## 量子場と波動関数の違いの分かる問題6

[問題6]-----------------------------
$\Lambda (x-{x}')= \Lambda ({x}'-x)$
を満たす２体ポテンシャル項を含んだ下記のハミルトニアンを考える」。
$\hat{H}= \int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi }(x)dx$
$+\frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}')dxd{x}'$
このときの量子場のハイゼンベルグ演算子は下記の非線形シュレディンガー方程式を満たすことを示せ。
$i\hbar\frac{\partial \hat{\Psi }(t,x)}{\partial t}$
$= \left (-\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2}+V(x) +\int_{-\infty }^{\infty }\Lambda (x-{x}')\hat{\Psi }^{\dagger }(t,{x}')\hat{\Psi }(t,{x}')d{x}' \right )\hat{\Psi }(t,x)$
----------------------------------

ハイゼンベルグの運動方程式の右辺を考えると

$[\hat{\Psi }(t,x),\hat{H}]= e^{it\hat{H}/\hbar}\: [\hat{\Psi }(x),\hat{H}]\: e^{-it\hat{H}/\hbar}$
$= e^{it\hat{H}/\hbar}\: \left [ \hat{\Psi }(x),\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }({x}')\left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial {x}'^2} +V({x}')\right )\hat{\Psi }({x}')d{x}'\right ] \: e^{-it\hat{H}/\hbar}$
$+e^{it\hat{H}/\hbar}\: \left [ \hat{\Psi }(x),\frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }^{\dagger }({x}')\Lambda ({x}''-{x}')\hat{\Psi }({x}'')\hat{\Psi }({x}')d{x}''d{x}'\right ] \: e^{-it\hat{H}/\hbar}$
$= \left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi }(t,x)$
$+e^{it\hat{H}/\hbar}\: \left [ \hat{\Psi }(x),\frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }^{\dagger }({x}')\Lambda ({x}''-{x}')\hat{\Psi }({x}'')\hat{\Psi }({x}')d{x}''d{x}'\right ] \: e^{-it\hat{H}/\hbar}$

ここで、

$\left [ \hat{\Psi }(x),\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }^{\dagger }({x}')\hat{\Psi }({x}'')\hat{\Psi }({x}')\right ]$
$= \left [ \hat{\Psi }(x),\hat{\Psi }^{\dagger }({x}'')\right ]\hat{\Psi }^{\dagger }({x}')\hat{\Psi }({x}'')\hat{\Psi }({x}') +\hat{\Psi }^{\dagger }({x}'') \left [ \hat{\Psi }(x),\hat{\Psi }^{\dagger }({x}')\right ]\hat{\Psi }({x}'')\hat{\Psi }({x}')$
$+ \hat{\Psi }^{\dagger }({x}'')\hat{\Psi }^{\dagger }({x}')\left [ \hat{\Psi }(x),\hat{\Psi }({x}'')\right ] \hat{\Psi }({x}') + \hat{\Psi }^{\dagger }({x}'')\hat{\Psi }^{\dagger }({x}')\hat{\Psi }({x}'') \left [ \hat{\Psi }(x),\hat{\Psi }({x}')\right ]$
$= \delta (x-{x}'')\hat{\Psi }^{\dagger }({x}')\hat{\Psi }({x}'')\hat{\Psi }({x}') +\hat{\Psi }^{\dagger }({x}'') \delta (x-{x}')\hat{\Psi }({x}'')\hat{\Psi }({x}')$

したがって、

$[\hat{\Psi }(t,x),\hat{H}]$
$= \left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi }(t,x)$
$+ e^{it\hat{H}/\hbar}\left ( \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\Lambda ({x}''-{x}')\delta (x-{x}'')\hat{\Psi }^{\dagger }({x}')\hat{\Psi }({x}'')\hat{\Psi }({x}') d{x}''d{x}'\right ) e^{-it\hat{H}/\hbar}$
$+e^{it\hat{H}/\hbar}\left ( \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty } \Lambda ({x}''-{x}')\hat{\Psi }^{\dagger }({x}'') \delta (x-{x}')\hat{\Psi }({x}'')\hat{\Psi }({x}')d{x}''d{x}' \right ) e^{-it\hat{H}/\hbar}$
$= \left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi }(t,x)$
$+ e^{it\hat{H}/\hbar}\left ( \frac{1}{2}\int_{-\infty }^{\infty }\Lambda (x-{x}')\hat{\Psi }^{\dagger }({x}')\hat{\Psi }({x}') d{x}' \:\hat{\Psi }(x) \right ) e^{-it\hat{H}/\hbar}$
$+e^{it\hat{H}/\hbar}\left ( \frac{1}{2}\int_{-\infty }^{\infty } \Lambda ({x}''-x)\hat{\Psi }^{\dagger }({x}'') \hat{\Psi }({x}'')d{x}'' \; \hat{\Psi }(x)\right ) e^{-it\hat{H}/\hbar}$
$= \left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi }(t,x)$
$+ e^{it\hat{H}/\hbar}\left ( \int_{-\infty }^{\infty }\Lambda (x-{x}')\hat{\Psi }^{\dagger }({x}')\hat{\Psi }({x}') d{x}' \:\hat{\Psi }(x) \right ) e^{-it\hat{H}/\hbar}$
$= \left (-\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2}+V(x) +\int_{-\infty }^{\infty }\Lambda (x-{x}')\hat{\Psi }^{\dagger }(t,{x}')\hat{\Psi }(t,{x}')d{x}' \right )\hat{\Psi }(t,x)$

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