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?

  • $\begingroup$ for reference, you seek the eigenvalues of an infinite matrix of the form $$\left( \begin{array}{ccccc} 0 & 3 & 0 & 0 & 0 \\ 3 & 0 & 5 & 0 & 0 \\ 0 & 5 & 0 & 7 & 0 \\ 0 & 0 & 7 & 0 & 9 \\ 0 & 0 & 0 & 9 & 0 \\ \end{array} \right)$$ $\endgroup$ Nov 22, 2022 at 21:54
  • 1
    $\begingroup$ 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). $\endgroup$ Nov 22, 2022 at 22:50
  • $\begingroup$ Also, your operator is not symmetric, you perhaps meant the sum of the two terms (rather than the difference). $\endgroup$ Nov 22, 2022 at 22:53
  • $\begingroup$ 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. $\endgroup$ Nov 22, 2022 at 23:02
  • $\begingroup$ @ChristianRemling Thank you for your comments. I forgot to multiply by $i$. I am going to see Teschl's book. $\endgroup$
    – Maklen
    Nov 23, 2022 at 2:10

1 Answer 1


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 $u_{\lambda,\pm}(t)$.

NB: In the comments, Giorgio Metafune outlined essentially the same argument.

  • $\begingroup$ Nice! I wonder if there is an argument to show that it is essentially selfadjoint without computing the resolvent. $\endgroup$ Nov 23, 2022 at 21:53
  • $\begingroup$ @GiorgioMetafune All indirect arguments that I can think of will have to imply something about the resolvent anyway. Since we know the formula for it from ODE theory, why not just compute it and check that it is well-defined. I think that is the most straightforward way to go about it. $\endgroup$ Nov 23, 2022 at 22:35
  • $\begingroup$ @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 $\sigma(\overline{J})$. $\endgroup$
    – Maklen
    Nov 24, 2022 at 15:50
  • $\begingroup$ @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 $\bar{J}-\lambda$. $\endgroup$ Nov 24, 2022 at 22:08
  • $\begingroup$ By the way I think that @Mainkit has an argument to prove essential selfadjointness on the original operator. Could you explain It? $\endgroup$ Nov 24, 2022 at 22:20

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