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  Self-adjointness of the components of the magnetic derivative

+ 1 like - 0 dislike

On $L^{2}(\mathbb{R}^{n})$ define the operator $\Pi_{j} u := (-i\partial/\partial x_{j} - A_{j})u$, where $A_{j} \in L^{2}_{loc}(\mathbb{R}^{n})$ represents the $j$-th component of the magnetic potential on $\mathbb{R}^{n}$ and $u \in Dom\ \Pi_{j} := \{ u \in L^{2}(\mathbb{R}^{n}) \ \vert\ -i\partial u/\partial x_{j} - A_{j}u \in L^{2}(\mathbb{R}^{2}) \}$, where $\partial u/\partial x_{j}$ is the weak derivative of $u$.

I wish to understand what sort of conditions on $A_{j}$ ensure that this operator is self-adjoint.

This post imported from StackExchange MathOverflow at 2015-02-20 16:54 (UTC), posted by SE-user Geno Whirl
asked Oct 13, 2014 in Mathematics by Geno Whirl (5 points) [ no revision ]
retagged Feb 20, 2015
You cannot use positivity, hence a result like the Kato inequality for the laplacian is not possible. With perturbation theory you can prove self-adjointness, i.e. if there is $a<1$ s.t. $$\lVert A_j \psi\rVert\leq a \lVert \partial/\partial x_j \psi\rVert +b\lVert \psi\rVert\; .$$

This post imported from StackExchange MathOverflow at 2015-02-20 16:54 (UTC), posted by SE-user yuggib

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