I've been learning about self-adjoint extensions from volume 2 of the Reed and Simon book. I'm still quite a beginner at the material and I'm having trouble with even a simple example. To test my knowledge, I decided to try looking at the operator: \begin{equation*} H_0 = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \frac{d}{dx} + \left( \begin{array}{cc} m & 0 \\ 0 & -m \end{array} \right) \end{equation*} With domain $\mathcal{D}(H_0) = \left\{ \phi \in C^\infty_0(\mathbb{R} \backslash \left\{0 \right\}: \phi(0^-)=a, \phi(0^+) = b) \right\}$ and $\phi:\mathbb{R} \backslash \left\{ 0 \right\} \rightarrow \mathbb{C}^2$. I'd like to calculate all self-adjoint extensions of this operator via two different methods. One is via direct calculation and the other is by utilizing isometries. Here's my attempt with direct calculation: \
Suppose $\hat{H}$ is a symmetric extension of $H_0$. Then we have for all $\phi \in \mathcal{D}(\hat{H})$ and $\psi \in \mathcal{D}(\hat{H}^*)$, \begin{equation*} \langle \hat{H} \phi, \psi \rangle = \langle \phi, \hat{H}^* \psi \rangle \end{equation*} Recalling the definition of $H_0$ and simplifying results in: \begin{equation*} \left\langle \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \phi_x , \psi \right\rangle = \left\langle \phi, \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \psi_x \right\rangle \end{equation*} Appealing to integration by parts and canceling equivalent terms on either side of the equality results in: \begin{equation*} \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \phi \cdot \psi \bigg|_{0^-}^{0^+} =0 \end{equation*} Since $\mathcal{D}(\hat{H}) \subseteq \mathcal{D}(\hat{H}^*)$ we must have $\forall \phi = \left( \begin{array}{c} \phi_1 \\ \phi_2 \end{array} \right) \in \mathcal{D}(\hat{H})$, \begin{equation*} \left( \begin{array}{c} \phi_2 \\ -\phi_1 \end{array} \right) \cdot \left( \begin{array}{c} \phi_1 \\ \phi_2 \end{array} \right) \bigg|_{0^-}^{0^+} = 0 \end{equation*} This holds trivially for any choice of $\phi$ with defined limits at $0$. Thus this tells us no additional restrictions on the boundary conditions so I'm not sure what to do in this case. Thus I will look at the second method via isometries:
This method begins by determining the deficiency subspaces $\mathcal{H}_+(H_0)$ and $\mathcal{H}_-(H_0)$. Recall the definition $\mathcal{H}_-(H_0) = \ker(i+H_0^*)$. Thus we are interested in solving the following equation for $\phi \in \mathcal{D}(H_0)$: \begin{equation*} \begin{split} 0 =& \left( i + H_0^* \right) \left( \begin{array}{c} \phi_1 \\ \phi_2 \end{array} \right) \\ =& \left( \begin{array}{c} m+i \\ -m + i \end{array} \right) \left( \begin{array}{c} \phi_1 \\ \phi_2 \end{array} \right) + \left( \begin{array}{c} \phi_2' \\ - \phi_1' \end{array} \right) \end{split} \end{equation*} This system is easily solved via the substitution method. We get \begin{equation*} \mathcal{H}_-(H_0) = \left\{ \left( \begin{array}{c} c_1 e^{\sqrt{m^2+1} x} + c_2 e^{-\sqrt{m^2+1} x} \\ \frac{c_1 \sqrt{m^2+1}}{-m+i} e^{\sqrt{m^2+1} x} + \frac{-c_2 \sqrt{m^2+1}}{-m+i} e^{- \sqrt{m^2+1} x} \end{array} \right) : c_1, c_2 \in \mathbb{C} \right\} \end{equation*} Similar calculations will show that \begin{equation*} \mathcal{H}_+(H_0) = \left\{ \left( \begin{array}{c} c_1 e^{\sqrt{m^2+1} x} + c_2 e^{-\sqrt{m^2+1} x} \\\frac{c_1 \sqrt{m^2+1}}{-(m+i)} e^{\sqrt{m^2+1} x} + \frac{-c_2 \sqrt{m^2+1}}{-(m+i)} e^{- \sqrt{m^2+1} x} \end{array} \right) : c_1, c_2 \in \mathbb{C} \right\} \end{equation*} Notice $\dim(\mathcal{H}_-(H_0))=\dim(\mathcal{H}_+(H_0)) = \dim(\mathbb{C} \times \mathbb{C}) = 4$. By Theorem X.2 (Reed Simon Vol. 2) if we can find a partial isometry $U$ from $\mathcal{H}_+$ into $\mathcal{H}_-$ then the corresponding closed symmetric extension $\hat{H}$ of $H_0$ has the following properties: \begin{equation*} \begin{split} &\mathcal{D}(\hat{H}) = \left\{ \phi +\phi_+ + U\phi_+ : \phi \in \mathcal{D}(H_0), \phi_+ \in I(U) \right\} \\ &\hat{H}(\phi + \phi_+ + U\phi_+) = H_0(\phi) + i \phi_+ - i U \phi_+ \end{split} \end{equation*} where $I(U)$ is the initial space of a partial isometry $U$. Furthermore the deficiency indices of $\hat{H}$ are \begin{equation*} n_{\pm}(\hat{H}) = n_{\pm}(H_0) - \dim[I(U)] \end{equation*} For $\hat{H}$ to be self-adjoint, we necessarily need $n_{\pm}(\hat{H}) =0$ and consequently we are looking for partial isometries $U$ with $\dim[I(U)] = n_{\pm}(H_0) = 4$. I'd like to know if I've made any mistakes up to here and how to find all such possible isometries. Thanks a lot for the help.