# 埃爾米特多項式與量子諧振子
###### tags: `應用數學三`
教學影片（@NTU COOL）：[Chapter 3-9](https://cool.ntu.edu.tw/courses/1649/modules/items/79879)

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一維量子諧振子的薛丁格方程式為 $$\hat{H}|\psi\rangle=\underbrace{\left(-\dfrac{\hbar^2}{2m}\dfrac{\text{d}^2}{\text{d}x^2}+\dfrac{1}{2}m\omega^2\hat{x}^2\right)}_{\equiv\ \hat{H}}|\psi\rangle=E|\psi\rangle\tag{1}$$ 其中 $\hat{H}$ 是哈密頓算子，它（投影在位置空間中）的固有函數 $\psi_n(x)$ 與固有值 $E_n$ 滿足 $$\hat{H}\psi_n(x)=-\dfrac{\hbar^2}{2m}\dfrac{\text{d}^2\psi_n(x)}{\text{d}x^2}+\dfrac{1}{2}m\omega^2x^2\psi_n=E_n\psi_n(x)\tag{2}$$

## 1. 基態波函數
若用約化位置坐標 $$\rho=\sqrt{\dfrac{m\omega}{\hbar}}x\tag{3}$$ 取代一般位置坐標 $x$，則 $$\hat{H}=\dfrac{\hbar\omega}{2}\left(-\dfrac{\hbar}{m\omega}\dfrac{\text{d}^2}{\text{d}x^2}+\dfrac{m\omega}{\hbar}\hat{x}^2\right)=\dfrac{\hbar\omega}{2}\left(-\dfrac{\text{d}^2}{\text{d}\rho^2}+\rho^2\right)。\tag{4}$$

### 定義1：創生與消滅算符
***消滅算符（annihilation operator）*** $\hat{a}$ $$\boxed{\hat{a}\equiv\dfrac{1}{\sqrt{2}}\left(\dfrac{\text{d}}{\text{d}\rho}+\rho\right)}\tag{5}$$

***創生算符（creation operator）*** $\hat{a}^\dagger$ $$\boxed{\hat{a}^\dagger\equiv\dfrac{1}{\sqrt{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)}\tag{6}$$

（我們之後就會知道它們為什麼被命名成這樣了。）
### 性質1：創生與消滅算符的伴隨關係
$\hat{a}^\dagger$ 右上角的 $\dagger$（匕首，dagger）指出它是 $\hat{a}$ 的[埃爾米特伴隨](https://reurl.cc/kdQEXn)（Hermitian adjoint），但前提如下：給定任兩個平方可積函數 $f$、$g$，把它們的內積定義成 $$\langle g|f\rangle\equiv\int^\infty_{-\infty}g^*(\rho)f(\rho)\,\text{d}x=\sqrt{\dfrac{\hbar}{m\omega}}\int^\infty_{-\infty}g^*(\rho)f(\rho)\,\text{d}\rho，\tag{7}$$ 如此一來我們可以驗證 $\hat{a}^\dagger$ 的確是$\hat{a}$ 的埃爾米特伴隨：$$\begin{align}\langle g|\hat{a}|f\rangle&=\int^\infty_{-\infty}g^*(\rho)\dfrac{1}{\sqrt{2}}\left(\dfrac{\text{d}}{\text{d}\rho}+\rho\right)f(\rho)\,\text{d}x\\&=\int^\infty_{-\infty}\left[\dfrac{1}{\sqrt{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)g(\rho)\right]^*f(\rho)\,\text{d}x=\langle \hat{a}^\dagger g|f\rangle\end{align}$$（關於 $\text{d}/\text{d}x$ 的埃爾米特伴隨是 $-\text{d}/\text{d}x$，可參考[這份文件](https://reurl.cc/0oyXGl)的 (5) 式）。

