Weak perturbations and degenerate subspaces
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| first-order energy shift | 一阶能量修正 | yī jiē néng liàng xiū zhèng |
| degenerate perturbation theory | 简并微扰理论 | jiǎn bìng wēi rǎo lǐ lùn |
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:
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:
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.
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.
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.
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.