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If $c_1,\dots ,c_n$ are the columns of a matrix, the matrix has the columns property when $[n]$ has an ordered partition $B_1\bigsqcup \cdots \bigsqcup B_s$ such thatand, for $j>1$, ${\sum }_{i\in B_j}{c}_i$ belongs to the linear span of the columns indexed by $B_1\cup \cdots \cup B_{j-1}$.

A scattering resonance is a metastable state represented by a pole of the analytically continued scattering matrix. Near an isolated narrow resonance, a cross-section has a Breit-Wigner peak whose energy width is the inverse lifetime up to $\hslash$.

A pole at $k=i\kappa$ with $\kappa >0$ represents a normalizable bound state of energy $-{\hslash }^2{\kappa }^2/(2m)$. Resonance poles instead lie away from the imaginary axis on the analytically continued unphysical sheet.

For elastic S-wave scattering by a sufficiently short-range central potential, the phase shift ${\delta }_0(k)$ defines a finite scattering lengthThe corresponding Partial-wave S-matrix is $S_0=e^{2i{\delta }_0}$, so its contribution to the scattering amplitude isSince $k\cot {\delta }_0\to -1/a_s$, a finite nonzero $a_s$ gives $f_0\to -a_s$. The zero-length case is understood by its own continuous low-energy limit.

For distinguishable particles in a single elastic channel, dominant S-wave scattering is isotropic, yieldingThe formula assumes a finite scattering length before taking the zero-energy limit. Near a zero-energy resonance, $a_s$ can diverge and the finite-$k$ denominator must be kept; elastic s-wave unitarity bounds the cross-section by $4\pi /k^2$.

Identical-particle symmetry also matters. Identical bosons have an exchange contribution, while identical fermions in the same internal state cannot scatter in an even-orbital channel. The distinguishable-particle prefactor must not be transferred unchanged to those cases.