### 性質2：創生與消滅算符的交換子
把 $\hat{a}$ 和 $\hat{a}^\dagger$ 的交換子 $\left[\hat{a},\hat{a}^\dagger\right]$ 對任意一個可微函數 $f(\rho)$ 作用，我們得到 $$\require{cancel}\begin{align}\left[\hat{a},\hat{a}^\dagger\right]\,f(\rho)&=\hat{a}\hat{a}^\dagger f(\rho)-\hat{a}^\dagger\hat{a} f(\rho)\\&=\dfrac{1}{2}\left(\small\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\left[-f'(\rho)+\rho f(\rho)\right]-\dfrac{1}{2}\left(\small-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\left[f'(\rho)+\rho f(\rho)\right]\\
&=\dfrac{1}{2}\left[-\bcancel{f''(\rho)}-\cancel{\rho f'(\rho)}+f(\rho)+\cancel{\rho f'(\rho)}+\bcancel{\rho^2f(\rho)}\right]\\
&\quad-\dfrac{1}{2}\left[-\bcancel{f''(\rho)}+\cancel{\rho f'(\rho)}-f(\rho)-\cancel{\rho f'(\rho)}+\bcancel{\rho^2f(\rho)}\right]\\&=f(\rho)，\tag{8}\end{align}$$ 或者我們也可以用一般的算符 $\hat{A}$、$\hat{B}$，觀察到 $$\begin{align}&\quad\left[\hat{A}+\hat{B},-\hat{A}+\hat{B}\right]\\&=\left(\hat{A}+\hat{B}\right)\left(-\hat{A}+\hat{B}\right)-\left(-\hat{A}+\hat{B}\right)\left(\hat{A}+\hat{B}\right)\\&=\left(-\cancel{\hat{A}^2}-\hat{B}\hat{A}+\hat{A}\hat{B}+\cancel{\hat{B}^2}\right)-\left(-\cancel{\hat{A}^2}+\hat{B}\hat{A}-\hat{A}\hat{B}+\cancel{\hat{B}^2}\right)\\&=2\left(\hat{A}\hat{B}-\hat{B}\hat{A}\right)=2\left[\hat{A},\hat{B}\right]，\end{align}$$ 並將此用於計算 $$\begin{align}\left[\hat{a},\hat{a}^\dagger\right]\,f(\rho)&=\left[\dfrac{1}{\sqrt{2}}\left(\dfrac{\text{d}}{\text{d}\rho}+\rho\right),\dfrac{1}{\sqrt{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\right]f(\rho)\\&=2\left[\dfrac{1}{\sqrt{2}}\dfrac{\text{d}}{\text{d}\rho},\dfrac{1}{\sqrt{2}}\rho\right]f(\rho)=\left[\dfrac{\text{d}}{\text{d}\rho},\rho\right]f(\rho)\\&=\dfrac{\text{d}[\rho f(\rho)]}{\text{d}\rho}-\rho\dfrac{\text{d} f(\rho)}{\text{d}\rho}=f(\rho)+\cancel{\rho\dfrac{\text{d} f(\rho)}{\text{d}\rho}}-\cancel{\rho\dfrac{\text{d} f(\rho)}{\text{d}\rho}}\\&=f(\rho)。\tag{9}\end{align}$$ 因此，(8)、(9)式都證明以下關係： $$\boxed{\left[\hat{a},\hat{a}^\dagger\right]=\hat{1}}\tag{10}$$

### 定義2：粒子數算符
我們可證明$$\boxed{\hat{H}=\hbar\omega\left(\hat{a}^\dagger\hat{a}+\dfrac{1}{2}\right)}，\tag{11}$$ 其方法是對創生及消滅算符做如下運算 $$\begin{align}\hat{a}^\dagger\hat{a}&=\dfrac{1}{2}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\left(\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\\&=\dfrac{1}{2}\left(-\dfrac{\text{d}^2}{\text{d}\rho^2}+\rho^2\right)-\dfrac{1}{2}\left[\dfrac{\text{d}}{\text{d}\rho},\rho\right]\\&=\dfrac{\hat{H}}{\hbar\omega}-\dfrac{1}{2}，\tag{12}\end{align}$$ 其中我們使用了 (4) 式以得到 $\frac{1}{2}(-\text{d}^2/\text{d}\rho^2+\rho^2)=\hat{H}/\hbar\omega$，又利用 (9) 式推導過程而得到 $[\text{d}/\text{d}\rho,\rho]=\hat{1}$。

