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Operator Solution of the Differential Equation Satisfied by Hermite Functions 📂Mathematical Physics

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.

$$ y_{n}=(D-x)^{n}y_{0}=(D-x)^{n}e^{-\frac{x^{2}}{2}} $$


  1. n이 양자역학에서 에너지 준위로 나타나기 때문에 이러한 아이디어로 미분 방정식을 풀려고 하는 것이다. ↩︎

  2. 양자역학에서 바닥상태에 내림 연산자를 적용하면 0이라고 둔다 ↩︎