I was wondering about the following: Let $T : dom(T) \subset H \rightarrow H$ be a self-adjoint operator, does this mean that the domain of $T$ is uniquely defined or is it possible to make the same operator self-adjoint on two different domains?

If anything is unclear, please let me know.

• With an operator $T$ comes its domain (just as for functions) $\text{dom}\, T$. On the other hand, one usually talks about (for example) the Laplace operator with Dirichlet boundary conditions and the Laplace operator with Neumann boundary conditions. These can both be defined as self-adjoint realisations, but they do not have the same domain, and thus they are not the same operator. – mickep Jan 23 '15 at 16:00

Just as an example, consider the operator $\frac{d^{2}}{dx^{2}}$ on the set $\mathscr{D}$ of all twice absolutely continuous functions $f \in L^{2}[0,2\pi]$ with $f'' \in L^{2}[0,2\pi]$. Then $T_{\alpha,\beta}=\frac{d^{2}}{dx^{2}}$ is selfadjoint on the domain $\mathcal{D}(T_{\alpha,\beta})$, $0 \le \alpha,\beta < \pi$ consisting of all $f \in \mathscr{D}$ for which $$\cos\alpha f(0)+\sin\alpha f'(0)=0,\\ \cos\beta f(2\pi)+\sin\beta f'(2\pi)=0.$$ That's a two parameter family of domains on which the same operator $\frac{d^{2}}{dx^{2}}$ is selfadjoint. It's also selfadjoint on the domain where $$f(0)=f(2\pi),\;\;\; f'(0)=f'(2\pi).$$ (There are other periodic types of conditions that work, too.) These are different operators, with a lot of spectral variation.

Is that what you had in mind?

For some reasons I read that you wanted to have $T$ self-adjoint on one domain and not-selfadjoint on the other.

If $T$ is densely defined (its domain is dense in $H$), then its adjoint exists. If moreover the adjoint is densely defined then $T$ is closable (there exists a closed operator which graph contains the graph of $T$). It may happen that its closure $\overline{T}$ (the smallest closed extension of $T$) is self-adjoint. From the inclusion of graphs we have that $\mathrm{dom}(T) \subseteq \mathrm{dom}(\overline{T})$ and $Tx=\overline{T}x$ for all $x \in \mathrm{dom}(T)$.

So $\overline{T}$ restricted to $\mathrm{dom}(T)$ equals $T$, and therefore on one domain is self-adjoint, but on the other (that is $\mathrm{dom}(T)$) it is just symmetric.

I hope that it answers your question. I can give you an example if you want to.

• no, i want the operator to be self-adjoint on BOTH domains! – user66906 Jan 23 '15 at 15:35
• Check Krein's theorem on non-negative self-adjoint extensions en.wikipedia.org/wiki/Friedrichs_extension – m_gnacik Jan 23 '15 at 15:54