$$[\hat{a},\hat{a}^†]=1・・・(1)$$$$[\hat{n},\hat{a}]=-\hat{a}・・・(2)$$

$$[\hat{n},\hat{a}^†]=\hat{a}^†・・・(3)$$

## 消滅演算子・生成演算子の復習

$$\hat{a}^†=\sqrt{ \frac{mω}{2ħ} }\left( \hat{x}-\frac{i\hat{p}}{mω} \right)$$

$$\hat{N}=\hat{a}^†\hat{a}$$

## 式(1)の導出

\begin{eqnarray} [\hat{a},\hat{a}^†]&=&\hat{a}\hat{a}^†-\hat{a}^†\hat{a}\\&=&\frac{mω}{2ħ}\left( \hat{x}+\frac{i\hat{p}}{mω} \right)\left( \hat{x}-\frac{i\hat{p}}{mω} \right)-\frac{mω}{2ħ}\left( \hat{x}-\frac{i\hat{p}}{mω} \right)\left( \hat{x}+\frac{i\hat{p}}{mω} \right)\\&=&\frac{mω}{2ħ} \left( \hat{x}^2+\left( \frac{\hat{p}}{mω} \right)^2 +\frac{i}{mω}(\hat{p}\hat{x}-\hat{x}\hat{p})\right)-\frac{mω}{2ħ} \left( \hat{x}^2+ \left( \frac{\hat{p}}{mω} \right)^2 +\frac{i}{mω}(\hat{x}\hat{p}-\hat{p}\hat{x}) \right)\\&=&\frac{mω}{2ħ}\frac{i}{mω}×2(\hat{p}\hat{x}-\hat{x}\hat{p})\\&=&-\frac{i}{ħ}[\hat{x},\hat{p}]\\&=&-\frac{i}{ħ}･iħ\\&=&1 \end{eqnarray}

## 式(2)の導出

\begin{eqnarray} [\hat{n},\hat{a}]&=&\hat{n}\hat{a}-\hat{a}\hat{n}\\&=&\hat{a}^†\hat{a}\hat{a}-\hat{a}\hat{a}^†\hat{a}\\&=&-(\hat{a}\hat{a}^†-\hat{a}^†\hat{a})\hat{a}\\&=&-[\hat{a},\hat{a}^†]\hat{a}\\&=&-\hat{a} \end{eqnarray}

## 式(3)の導出

\begin{eqnarray} [\hat{n},\hat{a}^†]&=&\hat{n}\hat{a}^†-\hat{a}^†\hat{n}\\&=&\hat{a}^†\hat{a}\hat{a}^†-\hat{a}^†\hat{a}^†\hat{a}\\&=&\hat{a}^†(\hat{a}\hat{a}^†-\hat{a}^†\hat{a})\\&=&\hat{a}^†[\hat{a},\hat{a}^†]\\&=&\hat{a}^† \end{eqnarray}

## 参考文献

・・猪木慶治・川合光(1994)『量子力学I』,講談社.