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A relatively well-known example of (continuous) Schrödinger operator with empty spectrum is the complex Airy operator on the line, i.e., the operator acting on $L^{2}(\mathbb{R})$ given by the differential expression $$ -\frac{d^{2}}{dx^{2}}+i x, $$ where $i$ is the imaginary unit; see for example Sec. 3.1 in this paper.

The question is whether there is a discrete counterpart of this phenomenon. More precisely, I would like to know an example (if any) of the operator $T_{v}:Dom\,T_{v}\subset\ell^{2}(\mathbb{Z})\to\ell^{2}(\mathbb{Z})$ with the maximal domain $Dom\,T_{v}=\{u\in\ell^{2}(\mathbb{Z}) \mid T_{v}u\in\ell^{2}(\mathbb{Z})\}$ and acting as $$ (T_{v}u)_{n}:=u_{n-1}+v_{n}u_{n}+u_{n+1}, \quad n\in\mathbb{Z}, $$ where $v=\{v_{n}\}_{n\in\mathbb{Z}}\subset\mathbb{C}$, such that the spectrum of $T_{v}$ is empty. The operator $T_{v}$ is refered to as the discrete Schrödinger operator with potential sequence $v$. Of course, such an example (if any) must have $v$ an unbounded and non-real sequence.

Remark: The discrete analogue of the complex Airy operator is the discrete Schrödinger operator $T_{v}$ with $v_{n}= in$. This is not an operator with empty spectrum. Actually, the spectrum of this operator can be computed fully explicitly and equals $i\mathbb{Z}$.

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    $\begingroup$ The discrete Schrödinger operator is a bounded perturbation of the multiplication operator $(u_n) \mapsto (v_n u_n)$. I never did perturbation theory, but I think in this case there are some continuity results for the spectrum which should imply that the spectrum is non-empty. Have you checked available results for bounded perturbations of normal operators? $\endgroup$ Commented Feb 12, 2021 at 17:06
  • $\begingroup$ @MateuszKwaśnicki I haven't checked that. Thanks for pointing this out! $\endgroup$
    – Twi
    Commented Feb 13, 2021 at 8:38

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