I am currently working with an operator fulfilling $$T:\mathcal{D} \to \mathcal{D}^\perp.$$ That means, it is "trivially" symmetric, i.e., $\langle \psi, T \phi \rangle = 0$ for $\psi,\phi \in \mathcal{D}$.
The first question can be answered without much detail:
- Is it a fact that $\mathcal{D}^\perp$ is not dense? And does it in turn imply that $T$ is not closable?
Now to add more detail, the operator acts on the symmtrized Fock space $\mathcal{H} = \bigoplus_{n=0}^\infty P_n(L^2(\mathbb{R})^{\otimes n})$ with $P_n$ symmetrization, and increases and decreases particle number (i.e., $(T\psi_n) = (T\psi)_{n-1} + (T\psi)_{n+1}$) by acting as multiplication with a dirac delta:
$$ (T\psi)_{n-1}(x_1, .., x_{n-1}) = \psi_n(x_1, x_1, x_2, .., x_{n-1}) $$ and $$ (T\psi)_{n+1}(x_1, .., x_{n+1}) = \delta(x_1 - x_2) \psi_n(x_1, x_3, .., x_{n+1}).$$ The domain are all $L^2$ functions with $\psi(x_1, .., x_n) =0$ if $x_i = x_j$ for $i\neq j$.
Now to my second and third question:
- Is it correct that $\mathcal{D}$ is dense in $\mathcal{H}$?
And most importantly:
- Can $T$ even be defined as an operator on $\mathcal{H}$, as its Dirac delta part does not map into $\mathcal{H}$? Maybe through expectation values of $T$ which are at least well-defined finite.
I know this question is a bit fuzzy, but I tried to formulate some clear questions.
Apart from that, I would be happy to know what if $T$ could be defined as an operator (under certain assumptions) and whether something of that sort has been studied already. Thanks for any help!