## 量子場と波動関数の違いの分かる問題7

[問題7]-----------------------------
$\hat{N}= \int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\hat{\Psi }(x)dx$
で定義される粒子数演算子は、２体系のポテンシャルを含んだハミルトニアンとも可換であることを示せ。
$\left [ \hat{H} ,\hat{N} \right ]= 0$
----------------------------------

$\hat{H}_{1}\equiv \int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi }(x)dx$
$\hat{H}_{2}\equiv \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}')dxd{x}'$
つまり
$\hat{H}= \hat{H}_{1}+\hat{H}_{2}\; \to\; [\hat{H}_{1},\hat{N}]+[\hat{H}_{2},\hat{N}]= [\hat{H},\hat{N}]$
ということです。
さらに、
$[\hat{\Psi}^{\dagger } (x)\hat{\Psi}(x),\hat{\Psi}^{\dagger } ({x}')\hat{\Psi}({x}')]$
$= \hat{\Psi}^{\dagger } (x)[\hat{\Psi}(x),\hat{\Psi}^{\dagger } ({x}')]\hat{\Psi}({x}') - \hat{\Psi}^{\dagger } ({x}')[\hat{\Psi}({x}'),\hat{\Psi}^{\dagger } (x)]\hat{\Psi}(x)$
$= \hat{\Psi}^{\dagger } (x) \delta (x-{x}')\hat{\Psi}({x}') - \hat{\Psi}^{\dagger } ({x}')\delta ({x}'-x)\hat{\Psi}(x)$
から
$[\hat{H}_{1},\hat{N}] =\int_{-\infty }^{\infty } \int_{-\infty }^{\infty }\delta (x-{x}')\hat{\Psi}^{\dagger } (x) \left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial {x}'^2} +V({x}')\right )\hat{\Psi}({x}')dxd{x}'$
$-\int_{-\infty }^{\infty } \int_{-\infty }^{\infty } \delta ({x}'-x)\hat{\Psi}^{\dagger } ({x}')\left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi}(x)dxd{x}'$
$= \int_{-\infty }^{\infty } \hat{\Psi}^{\dagger } (x)\left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi}(x)dx$
$- \int_{-\infty }^{\infty } \hat{\Psi}^{\dagger } (x)\left ( -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2} +V(x)\right )\hat{\Psi}(x)dx=0$
また
$[\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x)\hat{\Psi }({x}'), \hat{\Psi }^{\dagger }({x}'')\hat{\Psi }({x}'') ]$
$= \hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x)\hat{\Psi }({x}')\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }({x}'')$
$-\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }({x}'') \hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x)\hat{\Psi }({x}')$
で、第１項は
$\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x)\left \{ [\hat{\Psi }({x}'),\hat{\Psi }^{\dagger }({x}'')]+\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }({x}') \right \}\hat{\Psi }({x}'')$
$= \delta ({x}'-{x}'')\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x) \hat{\Psi }({x}'')$
$+\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x) \hat{\Psi }^{\dagger }({x}'')\hat{\Psi }({x}') \hat{\Psi }({x}'')$
$= \delta ({x}'-{x}'')\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x) \hat{\Psi }({x}'')$
$+\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}') \left \{ [\hat{\Psi }(x) , \hat{\Psi }^{\dagger }({x}'') ]+\hat{\Psi }^{\dagger }({x}'') \hat{\Psi }(x) \right \} \hat{\Psi }({x}') \hat{\Psi }({x}'')$
$= \delta ({x}'-{x}'')\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\hat{\Psi }(x) \hat{\Psi }({x}'')$
$+\delta (x-{x}'')\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}') \hat{\Psi }({x}') \hat{\Psi }({x}'')$
$+\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}') \hat{\Psi }^{\dagger }({x}'') \hat{\Psi }(x) \hat{\Psi }({x}') \hat{\Psi }({x}'')$

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

したがって
$[\hat{H}_{2},\hat{N}]$
$= \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\delta ({x}'-{x}'')\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}'')dxd{x}'d{x}''$
$+ \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\delta (x-{x}'')\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }({x}')\hat{\Psi }({x}'')dxd{x}'d{x}''$
$- \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\delta ({x}''-x)\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}')dxd{x}'d{x}''$
$- \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\delta ({x}''-{x}')\hat{\Psi }^{\dagger }({x}'')\hat{\Psi }^{\dagger }(x)\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}')dxd{x}'d{x}''$
$= \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}')dxd{x}'$
$+ \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }({x}')\hat{\Psi }(x)dxd{x}'$
$- \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }(x)\hat{\Psi }^{\dagger }({x}')\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}')dxd{x}'$
$- \frac{1}{2}\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }\hat{\Psi }^{\dagger }({x}')\hat{\Psi }^{\dagger }(x)\Lambda (x-{x}')\hat{\Psi }(x)\hat{\Psi }({x}')dxd{x}'=0$

つまり
$[\hat{H}_{1},\hat{N}]+[\hat{H}_{2},\hat{N}]= [\hat{H},\hat{N}]=0$

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