***粒子數算符*** 或***佔有數算符（occupation number operator）*** 定義為 $$\boxed{\hat{N}=\hat{a}^\dagger\hat{a}}\tag{13}$$ （我們之後就會看出為什麼如此命名），則 (11) 式也可以寫成 $$\boxed{\hat{H}=\hbar\omega\left(\hat{N}+\dfrac{1}{2}\right)}\tag{14}$$

### 定義3：基態
由 $$\int^\infty_{-\infty}\psi^*(\rho)\,\hat{a}^\dagger\hat{a} \psi(\rho)\, \text{d}x=\int^\infty_{-\infty}\left[\hat{a}\psi(\rho)\right]^* \left[\hat{a}\psi(\rho)\right] \text{d}x$$ 或者由更簡潔的 $$\langle \psi|\hat{a}^\dagger\hat{a}|\psi\rangle=\langle \hat{a}\psi|\hat{a}\psi\rangle=\|\hat{a}\psi\|^2\tag{15}$$ 都可看出 (14)、(15) 式都是 $$\hat{a}\psi(\rho)=\dfrac{1}{\sqrt{2}}\left(\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\psi(\rho)\tag{16}$$ 的範數（norm）。

若有一狀態 $\psi_0(\rho)$ 的範數為零，亦即 $\psi_0(\rho)$ 滿足 $$\boxed{\hat{a}\psi_0(\rho)=\dfrac{1}{\sqrt{2}}\left(\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\psi_0(\rho)=0}，\tag{17}$$ 則 $\psi_0(\rho)$ 也會滿足 $$\begin{align}\hat{H}\psi_0(\rho)&=\hbar\omega\left(\hat{a}^\dagger\hat{a}+\dfrac{1}{2}\right)\psi_0(\rho)\\&=\hbar\omega\hat{a}^\dagger\cancelto{0}{\left(\hat{a}\psi_0(\rho)\right)}+\dfrac{1}{2}\hbar\omega\psi_0(\rho)\\&=\dfrac{1}{2}\hbar\omega\psi_0(\rho)。\end{align}$$ 我們陳述（而不證明） $E_0\equiv \frac{1}{2}\hbar\omega$ 是 $\hat{H}$ 所有固有值中最小的一個，代表最低能量；此時的狀態 $\psi_0$ 稱為**基態**（ground state），下標記為 $0$ 的原因在此。

### 解出基態波函數
::: spoiler
<font size=2>一階線性微分方程式</font> $$y'(x)+P(x)y(x)=Q(x)$$ <font size=2>的解析解為</font> $$y(x)=\dfrac{1}{u(x)}\int Q(x)u(x)\,\text{d}x，$$ <font size=2>其中</font> $u(x)=e^{\int P(x)\,\text{d}x}$ <font size=2>為積分因子。</font>
:::

我們可解 (17) 式（一階線性微分方程式），得到基態波函數 $$\psi_0(\rho)=\dfrac{1}{e^{\rho^2/2}}\underbrace{\int_0^{\rho}\left[0\cdot e^{\rho'^2/2}\right]\text{d}\rho'}_{\equiv\ A}=Ae^{-\rho^2/2}，$$ 或換回一般位置坐標 $x$，就是 $$\psi_0(x)=Ae^{-\frac{m\omega}{2\hbar}x^2}，$$其中常數 $A$ 要由歸一化條件決定：$$\begin{align}1&=\int^\infty_{-\infty}\left[\psi_0(x)\right]^2\text{d}x=|A|^2\int^\infty_{-\infty}e^{-\frac{m\omega}{\hbar}x^2}\text{d}x=|A|^2\sqrt{\dfrac{\pi\hbar}{m\omega}}\\\implies A&=\left(\dfrac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}\tag{18}\end{align}$$ 於是，量子諧振子（歸一化後）的基態波函數為$$\boxed{\psi_0(x)=\left(\dfrac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}e^{-\frac{m\omega}{2\hbar}x^2}}。\tag{19}$$

