# 量子諧振子

（重定向自量子谐振子

## 一維諧振子

### 哈密頓算符與能量本徵態

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

${\displaystyle H\left|\psi \right\rangle =E\left|\psi \right\rangle }$ .

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

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

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

### 階梯算符方法

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

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

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

xp算符遵守下面的等式，稱之為正則對易關係

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

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

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

${\displaystyle \left(a\left|\psi _{E}\right\rangle ,a\left|\psi _{E}\right\rangle \right)=\left\langle \psi _{E}\right|a^{\dagger }a\left|\psi _{E}\right\rangle \geq 0}$

aa以哈密頓算符表示：

${\displaystyle \left\langle \psi _{E}\right|{H \over \hbar \omega }-{1 \over 2}\left|\psi _{E}\right\rangle =\left({E \over \hbar \omega }-{1 \over 2}\right)\geq 0}$

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

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

${\displaystyle H(a^{\dagger }\left|\psi _{E}\right\rangle )=(E+\hbar \omega )(a^{\dagger }\left|\psi _{E}\right\rangle )}$ .

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

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

${\displaystyle H\left|n\right\rangle =\hbar \omega (n+1/2)\left|n\right\rangle }$ ，這與前段所給的譜相符合。

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

${\displaystyle {\frac {d\ln \psi _{0}(x)}{dx}}=-{\frac {\hslash }{m\omega }}x+{\text{ Constant}}}$

${\displaystyle \psi _{0}(x)=\left({m\omega \over \pi \hbar }\right)^{1 \over 4}e^{-m\omega x^{2}/2\hbar }}$

### 自然長度與能量尺度

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

${\displaystyle \left\langle x|\psi _{n}\right\rangle ={1 \over {\sqrt {2^{n}n!}}}\pi ^{-1/4}{\hbox{exp}}(-u^{2}/2)H_{n}(u)}$
${\displaystyle E_{n}=n+{1 \over 2}}$ .

### 案例：雙原子分子

${\displaystyle \omega ={\sqrt {\frac {k}{m_{r}}}}}$

${\displaystyle \omega =2\pi f}$ 為角頻率，
k共價鍵勁度係數
${\displaystyle m_{r}}$ 約化質量

## N維諧振子

${\displaystyle {\begin{matrix}\left[x_{i},p_{j}\right]&=&i\hbar \delta _{i,j}\\\left[x_{i},x_{j}\right]&=&0\\\left[p_{i},p_{j}\right]&=&0\end{matrix}}}$ .

${\displaystyle H=\sum _{i=1}^{N}\left({p_{i}^{2} \over 2m}+{1 \over 2}m\omega ^{2}x_{i}^{2}\right)}$

${\displaystyle \langle \mathbf {x} |\psi _{\{n\}}\rangle =\prod _{i=1}^{N}\langle x_{i}|\psi _{n_{i}}\rangle }$

${\displaystyle a_{i}={\sqrt {m\omega \over 2\hbar }}\left(x_{i}+{i \over m\omega }p_{i}\right)}$
${\displaystyle a_{i}^{\dagger }={\sqrt {m\omega \over 2\hbar }}\left(x_{i}-{i \over m\omega }p_{i}\right)}$

${\displaystyle H=\hbar \omega \,\sum _{i=1}^{N}\left(a_{i}^{\dagger }\,a_{i}+{\frac {1}{2}}\right)}$

${\displaystyle E=\hbar \omega \left[(n_{1}+\cdots +n_{N})+{N \over 2}\right]}$

${\displaystyle d_{n}=\sum _{n_{1}=0}^{n}n-n_{1}+1={\frac {(n+1)(n+2)}{2}}}$

${\displaystyle d_{n}={\binom {N+n-1}{n}}}$

### 案例：三維均向諧振子

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

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

${\displaystyle \psi _{klm}(r,\theta ,\phi )=N_{kl}r^{l}e^{-\nu r^{2}}{L_{k}}^{(l+{1 \over 2})}(2\nu r^{2})Y_{lm}(\theta ,\phi )}$

${\displaystyle N_{kl}={\sqrt {{\sqrt {\frac {2\nu ^{3}}{\pi }}}{\frac {2^{k+2l+3}\;k!\;\nu ^{l}}{(2k+2l+1)!!}}}}}$ 是歸一常數，
${\displaystyle \nu \equiv {\mu \omega \over 2\hbar }}$
${\displaystyle {L_{k}}^{(l+{1 \over 2})}(2\nu r^{2})}$ ${\displaystyle k}$ 广义拉盖尔多项式 (generalized Laguerre polynomials)，${\displaystyle k}$ 是個正整數，
${\displaystyle Y_{lm}(\theta ,\phi )\,}$ 球諧函數
${\displaystyle \hbar }$ 約化普朗克常數

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

${\displaystyle n\equiv 2k+l}$

${\displaystyle l=0,\,2,\,\dots ,\,n-2,\,n}$

${\displaystyle l=1,\,3,\,\dots ,\,n-2,\,n}$

${\displaystyle -l\leq m\leq l}$

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

## 耦合諧振子

${\displaystyle H=\sum _{i=1}^{N}{p_{i}^{2} \over 2m}+{1 \over 2}m\omega ^{2}\sum _{1\leq i\leq N}(x_{i}-x_{i-1})^{2}}$