QM.2 · Weak perturbations and degenerate subspaces

일차 에너지 이동이 0인 경우가 상태의 변화와 비영장 고차 이동과 공존할 수 있습니다.

선행 지식: 5, 24.

• 비중첩 일차 에너지 이동을 정규화 기대값으로 계산합니다.
• 대칭성과 행렬 요소를 사용하여 약한 결합 한계를 평가합니다.
• 정확히 중첩된 하위 공간 내에서 섭동을 대각화합니다.

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.

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.

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.

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.

중첩된 하위 공간은 일차 이동 전에 행렬 대각화가 필요합니다. 두 상태에 제한된 섭동이 $\epsilon\begin{pmatrix}2&1\\1&2\end{pmatrix}$라고 가정하자.

영향은 epsilon과 3 epsilon이다. 정규화된 상태는 $(1,-1)/\sqrt2$과 $(1,1)/\sqrt2$이다. 대각 성분만 읽으면 분열을 놓치게 된다. 다른 준위와의 결합 에너지 간격에 비해 약하게 남아야 한다.

정규화된 상태 가중치를 사용하고, perturbation 단위를 유지하며, 중첩 블록을 대각화하라. 1차 소거와 정확히 불변임은 서로 다른 주장이다.

1차 에너지 영향: 비섭동 상태의 기댓값에 의해 주어지는 선형 근사 보정항.

비분리 섭동 이론: 비분리 비분리 하위 공간 내에서 섭동을 대각화하는 방법.