The Resonance Problem Woods-Saxon Potential

1 The Resonance Problem Woods-Saxon Potential. - Consider the numerical solution of the Schrodinger equation  

In the case of positive energies E = k2 the potential dies away faster than the term /(/ + l)/x2 and the Schrodinger equation effectively reduces to 

The above equation has linearly independent solutions foe/,(for) and ), where ,-(for), w,(fot) are the spherical Bessel and Neumann functions respectively. Thus the solution of equation (1) has (when x — 0) the asymptotic form 

Since the problem is treated as an initial-value problem, we need yo and y before starting a two-step method. From the initial condition, yo = 0, it can be shown that, for values of x close to the origin, the solution behaves like y(x) cxf+x as x — 0, where c is an independent constant. In view of this we take y = A/+1. With these starting values we evaluate at jcj of the asymptotic 

For positive energies one has the so-called resonance problem. This problem consists either of finding the phase-shift S(E) = 5/ or finding those E, for [1,1000], at which 3 equals n/2. We actually solve the latter problem, known as the resonance problem when the positive eigenenergies lie under the potential barrier. 

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