(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
![](https://www.sarthaks.com/?qa=blob&qa_blobid=7021685205477189116)
where A and B are normalization constants. The continuity of \(\psi\) and \(\psi'\) at r = R, or equivalently
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5815019165356069422)
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
![](https://www.sarthaks.com/?qa=blob&qa_blobid=18344335563973283841)
(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
![](https://www.sarthaks.com/?qa=blob&qa_blobid=241156500959609673)
For bound states, we require 0 > E > -V0. As V(x) = V(-z), the equation has solution
![](https://www.sarthaks.com/?qa=blob&qa_blobid=13162432387689576702)
where k, k’ have the same definitions as in (a). The continuity of \(\psi\) and \(\psi'\) at x = R/2 gives
![](https://www.sarthaks.com/?qa=blob&qa_blobid=10741884963928472256)
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 Vs(x) such that
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5497932366416133586)
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 Vs(x) for which
![](https://www.sarthaks.com/?qa=blob&qa_blobid=3560846416103936676)
This means that there is always a bound state for a one-dimensional well of any shape.