# Schrodinger operator with magnetic field: eigenvalues

Consider the self-adjoint operator on $$L^{2}(\mathbb{R}^{N})$$,

$$H=-\frac{1}{2}(\nabla-iA)^{2}+V,$$ where $$A\in C^{\infty}(\mathbb{R}^{N}, \mathbb{R}^{N} )$$, $$V\in C^{\infty}(\mathbb{R}^{N})$$, $$V\geq 0$$ and $$V(x)\rightarrow\infty$$ as $$|x|\rightarrow\infty$$.

Does H have a purely discrete spectrum?

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• In the examples that immediately come to my mind, say $N=2$, $A=(y,-x)B/2$ (constant magnetic field), and $V=C(x^2 +y^2 )$ (harmonic oscillator), the spectrum is indeed discrete. I don't know where to point you for a general statement, though. What if $A$ is strong enough to overwhelm $V$ for $|x|\rightarrow \infty$? That might be a way to get a continuous part of the spectrum. It could be you need more conditions in that respect. – Michael Engelhardt Feb 14 at 22:48