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Physics · Advanced university

Perturbation theory and approximation checks

Write H=H0+λV and expand observables or states about a solvable reference problem. Small λ is useful only relative to relevant spectral scales.

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Prerequisites

Eigenstates, inner products and nondegenerate quantum mechanics.

Study notes

Setup

Write H=H0+λV and expand observables or states about a solvable reference problem. Small λ is useful only relative to relevant spectral scales.

First order

For a normalized nondegenerate state, the first-order energy coefficient is ⟨n∣V∣n⟩. Include λ when calculating the first-order energy shift.

Degeneracy

Degenerate levels require diagonalizing the perturbation in the relevant subspace rather than applying a single arbitrary diagonal element.

Validation

Check dimensions, symmetry, limiting behaviour and the size of corrections. Near-degeneracy can spoil a naive small-coupling expansion.

Worked examples

For λ=0.1 and ⟨n∣V∣n⟩=3 eV, find the first-order energy shift.

Multiply the matrix element by λ.

0.3 eV

Which quantity gives the nondegenerate first-order energy coefficient?

The diagonal expectation in the unperturbed state is the first-order coefficient.

⟨n∣V∣n⟩

Check your understanding

  1. An unperturbed energy is 5 eV and first-order shift is −0.2 eV. Find the first-order corrected energy.

  2. What should be done first within a degenerate subspace?

    1. Divide by zero energy gaps
    2. Diagonalize the perturbation restricted to that subspace
    3. Discard normalization
    4. Choose any one basis vector and ignore mixing

  3. Why can a tiny coupling produce a significant change near two nearly degenerate levels?

  4. For a normalized state with components (1/√2,1/√2) and V=diag(2,4) eV, find ⟨V⟩.

Answers and marking guidance

Answers and marking notes
  1. 4.8 eV

    Add the correction to the unperturbed value.

  2. Diagonalize the perturbation restricted to that subspace

    The appropriate combinations resolve first-order splitting.

  3. State-mixing corrections involve energy-gap denominators. A small gap can make the ratio large despite a small numerator.

    Credit relative scales rather than coupling size alone.

  4. 3 eV

    The weighted diagonal mean is 2/2+4/2.

Further reading and review references

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