Jost function Created 2026-10-14 Updated 2026-11-05
For radial quantum scattering, the incoming coefficient of the regular solution defines the Jost function, with convention . A zero at , , removes the exponentially growing component and produces a bound state, subject to regularity and normalizability. Poles of the outgoing numerator alone need not produce bound states.
Past exam of the mathematics course of the University of Cambridge 2007 ii Paper 1 33A Solution Created 2026-09-24 Updated 2026-10-14
The incoming and outgoing coefficients give the scattering matrixIt obeys and , so for real , as required by elastic unitarity. With , the scattering phase shift satisfies . The inverse tangent needs a continuous branch: taking gives . Equivalently for .
The scattering length follows from :For a putative bound state put , . The incoming term grows as , so its coefficient must vanish. The incoming Jost function is ; its only upper-half-plane zero is . Under the regular-potential assumption this gives one bound state, with The pole at in instead comes from the outgoing numerator: it is a redundant pole of a scattering matrix, not a second normalizable state. The S-matrix alone, without its incoming/outgoing analytic structure, would tempt an incorrect count of two states.
For completeness, the nonsingular Eckart realization of this Jost function requires , as discussed in dipot.ulb.ac.be/dspace/bitstream/2013/373712/4/2306.12216.pdf. The supplied positive constants alone do not establish such a realization in every ordering; a coincident pole is likewise not evidence for two states.
There is no positive-energy scattering resonance. Indeed , and the s-wave total partial-wave total scattering cross-section is , strictly decreasing for . All poles are on the imaginary axis, with no oscillatory decaying resonance pole.
Redundant pole of a scattering matrix Created 2026-10-14 Updated 2026-10-25
A pole of a scattering matrix arising from the numerator rather than a zero of the incoming Jost function can fail to correspond to a bound state. For , the pole at is associated with the incoming zero; the pole at comes from the numerator. The regular Eckart realization has . The distinction and this explicit example are discussed in dipot.ulb.ac.be/dspace/bitstream/2013/373712/4/2306.12216.pdf.