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=== If the fundamental phase loop is half the geometric circumference (i.e. the physical boundary that the wave accumulates phase on has radius reff=r/2r_{\rm eff}=r/2reff=r/2), then the allowed angular‑momentum eigenvalues for the full geometric circle become half‑integers: === Lz=(s+12)ℏ,s∈Z,L_z = \left(s+\tfrac12\right)\hbar,\qquad s\in\mathbb{Z},Lz=(s+21)ℏ,s∈Z, which is the same discrete ladder structure as spin-12\tfrac1221. ==== Use the geometric angle φ∈[0,2π)\varphi\in[0,2\pi)φ∈[0,2π) around the full circle of radius rrr. The angular‑momentum operator is ==== L^z = −iℏ ∂∂φ.\hat L_z \;=\; -i\hbar\,\frac{\partial}{\partial\varphi}.L^z=−iℏ∂φ∂. With the usual single‑valued boundary condition ψ(φ+2π)=ψ(φ),\psi(\varphi+2\pi)=\psi(\varphi),ψ(φ+2π)=ψ(φ), the normalized eigenfunctions are ψm(φ)=12πeimφ,m∈Z,\psi_m(\varphi)=\dfrac{1}{\sqrt{2\pi}}e^{im\varphi},\qquad m\in\mathbb{Z},ψm(φ)=2π1eimφ,m∈Z, and the eigenvalues are L^zψm=mℏψm⇒Lz=mℏ, m∈Z.\hat L_z\psi_m = m\hbar \psi_m \qquad\Rightarrow\qquad L_z = m\hbar,\ m\in\mathbb{Z}.L^zψm=mℏψm⇒Lz=mℏ, m∈Z. So integer spectrum arises from 2π2\pi2π-periodicity of ψ(φ)\psi(\varphi)ψ(φ). ==== QAT hypothesis: the elementary phase domain for the microscopic boundary is a loop of radius reff=r/2r_{\rm eff}=r/2reff=r/2. Equivalently, the phase unit is accumulated along a path whose geometric angular advance corresponds to half the full geometric circle. ==== Mathematically implement this by introducing an elementary angular coordinate θ\thetaθ that parameterizes the elementary loop and runs 0≤θ<2π0\le\theta<2\pi0≤θ<2π. The geometric angle φ\varphiφ (full circle) is related to θ\thetaθ by φ=2θ.\varphi = 2\theta.φ=2θ. Thus advancing θ\thetaθ by 2π2\pi2π advances φ\varphiφ by 4π4\pi4π. In words: one elementary loop equals half the geometric circumference; two elementary loops equal one full geometric 2π2\pi2π rotation. Because the physical microscopic wave is defined on the elementary loop, impose single‑valuedness on θ\thetaθ: Ψ(θ+2π)=Ψ(θ).\Psi(\theta+2\pi)=\Psi(\theta).Ψ(θ+2π)=Ψ(θ). Rewrite that condition in terms of φ\varphiφ: since θ=φ/2\theta=\varphi/2θ=φ/2, Ψ (φ+4π2)=Ψ (φ2)⟹Ψ(φ+4π)=Ψ(φ).\Psi\!\big(\tfrac{\varphi+4\pi}{2}\big)=\Psi\!\big(\tfrac{\varphi}{2}\big) \quad\Longrightarrow\quad \Psi(\varphi+4\pi)=\Psi(\varphi).Ψ(2φ+4π)=Ψ(2φ)⟹Ψ(φ+4π)=Ψ(φ). So the wavefunction is 4π4\pi4π-periodic in the geometric angle φ\varphiφ. That allows two possibilities for the behavior under a geometric 2π2\pi2π rotation: Ψ(φ+2π)=± Ψ(φ).\Psi(\varphi+2\pi)=\pm\,\Psi(\varphi).Ψ(φ+2π)=±Ψ(φ). The physically interesting case giving half‑integers is the antiperiodic choice Ψ(φ+2π)=−Ψ(φ),\Psi(\varphi+2\pi)=-\Psi(\varphi),Ψ(φ+2π)=−Ψ(φ), because this encodes the fact that one geometric 2π2\pi2π rotation equals two elementary loops (so the wave picks up a minus sign after one geometric 2π2\pi2π but returns after 4π4\pi4π) — exactly the spinor double‑cover behaviour. ==== Seek eigenfunctions of L^z\hat L_zL^z of the form Ψs(φ)=exp(isφ)\Psi_s(\varphi)=\exp(i s \varphi)Ψs(φ)=exp(isφ). Impose antiperiodicity: ==== Ψs(φ+2π)=eis(φ+2π)=eisφei2πs=!