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Self-adjointness of the bounded-coefficient magnetic Schrodinger operator

Statement

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix $ℏ > 0$ , $m > 0$ , $q \in R$ and an admissible pair $(A, Φ)$ in the sense of Bounded-coefficient magnetic Schrodinger operator, and let $P_{A, j}$ , $K_{A, Φ}$ and $K_{0}$ be the operators defined there on $L^{2} (R^{3}; C)$ . Then:

(a) for every $j$ , $P_{A, j} = P_{j} - q A_{j}$ is self-adjoint with domain $D (P_{A, j}) = D (P_{j}) = {f : D_{j} f \in L^{2}}$ ; and for every $u \in W^{2, 2} (R^{3}; C)$ one has $P_{A, j} u \in W^{1, 2}$ and $P_{A, j} u = P_{j} u - q A_{j} u$ ;

(b) for every $u \in W^{2, 2} (R^{3}; C)$ the expansion holds: $K_{A, Φ} u = \frac{1}{2 m} P^{2} u - \frac{q}{2 m} \sum_{j = 1}^{3} (P_{j} A_{j} + A_{j} P_{j}) u + \frac{q^{2}}{2 m} ∣ A ∣^{2} u + q Φ u, P^{2} u : = \sum_{j = 1}^{3} P_{j} (P_{j} u) = - ℏ^{2} \sum_{j = 1}^{3} D_{j} D_{j} u,$ where $(P_{j} A_{j} + A_{j} P_{j}) u : = P_{j} (A_{j} u) + A_{j} P_{j} u$ , and each term lies in $L^{2}$ ;

(d) for every $a \in (0, 1)$ there is $b \in [0, \infty)$ with $∥ (K_{A, Φ} - K_{0}) u ∥ \leq a ∥ K_{0} u ∥ + b ∥ u ∥ (u \in W^{2, 2} (R^{3}; C)),$ so $K_{A, Φ} - K_{0}$ has $K_{0}$ -bound strictly below one (Relative boundedness with respect to an operator); consequently $K_{A, Φ}$ is bounded below.

The assertion is a statement about the displayed differential expression with its stated coefficient hypotheses; it identifies no physical field, particle or measured Hamiltonian.

Facts & Assumptions

Given: $ℏ > 0$ , $m > 0$ , $q \in R$ ; an admissible pair $(A, Φ)$ with $∥ A_{j} ∥_{\infty}, ∥ D_{k} A_{j} ∥_{\infty}, ∥ Φ ∥_{\infty} < \infty$ ; the operators $P_{j}$ , $Q_{j}$ , $P_{A, j}$ , $K_{A, Φ}$ , $K_{0}$ of the definition.

[F1]

$P_{j}$ is self-adjoint with $D (P_{j}) = {f : ξ_{j} F_{2} f \in L^{2}}$ , and $F_{2} (P_{j} f) = 2 π ℏ ξ_{j} F_{2} f$ for $f \in D (P_{j})$ ; also $D_{j} f \in L^{2}$ and $P_{j} f = - i ℏ D_{j} f$ for $f \in D_{j}$ in this domain (Self-adjointness, Fourier representation and translation identities of position and momentum).

[F2]

Multiplication by a real bounded measurable class is a bounded self-adjoint operator, and $q A_{j}$ , $Φ$ , $∣ A ∣^{2}$ are real bounded classes (Real multiplication operators and free plus bounded-potential Schrödinger operators, clause (a); Bounded-coefficient magnetic Schrodinger operator).

[F3]

Kato-Rellich: if $S$ is self-adjoint and an operator $B$ is symmetric with $D (S) \subseteq D (B)$ and $S$ -bound less than one, then $S + B$ with domain $D (S)$ is self-adjoint; moreover if $S \geq γ$ and $(a, b)$ is an admissible pair for $B$ with $a < 1$ , then $S + B \geq γ - \max {a ∣ γ ∣ + b, b / (1 - a)}$ (Kato-Rellich theorem, Relative boundedness with respect to an operator).

[F4]

For $u \in W^{1, 2}$ , $A_{j} u \in W^{1, 2}$ and $D_{k} (A_{j} u) = A_{j} D_{k} u + (D_{k} A_{j}) u$ ; hence $A_{j} u \in D (P_{j})$ with $P_{j} (A_{j} u) = A_{j} P_{j} u - i ℏ (D_{j} A_{j}) u$ (Weak product rule for a bounded coefficient with bounded weak derivative).

[F5]

For $u \in W^{2, 2}$ and admissible $A$ , $P_{A, j} u = P_{j} u - q A_{j} u \in W^{1, 2}$ : indeed $P_{j} u = - i ℏ D_{j} u \in W^{1, 2}$ because $D_{j} u \in W^{1, 2}$ and $A_{j} u \in W^{1, 2}$ by [F4] (Bounded-coefficient magnetic Schrodinger operator, Integer-order Sobolev spaces and their norms).

