Question

(a) Consider a particle of mass m moving in a three-dimensional square-well potential V(|r|). Show that for a well of fixed radius R, a bound state

exists only if the depth of the well has at least a certain minimum value. Calculate that minimum value.

(b) The analogous problem in one dimension leads to a different answer. What is that answer?

(c) Can you show that the general nature of the answers to (a) and (b) above remains the same for a well of arbitrary shape? For example, in the one-dimensional case (b)

V(x) = \(\lambda\)f(x) < 0, a ≤ x ≤ b,

V(x) = 0, x < a or x > b,

consider various values of \(\lambda\) while keeping f(x) unchanged.

Answer 1

(a) Suppose that there is a bound state \(\psi(r)\) and that it is the ground state (l = 0), so that \(\psi(r)\) = \(\psi(r)\). The eigenequation is

where A and B are normalization constants. The continuity of \(\psi\) and \(\psi'\) at r = R, or equivalently

These equations can be solved graphically. In a similar way, we can show that for there to be at least a bound state we require

(b) If the potential is a one-dimensional rectangular well potential, no matter how deep the well is, there is always a bound state. The ground state is always symmetric about the origin which is the center of the well. The eigenequation is

For bound states, we require 0 > E > -V₀. As V(x) = V(-z), the equation has solution

where k, k’ have the same definitions as in (a). The continuity of \(\psi\) and \(\psi'\) at x = R/2 gives

Since V₀ > -E > 0, there is always a bound-state solution for any V₀.

(c) For a one-dimensional potential well of arbitrary shape, we can always define a rectangular potential well V_s(x) such that

and -V₀ ≥ V(x) always (see Fig.). From (b) we see that there always exists a \(|\psi_0(x) \rangle\) which is a bound eigenstate of V_s(x) for which

This means that there is always a bound state for a one-dimensional well of any shape.