Absolute Continuity of the Spectrum of a Periodic Schrödinger Operator
Absolute Continuity of the Spectrum of a Periodic Schrödinger Operator
复制标题
周期性薛定谔算子谱的绝对连续性
DOI：
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发表时间：
2003
期刊：
影响因子：
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通讯作者：
L. I. Danilov
中科院分区：
文献类型：
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作者：
L. I. Danilov
摘要证明了中Schrödinger算子谱的绝对连续性
$$L^2 ({mathbb{R}}^n )$$
，
$$n geqslant 3$$
，具有周期格（具有共同周期格）
$$Lambda$$
)标量
$$V$$
矢量
$$A in C^1 ({mathbb{R}}^n ,{mathbb{R}}^n )$$
两者的势能
$$user1{A} in H_{user2{loc}}^user1{q} user2{(}mathbb{R}^user1{n} user2{;}mathbb{R}^user1{n} user2{)}$$
，
$$2q > n - 2$$
或者向量势的傅里叶级数
$$A$$
绝对收敛，
$$V in L_w^{p(n)} (K)$$
，其中
$$K$$
是晶格的基本单元吗
$$Lambda$$
，
$$p(n) = n/2$$
为了
$$n = 3, 4, 5, 6$$
，和
$$p(n) = n - 3$$
为了
$$n geqslant 7$$
的值
$$user2{lim}_{user1{t} o user2{ + }infty } left| { heta _user1{t} V}
ight|_{user1{L}_user1{w}^{user1{p}user2{(}user1{n}user2{)}} user2{(}user1{K}user2{)}} $$
足够小，在哪里
$$ heta _user1{t} user2{(}user1{x}user2{) = 0 if }left| {Vuser2{(}user1{x}user2{)}}
ight| leqslant user1{t}$$
和
$$ heta _t (x) = 1$$
否则，
$$x in K$$
，和
$$t > 0$$
．
AbstractWe prove the absolute continuity of the spectrum of the Schrödinger operator in
$$L^2 ({mathbb{R}}^n )$$
,
$$n geqslant 3$$
, with periodic (with a common period lattice
$$Lambda$$
) scalar
$$V$$
and vector
$$A in C^1 ({mathbb{R}}^n ,{mathbb{R}}^n )$$
potentials for which either
$$user1{A} in H_{user2{loc}}^user1{q} user2{(}mathbb{R}^user1{n} user2{;}mathbb{R}^user1{n} user2{)}$$
,
$$2q > n - 2$$
, or the Fourier series of the vector potential
$$A$$
converges absolutely,
$$V in L_w^{p(n)} (K)$$
, where
$$K$$
is an elementary cell of the lattice
$$Lambda$$
,
$$p(n) = n/2$$
for
$$n = 3, 4, 5, 6$$
, and
$$p(n) = n - 3$$
for
$$n geqslant 7$$
, and the value of
$$user2{lim}_{user1{t} o user2{ + }infty } left| { heta _user1{t} V}
ight|_{user1{L}_user1{w}^{user1{p}user2{(}user1{n}user2{)}} user2{(}user1{K}user2{)}} $$
is sufficiently small, where
$$ heta _user1{t} user2{(}user1{x}user2{) = 0 if }left| {Vuser2{(}user1{x}user2{)}}
ight| leqslant user1{t}$$
and
$$ heta _t (x) = 1$$
otherwise,
$$x in K$$
, and
$$t > 0$$
.