## 2. 一般狀態波函數
對於一般狀態的波函數 $\psi_n(x)$，我們可以用 (15) 式把 (2) 式改寫成 $$\hat{H}\psi_n(x)=\hbar\omega\left(\hat{N}+\dfrac{1}{2}\right)\psi_n(x)=E_n\psi_n(x)，$$ 更可以改寫成 $$\boxed{\hat{N}\psi_n(x)=\lambda_n\psi_n(x)}，\tag{20}$$ 其中$\hat{N}$ 的固有值是 $$\boxed{\lambda_n=\dfrac{E_n}{\hbar\omega}-\dfrac{1}{2}}，\tag{21}$$伴隨的固有函數仍是 $\psi_n(x)$。

為了解出 $\psi_n(x)$，我們不妨使用擬設（ansatz）$$\psi(x)=y(x)\psi_0(x)，$$ 把 $\psi_n(x)$ 假定為基態波函數（(19)式）與某函數 $y(x)$ 的積（我們將會證明 $y(x)$ 其實是個多項式）。則可藉由 $\hat{N}=\hat{a}^\dagger\hat{a}$ 的關係（(13)式）將 (20) 式寫成 $$\begin{align}\lambda_n y\psi_0&=\hat{a}^\dagger\hat{a}(y\psi_0)=\dfrac{1}{2}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\left(\dfrac{\text{d}}{\text{d}\rho}+\rho\right)(y\psi_0)\\&=\dfrac{1}{2}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\bigg(\psi_0\dfrac{\text{d}y}{\text{d}\rho}+\cancelto{0}{y\dfrac{\text{d}\psi_0}{\text{d}\rho}+y\rho\psi_0}\bigg)\\&=\dfrac{1}{2}\left(-\dfrac{\text{d}\psi_0}{\text{d}\rho}\dfrac{\text{d}y}{\text{d}\rho}-\psi_0\dfrac{\text{d}^2y}{\text{d}\rho^2}+\rho\psi_0\dfrac{\text{d}y}{\text{d}\rho}\right)\\&=\dfrac{1}{2}\left[-\dfrac{\text{d}y}{\text{d}\rho}\cancelto{0}{\left(\dfrac{\text{d}\psi_0}{\text{d}\rho}+\rho\psi_0\right)}-\psi_0\dfrac{\text{d}^2y}{\text{d}\rho^2}+2\rho\psi_0\dfrac{\text{d}y}{\text{d}\rho}\right]\\&=\left(\rho\dfrac{\text{d}y}{\text{d}\rho}-\dfrac{1}{2}\dfrac{\text{d}^2y}{\text{d}\rho^2}\right)\psi_0，\end{align}$$ 其中我們使用 $\frac{\text{d}\psi_0}{\text{d}\rho}+\rho\psi_0=0$（(17) 式）兩次。由此推得 $$\boxed{\dfrac{\text{d}^2y}{\text{d}\rho^2}-2\rho\dfrac{\text{d}y}{\text{d}\rho}+2\lambda_ny=0}。\tag{22}$$ (22) 式是二階線性微分方程式，叫做***埃爾米特方程式（Hermite equation）***。

### 定義1：階梯算符
利用 (10) 式的關係和 (13) 式的定義，我們能計算創生算符 $\hat{a}^\dagger$ 和 粒子數算符 $\hat{N}$ 的交換子，$$\boxed{\left[\hat{N},\hat{a}^\dagger\right]=\left[\hat{a}^\dagger\hat{a},\hat{a}^\dagger\right]=\hat{a}^\dagger}\tag{23}$$ 這是因為由[交換子的運算規則](https://reurl.cc/L3Z6ax) ，對任意三算符 $A$、$B$、$C$，我們有$$\left[\hat{A}\hat{B},\hat{C}\right]=\left[\hat{A},\hat{C}\right]\hat{B}+\hat{A}\left[\hat{B},\hat{C}\right]$$ 和 $$\left[\hat{A},\hat{A}\right]=0，$$ 所以 $\left[\hat{a}^\dagger\hat{a},\hat{a}^\dagger\right]=\cancelto{0}{\left[\hat{a}^\dagger,\hat{a}^\dagger\right]}\hat{a}+\hat{a}^\dagger\cancelto{1}{\left[\hat{a},\hat{a}^\dagger\right]}$，於是就得到 (23) 式。

