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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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    $\begingroup$ 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. $\endgroup$
    – Michael Engelhardt
    Commented Feb 14, 2020 at 22:48

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