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I am confused about why the following spectrum problem is self-adjoint:

$\begin{matrix}y'(x) = \mathbf{M}(k,x)y(x)&(y:\mathbf{R}\rightarrow\mathbf{C}^2,x\in\mathbf{R},k\in\mathbf{C})\end{matrix}$

and the operator

$\begin{matrix}\mathbf{M}(k,x) = \begin{bmatrix} -ik & q(x) \\ q(x)^* & ik \end{bmatrix} &(q:\mathbf{R}\rightarrow\mathbf{C})\end{matrix}$

is Hermitian, so that the discrete spectra lie on the real axis. I am a novice on functional analysis. The only explanation I have figured out is ugly: computing the eigensystem of $\mathbf{M}$ gives $\pm \sqrt{-\operatorname{Re}(k)^2-2i\operatorname{Re}(k)\operatorname{Im}(k)+\operatorname{Im}(k)^2+|q(x)|^2}$ as its eigenvalues. To make $\mathbf{M}(k,x)$ singular, it is necessary to set $\operatorname{Im}(k)$ zero. But this is irrelevant to the Hermitianess of $\mathbf{M}(k,x)$, which seems to me that

$\mathbf{M}(k,x)^* = \begin{bmatrix} ik^* & q(x)^* \\ q(x) & -ik^* \end{bmatrix} \ne \!\ \mathbf{M}(k,x)^T$

Is there a generalization of this "self-adjoint" problem in dimensions more than 2?

3x in advance!

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    This has me confused, and I believe your notation is a hindrance. Let the dagger, $\dagger$, denote the conjugate-transpose, or adjoint, i.e. $A^\dagger = (A^\ast)^T$, where $\ast$ denotes complex conjugation. Now, if $M$ is Hermitian (or, self-adjoint), then $M^\dagger = M$. This implies that the diagonal of $M$ must be real, so $k = i c$ where $c \in \mathbb{R}$. Then your eigenvalues are $\pm \sqrt{c^2 + |q(x)|^2}$. So, with the notation changes could you clarify why your confused?2011-04-18
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    Sincere 3x to rcollyer. I encountered this problem when studying nonlinear Schrödinger equations, as more than one reference state simply that the system $y'(x)=\mathbf{M}(k,x)y(x)$ is Hermitian. Hermitianess is not a known condition, but inferred directly from the system, without specifying $k$ to be a real number. I guess that computing the eigensystem is not the "right" (concise) way. To make clear I'll cite a paragraph:2011-04-18
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    The AKNS spectral problem reads $v_x = \begin{bmatrix} -ik & q \\ r & ik \end{bmatrix}v$ (1), for which there is a special case of the system under the symmetry reduction $r = \mp q^*$. "When $r = q^*$, the operator (1) is Hermitian. _In this case, the spectrum lies on the real axis._ ... Moveover, the problem is self-adjoint."2011-04-18

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