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Problem: I have a self-adjoint operator in $\ell^2(\mathbb{Z})$ which acts as $$T g(x)=q^{-2 x -3/2} g(x+1)+(1+q) q^{-2 x-1} g(x)+q^{-2 x +1/2} g(x-1),$$ and I am looking to diagonalize it. The spectrum should be positive, but I think I need eigenfunctions, too. The parameter $q$ is real and (say) between $0$ and $1$.

Motivation: This operator describes a limit of the correlation kernel in a certain dimer model.

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    $\begingroup$ Is there any reason why you expect the equation $Tg=\lambda g$ to have explicit solutions? It does not look plausible to me. Also, since you are dealing with an operator on an infinite-dimensional Hilbert space, existence of point spectrum is not guaranteed. $\endgroup$
    – Christian Remling
    Commented Dec 18, 2023 at 18:49
  • $\begingroup$ @ChristianRemling I know, but I would expect some Gaussian-like solutions, where g(x) contains a factor of q^(-x^2) which allows the eigenfunctions to belong to $\ell^2$. However, so far I failed to find any examples. $\endgroup$
    – Leonid Petrov
    Commented Dec 18, 2023 at 18:52
  • $\begingroup$ @ChristianRemling One other reason is that the operator is a limit of a tridiagonal operator for certain orthogonal polynomials $\endgroup$
    – Leonid Petrov
    Commented Dec 18, 2023 at 18:52

2 Answers 2

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Your (unbounded) operator is in the limit circle case at $+\infty$. This means that self-adjoint realizations are obtained only when a boundary condition at $\infty$ is imposed, and the eigenvalues will depend on this boundary condition. (In fact, for any $\lambda\in\mathbb R$, there is exactly one boundary condition that makes this $\lambda$ an eigenvalue.)

To see this, let's make the (obvious) try $u_n=q^{-2n}g_n$. Then $Tg=\lambda g$ (viewed as a difference equation) is equivalent to $$ q^{1/2}u_{n+1}+q^{-3/2}u_{n-1}+\frac{1+q}{q}u_n = \lambda q^{2n} u_n . $$ The coefficient $\lambda q^{2n}$ is summable on $n\ge 1$, so can be ignored as far as the asymptotic behavior of $u_n$ is concerned, and we are in control. (To prove this carefully, we would have to refer to a discrete version of Levinson's theorem, but in fact this is unnecessary for our current purposes since it's enough to observe that all solutions are square summable when $\lambda =0$.)

Two basis solutions for $\lambda =0$ are $u_n=(-q)^{-n/2}$ and $u_n=(-q)^{-3n/2}$. Going back to $g$, we thus have the two solutions $|g_n|=q^{3n/2}$ and $|g_n|= q^{n/2}$, both of which are exponentially decaying and in $\ell^2(\mathbb Z_+)$. All solutions are square integrable, as claimed.

We can also observe that the coefficients of the left half line operator $T_-$ go to zero, so $\sigma_{ess}(T_-)=\{ 0\}$. This shows that (for any boundary condition at $\infty$) $\sigma_{ess}(T)=\{ 0\}$, so the eigenvalues accumulate at $0$ and the spectrum is purely discrete everywhere else.

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  • $\begingroup$ I don't think the function $q^{n/2}$ is exponentially decaying for $n\in \mathbb{Z}$, so it is not summable. I do not think I understand the construction $\endgroup$
    – Leonid Petrov
    Commented Dec 18, 2023 at 20:53
  • $\begingroup$ @LeonidPetrov: I consider the two half lines separately and show that all solutions of $Tg=0$ are in $\ell^2(\mathbb Z_+)$, for $n\ge 1$, so we have limit point case at $+\infty$ (the limit circle/limit point alternative refers to endpoints, so no other set-up makes sense if this is what I want to discuss). $\endgroup$
    – Christian Remling
    Commented Dec 18, 2023 at 20:59
  • $\begingroup$ I was assuming some material that is basic in this subject but of course not necessarily well known generally. An attempt at a summary is as follows: your operator is unbounded, so needs a domain. One can define it on finitely supported $g$'s and then take the closure $T_0$ of this operator. $T_0$ is symmetric, but not necessarily self-adjoint. Whether or not this is the case depends on the lc/lp alternative that I discuss in my answer. In your case, it has deficiency $(1,1)$ (one endpoint lp, the other lc), and there is a one parameter family of self-adjoint extensions (cont'd) $\endgroup$
    – Christian Remling
    Commented Dec 18, 2023 at 21:05
  • $\begingroup$ corresponding to boundary conditions imposed on elements of $D(T_0^*)$. $\endgroup$
    – Christian Remling
    Commented Dec 18, 2023 at 21:06
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You can find a hint in this article, where a tridiagonal operator with similar properties is diagonalized.

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