# Mapping Issues with Unbounded Operators

Consider the operator-valued generalized function $\phi^{(k)}_{l}:=\phi^{(k)}_{l}$ on space-time $\mathcal{M}$. Now, smooth the operator-valued generalized function with test function $f(x)$ s.t.$$\phi^{(k)}_{l}(f):=\int \!\mathrm{d}^4x\, \phi^{(k)}_{l}(x)f(x)$$ are linear operators in the physical Hilbert space $\mathcal{H}$.

Note: These operators are not assumed to be bounded.

Instead, we suppose that all the operators $\phi^{(k)}_{l}$ have a common domain of definition $\mathcal{D}$, which is a dense linear subspace of $\mathcal{H}$. In order for algebraic operations on the operators to make sense, it is usually further supposed that the operator-valued generalized functions map $\mathcal{D}$ into itself.

It is the last assumption that I will be asking about. In particular what goes awry if the map doesn't take $\mathcal{D}$ into itself? I am interested in both mathematical reasoning, as well as any physical motivation behind this assumption.

• You could not apply the operators in succession if it were not guaranteed that they produce states that lie within their mutual domain of definition. Commented Sep 14, 2014 at 15:17

The assumption that $\mathcal{D}$ is invariant under $\phi(f)$ for each $f\in \mathcal{S}(\mathbb{R^4})$, the Schwartz space of functions of rapid decrease is one of the Wightman axioms. Its main use is for the vacuum expextation values $(\psi_0,\phi(f_1)...\phi(f_n)\psi_0)$ to make sense (where $\psi_0$ is the vacuum state), what would not happen in general if the domain wasn't invariant. See the section IX.8 of Reed and Simon for more details.