\right] = -\left[\hat H,\,\hat a^\dagger(\vec q)\right]\\ \left[\hat P^i,\, \hat a(\vec q)\right]^\dagger = \left[\hat a^\dagger(\vec q),\,\hat P^{i\dagger}\right] &= \left[\hat a^\dagger(\vec q),\,\hat P^i\right] = -\left[\hat P^i,\,\hat a^\dagger(\vec q)\right] \end{split} $$ on account of $\hat H, \hat P^i$ being self-adjoint.
So from $(3), (4)$, $$ \begin{split} -\left[\hat H,\,\hat a^\dagger(\vec q)\right] &= (-E(\vec q)\hat a(\vec q))^\dagger =-E(\vec q)\hat a^\dagger(\vec q)\\ -\left[\hat P^i,\,\hat a^\dagger(\vec q)\right] &= (-q^i\hat a(\vec q))^\dagger=-q^i\hat a^\dagger(\vec q) \end{split} $$ which is $(B)$ after multiplying by $-1$.