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.
What a learner can do afterwards
- 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
1 · Read
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.
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=<phi|H|phi>/<phi|phi>, 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.
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.
2 · Watch
به انگلیسیTake it off screen
Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.