What does it mean to expand an operator into a Fourier series? In quantum field theory it is quite common to talk about expanding an operator into a Fourier series. When this is done, at least in the cases I've seen, the "Fourier coefficients" are the creation and annihilation operators.
Now, this is something I find quite confusing. We know about expanding functions into Fourier series. That is, we consider functions $f\in L^2(S^1)$ and we can expand them into Fourier series.
Even for those functions the theory is quite complicated, if one goes into detail into analyzing the convergence of the series and the relation between the series and the original function. Nonetheless, at least in the $L^2(S^1)$ norm everything works fine.
Now, given a Hilbert space $\mathcal{E}$ that we use in QM, what does it mean to expand an operator into a Fourier series? How is this rigorously formulated?
 A: It is straightforwardly possible to generalize the notion of a Fourier transform of scalar functions $\mathbb{R}^n\to\mathbb{C}$ to Fourier transforms of functions $\mathbb{R}^n\to E$, where $E$ is a Banach space. The notion of integration chosen should be the Bochner integral or something equivalent. However, this is not the correct way to do it in this case, since rigorously the expression $[\phi(x),\pi(y)] = \mathrm{i}\delta(x-y)$ doesn't make any sense, since $\delta$ is not a function, but merely a distribution. Moreover, we don't have the guarantee that the field operators are bounded, so we are not guaranteed that the $E$ would actually be a Banach space. So we first need to fix the definition of the quantum field:
A (Wightman) quantum field is an operator-valued distrbution on Schwartz space $\mathscr{S}(\mathbb{R}^d)$, i.e. a linear map $\phi : \mathscr{S}(\mathbb{R}^d)\to\mathcal{O}(\mathcal{H})$ where $\mathcal{O}(\mathcal{H})$ are the operators on our Hilbert space of states. Now we simply apply the usual definition that the Fourier transform of a distribution is the distribution applied to the Fourier transform of the test function, that is $\mathscr{F}(\phi(f)) := \phi(\mathscr{F}(f))$ for $\mathscr{F}$ the Fourier transform operator on $\mathscr{S}(\mathbb{R}^d)$. Since $f$ is an ordinary function, its Fourier transform is perfectly well-defined.
