--- title: "Degenerate Perturbation Theory and the Variational Method" description: "For an isolated degenerate energy, diagonalise the perturbation within its eigenspace to find first-order shifts. The variational energy of an admissible nonzero trial state gives an upper bound on th" canonical: https://lightmysky.com/learn/science/degenerate-perturbation-theory-and-the-variational-method-mt_uov0PZ8vqy source: https://lightmysky.com/learn/science/degenerate-perturbation-theory-and-the-variational-method-mt_uov0PZ8vqy.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Degenerate Perturbation Theory and the Variational Method For an isolated degenerate energy, diagonalise the perturbation within its eigenspace to find first-order shifts. The variational energy of an admissible nonzero trial state gives an upper bound on the ground-state energy. Subject: Science · Area: Quantum & Modern Physics · Ages 23 to 24 Page: https://lightmysky.com/learn/science/degenerate-perturbation-theory-and-the-variational-method-mt_uov0PZ8vqy ## Ready when they can - Explains why an arbitrary degenerate basis can give incorrect first-order shifts and why the restricted perturbation must be diagonalised - Diagonalises a small perturbation matrix to get the split levels - Uses a trial function to bound a ground state energy and judges how tight the bound is ## Lesson: When levels split and how to bound them Exchange states used normalized combinations and energy expectations. Write H=H0+lambda V with small dimensionless lambda and Hermitian V. Let H0 have an isolated degenerate energy E_d. Diagonal averages can miss mixing within its eigenspace; changing basis alone creates no splitting. In non-degenerate theory, energy denominators enter state corrections and second-order energies, not first-order energy expectations. Coupling to outside levels must be small compared with their gaps to E_d. *(drawing: W is Hermitian. Its eigenvalues w give first-order shifts lambda w; its eigenvectors define the basis within the degenerate group.)* **Example.** For W with rows (0, 2) and (2, 0), solve det(W-wI)=w²-4=0. The corrections w are -2 and 2. The corresponding normalized vectors are (1, -1)/sqrt(2) and (1, 1)/sqrt(2). For rows (a, b) and (b, a), the same calculation gives a-b and a+b. A nonzero diagonal term can move both energies in the same direction. Add each shift to the shared energy: E=E_d+lambda w+O(lambda²). These are first-order energies, not generally exact ones. If E_d=10, lambda=0.1 and w=-2, 2, the first-order levels are 9.8 and 10.2. Distinct w values split the level at first order; equal w values leave a degeneracy that higher orders may resolve. For a self-adjoint H bounded below, use a nonzero trial vector with finite energy expectation. Its Rayleigh quotient is E_trial=/, and E_trial>=E0. For a normalized vector the denominator is 1. In an orthonormal energy basis, the quotient is sum(|c_n|² E_n)/sum(|c_n|²), a weighted average of energies no lower than E0. For the same Hamiltonian, the smaller of two valid trial energies is the tighter upper bound. If trials give 6.4 and 5.9, then E0<=5.9. This does not show how close 5.9 is to E0. Varying parameters and minimizing the quotient can improve the bound. Rescaling one trial vector leaves its quotient unchanged. A trial quotient below a known exact E0 signals a problem with the calculation or its assumptions. Check the Hamiltonian, admissibility and normalization. For H=diag(3,7) and phi=(2,0), the numerator is 12 and the norm squared is 4, so the quotient is 3. The numerator alone is not a trial energy unless the vector is normalized. **Recap.** Diagonalise the perturbation within the degenerate eigenspace for first-order shifts. An admissible trial vector's Rayleigh quotient bounds E0 from above; a lower valid quotient is tighter. *Original lesson, questions and flashcards written for this dataset about degenerate perturbation theory and variational energy bounds.* ## Practice 8 questions on this page, each with its working shown. ## Needs first - [Identical Particles and the Symmetry of a Many-Body State](https://lightmysky.com/learn/science/identical-particles-and-the-symmetry-of-a-many-body-state-mt_hns-eEGeHU) - [First-Order Perturbation Theory](https://lightmysky.com/learn/science/first-order-perturbation-theory-mt_UsbiiQFOAt) ## Optional connections - [Symmetric Matrices and the Spectral Theorem](https://lightmysky.com/learn/mathematics/symmetric-matrices-and-the-spectral-theorem-mt_sdQv4m7Nbk) ## Opens up - [Time-Dependent Perturbation Theory and Fermi's Golden Rule](https://lightmysky.com/learn/science/time-dependent-perturbation-theory-and-fermis-golden-rule-mt_zDSASGF7SC)