# Real eigenvalues of continuum spectrum of a self-adjoint operator

Is my understanding that if you assume eigenvectors of a self-adjoint operator are in Hilbert space, then is easy to prove that the eigenvalues must be real. However, it could happen that the "eigenvectors and eigenvalues" (as we called them in physics) form a continuous spectrum, in which case this assumption does not hold. I would like to know if even if the spectrum is continuous, eigenvalues of a self-adjoint operator are real.

• $\uparrow$ Yes. – Qmechanic Mar 23 '18 at 22:02
• Would Mathematics be a better home for this question? – Qmechanic Mar 23 '18 at 22:04
• I dont think the proof assumes a discrete spectrum , so it should hold for continuous eigenvalues as well. – cobra121 Mar 24 '18 at 16:28
• which theorem guarantees it? Is it the spectral theorem for unbounded self-adjoint operators? – angel leonardo Mar 24 '18 at 21:46

Suppose $\mathcal{H}$ is a complex Hilbert space. A general selfadjoint operator $A : \mathcal{D}(A)\subset\mathcal{H}\rightarrow\mathcal{H}$ has a linear domain $\mathcal{D}(A)$ that is dense in $\mathcal{H}$. The resolvent set $\rho(A)$ consists of all $\lambda\in\mathbb{C}$ for which $A-\lambda I$ has range $\mathcal{H}$ and null space $\{0\}$. The spectrum $\sigma(A)$ consists of all other $\lambda\in\mathbb{C}$.
A given $\lambda\in\mathbb{R}$ is in the spectrum of $A$ if (a) $Ax=\lambda x$ for some non-zero $x\in\mathcal{H}$ or (b) there exists a sequence of unit vectors $\{ x_n \} \subseteq\mathcal{D}(A)$ such that $(A-\lambda I)x_n \rightarrow 0$ as $n\rightarrow\infty$. Every point in the spectrum fits into one (or both) of these two cases (a) or (b). Case (a) is the point spectrum of $A$ (a.k.a. discrete spectrum,) and case (b) is defined as the approximate point spectrum (a.k.a. continuous spectrum.)
You can show that the continuous spectrum is real by assuming $\{ x_n \}$ is a sequence of unit vectors such that $(A-\lambda I)x_n\rightarrow 0$ for some $\lambda$, and using a modified classical argument (excuse the Math notation as opposed to bra-ket): $$\langle Ax_n,x_n\rangle-\lambda = \langle (A-\lambda I)x_n,x_n\rangle\rightarrow 0\;\;\;\mbox{ as } n\rightarrow\infty.$$ Because $\langle Ax_n,x_n\rangle$ is real for all $n$, then $\lambda$ must also be real. So an approximate eigenvalue of $A$ is seen to be real by basically the same argument used to show that eigenvalues must be real.