Ladder Operators
The following is essentially a elaboration on
Chebyshev polynomials.
The big trick is to write the Chebyshev differential equation in a slightly
different form, multiplied namely with a factor $(1-x^2)$. Then the very same
differential equation named after Pafnuty Chebyshev becomes:
$$
(1-x^2)^2\frac{d^2T_n(x)}{dx^2} - (1-x^2)x\frac{dT_n(x)}{dx}
+ (1-x^2)n^2 T_n(x) = 0
$$
When cast in Operator Calculus format, it reads:
$$
\left[ (1-x^2)^2\left(\frac{d}{dx}\right)^2 - (1-x^2)x\frac{d}{dx}
+ (1-x^2)n^2 \right] T_n(x) = 0
$$
We shall try to find a factorization again. Be careful:
$$
\left[ (1-x^2)^2\left(\frac{d}{dx}\right)^2 - (1-x^2)x\frac{d}{dx}
+ (1-x^2)n^2 \right] = \\
\left[ (1-x^2) \frac{d}{dx} + \alpha\,x \right]
\left[ (1-x^2) \frac{d}{dx} + \beta\,x \right] = \\
(1-x^2) \frac{d}{dx} (1-x^2) \frac{d}{dx} + \alpha\,x (1-x^2) \frac{d}{dx}
+ (1-x^2) \frac{d}{dx} \beta\,x + \alpha\,x \beta\,x = \\
(1-x^2)^2 \frac{d^2}{dx^2} + \\ \left[- 2 x (1-x^2)
+ \alpha\,x (1-x^2) + (1-x^2) \beta\,x \right] \frac{d}{dx} +
\\ (1-x^2) \beta + \alpha\beta\,x^2
$$
Where:
$$
- 2 x (1-x^2) + \alpha\,x (1-x^2) + (1-x^2) \beta\,x = - x (1-x^2) \\
\Longrightarrow \quad - 2 + \alpha + \beta = - 1 \quad \Longrightarrow \quad \beta = 1-\alpha
\quad \wedge \quad \alpha=1-\beta
$$
And:
$$
(1-x^2) \beta + \alpha\beta\,x^2 =
(1-x^2) \beta - \alpha\beta(1-x^2) + \alpha\beta = \\
(1-x^2) \beta(1-\alpha) + \alpha\beta =
(1-x^2) \beta^2 + \alpha\beta = (1-x^2) n^2
$$
If we put $\beta = + n$ then $\alpha = -(n-1)$ and a factor $-\alpha\beta =
n(n-1)$ must be added to the the left and to the right hand side:
$$
\left[ (1-x^2) \frac{d}{dx} - (n-1)\,x \right]
\left[ (1-x^2) \frac{d}{dx} + n\,x \right] T_n(x) = n(n-1) T_n(x)
$$
If we put $\beta = - n$ then $\alpha = (n+1)$ and a factor $-\alpha\beta =
n(n+1)$ must be added to the the left and to the right hand side:
$$
\left[ (1-x^2) \frac{d}{dx} + (n+1)\,x \right]
\left[ (1-x^2) \frac{d}{dx} - n\,x \right] T_n(x) = n(n+1) T_n(x)
$$
Now define the following operator:
$$
O_m = \left[ (1-x^2) \frac{d}{dx} + m\,x \right]
$$
Then the differential equation by Chebyshev assumes one of the following forms:
\begin{eqnarray*}
O_{-(n-1)}\; O_{+n}\; T_n(x) &=& n(n-1)\; T_n(x) \\
O_{+(n+1)}\; O_{-n}\; T_n(x) &=& n(n+1)\; T_n(x)
\end{eqnarray*}
Multiply the first of these two equations on the left with the operator
$O_{+n}$, times an arbitrary constant $c_1$ eventually:
$$
c_1\;O_{+n} \left[ O_{-(n-1)}\; O_{+n} \right] T_n(x) =
c_1\;O_{+n}\; n(n-1) T_n(x) \\
\left[ O_{+(n-1)+1}\; O_{-(n-1)} \right] c_1\;O_{+n}\; T_n(x) =
n(n-1)\; c_1\;O_{+n}\; T_n(x)
$$
Which demonstrates that $c_1\;O_{+n}\;T_n(x) $ is a solution of the differential
equation for $ n := n-1$. This means that, effectively:
$$
c_1\;O_{+n}\;T_n(x) = T_{n-1}(x) =
c_1\;\left[ (1-x^2) \frac{d}{dx}+n\,x \right] T_n(x)
$$
For this reason, the operator $O_{+n}$ is called a lowering operator.
Multiply the second of the two equations on the left with the operator
$O_{-n}$, times an arbitrary constant $c_2$ eventually:
$$
c_2\;O_{-n} \left[ O_{+(n+1)}\; O_{-n} \right] T_n(x) =
c_2\;O_{-n}\; n(n+1) T_n(x) \\
\left[ O_{-(n+1)+1}\; O_{+(n+1)} \right] c_2\;O_{-n}\; T_n(x) =
n(n+1)\; c_2\;O_{-n}\; T_n(x)
$$
Which demonstrates that $c_2\;O_{-n}\;T_n(x) $ is a solution of the differential
equation for $ n := n+1$. This means that, effectively:
$$
c_2\;O_{-n}\;T_n(x) = T_{n+1}(x) =
c_2\;\left[ (1-x^2) \frac{d}{dx}-n\,x \right] T_n(x)
$$
For this reason, the operator $O_{-n}$ is called a raising operator.
The raising and lowering operators together are called ladder operators,
because they enable us to construct a whole sequence of solutions, once we have
found only one of the possible ones. Let's repeat the result:
$$
c_1\;\left[ (1-x^2) \frac{d}{dx} - n\,x \right] T_n(x) = T_{n+1}(x) \\
c_2\;\left[ (1-x^2) \frac{d}{dx} + n\,x \right] T_n(x) = T_{n-1}(x)
$$
If we add these equations together, then:
$$
T_{n+1}(x) + T_{n-1}(x) = \left[ -n c_1 + n c_2 \right] x\; T_n(x)
+ \left[ c_1 + c_2 \right] (1-x^2) \frac{dT_n}{dx}
$$
While the well known recursion relation for Chebyshev Polynomials is:
$$
T_{n+1}(x) + T_{n-1}(x) = 2 x\; T_n(x)
$$
So if we put the arbitrary constants $c_{1,2}$ to well defined values, namely
$c_1 = -1/n$ and $c_2 = +1/n$, then the recursion relation for solutions of
the Chebyshev differential equation becomes the same as the one for Chebyshev
polynomials:
$$
T_{n+1}(x) + T_{n-1}(x) = 2 x\; T_n(x)
$$
And the ladder relations become:
$$
\left[ (1-x^2) \frac{d}{dx} - n\,x \right] T_n(x) = -n\; T_{n+1}(x) \\
\left[ (1-x^2) \frac{d}{dx} + n\,x \right] T_n(x) = +n\; T_{n-1}(x)
$$