# Difference in essential spectrum between Schrodinger operators

I am considering two Schrodinger operators on $$\mathbb{Z}^2$$ and compare their essential spectrum. The operators are both of the form $$H=A+V$$ where $$A$$ is the adjacency operator on the $$\mathbb{Z}^2$$-lattice, and $$V$$ is a diagonal operator of the form $$[ V\psi ](n,m)=u(n,m)\cdot \psi(n,m)$$ for any function $$\psi\in \ell^2(\mathbb{Z}^2)$$ and $$(n,m)\in \mathbb{Z}^2$$, where $$u:\mathbb{Z}^2\to \mathbb{R}$$ is some function.

Concretely, I am trying to compare $$\sigma_{ess}(H_1)$$ and $$\sigma_{ess}(H_2)$$, where $$u_1\equiv 0$$ and

$$u_2(m_1,m_2)= \begin{cases} 16 &;m_1=0, m_2\geq 0 \\ 0 &;\text{else} \end{cases} \; .$$

It is easy to show that $$\Vert H_1 \Vert= 4$$ and $$\sqrt{20} \leq \Vert H_2\Vert\leq 20$$. Since the operators are self-adjoint, we know that $$\lambda'\in \sigma(H_2)\setminus \sigma(H_1)$$ for $$\lambda':= \Vert H_2\Vert$$. I am now trying to show that there exists $$\lambda \in \sigma_{ess}(H_2)\setminus \sigma_{ess}(H_1)$$ with $$\lambda\geq 12$$, hopefully $$\lambda'=\lambda$$ itself works. I want to find a sequence of normalized functions $$\psi_n\in \ell^2(\mathbb{Z}^2)$$ such that $$\Vert (H_2-\lambda I)\psi_n\Vert \to 0$$ and $$\psi_n\to 0$$ weakly. It seems like the spectral projection onto $$\chi_{[12,20]}(H_2)$$ should be infinite dimensional, but I can't manage to show this directly. I thought to look at increasing finite sections around the origin which would each have a strictly growing dimension of projection onto $$[12,20]$$ using a Gershgorin circle argument, but this hasn't worked so far.

It feels like there should be a simple argument I am missing that someone well versed in spectral theory would notice. I would appreciate any input on this question.

What is clear here is that $$H_2$$ has spectrum above $$4$$, and that $$\lambda\in\sigma_{\textrm{ess}}(H_2)$$ also, with $$\lambda:=\max\sigma(H_2)=\|H_2\|>4$$. What $$\lambda$$ is actually equal to seems a rather delicate question. Certainly $$\lambda<20$$, and I'm also not at all sure that $$\lambda\ge 12$$; that may already be too ambitious.
To discuss these claims, let me switch to the continuous counterpart of your problem, the operator $$H=-\Delta-V(x,y)$$ on $$L^2(\mathbb R^2)$$. Everything I'm going to say has a precise analog in your (discrete) setting, but the details become considerably less tedious in the continuous case. Even so, I will only sketch most of them.
We now assume that $$V(x,y)=16$$ on $$|x|\le 1$$, $$y\ge 0$$, and $$V=0$$ otherwise. We are interested in the part of the spectrum below zero. (We clearly have $$[0,\infty)\subseteq\sigma (H)$$.)
First of all, there actually is negative spectrum. This would not be clear in dimension $$3$$ or higher, but here it still works. Compare my answer here. We can find $$\lambda=\min\sigma (H)$$ as the infimum of the quadratic form $$\lambda=\inf Q(f)$$, $$Q(f)=\int_{\mathbb R^2} \left(|\nabla f|^2-V|f|^2\right) , \tag{1}$$ taken over $$f\in C_0^{\infty}$$ (say) with $$\|f\|_{L^2}=1$$. Clearly, we can only hope to make $$Q(f)$$ small if we concentrate $$f$$ where $$V=16$$, but then $$Q(f)>-16$$ because we can not avoid paying a (potentially steep) price in the first term. A more chunky potential that is equal to $$16$$ on a large disk rather than a thin strip would be more effective at pulling down the bottom of the spectrum. In any event, we can try concrete test functions in (1) and get upper bounds on $$\lambda$$, but finding $$\lambda$$ exactly seems a formidable task.
Finally, there are $$f\in L^2$$ that make $$\|(H-\lambda)f\|$$ arbitrarily small. A carefully cut off version (compare the discussion in the comments to the linked answer) of $$f$$ will still keep $$(H-\lambda)f$$ small, and then we obtain a Weyl sequence by also shifting and considering $$g(x,y)=f(x,y-L)$$, $$L\gg 1$$.
• Thank you for (another) informative answer. I am unclear about the last part of your answer, which I think deals with why $\lambda$ is in the essential spectrum. You are essentially pushing the support of the function upwards and using the periodicity in that direction. But if the support intersects $y<0$, how can you be sure that the shifted function also makes $\Vert (H-\lambda)f\Vert$ small as well? Feb 25 at 11:24
• @Keen-ameteur: We have $f\in L^2$ (or $\ell^2$ in your actual setting), so $f$ decays and $y<0$ contributes only a small part to its norm. Feb 25 at 17:42