What is an example of a symmetric operator with non-real spectrum? I've been wracking my brains but all the examples I can think of with differential operators end up having either real eigen-values and/or intractable (possibly) complex eigen-values.
For example, take
\begin{align}
D(T)&=\{u\in L^2(0,\pi)~|~u''\in L^2(0,\pi),iu(0)=u'(0),u(\pi)=0\}, \\
Tu&=-u''.
\end{align}
Then integrating by parts shows it's symmetric, and the function $\sin(\kappa x)-i\kappa\cos(\kappa x)$ is an eigen-function for the eigen-value $\kappa^2$ which must satisfy $\tan(\kappa\pi)=i\kappa$. From here I must show that valid $\kappa$'s must be non-real (we exclude $0$ since $T$ has kernel $\{0\}$). I am not sure how to proceed.
Is there a better way? Is it too much of me to ask to be able to compute explicitly a complex eigen-value, as opposed to simply establishing the existence of some other piece of the spectrum, e.g., continuous, essential, etc.?
 A: In line with what you wanted to do, let $Tu=u''$ be defined on the domain $\mathcal{D}(T)$ consisting of all twice absolutely continuous $u \in L^2[0,1]$ for which $u''\in L^2[0,1]$ and
$$
        u(0)=u'(0)=u(1)=u'(1)=0.
$$
The operator $T$ is closed, densely-defined, and symmetric on its domain.
The adjoint $T^*$ has a domain consisting of all twice absolutely continuous $u\in L^2[0,1]$ for which $u''\in L^2[0,1]$, but with no restrictions at the endpoints. And $T^*u=u''$. The operator $T^*$ has every $\lambda$ as an eigenvalue because all classical solutions $u(x)=Ae^{\sqrt{\lambda}x}+Be^{-\sqrt{\lambda}x}$ of
$$
                 u''=\lambda u
$$
are in the domain of $T^*$. Consequently, every $\lambda\in\mathbb{C}$ is in the spectrum of $T$, as the range of $T-\lambda I$ is never dense in $L^2$ because
$$      \mathcal{R}(T-\lambda I)^{\perp}=\mathcal{N}(T^*-\overline{\lambda}I)\ne \{0\}.
$$
Note: You may also accomplish the same with a first order operator
$T$, but you were trying to deal with the second order, which is why I gave you this example. The first order $Tu=iu'$ with $u(0)=u(1)=0$ also works. In fact, the previous $T$ is the negative square of this operator.
A: $i\frac{d}{dx}$ defined on functions of compact support in $(0,\infty)$ (i.e. with no boundary condition at $0$).
The intuition is that $i\frac{d}{dx}$ is the infinitesimal generator of translations on $(-\infty,\infty)$ but this can't work on $(0,\infty)$ .
