Section 22.16 Power Series Solution of the Radial Equation
In this (optional section), we present the messy details of solving (22.15.12) using power series methods from
[cross-reference to target(s) "odepower" missing or not unique]. If you don’t want to go through the details now, you can skip to Section 22.17. The equation we are trying to solve is
\begin{equation}
\rho\frac{d^2H}{d\rho^2} + (2\ell+2-\rho)\frac{dH}{d\rho}
+ (\lambda-\ell-1)H(\rho) = 0\text{.}\tag{22.16.1}
\end{equation}
Start by writing the function we are trying to find as a power series
\begin{equation}
H(\rho) = \sum_{j=0}^\infty c_j\rho^j\text{.}\tag{22.16.2}
\end{equation}
Taking derivatives, we obtain
\begin{align}
H'(\rho) &= \sum_{j=1}^\infty jc_j\rho^j-1\tag{22.16.3}\\
H''(\rho) &= \sum_{j=2}^\infty j(j-1)c_j\rho^{j-2}\tag{22.16.4}
\end{align}
and substituting into (22.15.12) we find
\begin{align}
& \sum_{j=2}^\infty j(j-1)c_j\rho^{j-1} +
(2\ell+2)\sum_{j=1}^\infty jc_j\rho^{j-1}\notag\\
& - \sum_{j=1}^\infty jc_j\rho^j
+ (\lambda-\ell-1) \sum_{j=0}^\infty c_j\rho^j
= 0\tag{22.16.5}
\end{align}
Expanding each of the sums gives a power series in \(\rho\text{.}\) For all terms of the series to add to zero for any and all values of \(\rho\text{,}\) the coefficient of each power of \(\rho\) must be zero. Before we develop the solution, let’s write out the first few terms of (22.16.5):
\begin{align}
& \bigl[(2\ell+2)c_1+(\lambda-\ell-1)c_0\bigr]\rho^0
+ \bigl[(4\ell+6)c_2+(\lambda-\ell-2)c_1\bigr]\rho^1\notag\\
& + \bigl[6\ell+12)c_3+(\lambda-\ell-3)c_2\bigr]\rho^2 + ...
= 0\tag{22.16.6}
\end{align}
Each term in square brackets must be zero for this equation to be valid for any value of \(\rho\text{.}\) From the first term in square brackets, we obtain a relationship between \(c_1\) and \(c_0\text{,}\) from the second term in square brackets, we obtain a relationship between \(c_2\) and \(c_1\text{,}\) from the third term in square brackets, we obtain a relationship between \(c_3\) and \(c_2\text{,}\) and so forth. The value of \(c_0\) thus determines the values of all of the remaining \(c_j\text{.}\)
Now let’s find the complete solution to (22.16.5). It is convenient to begin by rewriting (22.16.5) so that all sums are given in terms of powers of \(\rho^j\text{.}\) We can accomplish this by replacing \(j\) with \(j+1\) in the first two sums
\begin{align}
& \sum_{j=1}^\infty (j+1)(j)c_{j+1}\rho^{j-1}
+ (2\ell+2)\sum_{j=0}^\infty (j+1)c_{j+1}\rho^{j-1}\notag\\
& - \sum_{j=1}^\infty jc_j\rho^j
+ (\lambda-\ell-1) \sum_{j=0}^\infty c_j\rho^j
= 0\tag{22.16.7}
\end{align}
Once again, the coefficient of each term in the sum must vanish. The coefficient of the general term \(\rho^j\) is
\begin{equation}
(j+1)(j)c_{j+1} + (2\ell+2)(j+1)c_{j+1}
- jc_j + (\lambda-\ell-1)c_j
= 0\tag{22.16.8}
\end{equation}
This leads to the recursion relation
\begin{equation}
c_{j+1} = \frac{j-\lambda+\ell+1}{(j+1)(j+2\ell+2)} c_j\tag{22.16.9}
\end{equation}
which agrees with the results we obtained from the first few terms in the expansion of (22.16.5). As we deduced from that analysis, choosing \(c_0\) determines all of the remaining expansion coefficients.
Subsection Terminating the series
So far we have assumed that the upper limit of the series expansion of \(H(\rho)\) is \(\infty\text{.}\) However, our analysis for the expansion coefficients does not depend on whether the series is finite or infinite. Let’s see how we can use the properties of \(R(r)\) to make this determination.