對於滿足 (23) 式這種對易關係的算符 $\hat{a}^\dagger$，我們稱之為***階梯算符（ladder operator）***，其一般定義如下：

[階梯算符](https://reurl.cc/0oyl0A)（ladder operator）
: 算符 $\hat{X}$ 是階梯算符，若且唯若：
1. $\hat{X}$ 與一任意算符 $\hat{N}$ 有對易關係如下：$$\left[\hat{N},\hat{X}\right]=c\hat{X}，$$ 其中 $c$ 為純量；而且
2. 如果 $|n\rangle$ 是上述算符 $\hat{N}$ 的固有態，對應到固有值 $n$， 即$\hat{N}|n\rangle=n|n\rangle$，則算符 $\hat{X}$ 作用在 $|n\rangle$ 後，對應到 $\hat{N}$ 的固有值 $n$ 移動了 $c$，也就是 $$\begin{align}\hat{N}\hat{X}|n\rangle&=\left(\hat{X}\hat{N}+\left[\hat{N},\hat{X}\right]\right)|n\rangle\\&=\left(\hat{X}\hat{N}+c\hat{X}\right)|n\rangle\\&=\hat{X}\hat{N}|n\rangle+c\hat{X}|n\rangle\\&=(n+c)\hat{X}|n\rangle\end{align}$$
當 $c>0$ 時，稱為***升算符（raising operator）***；當 $c<0$ 時，稱為***降算符（lowering operator）***。

按照以上定義，
- 消滅算符 $\hat{a}$ 是一種降算符，且 $c=-1$
- 創生算符 $\hat{a}^\dagger$ 是一種升算符，且 $c=1$。

