I have the following question regarding self-adjoint operator and I am not quite sure about its answer. Let $A_{n}: \operatorname{Dom}(A_{n})\subset H \rightarrow H$ be symmetric operator densely defined on a Hilbert space $H$ that has some self-adjoint extension due to the so-called Von Neumann theorem. Consider another operator $B_{c}: \operatorname{Dom}(B_{c}) \subset H \rightarrow H$ be symmetric operator densely defined on a Hilbert space $H$ which also has some self-adjoint extension, again due to Von Neumann theorem. In both cases, the dimension of the deficiency spaces is $n_{+}=n_{-}=1$. On the other hand, we can also write $B_{c}=A_{n}+ \delta B_{c}$ or equivalently, $\delta B_{c}=B_{c}-A_{n}$. I know that in principle $\delta B_{c}$ does not necessarily have a self-adjoint extension. Questions that I am interested in :
1-How can I prove that $\delta B_{c}$ is not self-adjoint? Or what are the conditions to become self-adjoint?
2-Let us suppose that $\delta B_{c} \phi=0$ with $\phi \in \operatorname{Dom}(B_{c})$. Can I conclude that the spectrum of the operator $B_{c}$ must be equal to the spectrum of $A_{n}$? So, in the end, both problems are equivalent. Is this correct?
In order to be much clear, I will re-write this problem in another way. Let us consider that we have two problems associated with Schrodinger-like equations in different patches of coordinates:
1-The first problem is given by an operator $A_{n}: Dom (A_{n}) \subset H \rightarrow H$ densely defined on a Hilbert space and its domain is given by $Dom(A_{n})= \{ \phi \in H^{2}(0, \infty)/ \phi(0)=\phi'(0)=\phi(\infty)=\phi'(\infty)=0 \}$ being $H^{2}$ the usual sobolev space and the prime stands for derivative with respect to the coordinate $x_{n} \in [0, \infty)$.
The first operator is given by $A_{n}= -\frac{d^{2}}{dx^{2}_{n}} + V_{n}[x_{n}]$ and the potential energy operator is
$V=\frac{e^{x_{n}/r_{0}}(9e^{x_{n}/r_{0}} -10)}{4r^{2}_{0}(9e^{x_{n}/r_{0}} -1)^{2}}$.
Here $r_{0}$ is just a positive constant. This operator has some self-adjoint extensions due to Von Neumann theorem. The dimension of the deficiency subspaces is $n_{+}=n_{-}=1$. The map between both subspaces $N_{+}$ and $N_{-}$ is given by an isometry and can be parametrized by a complex number.
2-The second problem is given by another operator $B_{c}: Dom (B_{c}) \subset H \rightarrow H$ densely defined on a Hilbert space and its domain is given by $Dom(B_{c})= \{ \psi \in H^{2}(-\infty, 0)/ \psi(0)=\psi(-\infty)=\psi'(-\infty)=0 \}$ being $H^{2}$ the usual sobolev space again and the prime stands for derivative with respect to another coordinate $x_{c} \in (-\infty, 0)$.
The second operator is given by $B_{c}= -\frac{(e^{-\mid b \mid x_{c}}-1)}{r^{2}_{0}b^{2}}\frac{d^{2}}{dx^{2}_{c}} + V_{c}[x_{c}]$, where the potential energy operator is $V=2\frac{e^{-\mid b\mid x_{c}}}{r^{2}_{0}}$. Here $b<0$. Besides, the second operator also has some self-adjoint extensions due to Von Neumann theorem. The dimension of the deficiency subspaces is $n_{+}=n_{-}=1$. The map between both subspaces $N_{+}$ and $N_{-}$ is given by an isometry and can be paramtrized by a complex number.
3-The map between the $x_{n}$ coordinate and the $x_{c}$ coodinate is given by $x_{c}=\frac{1}{\mid b \mid} \log (1- e^{-\frac{x_{n}}{r_{0}}})$, where $x_{n} \in [0, \infty)$ and $x_{c} \in (-\infty,0)$.
4-I started from the $B_{c}$ operator and used the chain rule to connect with the $A_n$ operator. I also had to add and substruct a the term $V_{n}$.
5-The rest of the problem is the same thing that I posted before.
Any help with these issues would be much appreciated.