[F6]

$F_{2}$ is unitary and $W^{2, 2} (R^{3}; C) = {f : ∣ ξ ∣^{2} F_{2} f \in L^{2}}$ with equivalent norms (Plancherel theorem, Integer-order W^{k,2} and H^k agree with equivalent norms).

Proof

1.1givenF1F2F3

[Self-adjoint magnetic momenta] For each $j$ , $q A_{j}$ is a bounded symmetric operator by [F2] with $∥ q A_{j} f ∥ \leq ∣ q ∣ ∥ A_{j} ∥_{\infty} ∥ f ∥$ for every $f \in L^{2}$ , so its $P_{j}$ -bound is $0$ : the pair $(a, b) = (0, ∣ q ∣ ∥ A_{j} ∥_{\infty})$ is admissible in the sense of [F3] since $D (P_{j}) \subseteq L^{2} = D (q A_{j})$ . By [F3] with $S = P_{j}$ and $B = - q A_{j}$ (symmetric), $P_{A, j} = P_{j} - q A_{j}$ is self-adjoint with domain $D (P_{j})$ . This is the first assertion of (a).

1.2givenF1F4F5

[Membership of the shifted momentum] Let $u \in W^{2, 2}$ . Then $D_{j} u \in W^{1, 2}$ , so $P_{j} u = - i ℏ D_{j} u \in W^{1, 2}$ , and $A_{j} u \in W^{1, 2}$ by [F4]; hence $P_{A, j} u = P_{j} u - q A_{j} u \in W^{1, 2}$ and $P_{A, j} u = P_{j} u - q A_{j} u$ holds as classes. This completes (a).

1.3givenF1F6algebra

[Momentum bound by the free operator] For $u \in W^{2, 2}$ , $P_{j} u \in D (P_{j})$ and $F_{2} (P_{j} u) = 2 π ℏ ξ_{j} F_{2} u$ by [F1], while $K_{0} u = F_{2}^{*} (c ∣ ξ ∣^{2} F_{2} u)$ with $c : = (2 π ℏ)^{2} / (2 m)$ by [F1] and the definition of $K_{0}$ : iterating the multiplier identity gives $F_{2} (\frac{1}{2 m} \sum_{j} P_{j} (P_{j} u)) = \frac{(2 π ℏ)^{2}}{2 m} ∣ ξ ∣^{2} F_{2} u$ . Hence $∥ P_{j} u ∥ = ∥ 2 π ℏ ξ_{j} F_{2} u ∥$ and $∥ K_{0} u ∥ = ∥ c ∣ ξ ∣^{2} F_{2} u ∥$ by unitarity of $F_{2}$ [F6]. Given $ε > 0$ , the elementary inequality $2 π ℏ ∣ t ∣ \leq ε c t^{2} + C_{ε}$ (with $t = ∣ ξ ∣$ , $C_{ε} = (2 π ℏ)^{2} / (4 ε c)$ ) gives $∥ P_{j} u ∥ \leq ε ∥ K_{0} u ∥ + C_{ε} ∥ u ∥$ .

2.1givenstep 1.1step 1.2algebra

[Symmetry of the individual magnetic squares] For $u, v \in W^{2, 2}$ and each $j$ , step 1.2 gives $P_{A, j} u, P_{A, j} v \in W^{1, 2} \subseteq D (P_{j}) = D (P_{A, j})$ by step 1.1, and $P_{j} u, P_{j} v \in W^{1, 2} \subseteq D (P_{j})$ (Bounded-coefficient magnetic Schrodinger operator, Integer-order Sobolev spaces and their norms). Using self-adjointness of $P_{A, j}$ and of $P_{j}$ (step 1.1, [F1]) on their domains, $⟨ P_{A, j} (P_{A, j} u), v ⟩ = ⟨ P_{A, j} u, P_{A, j} v ⟩ = ⟨ u, P_{A, j} (P_{A, j} v) ⟩$ and $⟨ P_{j} (P_{j} u), v ⟩ = ⟨ P_{j} u, P_{j} v ⟩ = ⟨ u, P_{j} (P_{j} v) ⟩$ . Subtracting, the operator $B_{j} : = \frac{1}{2 m} (P_{A, j}^{2} - P_{j}^{2})$ with $B_{j} u : = \frac{1}{2 m} (P_{A, j} (P_{A, j} u) - P_{j} (P_{j} u))$ on $W^{2, 2}$ is symmetric.

