LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingjudge pass (gpt-6-luna)

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Positive-energy eigenspaces pair under an odd self-adjoint supercharge

Statement

Assume Countable Choice. Let $H = H^{+} \oplus H^{-}$ be a complex Hilbert space with grading operator $Γ = + I$ on $H^{+}$ and $Γ = - I$ on $H^{-}$ . Let $Q$ be a densely defined self-adjoint operator with domain invariant under $Γ$ and $Q Γ ψ = - Γ Q ψ$ for every $ψ \in D (Q)$ . Set $H = Q^{2}$ on the composition domain $D (H) = {ψ \in D (Q) : Q ψ \in D (Q)}$ . For each $λ > 0$ , let $E_{λ} = {ψ \in D (H) : H ψ = λ ψ}$ be finite-dimensional, and put $E_{λ}^{\pm} = E_{λ} \cap H^{\pm}$ . Then $Q$ maps $E_{λ}$ into itself, and

$U_{λ} : = (Q / \sqrt{λ}) ∣_{E_{λ}^{+}} : E_{λ}^{+} ⟶ E_{λ}^{-}$

is a unitary isomorphism, with inverse $(Q / \sqrt{λ}) ∣_{E_{λ}^{-}}$ . In particular $\dim E_{λ}^{+} = \dim E_{λ}^{-}$ . Also $Q ψ = 0$ for every zero-energy vector $ψ \in \ker H$ . No discrete spectrum or finite-multiplicity assumption is made about energies other than the selected $λ$ .

Statement references

Martin, §7.1.

Gates et al., §8.4.

Hilbert space: Hilbert space.

Real and complex inner-product spaces and their induced length: Real and complex inner-product spaces and their induced length.

Unbounded linear operators: domain, graph and extension: Unbounded linear operators: domain, graph and extension.

Densely defined, closed and closable operators: Densely defined, closed and closable operators, and cores.

Adjoint of a densely defined operator: Adjoint of a densely defined operator.

Symmetric, self-adjoint and essentially self-adjoint operators: Symmetric, self-adjoint and essentially self-adjoint operators.

Complex arithmetic: The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ .

Complex numbers form a field: $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ .

Every nonnegative real number has a unique nonnegative square root: Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ .

Facts & Assumptions

Given: $H$ , $Γ$ , $Q$ , the composition operator $H = Q^{2}$ , and a finite-dimensional positive-energy eigenspace $E_{λ}$ as in the Statement. Assume Countable Choice for the imported adjoint and self-adjoint definitions.

[A1]

Countable Choice is used by the imported unbounded adjoint and self-adjoint operator definitions. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

Real and complex inner-product spaces and their induced length: the complex inner product is linear in its first argument, conjugate-linear in its second and positive definite.

[F2]

Hilbert space: $H$ is complete in the norm induced by its complex inner product.

[F3]

Unbounded linear operators: domain, graph and extension: the domain of a composition $Q^{2}$ is ${ψ \in D (Q) : Q ψ \in D (Q)}$ .

[F4]

Adjoint of a densely defined operator: self-adjointness gives $⟨ Q x, y ⟩ = ⟨ x, Q y ⟩$ whenever $x, y \in D (Q)$ .

[F5]

Symmetric, self-adjoint and essentially self-adjoint operators: a densely defined operator is self-adjoint when its domain agrees with the domain of its actual adjoint and the values agree.

[F6]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : every positive real $λ$ has a unique positive square root $\sqrt{λ}$ .

[F7]

$C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ : the nonzero complex scalar $\sqrt{λ}$ has a multiplicative inverse and scalar multiplication is associative.

Proof

Given: [A1], [F1]–[F7], and the domain, grading and anticommutation hypotheses in the Statement.

1.1F2F3givenalgebra

If $ψ \in E_{λ}$ , then $ψ \in D (Q)$ , $Q ψ \in D (Q)$ and $Q^{2} ψ = λ ψ$ by [F3]. Thus $Q ψ \in D (Q)$ and $Q (Q ψ) = λ ψ \in D (Q)$ , so a second application of [F3] gives $Q ψ \in D (H)$ . Moreover $H (Q ψ) = Q (Q^{2} ψ) = Q (λ ψ) = λ Q ψ$ . Therefore $Q$ maps $E_{λ}$ into itself.

2.1givenF3step 1.1algebra

The domain of $H$ is invariant under $Γ$ : if $ψ \in D (H)$ , then $Γ ψ \in D (Q)$ by hypothesis and $Q Γ ψ = - Γ Q ψ \in D (Q)$ because $Q ψ \in D (Q)$ and $D (Q)$ is $Γ$ -invariant. Applying the anticommutation relation twice gives $H Γ ψ = Γ H ψ$ . Hence $Γ E_{λ} = E_{λ}$ , and $Q Γ = - Γ Q$ shows that $Q$ maps $E_{λ}^{+}$ to $E_{λ}^{-}$ and $E_{λ}^{-}$ to $E_{λ}^{+}$ .

3.1A1F1F2F3F4F5F6F7step 1.1step 2.1algebra

For $ψ, φ \in E_{λ}$ , both vectors lie in $D (Q)$ and $Q ψ \in D (Q)$ , by [F2], the adjoint is taken in the complete Hilbert space; self-adjointness in [F5] and the identity in [F4] give $⟨ Q ψ, Q φ ⟩ = ⟨ Q^{2} ψ, φ ⟩ = λ ⟨ ψ, φ ⟩$ . By [F6], and since $\sqrt{λ}$ is nonzero in the complex field by [F7], division by $\sqrt{λ}$ therefore makes $Q$ an isometry on $E_{λ}$ . Also $(Q / \sqrt{λ})^{2} ψ = Q^{2} ψ / λ = ψ$ there. Together with the parity exchange in step 2.1, this proves that the restriction $U_{λ} : E_{λ}^{+} \to E_{λ}^{-}$ is an isometry with inverse the restriction from $E_{λ}^{-}$ to $E_{λ}^{+}$ ; it is consequently unitary and the finite dimensions agree.

4.1A1F1F2F4F5givenalgebra∎

If $ψ \in \ker H$ , then $ψ \in D (Q^{2})$ , so $Q ψ \in D (Q)$ . By [F2], [F5] and the adjoint identity [F4], $∥ Q ψ ∥^{2} = ⟨ Q ψ, Q ψ ⟩ = ⟨ Q^{2} ψ, ψ ⟩ = 0$ . Positive definiteness in [F1] forces $Q ψ = 0$ .

Positive-energy eigenspaces pair under an odd self-adjoint supercharge

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources