How far a finite-difference Hamiltonian's levels sit below n + 1/2

Adam Zettel · September 21, 2026 · written with an agent

H^= - \frac{1}{2} \partial_{x}^{2} + \frac{1}{2} x^{2}, E_{n} = n + \frac{1}{2}

Put $H = - \frac{1}{2} \partial_{x}^{2} + \frac{1}{2} x^{2}$ on a grid and diagonalize: the levels come out at $n + \frac{1}{2}$ , a little low. How low is predictable. First-order perturbation theory gives $- \frac{h ^{2}}{32} (2 n^{2} + 2 n + 1)$ , and for the six lowest levels the measured shift matches it to within half a percent (the test allows 4%).

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The three-point second difference is $\partial_{x}^{2} + \frac{h ^{2}}{12} \partial_{x}^{4} + \dots$ , so the grid Hamiltonian is the true one plus $- \frac{h ^{2}}{24} p^{4}$ . To first order the level moves by the expectation of that, and for the oscillator $⟨ n ∣ p^{4} ∣ n ⟩ = \frac{3}{4} (2 n^{2} + 2 n + 1)$ . The grid's error is a small perturbation of the Hamiltonian, so the perturbation theory used for the Stark shift predicts it.

Printed

H: 0.0993789

measured: [-0.000308821, -0.00154487, -0.00401889, -0.00773319, -0.0126901, -0.018892]

predicted: [-0.00030863, -0.00154315, -0.00401219, -0.00771575, -0.0126538, -0.0188264]

The three lowest eigenfunctions

Tests 3 of 3 pass

the ground state is at 1/2, to a part in a thousand (passes)
each of the first six levels is low, because the grid's p^4 term is negative (passes)
each shift is within 4% of what perturbation theory predicts (passes)