2.2givenstep 1.2F4algebra

[Expansion] Let $u \in W^{2, 2}$ . By [F4] applied to $D_{k} u$ (which lies in $W^{1, 2}$ ), or directly to $u$ , $P_{j} (A_{j} u) = A_{j} P_{j} u - i ℏ (D_{j} A_{j}) u$ ; hence $P_{A, j} (P_{A, j} u) = (P_{j} - q A_{j}) (P_{j} u - q A_{j} u) = P_{j} (P_{j} u) - q P_{j} (A_{j} u) - q A_{j} P_{j} u + q^{2} A_{j}^{2} u = P^{2} {-term}_{j} - q (P_{j} A_{j} + A_{j} P_{j}) u + q^{2} A_{j}^{2} u$ , where $(P_{j} A_{j} + A_{j} P_{j}) u : = P_{j} (A_{j} u) + A_{j} P_{j} u$ . Summing over $j$ and dividing by $2 m$ , and using $P^{2} u = \sum_{j} P_{j} (P_{j} u)$ and $∣ A ∣^{2} = \sum_{j} A_{j}^{2}$ , gives the displayed expansion; every term lies in $L^{2}$ by steps 1.2 and [F2]. This is (b).

3.1givenstep 2.1F2

[Symmetry of $K_{A, Φ}$ ] The zero-order term $q Φ$ is bounded symmetric by [F2]. Hence $K_{A, Φ} = K_{0} + \sum_{j} B_{j} + q Φ$ is symmetric on the common domain $W^{2, 2}$ , because $K_{0} u = \frac{1}{2 m} \sum_{j} P_{j} (P_{j} u)$ by definition and each summand is symmetric by step 2.1 and [F2].

3.2givenstep 2.2F2F4algebra

[The relative bound] Let $u \in W^{2, 2}$ . From (b), $∥ (K_{A, Φ} - K_{0}) u ∥ \leq \frac{∣ q ∣}{2 m} \sum_{j} ∥ (P_{j} A_{j} + A_{j} P_{j}) u ∥ + \frac{q^{2}}{2 m} ∥ ∣ A ∣^{2} u ∥ + ∣ q ∣ ∥ Φ u ∥$ . By [F4], $∥ (P_{j} A_{j} + A_{j} P_{j}) u ∥ \leq 2 ∥ A_{j} ∥_{\infty} ∥ P_{j} u ∥ + ℏ ∥ D_{j} A_{j} ∥_{\infty} ∥ u ∥$ ; and $∥ ∣ A ∣^{2} u ∥ \leq ∥ A ∥_{\infty}^{2} ∥ u ∥$ , $∥ Φ u ∥ \leq ∥ Φ ∥_{\infty} ∥ u ∥$ . So it suffices to bound $∥ P_{j} u ∥$ in terms of $∥ K_{0} u ∥$ and $∥ u ∥$ .

4.1step 3.2step 1.3algebra

[Conclusion of (d)] Fix $a \in (0, 1)$ and put $C : = \frac{∣ q ∣}{m} \sum_{j} ∥ A_{j} ∥_{\infty}$ and $B_{0} : = \frac{∣ q ∣ ℏ}{2 m} \sum_{j} ∥ D_{j} A_{j} ∥_{\infty} + \frac{q^{2}}{2 m} ∥ ∣ A ∣^{2} ∥_{\infty} + ∣ q ∣ ∥ Φ ∥_{\infty}$ . Steps 3.2 and 1.3 give $∥ (K_{A, Φ} - K_{0}) u ∥ \leq C ε ∥ K_{0} u ∥ + (C C_{ε} + B_{0}) ∥ u ∥$ . If $C > 0$ , take $ε = a / C$ and $b = C C_{ε} + B_{0}$ ; if $C = 0$ , take $b = B_{0}$ . In both cases the asserted inequality holds for this prescribed $a$ . The infimum of admissible relative constants is therefore zero, but an estimate with $a = 0$ is not asserted.

5.1step 3.1step 4.1F3∎

[Self-adjointness] By step 2.2 the operator $W : = K_{A, Φ} - K_{0}$ is defined on $D (K_{0}) = W^{2, 2}$ and is symmetric by step 3.1 (it is the sum of the symmetric $B_{j}$ and the bounded symmetric $q Φ$ ). By [F3] (Kato-Rellich) with $S = K_{0}$ and $B = W$ , whose $K_{0}$ -bound is below one by step 4.1, the operator $K_{0} + W = K_{A, Φ}$ is self-adjoint with domain $D (K_{0}) = W^{2, 2}$ . Moreover $K_{0} u = F_{2}^{*} (c ∣ ξ ∣^{2} F_{2} u)$ is positive (its quadratic form is $\int c ∣ ξ ∣^{2} ∣ F_{2} u ∣^{2} \geq 0$ ), so $K_{0} \geq 0$ and [F3] gives the lower bound $K_{A, Φ} \geq - \max {0, b / (1 - a)} = - b / (1 - a)$ for the admissible pair of step 4.1. This is (c) and the final clause of (d).

Self-adjointness of the bounded-coefficient magnetic Schrodinger operator

Statement

Facts & Assumptions

Proof

Depends on

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