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Physics · lesson 1 of 4

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.

Advanced university8 practice questionsNotes + worked examples
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Prerequisites: Eigenstates, inner products and nondegenerate quantum mechanics.

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

Example 1

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

Example 2

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

8 questions

Decimals or fractions are accepted for numeric answers. Proofs and written responses use sample answers and require human marking. This is practice, not a protected examination.

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

Show answer and explanation

0.3 eV

Multiply the matrix element by λ.

2 Which quantity gives the nondegenerate first-order energy coefficient?

Show answer and explanation

⟨n∣V∣n⟩

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

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

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4.8 eV

Add the correction to the unperturbed value.

4 What should be done first within a degenerate subspace?

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Diagonalize the perturbation restricted to that subspace

The appropriate combinations resolve first-order splitting.

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

Sample answer and marking guidance

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.

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

Show answer and explanation

3 eV

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

7 How can symmetry make a first-order correction vanish without making all higher corrections vanish?

Sample answer and marking guidance

A symmetry can force the diagonal expectation of V to zero, while allowed off-diagonal couplings can still contribute at higher order.

Credit the distinction between diagonal and off-diagonal terms.

8 State two checks on a claimed perturbative energy correction.

Sample answer and marking guidance

Its units must be energy, it should vanish as the perturbation strength tends to zero, and it should be small compared with relevant separation scales in the asserted regime.

Award credit for any two appropriate checks.

Further reading and review references

These are reference links, not copied exercises, institutional endorsement or proof of independent review.