### 解出一般狀態波函數
用 (23) 式，計算作用一次 $\hat{a}^\dagger$ 時得到$$\begin{align}\hat{N}\hat{a}^\dagger\psi_n&=\hat{a}^\dagger\hat{N}|\psi_n\rangle-\left[\hat{N},\hat{a}^\dagger\right]|\psi_n\rangle=\hat{a}^\dagger\left(\hat{N}+\hat{1}\right)|\psi_n\rangle\\&=(\lambda_n+1)\hat{a}^\dagger|\psi_n\rangle，\end{align}$$ 作用兩次 $\hat{a}^\dagger$ 時，會是$$\begin{align}\hat{N}\left(\hat{a}^\dagger\right)^2|\psi_n\rangle&=\hat{N}\hat{a}^\dagger\left(\hat{a}^\dagger|\psi_n\rangle\right)=\hat{a}^\dagger\left(\hat{N}+\hat{1}\right)\left(\hat{a}^\dagger|\psi_n\rangle\right)\\
&=\hat{a}^\dagger\left(\hat{N}\hat{a}^\dagger|\psi_n\rangle+\hat{a}^\dagger|\psi_n\rangle\right)=\hat{a}^\dagger\left[(\lambda_n+1)\hat{a}^\dagger|\psi_n\rangle+\hat{a}^\dagger|\psi_n\rangle\right]\\
&=(\lambda_n+2)\left(\hat{a}^\dagger\right)^2|\psi_n\rangle，\end{align}$$ 如此重複下去，可歸納出，作用 $k$ 次 $\hat{a}^\dagger$ 時，會是 $$\hat{N}\left(\hat{a}^\dagger\right)^k|\psi_n\rangle=(\lambda_n+k)\left(\hat{a}^\dagger\right)^k|\psi_n\rangle，\tag{24}$$ 這就驗證了階梯算符的意義——使固有值移動。對 (24) 式而言，比較特別的是 $$\hat{N}\left(\hat{a}^\dagger\right)^n|\psi_0\rangle=(\lambda_0+n)\left(\hat{a}^\dagger\right)^n|\psi_0\rangle=n\left(\hat{a}^\dagger\right)^n|\psi_0\rangle，\tag{25}$$ （由 (21) 式和 $E_0=\frac{1}{2}\hbar\omega$ 可知 $\lambda_0=0$）。但把 (25) 式與 $\hat{N}|\psi_n\rangle=\lambda_n|\psi_n\rangle$（(20) 式）比較，我們發現 $$\lambda_n=n，\tag{26}$$ 或者用 (21) 式解出能量固有值 $$\boxed{E_n=\hbar\omega\left(n+\dfrac{1}{2}\right)}；\tag{27}$$ 我們又發現$$|\psi_n\rangle=\left(\hat{a}^\dagger\right)^n|\psi_0\rangle。\tag{28}$$ 換句話說，==對 $|\psi_0\rangle$ 作用 $n$ 次 $\hat{a}^\dagger$ 就可以得到 $|\psi_n\rangle$==。但我們希望 $|\psi_n\rangle=\left|\left(\hat{a}^\dagger\right)^n\psi_0\right>$ 歸一化，所以我們計算$$\require{color}\begin{align}\langle\psi_n|\psi_n\rangle&=\left<\left(\hat{a}^\dagger\right)^{\color{darkcyan}n}\psi_0\middle|\left(\hat{a}^\dagger\right)^{\color{darkcyan}n}\psi_0\right>
\\&=\left<\psi_0\middle|\hat{a}^{\color{darkcyan}n}\left(\hat{a}^\dagger\right)^{\color{darkcyan}n}\middle|\psi_0\right>\\
&=\left<\psi_0\middle|\hat{a}^{\color{darkcyan}n-1}\left.\color{blue}\left(\hat{a}\hat{a}^\dagger\right)\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}\middle|\psi_0\right>\\
&=\left<\psi_0\middle|\hat{a}^{\color{darkcyan}n-1}\left.\color{blue}\left(\hat{N}+\left[\hat{a},\hat{a}^\dagger\right]\right)\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}\middle|\psi_0\right>\\
&=\left<\psi_0\middle|\hat{a}^{\color{darkcyan}n-1}\left.\color{blue}\left(\hat{N}+\hat{1}\right)\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}\middle|\psi_0\right>\\
&=\Big\langle\psi_0\hat{a}^{\color{darkcyan}n-1}\bigg|\left[\left.\color{blue}\hat{N}\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}|\psi_0\rangle+\left.\color{blue}\hat{1}\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}|\psi_0\rangle\right]\\
&=\Big\langle\psi_0\hat{a}^{\color{darkcyan}n-1}\bigg|\left[\left.\color{blue}(n-1)\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}|\psi_0\rangle+\left.\color{blue}1\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}|\psi_0\rangle\right]\\
&=\left<\psi_0\hat{a}^{\color{darkcyan}n-1}\middle|\left.\color{blue}n\right.\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}\psi_0\right>=\left.\color{blue}n\right.\left<\psi_0\middle|\hat{a}^{\color{darkcyan}n-1}\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-1}\middle|\psi_0\right>\\
&=\left.\color{blue}n(n-1)\right.\left<\psi_0\middle|\hat{a}^{\color{darkcyan}n-2}\left(\hat{a}^\dagger\right)^{\color{darkcyan}n-2}\middle|\psi_0\right>=\cdots\\
&=\left.\color{blue}n!\right.\left<\psi_0\middle|\hat{a}^{\color{darkcyan}0}\left(\hat{a}^\dagger\right)^{\color{darkcyan}0}\middle|\psi_0\right>=\left.\color{blue}n!\right.\left<\psi_0\middle|\psi_0\right>=\left.\color{blue}n!\right.。\end{align}$$ 所以如果把 $|\psi_n\rangle$ 除以它的範數 $\sqrt{n!}$ 等於是為 $|\psi_n\rangle$ 做歸一化，於是，量子諧振子（歸一化後）的固有態波函數為 $$\boxed{\psi_n(x)=\dfrac{1}{\sqrt{n!}}\left(\hat{a}^\dagger\right)^n\psi_0(x)}。\tag{29}$$

