# 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?

• Covergence of series of operators is defined exactly the same way as for functions. You need need a definition of distance. Then you make a sequence of partial sums of the series. Then you show that the sequence of distances between partial sums and the intended result has a limit of zero. What exactly do you want to know? Do you want to know how to define the distance between operators? In any case, this question might be better off in the Math site, since it doesn't seem like anything here is actually a physics question. – DanielSank Oct 17 '16 at 21:46
• @DanielSank, in the case, what confuses me most is that usually when we do Fourier series for functions we consider functions on the circle, that is, they are periodic. Also, for functions there are lots of complications, since one can consider the convergence of the partial sums, the convergence of the Cesàro means, the convergence of the Abel means, or the convergence in the $L^2$ norm. I wanted to know how this works out for operators, but specifically for the case we use in QFT. Perhaps a general theory, as I believe would be presented in Math, would not be directly related to QFT. – user1620696 Oct 17 '16 at 21:50
• The short answer is that one has to regard a quantum field as an operator-valued distribution on a certain space of test functions. Then Fourier transform is then defined by transposition. The long and well-detailed answer is in Streater-Wightman. – Phoenix87 Oct 17 '16 at 21:53
• I think this is one of those cases where the boundary between advanced theoretical physics and math is fuzzy enough that this is fine here. – David Z Oct 17 '16 at 21:53
• Let me remind everyone that answers should be posted as answers, and comments should be reserved for improving the question.... – David Z Oct 17 '16 at 21:54

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.