# Self-adjointness of the components of the magnetic derivative

+ 1 like - 0 dislike
163 views

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
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. $$\lVert A_j \psi\rVert\leq a \lVert \partial/\partial x_j \psi\rVert +b\lVert \psi\rVert\; .$$
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.