Given a massive free scalar field. We can define the quantum *-algebra of observables as a subset $\mathcal{E}\subseteq C^{\infty}(\mathcal{F})$ given by $$ \mathcal{E}:= \lbrace 1,a[f],\overline{a[g]},H \hspace{2mm}|\hspace{2mm} a:= \frac{1}{\sqrt{2\hbar \kappa}}(\sqrt{w}\phi-i\kappa\frac{1}{\sqrt{w}}\pi), \hspace{2mm} \forall f,g \in \mathcal{S(\mathbb{R}^D)}\rbrace, \hspace{1cm} (1) $$ where $\mathcal{F}$ corresponds to the phase space, $\mathcal{S}$ is the Schwarz space and $a[f]$ is defined by $a[f]:=\int d^Dx \overline{f}(x)a(x)$.
We can check that this is indeed an algebra. For the property of closeness under its symplectic structure we can calculate the following brackets: \begin{equation} \lbrace a[f],\overline{a[g]}\rbrace=\frac{1}{i\hbar}<f,g>_{L_2}. \hspace{2cm} (2) \end{equation}
The details of the calculation of this bracket are unimportant for my question. For the normal ordered Hamiltonian $:H:=\hbar c\int d^Dx\overline{a}(x)(\omega a)(x)$ where $\omega:=-\Delta+m^2$ we can calculate the bracket $$ \lbrace H,a[f]\rbrace = \hbar c\int d^Dx \lbrace \overline{a}(x),a[f]\rbrace(\omega a)(x) = ic \int d^Dx <\delta_x,f>(\omega a)(x)= ic\int d^Dx (\overline{f}\omega a)(x) = ic \int d^Dx (\overline{\omega f}a)(x)=ic <wf,a>_{L_2} = ic \hspace{1mm} a[\omega f]. \hspace{2cm} (3) $$
Now, using the canonical commutation relations (2) and getting inspired by the calculation (3) I would like to compute the following integral $$ \int d^Dx (\bar{a}^n\lbrace\bar{a}(x),a[f]\rbrace\bar{a}^m a^k)(x)= \frac{i}{\hbar}\int d^Dx (\bar{a}^n\bar{f}\bar{a}^m a^k)(x), $$
where $m,n,k >0$. Now I would like to pass the function to the left of the operators so I can write the integral in the form $<f,\bar{a}^n\bar{a}^m a^{k}>_{L_2}=\bar{a}^n\bar{a}^m a^{k}[f]$. My question is, can I commute $\bar{a}^n\bar{f}=\bar{f}\bar{a}^n$? Or do I need to do what I did for the commutation of $\bar{f}\omega$ in equation (2). I mean something like $\bar{a}^n\bar{f}=\overline{f a^n}$? If so, then I don't know how to get something like $\bar{a}^n\bar{a}^m a^{k}[f]$ from the integral.
Since $f$ is a function my guess is that I can commute with the operator without complex conjugation issues but I am not quite sure.
I hope you can help me to clarify this.