# Why this 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?

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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? – rcollyer Apr 18 '11 at 13:03
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: – Ansel Apr 18 '11 at 14:01
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." – Ansel Apr 18 '11 at 14:01

Based upon your comments, there must be additional conditions on $k$ that aren't listed. First of all, the reality of the eigenvalues of $M$ imposes two conditions on $k$, either $k = ic, \textrm{where }c \in \mathbb{R}$ or $k \in \mathbb{R}$. The former condition does not impose any additional constraints on $k$, but if $k \in \mathbb{R}$, then $|k| \leq |q(x)|$ for the eigenvalues to be real. However, the statement that $M$ is Hermitian (or, self-adjoint) implies that its diagonal elements (and eigenvalues) must be real, or more specifically, that $k$ is purely imaginary. So, the self-adjointness of $M$ relies directly on the nature of $k$. This suggests to me that you need to research how the problem was derived and see what assumptions were made with regards to $k$.

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Thanks very much to rcollyer for helpful comments, which make me keep on thinking about this problem. I have worked out an explanation consistent with the context. That the spectrum of a self-adjoint operator lies on the real axis can be found in a common textbook on functional analysis, so the primal issue is to make clear the self-adjointness.

Rewriting the system as

$\begin{matrix}\begin{bmatrix}iy_1\\-iy_2\end{bmatrix}_x-\begin{bmatrix}0&iq\\-ir&0\end{bmatrix} = \begin{bmatrix} ky_1\\ky_2 \end{bmatrix} &(1)\end{matrix}$

to emphasize that k is the spectrum. Then the operator we are interested in satisfies

$\begin{matrix}\operatorname{T}\begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}iy_1\\-iy_2\end{bmatrix}_x-\begin{bmatrix}0&iq\\-ir&0\end{bmatrix} &(2)\end{matrix}$

A bounded operator $\operatorname{T}$ defined everywhere is symmetric, or self-adjoint, or Hermitian if

$(\operatorname{T}f,g)=(f,\operatorname{T}g)$

The inner product of two vector-valued functions is defined here (and in common) to be

$\begin{matrix}(f,g)=\int_{-\infty}^{\infty}(f_1^*g_1+f_2^*g_2)~dx&(3)\end{matrix}$

Then, it is easy to show that if $r=q^*$,

$(\operatorname{T}f,g)=(f,\operatorname{T}f)$

based on the prerequisite that $\int_{-\infty}^{\infty}|x^jq|~dx<\infty$,$j=0,1$.

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