** For any self-generated unitary Û: [σ̂, Û] = 0.
Self-operations are unitary by definition. Non-unitary evolution requires coupling to external systems (Lindblad dynamics). If you claim internal non-unitarity, you've either (a) expanded the system to include external input (contradicting "self-generated") or (b) violated quantum mechanics. The theorem applies within standard quantum formalism.
When two Hermitian operators commute, they share eigenstates. Unitary evolution preserves the eigenvalue associated with each eigenstate. If σ̂ and Û commute, and |−⟩ is a σ̂ eigenstate with eigenvalue −1, then after Û acts, the state remains in the −1 eigenspace. This is not accidental—it's the spectral theorem.
For physically realistic Hamiltonians (bounded below, finite energy), the generated unitaries are well-defined. The sign operator acts on a 2-dimensional space (compact); domain issues don't arise for σ̂. The theorem holds for all relevant cases.
No. The theorem derives from: (1) Self-operations are generated by the system's own Hamiltonian H. (2) σ̂ represents a symmetry of the system (moral orientation is intrinsic). (3) Symmetries commute with Hamiltonians (Noether). (4) Therefore [σ̂, Û] = 0 for Û = exp(−iHt). The commutation is derived, not stipulated.
The soul has quantum structure ([[087_E10.1_Soul-Field-Equation|E10.1]]). The sign operator acts on the moral Hilbert space, which is part of the soul's full Hilbert space. Theophysics extends quantum formalism to include moral degrees of freedom. If you reject this extension, you must provide an alternative—but the quantum structure is the most rigorous available.
Noether's Theorem: Every continuous symmetry corresponds to a conserved quantity.
For sign:
Physical parallel: Charge conservation follows from U(1) symmetry. Sign conservation follows from Z₂ moral symmetry.
Theorem ([[070_T8.1_Sign-Invariance-Theorem|T8.1]]): For any self-generated unitary Û, [σ̂, Û] = 0.
Proof:
1. Let Û be self-generated: Û = exp(−iHt/ℏ) for system Hamiltonian H
2. σ̂ is a symmetry: [σ̂, H] = 0 (sign is conserved under internal dynamics)
3. Baker-Campbell-Hausdorff: [σ̂, exp(−iHt)] = Σ (−it)ⁿ/n! [σ̂, Hⁿ]
4. [σ̂, Hⁿ] = 0 for all n (since [σ̂, H] = 0)
5. Therefore: [σ̂, Û] = 0 ∎