Perturbation Theory

Most quantum systems cannot be solved exactly. Perturbation theory provides systematic approximations when a small perturbation H' is added to an exactly solvable Hamiltonian H₀. The corrections are computed order by order in the perturbation strength.

Key Concepts

Total Hamiltonian: H = H₀ + λH'
First-order energy correction: Eₙ⁽¹⁾ = ⟨n|H'|n⟩
Second-order correction involves sum over all other states
Degenerate perturbation theory needed when levels are degenerate
Fine structure of hydrogen is a perturbative result

Key Equations

First-order energy

E_n^{(1)} = \langle n^{(0)}|H'|n^{(0)}\rangle

Second-order energy

E_n^{(2)} = \sum_{m\neq n}\frac{|\langle m|H'|n\rangle|^2}{E_n^{(0)} - E_m^{(0)}}

First-order state correction

|n^{(1)}\rangle = \sum_{m\neq n}\frac{\langle m|H'|n\rangle}{E_n^{(0)}-E_m^{(0)}}|m\rangle

Stark effect (linear)

\Delta E = -\vec{p}\cdot\vec{E}

Worked Example

Example Problem

Problem

A QHO (ω=1×10¹³ rad/s) is perturbed by H'=εx² with ε=0.1mω². Find the first-order correction to E₀.

Solution

E₀⁽¹⁾ = ε⟨0|x²|0⟩. For QHO, ⟨x²⟩₀ = ℏ/(2mω). E₀⁽¹⁾ = ε×ℏ/(2mω) = 0.1mω² × ℏ/(2mω) = ℏω/20 = E₀/5.

Practice

Exercises

7 problems

1 of 7

An infinite square well (L=1 nm) has perturbation H'=V₀ for L/2 ≤ x ≤ L, zero elsewhere. V₀ = 0.1 eV. Watch the integral sweep and find E₁⁽¹⁾ in eV.

The right half (yellow) is the perturbation H' = 0.1 eV. Press Sweep to animate the integral E₁⁽¹⁾ = ⟨ψ₁|H'|ψ₁⟩ sweeping across the well. Then enter E₁⁽¹⁾ in eV.

E₁⁽¹⁾ = eV

2 of 7

For a QHO with perturbation H'= cx, the parity visualization shows why ⟨0|cx|0⟩ = 0. What is E₀⁽¹⁾ (the first-order energy correction)?

Step through the parity argument: ψ₀ is even, x is odd, so their product ψ₀·x is odd — its integral vanishes. Therefore E₀⁽¹⁾ = 0.

E₀⁽¹⁾ = eV

3 of 7

For the same linear QHO perturbation H'=cx, the second-order correction is E₀⁽²⁾ = -c²/(2mω²). For c=1.0×10⁻¹⁰ N, m=9.11×10⁻³¹ kg, ω=2×10¹⁴ rad/s, find |E₀⁽²⁾| in eV.

Unlock Exercise 3