1
$\begingroup$

I've not used singular before, so I hope this question is not silly or trivial.

I assume I have a finite nonempty real set $\mathbb{V}\subseteq \mathbb{R}$ and a potential function $V:\mathbb{Z}^2\to \mathbb{V}$. I consider the bounded self-adjoint operator

$[H\psi](\mathsf{n})= \sum_{|| \mathsf{m}-\mathsf{n} ||_1=1} \psi(\mathsf{m}) + V(\mathsf{n}) \psi(\mathsf{n}) \quad \text{for all} \quad \psi\in \ell^2(\mathbb{Z}^2) \quad\text{and} \quad \mathsf{n}\in \mathbb{Z}^2. $

I then consider the set $S_1=\{ -1,0 \}^2$ and $S_2=\big([-2,2] \times \{ -1,0 \} \big) \cup \big( \{ -1,0 \} \times [-2,2] \big) $. I denote the corresponding orthogonal projections on $\ell^2(\mathbb{Z}^2)$ by $\mathbf{1}_{S_1}$ and $\mathbf{1}_{S_2}$ accordingly. I want to associate finitely many values to the operator $H_{S_2,S_1} :=\mathbf{1}_{S_2} H \mathbf{1}_{S_1}$ by locally minimizing

$$ || (H_{S_2,S_1}-zI)\psi||_2 \quad \text{for} \quad \psi\in \text{span}\big(\{ \mathbf{1}_s: s\in S_1\} \big) \quad \text{with} \quad ||\psi ||_2=1, $$

and $z\in \mathbb{R}$. Reading a little on singular values, it seems to me that the minimizers should be $z$'s which minimize the smallest singular value of $H_{S_2,S_1}-zI$. Is there a way to compute this minimizers without finding an explicit form for the smallest singular value? Also, is the smallest singular value given by the smallest eigenvalue of $H_{S_2,S_1}H_{S_2,S_1}^*$ or of $H_{S_2,S_1}^*H_{S_2,S_1}$? Or does it not matter?

This seems like it should be a simple problem, but I am having a problem finding the right values.

I should say that my motivation is approximating points in $\text{spec}(H)$ ,using Weyl's criterion for spectrum of self adjoint operators. I believe that if I do this minimization problem for all $H_{\mathsf{m}+S_2,\mathsf{m}+ S_1}$, where $\mathsf{m}\in \mathbb{Z}^2$, then I should get some finite cover of $\text{spec}(H)$.

Since my understanding of singular value is basic, I would appreciate any input or corrections on this problem.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Following a more recent question I asked, I think I have an answer for this answer. I attach an attempt of an answer, in the hope that someone would correct it if it is wrong.

I denote $s_1=(-1,0)$, $s_2=(0,0)$, $s_3=(-1,-1)$ and $s_4=(0,-1)$. I am therefore considering functions $\psi:S_1\to \mathbb{R}$ such that

$$ \sum_{j=1}^4 |\psi(s_j)|^2=1. $$

Writing the size I want to minimize is given explicitly by

$$\mathcal{L}(\psi,z):= \sum_{j=1}^4 \Bigg[ \Big( \big( V(s_j)-z \big)\psi(s_j)+ A(\psi_j) \Big)^2 +2 | \psi(s_j) |^2 \Bigg], $$

where $A(\psi_1)=A(\psi_4)= \psi(s_2)+\psi(s_3)$ and $A(\psi_2)=A(\psi_3)= \psi(s_1)+\psi(s_4)$.

Since $\psi$ is normalized, I can rewrite

$$ \mathcal{L}(\psi,z)= \sum_{j=1}^4 \Big( \big( V(s_j)-z \big)\psi(s_j)+ A(\psi_j) \Big)^2 +2$$.

Let us denote the Dirichlet Schrodinger operator $H_{S_1,S_1}=\mathbf{1}_{S_1} H \mathbf{1}_{S_1}$. A short computation shows that

$$ \mathcal{L}(\psi,z)= || (H_{S_1,S_1}-zI)\psi ||^2+2. $$

The first term vanishes when $z$ is an eigenvalue of $H_{S_1,S_1}$, so $\mathcal{L}(\psi,z)$ is minimized when $z$ is an eigenvalue of the Dirichlet Schrodinger operator.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.