While doing QFT when we try to canonically quantize the Klein Gordon equation $\Box \phi =0$ we promote the $\phi $ to an operator field and impose the commutation rule $[\phi(x,t),\pi (y,t)]=i\hbar\delta^3(x-y)$.
So my doubt is following when we try to do expansion of $\phi$ in terms of $f(x,t)$ which are c-numbered scalar field and satisfy $\Box f=0$. All of sudden the creation and annihilation operator pops up i.e. $\phi= \Sigma (af(x,t) + a^\dagger f(x,t))$ I am right now going for box normalization so the $\Sigma$ sign is present instead of $\int$ sign. Since the $\phi$ field is a hermitian field therefore it has to be such that $\phi ^\dagger = \phi$ and since $\phi$ is also an operator it's expansion has to involve operator so if go for operators $a$ and $a^\dagger$ such that $[a,a^\dagger]=\frac{1}{2}$ then we retrieve the original $\phi$ and $\pi$ commutation relation.
So is this the reason that above expansion is the way it is i.e. we guessed the most simple and consistent expansion there possible and it worked out.
Also is there any text in the literature which tackles with the above idea of expanding an operator field in terms of its equivalent c-numbered field ?