## 場の理論入門（３０）

$\widehat{H}= \int_{-\infty }^{\infty }\widehat{a}^{\dagger }(x)\left [ -\frac{\hbar^{2}}{2m} \frac{\partial^2 }{\partial x^2}\right ]\widehat{a}(x)dx= -\frac{\hbar^{2}}{2m}\int_{-\infty }^{\infty }\widehat{a}^{\dagger }(x)\left [ \frac{\partial^2 }{\partial x^2}\: \widehat{a}(x)\right ]dx$
$= -\frac{\hbar^{2}}{2m}\int_{-\infty }^{\infty }\left [ \int_{-\infty }^{\infty } \frac{e^{-ikx}}{\sqrt{2\pi }}\widehat{a}^{\dagger }(k)dk\right ]\left [\frac{\partial^2 }{\partial x^2} \int_{-\infty }^{\infty } \frac{e^{i{k}'x}}{\sqrt{2\pi }}\: \widehat{a}({k}')d{k}'\right ]dx$
$= -\frac{\hbar^{2}}{2m}\int_{-\infty }^{\infty }\left [ \int_{-\infty }^{\infty } \frac{e^{-ikx}}{\sqrt{2\pi }}\widehat{a}^{\dagger }(k)dk\right ]\left [(-{k}'^{2}) \int_{-\infty }^{\infty } \frac{e^{i{k}'x}}{\sqrt{2\pi }}\: \widehat{a}({k}')d{k}'\right ]dx$
$=\frac{\hbar^{2}}{2m}\int_{-\infty }^{\infty }\left [ \int_{-\infty }^{\infty } \frac{e^{-ikx}}{\sqrt{2\pi }}\widehat{a}^{\dagger }(k)dk\right ]\left [{k}'^{2} \int_{-\infty }^{\infty } \frac{e^{i{k}'x}}{\sqrt{2\pi }}\: \widehat{a}({k}')d{k}'\right ]dx$
$=\frac{\hbar^{2}}{2m}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty } {k}'^{2}\widehat{a}^{\dagger }(k)\widehat{a}({k}')\left [ \int_{-\infty }^{\infty } \frac{e^{i({k}'-k)x}}{2\pi }dx \right ] d{k}'dk$
$=\frac{\hbar^{2}}{2m}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty } {k}'^{2}\widehat{a}^{\dagger }(k)\widehat{a}({k}')\delta ({k}'-k) d{k}'dk$
$=\frac{\hbar^{2}}{2m}\int_{-\infty }^{\infty } k^{2}\widehat{a}^{\dagger }(k)\widehat{a}(k)dk=\int_{-\infty }^{\infty }\frac{(\hbar k)^{2}}{2m}\: \widehat{a}^{\dagger }(k)\widehat{a}(k)dk$

$\widehat{H}\widehat{a}^{\dagger }({k}')|0\rangle=\left [ \int_{-\infty }^{\infty }\frac{(\hbar k)^{2}}{2m}\: \widehat{a}^{\dagger }(k)\widehat{a}(k)dk \right ]\widehat{a}^{\dagger }({k}')|0\rangle$
$=\left [ \int_{-\infty }^{\infty }\frac{(\hbar k)^{2}}{2m}\: \widehat{a}^{\dagger }(k)\left \{ \int_{-\infty }^{\infty }\frac{e^{-ikx}}{\sqrt{2\pi }} \widehat{a}(x)dx\right \}dk \right ] \int_{-\infty }^{\infty }\frac{e^{i{k}'{x}'}}{\sqrt{2\pi }} \: \widehat{a}^{\dagger }({x}')d{x}'|0\rangle$
$=\left [ \int_{-\infty }^{\infty }\frac{(\hbar k)^{2}}{2m}\: \widehat{a}^{\dagger }(k)\left \{ \int_{-\infty }^{\infty}\int_{-\infty }^{\infty }\frac{e^{-ikx+i{k}'{x}'}}{2\pi } \widehat{a}(x)\: \widehat{a}^{\dagger }({x}')d{x}'dx\right \}dk \right ] |0\rangle$

ここで、

$\widehat{a}(x)\: \widehat{a}^{\dagger }({x}')|0\rangle= \delta (x-{x}')|0\rangle$

という関係を使うと

$\widehat{H}\widehat{a}^{\dagger }({k}')|0\rangle =\left [ \int_{-\infty }^{\infty }\frac{(\hbar k)^{2}}{2m}\: \widehat{a}^{\dagger }(k)\left \{ \int_{-\infty }^{\infty}\int_{-\infty }^{\infty }\frac{e^{-ikx+i{k}'{x}'}}{2\pi } \delta (x-{x}') d{x}'dx\right \}dk \right ] |0\rangle$
$=\left [ \int_{-\infty }^{\infty }\frac{(\hbar k)^{2}}{2m}\: \widehat{a}^{\dagger }(k)\left \{ \int_{-\infty }^{\infty}\frac{e^{i({k}'-k)x}}{2\pi } dx\right \}dk \right ] |0\rangle$
$=\left [ \int_{-\infty }^{\infty }\frac{(\hbar k)^{2}}{2m}\: \widehat{a}^{\dagger }(k) \delta ({k}'-k)dk \right ] |0\rangle$
$= \frac{(\hbar {k}')^{2}}{2m}\: \widehat{a}^{\dagger }({k}') |0\rangle$

つまり、

$\widehat{H}\left ( \widehat{a}^{\dagger }({k}')|0\rangle \right )= \frac{(\hbar {k}')^{2}}{2m}\:\left ( \widehat{a}^{\dagger }({k}') |0\rangle \right )$

となり、状態 $\110dpi \widehat{a}^{\dagger }({k}')|0\rangle$ は演算子 $\110dpi \widehat{H}$ の固有状態であることがわかります。

## ブログ気持玉

クリックして気持ちを伝えよう！

ログインしてクリックすれば、自分のブログへのリンクが付きます。

→ログインへ