Question

(a) Given a one-dimensional potential

V = -V₀, |x| < a,

V = 0, |x| > a,

show that there is always at least one bound state for attractive potentials V₀ > 0. (You may solve the eigenvalue condition by graphical means.)

(b) Compare the Schrodinger equation for the above one-dimensional case with that for the radial part U(r) of the three-dimensional wave function when L = 0,

\(\psi(r) = r^{-1}U(r) Y_{LM} (\Omega),\)

where \(\psi(r) \) is the solution of the Schrodinger equation for the potential

V = -V₀, r < a,

V = 0, r > a.

Why is there not always a bound state for V₀ > 0 in the three-dimensional case?

Answer 1

(a) For the bound state of a particle, E < 0. For |x| > a, V = 0 and the Schrodinger equation

provided E > -V₀. Here we need only consider states of even parity which include the ground state.

The continuity of the wave function and its derivative at x = \(\pm a\) requires k tan ka = k'. Set ka = \(\xi, k' = \eta\). Then the following equations determine the energy levels of the bound states:

These equations must in general be solved by graphical means. In Fig.(a), curve 1 is a plot of \(\eta = \xi \tan \xi\), and curve 2 plots \(\xi^2 + \eta^2 = 1\). The dashed line 3 is the asymtotic curve for the former with \(\xi = \pi /2\). Since curve 1 goes through the origin, there is at least one solution no matter how small is the value of V₀a². Thus there is always at least one bound state for a one-dimensional symmetrical square-well potential.

(b) For r > a, the radial Schrodinger equation

and the solution that satisfies the boundary condition

The continuity of the wave function and its derivative at r = a, requires k cot ka = -k'. Setting ka = \(\xi\), ka' = \(\eta\) we get

These are again to be solved by graphical means. Fig.(b) shows curve 1 which is a plot of \(\xi\) cot \(\xi\) = -\(\eta\), and the dashed line 2 which is its asymptotic if \(\xi\) = π. It can be seen that only when

can the equations have a solution. Hence, unlike the one-dimensional case, only when V₀a² ≥ π²h²/8m can there be a bound state.