A spin-one multiplet in a uniform field: the exact Zeeman spectrum and degenerate first-order check

Adopt P1, P2, P4 and P6 for the massive spin-$s$ particle model of The massive spin-s particle model of the framework: state space, operators, units and frame readings (State and composition postulate P1: Hilbert spaces, preparations and composite systems, Observable and probability postulate P2: self-adjoint observables, spectral probability and finite unsharp outcomes, Autonomous dynamics postulate P4: self-adjoint generators and unitary evolution, Identical-particle and spin assignments P6: permutation sectors and finite-spin representations), together with the effective magnetic-moment coupling Effective magnetic-moment coupling of the finite-spin model: gyromagnetic parameter, magnetic moment and Zeeman energy operator with real gyromagnetic parameter $\gamma :={\gamma }_1$ and the spin-one internal factor ${\mathbb{C}}^3$ with spin operators $S_i=\mathrm{\hslash }{J}_i$ of Finite spin matrices at $s=1$; read energies in joules as in Dimensions, SI unit conventions and quantity types of the framework. Declare the finite-level Zeeman model by the perturbed-family model of The declared perturbed-family model: bounded self-adjoint perturbation, isolated gapped label and followed spectral cluster with $\mathfrak{H}:={\mathbb{C}}^3,H_0:=\epsilon I_3,V:=-\gamma B_z{S}_3,$ where $\epsilon \in \mathbb{R}$ is a declared multiplet energy, $B_z>0$ is the field magnitude in tesla of the uniform field $\mathbf{B}=(0,0,B_z)$, so that the family is $H(\lambda )=H_0+\lambda V$ with real coupling $\lambda$ (dimensionless, equal to $1$ for the physical field) and $v=\mathrm{\parallel }V\mathrm{\parallel }=\mathrm{\mid }\gamma \mathrm{\mid }{B}_z\mathrm{\hslash }$. For the physical value $\lambda =1$ the family is the model's Zeeman operator on the internal multiplet. Then:

(a) the perturbation preserves the multiplet: $[H_0,V]=0$, the first-order operator is $V_E=V$, and its eigenvalues counted with multiplicity are $-\mathrm{\hslash }\gamma B_z,0,+\mathrm{\hslash }\gamma B_z$ in joules. If $\gamma \mathrm{\ne }0$, these are three distinct simple eigenvalues; if $\gamma =0$, the only eigenvalue is $0$ with multiplicity three. Thus the first-order operator describes a triply degenerate level when the adopted magnetic coupling vanishes;

(b) the exact energy eigenvalues of the family, counted with multiplicity, are $\epsilon -\lambda \mathrm{\hslash }\gamma B_z,\epsilon ,\epsilon +\lambda \mathrm{\hslash }\gamma B_z\mathrm{.}$ They are three distinct simple levels exactly when $\lambda \gamma \mathrm{\ne }0$; when $\lambda =0$ or $\gamma =0$, the only energy is $\epsilon$ with multiplicity three. Each eigenvalue is affine in $\lambda$, so its second-order and higher Taylor coefficients vanish identically;

(c) when $v>0$, the page's degenerate theorem applies for every $g>0$ (the gap condition is vacuous because the spectrum of $H_0$ is the single value $\epsilon$) and every real $\lambda$ with $\mathrm{\mid }\lambda \mathrm{\mid }v<g\mathrm{/}4$; each eigenvalue $z$ then satisfies $\mathrm{dist}(z,\epsilon +\lambda \mathrm{spec}(V_E))\le 4{\lambda }^2{v}^2\mathrm{/}g$, and the exact spectrum in (b) shows that this distance is $0$. If $v=0$, equivalently $\gamma =0$, then $H(\lambda )=\epsilon I_3$ for every real $\lambda$, so the same distance is $0$ directly, without invoking the theorem's positive-norm hypothesis. In both cases the first-order eigenvalue set is exact, with multiplicity retained as in (a)-(b);

(d) for a unit preparation vector $u\in {\mathbb{C}}^3$, put $u(t)=e^{-itH(\lambda )\mathrm{/}\mathrm{\hslash }}u$ and $s_i(t)=⟨S_{i}u(t),u(t)⟩$. With $\omega =\lambda \gamma B_z$ (in ${\mathrm{s}}^{-1}$), $s_3(t)=s_3(0),s_1(t)=\cos (\omega t)s_1(0)+\sin (\omega t)s_2(0),s_2(t)=\cos (\omega t)s_2(0)-\sin (\omega t)s_1(0)\mathrm{.}$ Thus the transverse means rotate by $-\omega t$, with the signed Larmor parameter $\omega$ equal to $\gamma B_z$ at $\lambda =1$;

(e) no branch selection: the energy eigenspaces of (b) are spectral data of the family; a preparation in any eigenstate is stationary, and a superposition evolves with the relative phases of (b). The example asserts no instantaneous transition of any preparation to the lowest sublevel, no relaxation and no measurement; which sublevel a preparation occupies is fixed by the preparation and by P4.

• Degenerate perturbation theory is exact here. The perturbation commutes with $H_0$ and acts inside the multiplet, so the first-order operator is the whole perturbation and there is no second-order shift; when $\lambda \gamma \mathrm{\ne }0$ it gives three distinct levels, while zero coupling leaves one triply degenerate level. The example exhibits the degenerate theorem's first-order content on a clean exactly solvable family. This is the Zeeman side of the page's examples; the Stark side uses a nondegenerate truncation with a genuine second-order shift.
• No relaxation. When distinct levels occur, they do not force a preparation to move to the lowest one; the affine spectral shifts, with degeneracy retained at zero coupling, are statements about the spectrum, in agreement with the corrected reading of the textbook degenerate-perturbation conjecture recorded in What the controlled-approximations page adopts, proves and does not claim: gaps, branch populations, turning points and the static-Stark limitation.