약한 섭동 및 중첩 하위 공간
도입됨| English | 의미 |
|---|---|
| first-order energy shift/fɜːst ˈɔːdə ˈenədʒi ʃɪft/ | 1차 에너지 보정 |
| degenerate perturbation theory/dɪˈdʒenəreɪt pətəˈbeɪʃn ˈθɪəri/ | 축퇴 Perturbation theory: 축퇴 perturbation theory |
A decision before an answer
- A zero first-order energy shift can coexist with a changed state and a nonzero higher-order shift.
- Your goal: Calculate nondegenerate first-order energy shifts as normalised expectation values.
Weight the perturbation by probability
- Let H=H₀+λW, where λ is a small dimensionless parameter and the eigenstates of H₀ are known and normalised. For a nondegenerate level n, the first-order energy correction is λ⟨n|W|n⟩. In position space this is λ∫ψ_n*(x)W(x)ψ_n(x)dx for a multiplicative potential, with integration over the allowed domain.
- A potential value at one point is not an expectation value; the state’s probability density weights the whole domain. For an infinite well 0<x<L, ψ_n=√(2/L)sin(nπx/L). Reflection about its centre makes ⟨x⟩=L/2, so a weak added potential γx has shift γL/2 for every nondegenerate well level at first order. γ has units of energy per length.
In 0<x<L, a weak perturbation γ(x−L/2) acts on an unperturbed infinite-well eigenstate. Its first-order energy shift is:
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
The density is symmetric about L/2 while the perturbation is odd, so the integral cancels.
Use parity carefully
- For a perturbation odd about the centre, such as γ(x−L/2), the probability density of an unperturbed well eigenstate is even, so its diagonal expectation vanishes. A centred quadratic perturbation η(x−L/2)² instead gives shift ηL²[1/12−1/(2π²n²)], which is positive for η>0 and depends on n.
- Parity makes the first-order integral zero only for the specified state and operator symmetry. Off-diagonal matrix elements can still change the wavefunction. Do not turn a symmetry cancellation into a claim that every energy correction vanishes or that the original state remains exact.
A two-state degenerate block has perturbation ε[[0,1],[1,0]], ε>0. First-order shifts are:
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
The matrix eigenvalues are ±ε. Its vanishing diagonal does not mean vanishing splitting.
Compare coupling with level gaps
- For a nondegenerate state, the leading admixture of another unperturbed state m is proportional to λW_mn/(E_n⁰−E_m⁰). The useful smallness condition therefore compares coupling matrix elements with the relevant level separations, not just with an arbitrary absolute energy zero.
- To second order, the energy correction contains Σ_(m≠n)|λW_mn|²/(E_n⁰−E_m⁰). For the lowest nondegenerate state, all these denominators are negative, so the second-order correction is nonpositive in this model. Near a degeneracy, a small denominator defeats the nondegenerate expansion; use a coupled subspace instead of dividing by zero.
In a 2 nm infinite well, a weak γx perturbation with γ=0.05 eV/nm gives ΔE_n^(1)=γL/2=0.05 eV. Separately, η=0.03 eV/nm² multiplying (x−L/2)² gives the ground shift 0.03·4[1/12−1/(2π²)]=0.00392073 eV. These are first-order results conditional on weak coupling relative to gaps. A degenerate pair with ε[[2,1],[1,2]] and ε=0.004 eV splits into shifts 0.004 and 0.012 eV; the symmetric state gets the larger shift.
For a 4 nm well and weak γx with γ=0.02 eV/nm, first-order shift is ____ eV.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
γ⟨x⟩=0.02·4/2=0.04 eV.
Resolve a degenerate subspace
- If a level of H₀ is exactly degenerate, choose an orthonormal basis within that subspace and form the Hermitian matrix of the perturbation there. Its eigenvalues are the first-order energy shifts, and its eigenvectors specify the combinations that diagonalise the leading splitting.
- For a two-state subspace with perturbation ε[[2,1],[1,2]], the normalised symmetric and antisymmetric combinations have shifts 3ε and ε. Reading only the two diagonal entries would incorrectly predict two shifts of 2ε. A common scalar multiple of the identity shifts both states equally and does not split their degeneracy; off-subspace couplings can matter at higher order.
Use normalised state weights, keep perturbation units, and diagonalise a degenerate block. First-order cancellation and exact invariance are different claims.
Which answer fits this case?
Calculate nondegenerate first-order energy shifts as normalised expectation values
A zero diagonal first-order expectation guarantees all higher-order energy shifts are zero.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Off-diagonal couplings can contribute at second order even when the diagonal expectation cancels.
Keep the distinctions
- first-order energy shift 一阶能量修正 — Leading weak-perturbation correction given by the unperturbed state’s expectation value.
- degenerate perturbation theory 简并微扰理论 — Method that first diagonalises the perturbation within an unperturbed degenerate subspace.
- Calculate nondegenerate first-order energy shifts as normalised expectation values.
- Use symmetry and matrix elements to assess weak-coupling limits.
- Diagonalise the perturbation within an exactly degenerate subspace.
Match each term with its precise meaning in this lesson.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
난이도 3/5 · 200 코인 + 오늘 첫 완료 시 소액 보너스
The order follows the stated process; check each stage before the next.