Section 16.2 Asymptotic Solutions of the Radial Equation
At first glance there seems to be no obvious direct approach to solving (16.1.5), so let’s see if we can get some clues to the form of the solution by looking at the limiting behavior of the solutions for large \(r\) and for small \(r\text{.}\)
For large \(r\text{,}\) the terms in (16.1.5) involving \(r^{-1}\) and \(r^{-2}\) can be neglected, so (16.1.5) becomes approximately
\begin{equation}
\frac{d^2R}{dr^2} + \frac{2\mu E}{\hbar^2} R
= 0\tag{16.2.1}
\end{equation}
It is helpful to remind ourselves that \(E<0\) by writing (16.2.1) as
\begin{equation}
\frac{d^2R}{dr^2} = \frac{2\mu |E|}{\hbar^2} R\tag{16.2.2}
\end{equation}
which has the familiar exponential solutions \(R(r)=e^{\pm br}\) with \(b=\sqrt{2\mu|E|/\hbar^2}\text{.}\)
Note the \(\pm\) symbol in the exponential, which is there because (16.2.2) involves the second derivative of \(R\text{.}\) Can we eliminate one of these signs? As \(r\) goes to infinity, \(e^{+br}\) blows up. We will eventually want our solutions for the wave functions to give us reasonably behaved probability densities (that is, they must be finite everywhere), and we must therefore discard any solution that leads to an infinite probability. Our solution for the radial wave function in this limit then becomes:
\begin{equation}
R(r) \sim e^{-br} \qquad \text{(large $r$)}\tag{16.2.3}
\end{equation}
Now let’s look at the behavior of the solutions when \(r\) is small. In this limit, the \(r^{-2}\) term will dominate and we can neglect the other terms in the square brackets in (16.1.5). In this case, we obtain
\begin{equation}
\frac{d^2 R}{dr^2} + \frac{2}{r}\frac{dR}{dr}
- \frac{\ell(\ell+1)}{r^2} R
= 0\tag{16.2.4}
\end{equation}
Reflection.
The solution to this equation apparently doesn’t depend on \(\hbar\text{.}\) Don’t all quantum-mechanical results depend in some way on Planck’s constant?
By inspection, you can see that a solution of the form \(R(r)=r^q\) can be made to satisfy (16.2.4)—note that for this choice of \(R(r)\) each term in (16.2.4) will then depend on \(r^{q-2}\) and the three terms can sum to zero. With this substitution, we obtain
\begin{equation}
q(q-1) r^{q-2} + \frac{2}{r} q r^{q-1}
- \frac{\ell(\ell+1)}{r^2} r^q
= 0\tag{16.2.5}
\end{equation}
or
\begin{equation}
q(q+1)-\ell(\ell+1) = 0\tag{16.2.6}
\end{equation}
This quadratic equation for \(q\) yields two solutions: \(q=\ell\) and \(q=-(\ell-1)\text{.}\) For small \(r\text{,}\) \(r^{-\ell-1}\) blows up, so we discard this solution. We then have
\begin{equation}
R(r) \sim r^\ell \qquad \text{(small $r$)}\tag{16.2.7}
\end{equation}
Combining (16.2.3) and (16.2.7), we expect the radial solution to look something like \(R(r)\sim r^\ell e^{-br}\text{.}\) Note that we haven’t violated the proper behavior at the limits by combining these two solutions; \(R(r)\) remains well-behaved for \(r=0\) and \(r\to\infty\text{.}\) What else do we need to complete the solution? Perhaps an additional function \(H(r)\text{,}\) which gives the remaining radial dependence and is well-behaved by not blowing up at \(r=0\) (or blowing up more slowly than \(r^{-\ell}\)) nor as \(r\to\infty\) (or blowing up more slowly than \(e^{br}\)). Let us therefore seek solutions of the radial equation of the form
\begin{equation}
R(r) = r^\ell e^{-br} H(r)\tag{16.2.8}
\end{equation}
Our next goal is to determine \(H(r)\text{.}\)
Reflection.
Other than its asymptotic behavior, what other properties do you expect for \(H(r)\text{?}\) If we expect \(R(r)\) to be a radial wave function, what type of behavior is missing from the other parts of \(R(r)\) that \(H(r)\) must provide?
Before we begin substituting into (16.1.5), we’ll need the first two derivatives of \(R(r)\text{.}\)
\begin{equation}
\frac{dR}{dr}
= \ell r^{\ell-1} e^{-br} H(r)
- br^\ell e^{-br} H(r) + r^\ell e^{-br} H'(r)\tag{16.2.9}
\end{equation}
where \(H'(r)=dH/dr\text{,}\) and
\begin{align}
\frac{d^2R}{dr^2}
&=
\ell(\ell-1)r^{\ell-2} e^{-br} H(r) - 2b\ell r^{\ell-1} e^{-br} H(r)
+ 2\ell r^{\ell-1} e^{-br} H'(r)\notag\\
&\qquad + b^2 r^\ell e^{-br} H(r)
- 2br^\ell e^{-br} H'(r) + r^\ell e^{-br} H''(r)\tag{16.2.10}
\end{align}
\begin{equation}
rH'' + \bigl(2(\ell+1)-2br\bigr) H'
+ \left(-2b(\ell+1)+\frac{2\mu}{\hbar^2}\right) H = 0\tag{16.2.11}
\end{equation}
It is convenient at this point to re-write this equation in terms of the dimensionless variables \(\rho=2br\) and \(\lambda=\mu k/\hbar^2b\)
\begin{equation}
\rho\frac{d^2H}{d\rho^2} + (2\ell+2-\rho)\frac{dH}{d\rho}
+ (\lambda-\ell-1)H(\rho) = 0\tag{16.2.12}
\end{equation}
(16.2.12) is the equation that we now need to solve. Either you recognize that this is Laguerre’s Associated Equation and skip quickly to Section 16.4 or you can use power series techniques from Chapter 7 to solve this equation. The messy details are given in Section 16.3.
