# 量子諧振子

## 一維諧振子

### 哈密頓算符與能量本徵態

$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$

$H \left| \psi \right\rangle = E \left| \psi \right\rangle$.

$\left\langle x | \psi_n \right\rangle = \sqrt{\frac{1}{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \cdot \exp \left(- \frac{m\omega x^2}{2 \hbar} \right) \cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right)$
$n = 0, 1, 2, \ldots$

$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$

$E_n = \hbar \omega \left(n + {1\over 2}\right)$

### 階梯算符方法

$\begin{matrix} a &=& \sqrt{m\omega \over 2\hbar} \left(x + {i \over m \omega} p \right) \\ a^{\dagger} &=& \sqrt{m \omega \over 2\hbar} \left( x - {i \over m \omega} p \right) \end{matrix}$

$\begin{matrix} a \left| \phi _n \right\rangle &=& \sqrt{n} \left| \phi _{n-1} \right\rangle \\ a^{\dagger} \left| \phi _n \right\rangle &=& \sqrt{n+1} \left| \phi _{n+1} \right\rangle \end{matrix}$

$\begin{matrix} x &=& \sqrt{\hbar \over 2m\omega} \left( a^{\dagger} + a \right) \\ p &=& i \sqrt{{\hbar}m\omega \over 2} \left( a^{\dagger} - a \right) \end{matrix}$

xp 算符遵守下面的等式，稱之為正則對易關係

$\left[x , p \right] = i\hbar$.

$\left[A , B \right] \ \stackrel{\mathrm{def}}{=}\ AB - BA$.

$H = \hbar \omega \left(a^{\dagger}a + 1/2\right)$
$\left[a , a^{\dagger} \right] = 1$.

$\left(a \left|\psi_E \right\rangle, a \left|\psi_E \right\rangle\right) = \left\langle\psi_E \right| a^\dagger a \left| \psi_E \right\rangle \ge 0$

aa 以哈密頓算符表示：

$\left\langle\psi_E \right| {H \over \hbar \omega} - {1 \over 2} \left|\psi_E\right\rangle = \left({E \over \hbar \omega} - {1 \over 2} \right) \ge 0$,

$\begin{matrix} \left[H , a \right] &=& - \hbar \omega a \\ \left[H , a ^\dagger\right] &=& \hbar \omega a^\dagger \end{matrix}$.

$\begin{matrix} H (a \left| \psi_E \right\rangle) &=& (\left[H,a\right] + a H) \left|\psi_E\right\rangle \\ &=& (- \hbar\omega a + a E) \left|\psi_E\right\rangle \\ &=& (E - \hbar\omega) (a\left|\psi_E\right\rangle) \end{matrix}$.

$H (a^\dagger \left| \psi_E \right\rangle) = (E + \hbar\omega) (a^\dagger \left| \psi_E \right\rangle)$.

$a \left| 0 \right\rangle = 0$（即 a$\left| 0 \right\rangle$ 作用後產生零右括向量 (zero ket)）。

$H \left|0\right\rangle = (\hbar\omega/2) \left|0\right\rangle$

$H \left|n\right\rangle = \hbar\omega (n + 1/2) \left|n\right\rangle$

$x\psi_0(x) + \frac{\hslash}{m \omega} \frac{d \psi_0 (x)}{dx} = 0$

$\frac{d\ln \psi_0 (x)}{dx}= - \frac{\hslash}{m \omega} x + \text{ Constant}$

$\psi_0(x)= \left({m\omega \over \pi\hbar}\right)^{1 \over 4}e^{ - m\omega x^2/2\hbar}$

### 自然長度與能量尺度

$H = - {1\over2} {d^2 \over du^2 } + {1 \over 2} u^2$,

$\left\langle x | \psi_n \right\rangle = {1 \over \sqrt{2^n n!}} \pi^{-1/4} \hbox{exp} (-u^2 / 2) H_n(u)$
$E_n = n + {1\over 2}$.

### 案例：雙原子分子

$\omega = \sqrt{\frac{k}{m_r}}$

$\omega = 2 \pi f$ 為角頻率，
k共價鍵勁度係數
$m_r$約化質量

## $N$ 維諧振子

$\begin{matrix} \left[x_i , p_j \right] &=& i\hbar\delta_{i,j} \\ \left[x_i , x_j \right] &=& 0 \\ \left[p_i , p_j \right] &=& 0 \end{matrix}$.

$H = \sum_{i=1}^N \left( {p_i^2 \over 2m} + {1\over 2} m \omega^2 x_i^2 \right)$

$\langle \mathbf{x}|\psi_{\{n\}}\rangle=\prod_{i=1}^N\langle x_i|\psi_{n_i} \rangle$

$a_i = \sqrt{m\omega \over 2\hbar} \left(x_i + {i \over m \omega} p_i \right)$
$a^{\dagger}_i = \sqrt{m \omega \over 2\hbar} \left( x_i - {i \over m \omega} p_i \right)$

$H = \hbar \omega \, \sum_{i=1}^N \left(a_i^\dagger \,a_i + \frac{1}{2}\right)$

$E = \hbar \omega \left[(n_1 + \cdots + n_N) + {N\over 2}\right]$

$d_n = \sum_{n_1=0}^n n - n_1 + 1 = \frac{(n+1)(n+2)}{2}$

$d_n = \binom{N+n-1}{n}$

### 案例：三維均向諧振子

$V(r) = {1\over 2} \mu \omega^2 r^2$

$- \frac{\hbar^2}{2\mu}\nabla^2\psi + {1\over 2} \mu \omega^2 r^2\psi=E\psi$

$\psi_{klm}(r,\theta,\phi) = N_{kl} r^{l}e^{-\nu r^2}{L_k}^{(l+{1\over 2})}(2\nu r^2) Y_{lm}(\theta,\phi)$

$N_{kl}=\sqrt{\sqrt{\frac{2\nu ^{3}}{\pi }}\frac{2^{k+2l+3}\;k!\;\nu ^{l}}{ (2k+2l+1)!!}}$ 是歸一常數，
$\nu \equiv {\mu \omega \over 2 \hbar}$
${L_k}^{(l+{1\over 2})}(2\nu r^2)$$k$廣義拉格耳多項式 (generalized Laguerre polynomials) ，$k$ 是個正整數，
$Y_{lm}(\theta,\phi)\,$球諧函數
$\hbar$約化普朗克常數

$E=\hbar \omega (2k+l+{3\over 2})$

$n\equiv 2k+l$

$l=0,\,2,\,\dots,\,n - 2,\,n$

$l=1,\,3,\,\dots,\,n - 2,\,n$

$-l \le m \le l$

$\sum_{l=i,\,i+2,\,\ldots,\,n - 2,\,n} (2l+1) = {(n+1)(n+2)\over 2}$

## 耦合諧振子

$H = \sum_{i=1}^N {p_i^2 \over 2m} + {1\over 2} m \omega^2\sum_{1\le i\le N} (x_i - x_{i-1})^2$

## 參考文獻

• Griffiths, David J。. Introduction to Quantum Mechanics (2nd ed。). Prentice Hall. 2004. ISBN 0-13-111892-7.
• Liboff, Richard L。. Introductory Quantum Mechanics. Addison-Wesley. 2002. ISBN 0-8053-8714-5.