Self-adjoint operator- domain unique? 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.
 A: 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?
A: 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. 