How does the recursion relationship, (22.16.9), behave in the limit of large \(j\text{?}\) As \(j\to\infty\text{,}\) (22.16.9) becomes approximately \(c_{j+1}\approx c_j/(j+1)\text{.}\) That is, \(c_1=c_0\text{,}\) \(c_2=c_1/2=c_0/2\text{,}\) \(c_3=c_2/3=c_0/3!\text{,}\) and so forth. In general, then, for large \(j\text{,}\) \(c_j\approx c_0/j!\) and because \(\sum_{j=0}^\infty \rho^j/j! = e^\rho\) we have \(H(\rho)=c_0e^\rho\text{.}\) Recalling that \(\rho=2br\text{,}\) (22.15.8) gives the behavior of \(R(r)\) for large \(r\) as \(R(r)\approx r^\ell e^{-br} e^{2br}\text{,}\) which clearly blows up for large \(r\text{.}\) We cannot allow this to happen, and we can prevent it by arranging for the series expansion of \(H(\rho)\) to terminate at some value \(j_\text{max}\) such that the numerator of the recursion relationship, (22.16.9), goes to zero
\begin{equation}
j_\text{max} - \lambda+\ell+1 = 0\tag{22.16.10}
\end{equation}
Because \(j\) and \(\ell\) are integers, (22.16.10) can be satisfied only if \(\lambda\) is also an integer. Let’s call this integer \(n\text{:}\)
\begin{equation}
j_\text{max} = n-\ell-1\tag{22.16.11}
\end{equation}
Using our previous definition of \(\lambda=\mu k/\hbar^2b\) with \(\lambda=n\text{,}\) we have
\begin{equation}
n
= \frac{\mu k}{\hbar^2b}
= \frac{\mu(Ze^2/4\pi\epsilon_0)}{\hbar^2\sqrt{2\mu|E|/\hbar^2}}\tag{22.16.12}
\end{equation}
or
\begin{equation}
E_n = -\frac{Z^2e^4\mu}{32\pi^2\epsilon^2\hbar^2}\frac{1}{n^2}\tag{22.16.13}
\end{equation}
This is the quantization condition on the energy levels of the hydrogen atom, which agrees with the semi-classical condition of the Bohr model. Note that \(E\) depends only on \(n\) and not on \(\ell\) (even though \(R(r)\) depends on both \(n\) and \(\ell\)). For \(Z=1\) and \(\mu=m_e\) (the electron mass), the numerical factors in (22.16.13) evaluate to 13.6 eV.
Reflection.
What property did we impose on the radial wave functions that ultimately resulted in energy quantization?
It is also convenient at this point to define the quantity
\begin{equation}
a_0 = \frac{4\pi\epsilon_0\hbar^2}{\mu e^2}\tag{22.16.14}
\end{equation}
\(a_0\) has the dimension of length and provides a convenient length scale for atomic systems. When \(\mu=m_e\text{,}\) \(a_0\) is known as the Bohr radius and has the value 0.0529 nm = 52.9 pm. In terms of \(a_0\text{,}\) the parameter \(b\) can be written
\begin{equation}
b
= \sqrt{\frac{2m|E|}{\hbar^2}}
= \frac{\mu Ze^2}{4\pi\epsilon_0\hbar^2n}
= \frac{Z}{n a_0}\tag{22.16.15}
\end{equation}
Subsection Some radial wave functions
Before we discuss the general results for \(H(\rho)\text{,}\) let’s look at the form of some of the solutions.
\(n=1\).
In this case, (22.16.10) gives \(j_\text{max}=-\ell\text{,}\) and since neither \(j\) nor \(\ell\) can be negative, the only possible solution is \(j=\ell=0\text{.}\) In this case, \(H(\rho)=c_0\) and \(R(r)\) becomes
\begin{equation}
R_{10}(r) = c_0 e^{-Zra_0}\tag{22.16.16}
\end{equation}
We have labeled the radial wave functions as \(R_{n\ell}\) using the indices \(n\) and \(\ell\text{.}\) The constant \(c_0\) can be determined from the normalization condition.
\(n=2\).
In this case, \(j_\text{max}=1-\ell\text{.}\) The possible \(\ell\) values are \(\ell=0\) (in which case \(j_\text{max}=1\)) and \(\ell=1\) (in which case \(j_\text{max}=0\)). For \(\ell=1\text{,}\) only \(j=0\) contributes and \(H(\rho)=c_0\text{.}\) We then have
\begin{equation}
R_{21}(r) = c_0 r e^{-Zr/2a_0}\tag{22.16.17}
\end{equation}
For \(\ell=0\text{,}\) \(j_\text{max}=1\) and thus \(H(\rho)=c_0 + c_1\rho\text{.}\) We can use the recursion formula, (22.16.9), to determine \(c_1\) in terms of \(c_0\text{:}\)
\begin{equation}
c_1 = \frac{0-2+0+1}{(1)(0+0+2)}c_0 = -\frac12 c_0\tag{22.16.18}
\end{equation}
so that \(H(\rho)=c_0-\frac12c_0\rho=c_0(1-\frac12\rho)\) or \(H(r)=c_0(1-Zr/2a_0)\text{.}\) We thus have
\begin{equation}
R_{20}(r) = c_0 e^{-Zr/2a_0} \left(1-\frac{Zr}{2a_0}\right)\tag{22.16.19}
\end{equation}
Practice problems:.
Continue this process for \(n=3\) and show
\begin{align}
R_{30}(r)
&= c_0 \left[
1-\frac{2Zr}{3a_0}+\frac{2}{27}\left(\frac{Zr}{a_0}\right)^2
\right] e^{-Zr/3a_0}\tag{22.16.20}\\
R_{31}(r)
&= c_0 r \left( 1-\frac{Zr}{6a_0} \right) e^{-Zr/3a_0}\tag{22.16.21}\\
R_{32}(r) &= c_0 r^2 e^{-Zr/3a_0}\tag{22.16.22}
\end{align}
