朗道能级在$δ$相互作用下的命运
The fate of Landau levels under $δ$-interactions
摘要：
考虑$L^2(\mathbb{R}^2)$中的自伴Landau Hamilton $H_0$，其谱由无限退化的本征值$Λ_q$，$q \in \mathbb{Z}_+$和扰动算子$H_\upsilon = H_0 + \upsilonδ_Γ$组成，其中$Γ\subset \mathbb{R}^2$是正则的Jordan $C^{1,1}$ -曲线，而$\upsilon \in L^p(Γ;\mathbb{R})$，$p>1$具有常数符号.我们研究了${\rm Ker}(H_\upsilon -Λ_q)$，$q \in \mathbb{Z}_+$，并证明了一般的$$0 \leq {\rm dim \, Ker}(H_\upsilon -Λ_q) - {\rm dim \, Ker}(T_q(\upsilon δ_Γ)) < \infty,$$，其中$T_q(\upsilon δ_Γ) = p_q (\upsilon δ_Γ)p_q$，是Berezin-Toeplitz型算子，作用在$p_q L^2(\mathbb{R}^2)$中，$p_q$是${\rm Ker}\,(H_0 -Λ_q)$上的正交投影.如果$\upsilon \neq 0$和$q = 0$，我们证明了${\rm Ker}\,(T_0(\upsilon δ_Γ)) = \{0\}$ .如果$q \geq 1$，$Γ= \mathcal{C}_r$是一个圆的半径$r$，我们表明，${\rm dim \, Ker} (T_q(δ_{\mathcal{C}_r})) \leq q$，和一套$r \in (0,\infty)$的${\rm dim \, Ker}(T_q(δ_{\mathcal{C}_r})) \geq 1$，是无限的和离散的。
英文摘要：
We consider the self-adjoint Landau Hamiltonian $H_0$ in $L^2(\mathbb{R}^2)$ whose spectrum consists of infinitely degenerate eigenvalues $Λ_q$, $q \in \mathbb{Z}_+$, and the perturbed operator $H_\upsilon = H_0 + \upsilonδ_Γ$, where $Γ\subset \mathbb{R}^2$ is a regular Jordan $C^{1,1}$-curve, and $\upsilon \in L^p(Γ;\mathbb{R})$, $p>1$, has a constant sign. We investigate ${\rm Ker}(H_\upsilon -Λ_q)$, $q \in \mathbb{Z}_+$, and show that generically $$0 \leq {\rm dim \, Ker}(H_\upsilon -Λ_q) - {\rm dim \, Ker}(T_q(\upsilon δ_Γ)) < \infty,$$ where $T_q(\upsilon δ_Γ) = p_q (\upsilon δ_Γ)p_q$, is an operator of Berezin-Toeplitz type, acting in $p_q L^2(\mathbb{R}^2)$, and $p_q$ is the orthogonal projection on ${\rm Ker}\,(H_0 -Λ_q)$. If $\upsilon \neq 0$ and $q = 0$, we prove that ${\rm Ker}\,(T_0(\upsilon δ_Γ)) = \{0\}$. If $q \geq 1$, and $Γ= \mathcal{C}_r$ is a circle of radius $r$, we show that ${\rm dim \, Ker} (T_q(δ_{\mathcal{C}_r})) \leq q$, and the set of $r \in (0,\infty)$ for which ${\rm dim \, Ker}(T_q(δ_{\mathcal{C}_r})) \geq 1$, is infinite and discrete.
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