Example 6 Given the function $f(x)=x^{2}+x-1, \alpha 、 \beta$ $(\alpha>\beta)$ are the two roots of the equation $f(x)=0$, and $f^{\prime}(x)$ is the derivative function.
Let $a_{1}=1, a_{n+1}=a_{n}-\frac{f\left(a_{n}\right)}{f^{\prime}\left(a_{n}\right)}(n=1,2, \cdots)$.
(1) Find the values of $\alpha 、 \beta$;
(2) It is known that for any positive integer $n$, $a_{n}>\alpha$, and let $b_{n}=\ln \frac{a_{n}-\beta}{a_{n}-\alpha}(n=1,2, \cdots)$. Find the sum of the first $n$ terms $S_{n}$ of the sequence $\left\{b_{n}\right\}$.
(2007, National College Entrance Examination (Guangdong Paper))
(1) From $x^{2}+x-1=0$, we get $x=\frac{-1 \pm \sqrt{5}}{2}$. Therefore, $\alpha=\frac{-1+\sqrt{5}}{2}, \beta=\frac{-1-\sqrt{5}}{2}$.
(2) Note that
$$
\begin{array}{l}
f^{\prime}(x)=2 x+1, \\
a_{n+1}=a_{n}-\frac{a_{n}^{2}+a_{n}-1}{2 a_{n}+1}=\frac{a_{n}^{2}+1}{2 a_{n}+1}, \\
\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}=\frac{\frac{a_{n}^{2}+1}{2 a_{n}+1}+\frac{1+\sqrt{5}}{2}}{\frac{a_{n}^{2}+1}{2 a_{n}+1}+\frac{1-\sqrt{5}}{2}} \\
=\frac{a_{n}^{2}+(1+\sqrt{5}) a_{n}+\frac{3+\sqrt{5}}{2}}{a_{n}^{2}+(1-\sqrt{5}) a_{n}+\frac{3-\sqrt{5}}{2}} \\
=\left(\frac{a_{n}+\frac{1+\sqrt{5}}{2}}{a_{n}+\frac{1-\sqrt{5}}{2}}\right)^{2}=\left(\frac{a_{n}-\beta}{a_{n}-\alpha}\right)^{2}. \\
\end{array}
$$

Therefore, $b_{n+1}=2 b_{n}$.
$$
\text { Also, } b_{1}=\ln \frac{a_{1}-\beta}{a_{1}-\alpha}=\ln \frac{3+\sqrt{5}}{3-\sqrt{5}}=4 \ln \frac{1+\sqrt{5}}{2},
$$

so the sequence $\left\{b_{n}\right\}$ is a geometric sequence with the first term $4 \ln \frac{1+\sqrt{5}}{2}$ and common ratio 2.
Hence, $S_{n}=4\left(2^{n}-1\right) \ln \frac{1+\sqrt{5}}{2}$.