−eisφ.\Psi_s(\varphi+2\pi)=e^{i s(\varphi+2\pi)} = e^{i s\varphi} e^{i 2\pi s} \stackrel{!}{=} - e^{i s\varphi}.Ψs(φ+2π)=eis(φ+2π)=eisφei2πs=!−eisφ. Cancel eisφe^{i s\varphi}eisφ to get ei2πs=−1⇒2πs=(2k+1)π,k∈Z.e^{i2\pi s}=-1 \quad\Rightarrow\quad 2\pi s = (2k+1)\pi,\quad k\in\mathbb{Z}.ei2πs=−1⇒2πs=(2k+1)π,k∈Z. Therefore s=k+12,k∈Z.s = k+\tfrac12,\qquad k\in\mathbb{Z}.s=k+21,k∈Z. Acting with L^z\hat L_zL^z, L^zΨs=−iℏ∂∂φeisφ=sℏ eisφ.\hat L_z \Psi_s = -i\hbar \frac{\partial}{\partial\varphi} e^{i s \varphi} = s\hbar\, e^{i s\varphi}.L^zΨs=−iℏ∂φ∂eisφ=sℏeisφ. Hence the allowed eigenvalues are Lz=sℏ=(k+12)ℏ,k∈Z,L_z = s\hbar = \big(k+\tfrac12\big)\hbar,\qquad k\in\mathbb{Z},Lz=sℏ=(k+21)ℏ,k∈Z, i.e. half‑integer multiples of ℏ\hbarℏ. QED. ==== Classically L=pφ=mvrL = p_\varphi = m v rL=pφ=mvr. If the elementary momentum circulation is quantized around the smaller loop of radius reff=r/2r_{\rm eff}=r/2reff=r/2, ==== mv(reff)=nℏ,m v (r_{\rm eff}) = n\hbar,mv(reff)=nℏ, then for the full geometric radius rrr the corresponding angular momentum reads Lgeom=mvr=rreff nℏ=2nℏ.L_{\rm geom}=m v r = \frac{r}{r_{\rm eff}}\,n\hbar = 2 n\hbar.Lgeom=mvr=reffrnℏ=2nℏ. Counting the elementary quantum as n=12n=\tfrac12n=21 relative to the full-circle counting yields the half‑integer ladder for the full geometry. The precise quantum derivation above (antiperiodic wavefunctions) gives the correct operator result without ambiguity. ==== - Mathematically: if the microscopic boundary on which the wavefunction is defined has half the geometric circumference (or equivalently the physical single‑valuedness condition is that the wave is single only after a 4π4\pi4π geometric rotation), then the allowed eigenstates of L^z\hat L_zL^z on the full circle are half‑integers: Lz=(k+12)ℏL_z=(k+\tfrac12)\hbarLz=(k+21)ℏ. This follows directly from imposing antiperiodic boundary conditions Ψ(φ+2π)=−Ψ(φ)\Psi(\varphi+2\pi)=-\Psi(\varphi)Ψ(φ+2π)=−Ψ(φ). ==== * Physically: the step “reff=r/2r_{\rm eff}=r/2reff=r/2 ⇒ antiperiodic BC” must be justified by a microphysical boundary mechanism (why is the elementary phase domain half the geometric circumference?). If QAT provides that geometric/physical reason (boundary-layer, concave/convex charge partition, photon‑electron dipole geometry), then the above is a direct, rigorous one‑dimensional quantum demonstration of half‑integer quantization. * A caution: this is not a replacement for the full QFT explanation of spin (which arises from representation theory of the Lorentz group and field‑theory construction). What we have derived is a clear quantum‑mechanical mechanism that produces the same discrete eigenvalue pattern (half‑integers) from a geometric boundary condition — exactly the behavior you want to attribute geometrically in QAT. If you like, I’ll now: • produce a short, slide‑ready block of LaTeX you can paste into your Core Equation document, or • write the boundary‑action derivation (Lagrangian on the 2‑D sphere boundary leading to identical antiperiodic condition), which is the next step to make this fully rigorous. Which would you prefer next?
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