The \(1 / x^{2}\) potential. Suppose $$V(x)=\left\\{\begin{array}{ll}-\alpha /
x^{2}, & x>0, \\
\infty, & x \leq 0.\end{array}\right.$$
where \(a\) is some positive constant with the appropriate dimensions. We'd like
to find the bound states-solutions to the time-independent Schr枚dinger
equation
$$-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}-\frac{\alpha}{x^{2}}
\psi=E \psi$$
with negative energy \((E<0)\)
(a) Let's first go for the ground state energy, \(E_{0}\). Prove, on dimensional
grounds, that there is no possible formula for \(E_{0}\) -no way to construct
(from the available constants \(m, \hbar,\) and \(\alpha\) ) a quantity with the
units of energy. That's weird, but it gets worse \(\ldots .\)
(b) For convenience, rewrite Equation 2.190 as $$\frac{d^{2} \psi}{d
x^{2}}+\frac{\beta}{x^{2}} \psi=\kappa^{2} \psi, \text { where } \beta \equiv
\frac{2 m \alpha}{\hbar^{2}} \text { and } \kappa \equiv \frac{\sqrt{-2 m
E}}{\hbar}.$$
Show that if \(\psi(x)\) satisfies this equation with energy \(E\), then so too
does
\(\psi(\lambda x),\) with energy \(E^{\prime}=\lambda^{2} E,\) for any positive
number \(\lambda .[\) This is a catastrophe: if there exists any solution at
all, then there's a solution for every (negative) energy! Unlike the square
well, the harmonic oscillator, and every other potential well we have
encountered, there are no discrete allowed states - and no ground state. A
system with no ground state-no lowest allowed energy-would be wildly unstable,
cascading down to lower and lower levels, giving off an unlimited amount of
energy as it
falls. It might solve our energy problem, but we'd all be fried in the
process.] Well, perhaps there simply are no solutions at all \(\ldots .\) (c)
(Use a computer for the remainder of this problem.) Show that
$$\psi_{\kappa}(x)=A \sqrt{x} K_{i g}(\kappa x),$$ satisfies Equation 2.191
(here \(K_{i g}\) is the modified Bessel function of order \(i g,\) and \(g \equiv
\sqrt{\beta-1 / 4}\) ). Plot this function, for \(g=4\) (you might as well let
\(_{K}=1\) for the graph; this just sets the scale of length). Notice that it
goes to 0 as \(x \rightarrow 0\) and as \(x \rightarrow \infty\). And it's
normalizable: determine \(A\). How about the old rule that the number of nodes
counts the number of lower-energy states? This function has an infinite number
of nodes, regardless of the energy (i.e. of \(\kappa\) ). I guess that's
consistent, since for any \(E\) there are always an infinite number of states
with even lower energy.
(d) This potential confounds practically everything we have come to expect.
The problem is that it blows up too violently as \(x \rightarrow 0\). If you
move the "brick wall" over a hair,
$$V(x)=\left\\{\begin{array}{ll}-\alpha / x^{2}, & x>\epsilon>0, \\\\\infty, &
x \leq, \epsilon \end{array}\right.$$
it's suddenly perfectly normal. Plot the ground state wave function, for \(g=4\)
and \(\epsilon=1\) (you'll first need to determine the appropriate value of
\(\kappa\) ), from \(x=0\) to \(x=6 .\) Notice that we have introduced a new
parameter \((\epsilon),\) with the dimensions of length, so the argument in (a)
is out the window. Show that the ground state energy takes the form
$$E_{0}=-\frac{\alpha}{\epsilon^{2}} f(\beta),$$ for some function \(f\) of the
dimensionless quantity \(\beta,\)
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