14.5: Determination of Phase-Shifts
( \newcommand{\kernel}{\mathrm{null}\,}\)
Let us now consider how the phase-shifts, δl, in Equation ([e17.73]) can be evaluated. Consider a spherically symmetric potential, V(r), that vanishes for r>a, where a is termed the range of the potential. In the region r>a, the wavefunction ψ(r) satisfies the free-space Schrödinger equation ([e17.54]). The most general solution that is consistent with no incoming spherical-waves is ψ(r)=√n∑l=0,∞il(2l+1)Rl(r)Pl(cosθ), where Rl(r)=exp(iδl)[cosδljl(kr)−sinδlyl(kr)]. Note that yl(kr) functions are allowed to appear in the previous expression because its region of validity does not include the origin (where V≠0). The logarithmic derivative of the lth radial wavefunction, Rl(r), just outside the range of the potential is given by βl+=ka[cosδlj′l(ka)−sinδly′l(ka)cosδljl(ka)−sinδlyl(ka)], where j′l(x) denotes djl(x)/dx, et cetera. The previous equation can be inverted to give tanδl=kaj′l(ka)−βl+jl(ka)kay′l(ka)−βl+yl(ka). Thus, the problem of determining the phase-shift, δl, is equivalent to that of obtaining βl+.
The most general solution to Schrödinger’s equation inside the range of the potential (r<a) that does not depend on the azimuthal angle ϕ is ψ(r)=√n∑l=0,∞il(2l+1)Rl(r)Pl(cosθ), where Rl(r)=ul(r)r, and d2uldr2+[k2−l(l+1)r2−2mℏ2V]ul=0. The boundary condition ul(0)=0 ensures that the radial wavefunction is well behaved at the origin. We can launch a well-behaved solution of the previous equation from r=0, integrate out to r=a, and form the logarithmic derivative βl−=1(ul/r)d(ul/r)dr|r=a. Because ψ(r) and its first derivatives are necessarily continuous for physically acceptable wavefunctions, it follows that βl+=βl−. The phase-shift, δl, is then obtainable from Equation ([e17.82]).
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)