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Finite diagonal lift and its relative frequency window

Statement

Fix a spin label J≥1 and the spin blocks VJ and VJ−1. Let H0 and D be diagonal in their combined orthonormal spin basis, with real eigenvalues EKn and dKn=ΔKn, respectively, for K∈{J,J−1} and n=−K,−K+1,…,K, where Δ is real. Use units with ℏ=1, so directed frequencies are source energies minus destination energies.

For every formal triple with m=−J,−J+1,…,J and δ∈{−1,0,1}, define sJmδ=Δ[m−(J−1)δ]. Then ∣sJmδ∣≤(2J−1)∣Δ∣. The pair (J,m)→(J−1,m+δ) designates existing spin states only when m+δ∈{−(J−1),−(J−1)+1,…,J−1}. For such a pair, sJmδ is the change of its directed frequency on replacing H0 by H0+D. This is an energy-difference statement and does not assert that a transition matrix element is nonzero; for labels without a destination state, sJmδ remains only a formal algebraic expression.

If the transitions under consideration have unperturbed directed frequency EJm−EJ−1,m+δ=ΩJ=ω0hJ, where ω0,hJ>0, then ∣sJmδ/ΩJ∣≤rJ=(2J−1)∣Δ∣/(ω0hJ). For rJ<1, all their shifted frequencies ΩJ+sJmδ remain positive. The relative bound and positivity of the formal numbers ΩJ+sJmδ also hold for every listed formal triple.

Facts & Assumptions

[F1]

The spin convention supplies the blocks VK with orthonormal basis labels n=−K,−K+1,…,K. Thus a destination in VJ−1 must have its magnetic label in that block's range. Finite spin blocks, spherical tensors and Clebsch–Gordan phases

[F2]

If the unperturbed operator and lift are diagonal in the same specified basis, replacing the former by their sum changes a directed source-minus-destination energy difference by the source lift eigenvalue minus the destination lift eigenvalue. Calibrated diagonal lifts and transition phase differences

Proof

Given: The common diagonal spin basis, real eigenvalues, formal label ranges and positive unperturbed frequency specified in the Statement. Goal: Establish the algebraic window, its transition-energy interpretation on existing states, and positivity.

1.1givenalgebra

For every formal triple, ∣m∣≤J and ∣δ∣≤1. Since J−1≥0, the triangle inequality gives ∣sJmδ∣=∣Δ∣ ∣m−(J−1)δ∣≤∣Δ∣[J+(J−1)]=(2J−1)∣Δ∣. No destination state is needed for this algebraic inequality.

2.1givenF1F2step 1.1algebra

When m+δ is in the magnetic range of VJ−1, both states exist by [F1] and are common eigenvectors of H0 and D by hypothesis. Applying [F2] changes their directed frequency by dJm−dJ−1,m+δ=Δ[Jm−(J−1)(m+δ)]=sJmδ. A zero matrix element does not alter this energy-difference identity or establish a nonzero transition. Outside that destination range only the formal expression in step 1.1 is asserted.

3.1givenstep 1.1step 2.1algebra∎

Divide the inequality in step 1.1 by ΩJ=ω0hJ>0 to obtain ∣sJmδ/ΩJ∣≤rJ. If rJ<1, then ΩJ+sJmδ≥ΩJ−∣sJmδ∣≥ΩJ(1−rJ)>0 for every formal triple. For each existing-state pair whose unperturbed directed frequency is ΩJ, step 2.1 identifies this positive number as its shifted frequency.

Depends on

Used by · 1

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