A new proof shows the exact best constant in a basic spectral inequality for one‑dimensional quantum systems. The paper proves that, for a one‑dimensional Schrödinger operator, the sum of the absolute values of all negative energy levels is bounded by (4/(3√3 π)) times the integral of the positive part of the potential raised to the 3/2 power. The author shows this constant cannot be improved, confirming the Lieb–Thirring conjecture in this specific case.

The inequality at stake compares two ways of measuring how attractive a potential is. On the left is the spectral quantity: the sum of the negative eigenvalues of the Schrödinger operator, which counts the total binding energy. On the right is a simple integral of the potential, ∫ V_+(x)^{3/2} dx, where V_+ is the positive part of −V. The main theorem states that the sum of the negative eigenvalues is at most (4/(3√3 π)) times that integral. The constant is sharp, and equality is achieved by a potential that produces a single negative eigenvalue.

The proof builds on earlier one‑eigenvalue arguments and extends them to any finite orthonormal family of wave functions. The paper uses a standard duality between the spectral inequality and a related “kinetic” inequality for orthonormal functions. At a high level the method replaces a single‑function change of variables by a matrix version that tracks cumulative mass for a whole orthonormal system. An auxiliary function and a careful completion of squares control the terms that arise. This allows the author to carry the sharp constant from the scalar case over to the many‑state case.

Why this matters: Lieb–Thirring inequalities are central tools in mathematical physics. They connect the shape of a potential to how many and how deep bound states it can create. These bounds play a role in questions about stability of matter and in many spectral estimates. Confirming the conjectured optimal constant in one dimension removes a long‑standing gap in our understanding of this classical inequality.