QM.2 · Weak perturbations and degenerate subspaces

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.

A degenerate subspace requires matrix diagonalisation before first-order shifts. Suppose the perturbation restricted to two states is $\epsilon\begin{pmatrix}2&1\\1&2\end{pmatrix}$.

Shifts are epsilon and 3 epsilon. Normalised states are $(1,-1)/\sqrt2$ and $(1,1)/\sqrt2$. Reading only the diagonal entries would miss the splitting. Couplings to other levels must remain weak relative to their energy gaps.