QM.4 · Finite-well bound states and hydrogenic quantum numbers

A bound wavefunction can extend outside a finite well even though the particle’s energy is below the external potential.

필요 사전 지식: 5, 24, 37.

• 감쇠하는 유한 우물 상태 매칭 및 패리티 의존적 양자화 조건 도출
• 무차원 결속 상태 근해 해석 및 국소화와 산란 구분
• 수소형 양자수, 각운동량 및 반경 확률 측정값 사용

Declare a symmetric well V=−V₀ for |x|<a and V=0 outside, with V₀>0 and the same mass throughout. A bound energy lies between −V₀ and zero. Inside, k=√[2m(E+V₀)]/$\hbar$ gives oscillatory solutions; outside, κ=√(−2mE)/$\hbar$ gives decaying tails. Reject growing exponentials to obtain a normalisable state. Reflection symmetry permits even interior cos(kx) or odd sin(kx) states. At finite boundaries without delta interactions, both ψ and ψ′ are continuous. Even matching gives k tan(ka)=κ; odd matching gives −k cot(ka)=κ. A finite well does not require ψ to vanish at its edges as an infinite wall does.

Introduce z=ka and ρ=a√(2mV₀)/$\hbar$. Then κa=√(ρ²−z²) and 0<z<ρ. Solve z tan z=√(ρ²−z²) for an even state, or −z cot z=√(ρ²−z²) for an odd one, on intervals with the appropriate sign and away from tangent poles. Energy follows as E/V₀=z²/ρ²−1. For ρ=1, the ground root lies in 0<z<1 and the even equation is equivalent there to z=cos z. Bisection or a converged root finder gives z≈0.739085 and E/V₀≈−0.453753. Since ρ<π/2, no odd bound-state branch fits. In one dimension an attractive square well has an even bound ground state however shallow it is; an E=0 threshold tail is not square integrable.

For an even bound state write ψ=A cos(kx) inside and ψ=A cos(ka)exp[−κ(|x|−a)] outside. Continuity sets the relative tail amplitude; integrate |ψ|² over both the interior and tails to determine A. Probability outside the well is nonzero and depends on its depth, width and the selected state. Bound states have discrete energies and normalisable decaying asymptotes. A scattering state with E>0 has propagating external waves and is normalised or interpreted using a continuum/flux convention. A decaying tail is not evidence that the state violates energy conservation, and raw tail amplitude is not the outside probability until it has been integrated and normalised.

For a one-electron Coulomb ion with nuclear charge Ze, use the nonrelativistic central-potential model and a heavy nucleus approximation unless reduced mass is specified. E_n≈−13.6 Z²/n² eV; allowed orbital numbers are n≥1, l=0,…,n−1 and m_l=−l,…,l. L² has eigenvalue l(l+1)$\hbar$² and L_z=m_l$\hbar$; the ground 1s orbital has l=0, despite the historical Bohr circular-orbit rule. Spatial degeneracy at fixed n is Σ_l(2l+1)=n² before spin and fine-structure corrections. If ψ=R_nl(r)Y_lm with angular part normalised, radial probability in dr is r²|R|²dr, not just |R|²dr. For hydrogen 1s, the radial density is 4r²e^(−2r/a₀)/a₀³: it peaks at r=a₀ and has mean 3a₀/2, while the three-dimensional point density is largest at the origin. Distinguish these measures before locating a most probable radius.

유한 우트는 $|x|<a$에서 V = -V0이며, 그 외에서는 0입니다. 결합 에너지는 -V0와 0 사이에 엄격하게 위치합니다. 정의하십시오.

even 대칭성은 내부에서 코사인 함수와 감쇠 꼬리를 줍니다. a에서 값과 미분 계수를 일치시키면 odd 대칭성 대신 $-k\cot(ka)=\kappa$을 줍니다. 유효한 대칭성 구간 내에서만 풀이하십시오; tan z의 극점에서의 근은 결합 상태가 아닙니다.

Use the declared potential zero, match ψ′ as well as ψ, and include the radial shell measure. A bound tail and a continuum travelling wave have different energy regimes.

bound state: Normalisable stationary state with confined probability and discrete energy in the specified model.

반경 방향 확률 밀도: 각도 적분 후 구껍질 측도를 포함하는 반경 거리당 확률입니다.