### 定義2：埃爾米特多項式
(29) 式還留有 $n$ 個算符 $\hat{a}^\dagger$，我們讓它對 $\psi_0(x)$ （(19)式）作用，得到 $$\begin{align}\psi_n(x)&=\dfrac{1}{\sqrt{n!}}\left(\hat{a}^\dagger\right)^n\psi_0(x)=\dfrac{1}{\sqrt{n!}}\left[\dfrac{1}{\sqrt{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\right]^n\left[\left(\dfrac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}e^{-\frac{m\omega}{2\hbar}x^2}\right]\\
&=\dfrac{1}{\sqrt{2^nn!}}\left(\dfrac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)^ne^{-\frac{\rho^2}{2}}。\tag{30}\end{align}$$ 這裡出現了 $\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)^ne^{-\frac{\rho^2}{2}}$，我們注意到 $$\begin{align}e^{-\frac{\rho^2}{2}}&=\left.\color{darkgreen}1\right.e^{-\frac{\rho^2}{2}}\\\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)e^{-\frac{\rho^2}{2}}&=\left.\color{darkgreen}2\rho\right.e^{-\frac{\rho^2}{2}}\\\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)^2e^{-\frac{\rho^2}{2}}=\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)\left[(2\rho)e^{-\frac{\rho^2}{2}}\right]&=\left.\color{darkgreen}(4\rho^2-2)\right.e^{-\frac{\rho^2}{2}}\\\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)^3e^{-\frac{\rho^2}{2}}=\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)\left[(4\rho^2-2)e^{-\frac{\rho^2}{2}}\right]&=\left.\color{darkgreen}(8\rho^3-12\rho)\right.e^{-\frac{\rho^2}{2}}\\&\ \ \vdots\\\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)^ne^{-\frac{\rho^2}{2}}&=\left.\color{darkgreen}(n~次多項式)\right.e^{-\frac{\rho^2}{2}}\end{align}$$ 一直都是某個多項式乘上 $e^{-\frac{\rho^2}{2}}$，我們把它命名為***埃爾米特多項式（Hermite polynomial）***，記作 $H_n(\rho)$，它滿足 $$\boxed{H_n(\rho)=e^{\frac{\rho^2}{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)^ne^{-\frac{\rho^2}{2}}},\quad n=0,1,2,\ldots，\tag{31}$$ 其中前四個埃爾米特多項式是 $$H_0(\rho)=1,\quad H_1(\rho)=2\rho,\quad H_2(\rho)=4\rho^2-2,\quad H_3(\rho)=8\rho^3-12\rho。$$ 使用 (31) 式，可以把 (30) 式寫成 $$\psi_n(\rho)=\dfrac{1}{\sqrt{2^nn!}}\left(\dfrac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}H_n\left(\rho\right)e^{-\frac{\rho^2}{2}}$$ 或 $$\boxed{\psi_n(x)=\dfrac{1}{\sqrt{2^nn!}}\left(\dfrac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}H_n\left(\sqrt{\dfrac{m\omega}{\pi\hbar}}x\right)e^{-\frac{m\omega}{2\hbar}x^2}}\tag{32}$$
[](https://www.desmos.com/calculator/y5faaiesio)

