Bounded-coefficient magnetic Schrodinger operator

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Fix $\mathrm{\hslash }>0$, $m>0$ and $q\in \mathbb{R}$, and work on $L^2({\mathbb{R}}^3;\mathbb{C})$ with the position and momentum operators $Q_j,P_j$ of Position and momentum operators on Euclidean L2.

Admissible magnetic data. Call the pair $(A,\mathrm{\Phi })$ admissible when $A=(A_1,A_2,A_3)$ is a triple of real-valued classes $A_j\in L^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{R})$ having weak first derivatives $D_k{A}_j\in L^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{R})$ for all $j,k\in \{1,2,3\}$, and $\mathrm{\Phi }\in L^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{R})$ is a real-valued bounded class (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms). Write $\mathrm{\mid }A{\mathrm{\mid }}^2:=A_1^2+A_2^2+A_3^2$ and $\mathrm{div}A:=D_1{A}_1+D_2{A}_2+D_3{A}_3$ for chosen representatives; both are bounded classes.

Magnetic momentum operators. For admissible $(A,\mathrm{\Phi })$ and $j\in \{1,2,3\}$ define the $\mathbb{C}$-linear operator $P_{A,j}:=P_j-q{A}_j,D(P_{A,j}):=D(P_j)=\{f\in L^2:D_{j}f\in L^2\},$ where $A_j$ acts as multiplication; this is well defined into $L^2$ because $A_j\in L^{\mathrm{\infty }}$ and $f\in L^2$. The self-adjointness of $P_{A,j}$ is proved later on this page, not asserted here.

The operator. Put $D(K_{A,\mathrm{\Phi }}):=W^{2,2}({\mathbb{R}}^3;\mathbb{C})$ and $K_{A,\mathrm{\Phi }}u:=\frac{1}{2m}{\sum }_{j=1}^3{P}_{A,j}\left(P_{A,j}u\right)+q\mathrm{\Phi }u(u\in D(K_{A,\mathrm{\Phi }}))\mathrm{.}$ This is well defined: if $u\in W^{2,2}$ and $(A,\mathrm{\Phi })$ is admissible, then $P_{A,j}u=P_{j}u-q{A}_{j}u\in W^{1,2}({\mathbb{R}}^3;\mathbb{C})$ by Weak product rule for a bounded coefficient with bounded weak derivative together with $P_{j}u\in W^{1,2}$, so $P_{A,j}u\in D(P_{A,j})$ and every summand lies in $L^2$. Write $K_0:=K_{(0,0)}$ for the free operator, so that $K_{0}u=\frac{1}{2m}{\sum }_j{P}_j(P_{j}u)$ on $W^{2,2}$. The same formula shows $K_{A,\mathrm{\Phi }}$ is $\mathbb{C}$-linear. Its symmetry, self-adjointness, expansion in the magnetic momenta and relative boundedness with respect to $K_0$ are the content of the self-adjointness theorem on this page, not asserted here.