Finite-well bound states and hydrogenic quantum numbers
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| bound state | 束缚态 | shù fù tài |
| radial probability density | 径向概率密度 | jìng xiàng gài lǜ mì dù |
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:
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?
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 ____.
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.
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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.