- #1
jostpuur
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This is about quantum mechanics, but it is sufficiently difficult existence question dealing with DE and operator theory, that I think it fits the DE subforum the best:
Let [tex]E:\mathbb{N}\to\mathbb{R}[/tex] be an arbitrary map, but so that [tex]\textrm{Im}(E)[/tex] is bounded from below. Does there exist a measurable function [tex]V:\mathbb{R}\to\mathbb{R}[/tex] such that the eigenvalues of the operator
[tex]
H = -\frac{1}{2}\partial_x^2 + M_V
[/tex]
are the given [tex]E(n)[/tex]?
Here [tex]M_V:\mathbb{C}^{\mathbb{R}}\to \mathbb{C}^{\mathbb{R}}[/tex] is the multiplication operator [tex](M_V\psi)(x) = V(x)\psi(x)[/tex].
For the sake of rigor we can give the following definition for the domain of H,
[tex]
D(H) = \{\psi\in L^2(\mathbb{R},\mathbb{C})\;|\; \psi\;\textrm{is piece wisely}\;C^2\quad\textrm{and}\quad \int dx\;\Big|-\frac{1}{2}\partial_x^2\psi(x) + V(x)\psi(x)\Big|^2 < \infty\}
[/tex]
So H is a mapping [tex]H:D(H)\to L^2(\mathbb{R},\mathbb{C})[/tex].
Let [tex]E:\mathbb{N}\to\mathbb{R}[/tex] be an arbitrary map, but so that [tex]\textrm{Im}(E)[/tex] is bounded from below. Does there exist a measurable function [tex]V:\mathbb{R}\to\mathbb{R}[/tex] such that the eigenvalues of the operator
[tex]
H = -\frac{1}{2}\partial_x^2 + M_V
[/tex]
are the given [tex]E(n)[/tex]?
Here [tex]M_V:\mathbb{C}^{\mathbb{R}}\to \mathbb{C}^{\mathbb{R}}[/tex] is the multiplication operator [tex](M_V\psi)(x) = V(x)\psi(x)[/tex].
For the sake of rigor we can give the following definition for the domain of H,
[tex]
D(H) = \{\psi\in L^2(\mathbb{R},\mathbb{C})\;|\; \psi\;\textrm{is piece wisely}\;C^2\quad\textrm{and}\quad \int dx\;\Big|-\frac{1}{2}\partial_x^2\psi(x) + V(x)\psi(x)\Big|^2 < \infty\}
[/tex]
So H is a mapping [tex]H:D(H)\to L^2(\mathbb{R},\mathbb{C})[/tex].
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