Operator Solution of the Differential Equation Satisfied by Hermite Functions
Theorem
Given the differential equation
$$ y_{n}^{\prime \prime}-x^{2}y_{n}=-(2n+1)y_{n},\quad n=0,1,2,\cdots \tag{1} \label{eq1} $$
The solution to $(1)$ is as follows, known as the Hermite function.
$$ \begin{align*} y_{n} &= \left( D-x \right)^{n} e^{-\frac{x^{2}}{2}} \\ &= e^{\frac{x^{2}}{2}} D^{n} x^{-x^{2}} \end{align*} $$
Here, $D$ is the differential operator $D=\frac{ d }{ dx }$.
Explanation
The first equation of $y_{n}$ can be directly obtained by solving the differential equation. That the second equation is equivalent to the first can be proven through mathematical induction.
Proof
Given the differential equation can be represented as follows by the properties of differential operators $(e)$, $(f)$.
$$ \begin{align} (D-x)(D+x)y_{n} &= -2ny_{n} \\ (D+x)(D-x)y_{n} &=-2(n+1)y_{n} \end{align} $$
If we substitute $n-1$ for $n$ in $(2)$ and apply $(D-x)$ to both sides, we get the following.
$$ \begin{equation} (D-x)(D+x)(D-x)y_{n-1} = -2n(D-x)y_{n-1} \end{equation} $$
Substituting $n+1$ for $n$ in $(1)$ and applying $(D+x)$ to both sides gives the following.
$$ \begin{equation} (D+x)(D-x)(D+x)y_{n+1} = -2(n+1)(D+x)y_{n+1} \end{equation} $$
Let’s say $y_{n}$ satisfies the equation below.
$$ \begin{align*} (D-x)y_{n-1} &= y_{n} \\ (D+x)y_{n+1} &= y_{n} \end{align*} $$
Then, $(3)=(1)$ and, therefore, $(4)=(2)$ so, $y_{n}$ still satisfies the differential equation. Thus, let’s find a $y_{n}$ with such properties1.
$(D-x)$ changes $y_{n-1}$ to $y_{n}$, thus let’s call it the raising operator. Conversely, $(D+x)$ changes $y_{n+1}$ into $y_{n}$, so let’s call it the lowering operator. Now, if we find $y_{0}$ that satisfies $(3)$ and $(4)$, then we can express the solution $y_{n}$ of the differential equation through the raising operator. $y_{0}$ is the ground state, so applying the lowering operator results in $0$. This is a physical condition2. Therefore, we get the following equation.
$$ (D+x)y_{0}=0 $$
This equation is a simple separable differential equation. $$ \begin{align*} && \frac{ d y_{0}}{ d x } &=-xy_{0} \\ \implies && \frac{1}{y_{0}}dy_{0} &=-xdx \\ \implies && \ln y_{0} &= -\frac{x^{2}}{2} \\ \implies && y_{0} &=e^{-\frac{x^{2}}{2}} \end{align*} $$
Therefore, $y_{n}$ is as follows.
$$ y_{n}=(D-x)^{n}y_{0}=(D-x)^{n}e^{-\frac{x^{2}}{2}} $$
■