- #1
julian
Gold Member
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Could someone prove the following (if it is precisely correct):
Since Fock space is the closure of the finite linear span of finite excitations of the vacuum state $\Omega$, then the operator $\hat{O}$ is densely defined if and only if $\hat{O} \Omega$ has finite norm.
Or more generally, an operator is densely defined on a separable space if it is defined on a given single vector in that space.
thanks
Since Fock space is the closure of the finite linear span of finite excitations of the vacuum state $\Omega$, then the operator $\hat{O}$ is densely defined if and only if $\hat{O} \Omega$ has finite norm.
Or more generally, an operator is densely defined on a separable space if it is defined on a given single vector in that space.
thanks