Question

A particle of mass m moves with zero angular momentum in a spherically symmetric attractive potential V(r).

(a) Write down the differential equation of radial motion, defining your radial wave function carefully and specifying the boundary conditions on it for bound states. What is the WKB eigenvalue condition for s-states in such a potential? (Be careful to incorporate in your one-dimensional WKB analysis the constraints of radial motion (0 < r < \(\infty\)).

(b) For V(r) = -V₀exp(-r/a), use the WKB relation to estimate the minimum value of V₀ such that there will be one and only one bound state, just barely bound. Compare your value to the exact result for the exponential potential, 2mV₀a²/h² = 1.44.

Answer 1

(a) The wave function of the particle can be written as the product of a radial part and an angular part, \(\psi(r) = R(r) Y_{lm} (\theta, \phi)\). Here R(r) satisfies the equation

in which l = 0 for zero angular momentum has been incorporated. The boundary conditions for a bound state are R(r) finite for r → 0, R(r) → 0 for r → \(\infty\).

Let X(r) = R(r)/r, the above becomes

Thus the problem becomes that of the one-dimensional motion of a particle in a potential V(r) defined for r > 0 only. The WKB eigenvalue condition for s-state is

(b) Substituting V = -V₀ exp(-r/a) in the loop integral we have

Within the requirements that V₀ is finite and that there is one and only one bound state which is just barely bound, we can consider the limiting case where |E| \(\approx\) V₀. Then the integral on the left-hand side can be approximated by

If there is to be one and only one bound state, we require -E = |E| < V₀ for n = 0 but not for n = 1, or equivalently

The minimum V₀ that satisties this condition is given by

which is very close to the exact result of 1.44.