### 定義3：羅德里格公式
埃爾米特多項式的羅德里格公式（Rodrigues' formular）為$$\boxed{H_n(\rho)=e^{\rho^2}\left(-\frac{\text{d}}{\text{d}\rho}\right)^ne^{-\rho^2}},\quad n=0,1,2,\ldots。\tag{31}$$ (30) 式的定義和 (31) 式是等價的，證明如下：$$\begin{align}H_n(\rho)&=\left.\color{darkgoldenrod}e^{\frac{\rho^2}{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)^ne^{-\frac{\rho^2}{2}}\right.=e^{\frac{\rho^2}{2}}\overbrace{\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)\cdots\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)}^{n~個}e^{-\frac{\rho^2}{2}}\\
&=\overbrace{\Big[e^{\frac{\rho^2}{2}}\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)\underbrace{\color{red}e^{-\frac{\rho^2}{2}}\Big]\Big[e^{\frac{\rho^2}{2}}}_{=\ 1}\cdots\underbrace{\color{red}e^{-\frac{\rho^2}{2}}\Big]\Big[e^{\frac{\rho^2}{2}}}_{=\ 1}\left(-\tfrac{\text{d}}{\text{d}\rho}+\rho\right)e^{-\frac{\rho^2}{2}}\Big]}^{n~個}\\
&=\left.\color{darkgoldenrod}\left[e^{\frac{\rho^2}{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)e^{-\frac{\rho^2}{2}}\right]^n\right.=e^{\frac{\rho^2}{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)\Bigg\{e^{-\frac{\rho^2}{2}}\underbrace{\left[\small e^{\frac{\rho^2}{2}}\left(-\dfrac{\text{d}}{\text{d}\rho}+\rho\right)e^{-\frac{\rho^2}{2}}\right]^{n-1}}_{=\ H_{n-1}(\rho)}\Bigg\}\\
&=e^{\frac{\rho^2}{2}}\left\{-\frac{\text{d}}{\text{d}\rho}\left[e^{-\frac{\rho^2}{2}}H_{n-1}(\rho)\right]+\rho e^{-\frac{\rho^2}{2}}H_{n-1}(\rho)\right\}\\
&=\cancel{e^{\frac{\rho^2}{2}}}\rho\cancel{e^{-\frac{\rho^2}{2}}}H_{n-1}(\rho)-\cancel{e^{\frac{\rho^2}{2}}}\cancel{e^{-\frac{\rho^2}{2}}}\frac{\text{d}H_{n-1}(\rho)}{\text{d}\rho}+\cancel{e^{\frac{\rho^2}{2}}}\rho\cancel{e^{-\frac{\rho^2}{2}}}H_{n-1}(\rho)\\
&=2\rho H_{n-1}(\rho)-\dfrac{\text{d}H_{n-1}(\rho)}{\text{d}\rho}=\left.\color{red}e^{\rho^2}\right.\left[2\rho\left.\color{red}e^{-\rho^2}\right.H_{n-1}(\rho)-\left.\color{red}e^{-\rho^2}\right.\dfrac{\text{d}H_{n-1}(\rho)}{\text{d}\rho}\right]\\
&=e^{\rho^2}\left(-\frac{\text{d}}{\text{d}\rho}\right)\left[e^{-\rho^2}H_{n-1}(\rho)\right]=\color{darkgoldenrod}\left[e^{\rho^2}\left(-\frac{\text{d}}{\text{d}\rho}\right)e^{-\rho^2}\right]^n\\
&=\overbrace{\Big[e^{\rho^2}\left(-\tfrac{\text{d}}{\text{d}\rho}\right)\cancel{e^{-\rho^2}}\Big]\Big[\cancel{e^{\rho^2}}\cdots \cancel{e^{-\rho^2}}\Big]\Big[\cancel{e^{\rho^2}}\left(-\tfrac{\text{d}}{\text{d}\rho}\right)e^{-\rho^2}\Big]}^{n~個}=\left.\color{darkgoldenrod}e^{\rho^2}\left(-\frac{\text{d}}{\text{d}\rho}\right)^ne^{-\rho^2}\right.。\end{align}$$

### 性質2：埃爾米特多項式的領導係數
我們也觀察到 $H_n(\rho)$ 是領導係數（leading coefficient）為 $2^n$ 的 $n$ 次多項式。

### 性質4：埃爾米特多項式的正交歸一條件

### 性質4：埃爾米特多項式的生成函數
埃爾米特多項式 $H_n(x)$ 的生成函數是 $e^{2xt-t^2}$
因為 $$$$