有限ポテンシャル井戸の束縛状態および水素様量子数
導入済み| English | 意味 |
|---|---|
| bound state/baʊnd steɪt/ | 束縛状態 |
| radial probability density/ˈreɪdɪəl ˌprɒbəˈbɪlɪti ˈdensɪti/ | 放射確率密度 |
A decision before an answer
- A bound wavefunction can extend outside a finite well even though the particle’s energy is below the external potential.
- Your goal: Match decaying finite-well states and derive parity-dependent quantisation conditions.
Match bound-state asymptotes
- 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)]/ℏ gives oscillatory solutions; outside, κ=√(−2mE)/ℏ 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.
For a finite square well with equal mass on both sides and no delta interaction, boundary matching requires:
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Both wavefunction and first derivative are continuous at a finite step under these assumptions.
Solve a dimensionless root
- Introduce z=ka and ρ=a√(2mV₀)/ℏ. 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 a hydrogenic orbital n=4, which pair (l,m_l) is allowed?
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l can be 0 through 3, and |m_l| cannot exceed l.
Normalise the interior and tails
- 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 ρ=1, z≈0.739085 gives κa≈0.673612 and E≈−0.453753V₀; the even ground state leaks beyond ±a but decays at infinity. For a one-electron Z=2 ion in n=2, E≈−13.6·4/4=−13.6 eV. Its allowed l values are 0 and 1, giving four spatial states before spin. In a hydrogen 1s orbital, the radial density peaks at a₀ because the r² shell factor offsets the decreasing point density; integrating the radial density gives ⟨r⟩=3a₀/2.
For n=4 in the nonrelativistic Coulomb model, spatial degeneracy before spin is ____.
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Σ_l=0 to 3 (2l+1)=1+3+5+7=16.
Choose the hydrogenic probability measure
- 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)ℏ² and L_z=m_lℏ; 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.
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.
Which answer fits this case?
Match decaying finite-well states and derive parity-dependent quantisation conditions
A finite-well bound state can have nonzero probability outside the well while decaying at infinity.
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Its external exponential tails are nonzero near the edges but integrable for κ>0.
Keep the distinctions
- bound state 束缚态 — Normalisable stationary state with confined probability and discrete energy in the specified model.
- radial probability density 径向概率密度 — Probability per radial distance, including the spherical-shell measure after angular integration.
- Match decaying finite-well states and derive parity-dependent quantisation conditions.
- Interpret dimensionless bound-state roots and distinguish confinement from scattering.
- Use hydrogenic quantum numbers, angular momentum and radial probability measures.
Match each term with its precise meaning in this lesson.
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Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
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The order follows the stated process; check each stage before the next.