All Questions
8 questions
0
votes
1
answer
100
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Spatially localised solution to the Schrödinger equation with potential is a combination of eigenfunctions
In this Terry Tao's blog post, he claims that if one has a solution to the Schrödinger equation
$$i\,\partial_t u +\Delta u=Vu $$
with a "reasonably smooth and localised $V$", $u$ has ...
4
votes
1
answer
167
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Spectrum near zero of $-\partial^2_x + V : L^2(\mathbb{R}) \to L^2(\mathbb{R})$, where $V = O(|x|^{-2 - \delta})$
Let $H = -\partial^2_x + V(x) : L^2(\mathbb{R}) \to L^2(\mathbb{R})$ be a one dimensional Schrödinger operator, where the potential $V$ is real-valued, belongs to $L^\infty(\mathbb{R})$, and, as $|x| \...
- 24.2k
7
votes
2
answers
641
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Decay of solutions to Schrodinger equation with local minimum in potential
Consider the one-dimensional Schrodinger operator on the real line $\mathbb{R}$ given by
$$ L = - \partial_x^2 + V $$
where $V$ is a potential with the following properties:
$V$ is non-negative, and ...
- 9,007
2
votes
0
answers
145
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Are Weyl sequences polynomially bounded?
Look at the Hilbert space $l^2( \mathbb{Z}) $ and let $A$ be a translation invariant band operator. I.e. if $\{ e_n \}_{n \in \mathbb Z} $ is the standard basis for $l^2( \mathbb{Z}) $ then it holds ...
- 1,054
2
votes
0
answers
145
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When does a one-dimensional Schrödinger operator have a threshold resonance?
Consider the operator
$$ L = -\partial_x^2 + V(x),$$
for some bounded, decaying potential, i.e. $V(x)\to 0$ as $x\to \pm \infty$. I'm interested in the $L^2(\mathbb R)$ spectrum of $L$. We know that $...
- 12.9k
1
vote
0
answers
128
views
Angular excitations and Schrodinger operators with radial potential in N-dimensions
Can someone please explain the following in mathematical language?
"First of all, angular excitations only push the energy up, never down, so it is enough to analyze spherically symmetric s-waves....
- 141
3
votes
1
answer
206
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Bounded solutions for Schrödinger equation at the edge of the essential spectrum
Let $V:R^d\to R_+$ be with a compact support. The Schrödinger operator $H_a=-\Delta - a V$ acting in $L^2(R^d)$ has then (at most) finitely many negative eigenvalues. Denote the number of negative ...
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- 1
10
votes
0
answers
284
views
Comparing spectra of Laplacian and Schrödinger operator
Let $M$ be a closed (compact without boundary) Riemannian manifold. Is there a body of results that compares the eigenvalues of the Laplace-Beltrami operator with that of Schrödinger operators $-\...
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