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.

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    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Commented Sep 14, 2014 at 15:17

1 Answer 1


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.


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