# Spectrum of $(Jx)_n =i((2n+1)x_{n+1}-(2n-1)x_{n-1})$ on $\ell^2(\mathbb{Z})$

I've been working on the spectrum of the closure of the operator $$J: \mathcal{D}(J)= \mbox{span}\{ e_n: n \in \mathbb{Z}\} \subseteq \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$$ defined for $$x=(x_n)_{n \in \mathbb{Z}} \in \mathcal{D}(J)$$ by

$$(Jx)_{n} =i((2n+1)x_{n+1}-(2n-1)x_{n-1}).$$

I know that $$J$$ is essentially self-adjoint and I have shown that if $$\lambda$$ is an eigenvalue of $$\overline{J}$$, then $$-\lambda$$ is also an eigenvalue. But I don't know another approach to study $$\sigma(\overline{J})$$. Can you give me any help?

• for reference, you seek the eigenvalues of an infinite matrix of the form $$\left( \begin{array}{ccccc} 0 & 3 & 0 & 0 & 0 \\ 3 & 0 & 5 & 0 & 0 \\ 0 & 5 & 0 & 7 & 0 \\ 0 & 0 & 7 & 0 & 9 \\ 0 & 0 & 0 & 9 & 0 \\ \end{array} \right)$$ Nov 22 at 21:54
• If your question is mainly "what are the methods to study the spectrum of Jacobi matrices", that is a topic for a book, and one such book is Teschl's (available online for free). Nov 22 at 22:50
• Also, your operator is not symmetric, you perhaps meant the sum of the two terms (rather than the difference). Nov 22 at 22:53
• My guess would be that the spectrum is purely discrete (though the only thing that's clear right away (to me) is that there is no ac spectrum), and I wouldn't necessarily expect to be able to find the eigenvalues explicitly. Nov 22 at 23:02
• @ChristianRemling Thank you for your comments. I forgot to multiply by $i$. I am going to see Teschl's book. Nov 23 at 2:10

Under the Fourier series isomorphism $$\ell^2(\mathbb{Z}) \cong L^2(-\pi,\pi)$$, $$u(t) = \sum_{n\in\mathbb{Z}} x_n e^{int}$$, the operator becomes \begin{aligned} (Ju)(t) &= 4i\sin(t) u'(t) + 2i\cos(t) u(t) \\ &= \begin{cases} +4i\left|\sin(t)\right|^{1/2} \partial_t (\left|\sin(t)\right|^{1/2} u(t)) & t\in(0,\pi) \\ -4i\left|\sin(t)\right|^{1/2} \partial_t (\left|\sin(t)\right|^{1/2} u(t)) & t\in(-\pi,0) \end{cases} . \end{aligned} Solving the eigenvalue equation $$Ju = \lambda u$$ as an ODE, gives two independent weak solutions $$u_{\lambda,\pm}(t) = \frac{\left|\tan(t/2)\right|_\pm^{-i\lambda/4}}{\left|\sin(t)\right|_\pm^{1/2}} ,$$ where $$\left|A\right|_\pm = \left|A\right| \Theta(\pm A)$$. The explicit expressions tells us that $$u_{\lambda,\pm} \not\in L^2(-\pi,\pi)$$ for any complex $$\lambda$$. However, for $$\Im\lambda > 0$$ we have $$u_{\lambda,\pm}$$ in $$L^2_{\text{loc}}$$ near $$t=0$$, while for $$\Im\lambda < 0$$ we have $$u_{\lambda,\pm}$$ in $$L^2_{\text{loc}}$$ near $$t=\pi$$.
Thus, for $$\lambda \in \mathbb{C} \setminus \mathbb{R}$$ we can adapt the variation of constants formula to define the resolvents (hopefully getting all the factors correct) \begin{aligned} ((J-\lambda)^{-1}v)(t) = \begin{cases} \frac{\Theta(t)}{4i} \int_t^\pi u_{\lambda,+}(t) u_{-\lambda,+}(s) v(s)\, ds + \frac{\Theta(-t)}{4i} \int_{-\pi}^{t} u_{\lambda,-}(t) u_{-\lambda,-}(s) v(s)\, ds & \Im\lambda > 0 \\ -\frac{\Theta(t)}{4i} \int_0^t u_{\lambda,+}(t) u_{-\lambda,+}(s) v(s)\, ds - \frac{\Theta(-t)}{4i} \int_{t}^{0} u_{\lambda,-}(t) u_{-\lambda,-}(s) v(s)\, ds & \Im\lambda < 0 \end{cases} . \end{aligned} The resolvent is symmetric, $$((J-\lambda)^{-1})^* = (J-\bar{\lambda})^{-1}$$ and is well-defined for any $$v\in L^2(-\pi,\pi)$$. Moreover, it did not require any boundary conditions other than being defined from $$L^2$$ to $$L^2$$. Hence, $$J$$ is essentially self-adjoint, with the unique self-adjoint extension corresponding to the above resolvent. As a function of $$\lambda$$, $$(J-\lambda)^{-1}$$ is discontinuous across $$\mathbb{R}\subset \mathbb{C}$$ (the solutions $$u_{\lambda,\pm}$$ switch the location of their $$L^2_{\text{loc}}$$ behavior as $$\lambda$$ crosses $$\mathbb{R}$$), hence $$\sigma(J) = \mathbb{R}$$, with generalized eigenfunctions given by $$u_{\lambda,\pm}(t)$$.
• @IgorKhavkine Thank you for your answer. But, since that $J$ on the domain of finite support sequences $D(J)=\mbox{span}\{ e_n: n \in \mathbb{\mathbb{Z}}\}$ is not closed, we have that $\sigma(J)=\mathbb{C}$, doesn't we? That's why I'm interested in $\sigma(\overline{J})$. Nov 24 at 15:50
• @Mainkit Sorry, I didn't explicitly write $\bar{J}$ where I perhaps should have. Once a self-adjoint resolvent is defined it automatically gives rise to a closed self-adjoint realization of $J$ (the resolvent is bounded, so it's graph is closed, which is the reflection of the graph of the corresponding closed extension of $J-\lambda$). By essential self-adjointness (in this case the uniqueness of the resolvent), this closed extension is just $\bar{J}-\lambda$. Nov 24 at 22:08