线性谐振子哈密顿量 H ^ L H O : V → V {\displaystyle {\hat {H}}_{LHO}:\mathbb {V} \rightarrow \mathbb {V} } 是一个粒子 p ∈ P a r t i c l e S e t {\displaystyle p\in {\mathfrak {ParticleSet}}} 的定义是 H ^ ( | n > ) = p ^ 2 ( | n > ) 2 m + 1 2 m ω x ^ 2 ( | n > ) {\displaystyle {\hat {H}}(|n>)={\frac {{\hat {p}}^{2}(|n>)}{2m}}+{\frac {1}{2}}m\omega {\hat {x}}^{2}(|n>)} 其中算符 x ^ ( | n > ) , p ^ ( | n > ) : V → V {\displaystyle {\hat {x}}(|n>),{\hat {p}}(|n>):\mathbb {V} \rightarrow \mathbb {V} }
湮灭算符 a ^ : V → V {\displaystyle {\hat {a}}:\mathbb {V} \rightarrow \mathbb {V} } 的定义是 a ^ ( | n > ) = m ω 2 ℏ x ^ + i 1 2 ℏ m ω p ^ {\displaystyle {\hat {a}}(|n>)={\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}+i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}}
产生算符 a ^ + : V → V {\displaystyle {\hat {a}}^{+}:\mathbb {V} \rightarrow \mathbb {V} } 的定义是 a ^ + ( | n > ) = m ω 2 ℏ x ^ − i 1 2 ℏ m ω p ^ {\displaystyle {\hat {a}}^{+}(|n>)={\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}-i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}}
假设线性谐振子哈密顿量。然后
a ^ [ a ^ + ( | n > ) ] = H ^ ( | n > ) ℏ ω + 1 ^ ( | n > ) 2 , a ^ + [ a ^ ( | n > ) ] = H ^ ( | n > ) ℏ ω − 1 ^ ( | n > ) 2 , {\displaystyle {\hat {a}}[{\hat {a}}^{+}(|n>)]={\frac {{\hat {H}}(|n>)}{\hbar \omega }}+{\frac {{\hat {1}}(|n>)}{2}},\qquad {\hat {a}}^{+}[{\hat {a}}(|n>)]={\frac {{\hat {H}}(|n>)}{\hbar \omega }}-{\frac {{\hat {1}}(|n>)}{2}},}
证明直接从 a、a+ 定义，关于 LHO 哈密顿量定义和 [x,p] 算符对易关系得出
a [ a + ( | n > ) ] = d e f a , a + ( m ω 2 ℏ x ^ + i 1 2 ℏ m ω p ^ ) ( m ω 2 ℏ x ^ − i 1 2 ℏ m ω p ^ ) = {\displaystyle a[a^{+}(|n>)]{\overset {def\,a,a^{+}}{=}}\left({\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}+i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}\right)\left({\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}-i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}\right)=} = m ω 2 ℏ x ^ 2 ( | n > ) − i 2 1 2 ℏ m ω p ^ 2 ( | n >) − i m ω 2 ℏ 1 2 ℏ m ω x ^ [ p ^ ( | n > ) ] + i m ω 2 ℏ 1 2 ℏ m ω p ^ [ x ^ ( | n > ) ] = {\displaystyle ={\frac {m\omega }{2\hbar }}{\hat {x}}^{2}(|n>)-i^{2}{\frac {1}{2\hbar m\omega }}{\hat {p}}^{2}(|n>)-i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {x}}[{\hat {p}}(|n>)]+i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}[{\hat {x}}(|n>)]=} = | H ^ = p ^ 2 | n > m + 1 2 m ω x ^ 2 | n > | = H ^ 2 ℏ ω − i m ω 2 ℏ 1 2 ℏ m ω ( x ^ p ^ | n > − p ^ x ^ | n > ) = {\displaystyle =|{\hat {H}}={\frac {{\hat {p}}^{2}|n>}{m}}+{\frac {1}{2}}m\omega {\hat {x}}^{2}|n>|={\frac {\hat {H}}{2\hbar \omega }}-i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}({\hat {x}}{\hat {p}}|n>-{\hat {p}}{\hat {x}}|n>)=} = H ^ 2 ℏ ω − i 2 ℏ [ x ^ , p ^ ] | n >= H ^ 2 ℏ ω − i 2 ℏ ( i ℏ 1 ^ ) | n >= H ^ | n > 2 ℏ ω + 1 ^ | n > 2 {\displaystyle ={\frac {\hat {H}}{2\hbar \omega }}-{\frac {i}{2\hbar }}[{\hat {x}},{\hat {p}}]|n>={\frac {\hat {H}}{2\hbar \omega }}-{\frac {i}{2\hbar }}(i\hbar {\hat {1}})|n>={\frac {{\hat {H}}|n>}{2\hbar \omega }}+{\frac {{\hat {1}}|n>}{2}}}
[ a ^ , a ^ + ] ( | n > ) = 1 ^ ( | n > ) , [ H ^ , a ^ + ] ( | n > ) = ℏ ω a ^ + ( | n > ) [ H ^ , a ^ ] ( | n > ) = − ℏ ω a ^ ( | n > ) {\displaystyle [{\hat {a}},{\hat {a}}^{+}](|n>)={\hat {1}}(|n>),\qquad [{\hat {H}},{\hat {a}}^{+}](|n>)=\hbar \omega {\hat {a}}^{+}(|n>)\qquad [{\hat {H}},{\hat {a}}](|n>)=-\hbar \omega {\hat {a}}(|n>)}
证明第一个表达式可以通过算符定义直接从a, a+ 复合表达式定理来评估
[ a ^ , a ^ + ] = a ^ a ^ + − a ^ + a ^ = t h e o r e m H ^ | n > 2 ℏ ω + 1 ^ | n > 2 − H ^ | n > 2 ℏ ω − 1 ^ | n > 2 = 1 ^ | n > {\displaystyle [{\hat {a}},{\hat {a}}^{+}]={\hat {a}}{\hat {a}}^{+}-{\hat {a}}^{+}{\hat {a}}{\overset {theorem}{=}}{\frac {{\hat {H}}|n>}{2\hbar \omega }}+{\frac {{\hat {1}}|n>}{2}}-{\frac {{\hat {H}}|n>}{2\hbar \omega }}-{\frac {{\hat {1}}|n>}{2}}={\hat {1}}|n>}
第二个和第三个表达式需要 H ^ = H ^ ( a ^ , a ^ + ) {\displaystyle {\hat {H}}={\hat {H}}({\hat {a}},{\hat {a}}^{+})} 首先，这可以通过 a、a+ 复合表达式定理再次获得。
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![{\displaystyle {\hat {a}}[{\hat {a}}^{+}(|n>)]={\frac {{\hat {H}}(|n>)}{\hbar \omega }}+{\frac {{\hat {1}}(|n>)}{2}},\qquad {\hat {a}}^{+}[{\hat {a}}(|n>)]={\frac {{\hat {H}}(|n>)}{\hbar \omega }}-{\frac {{\hat {1}}(|n>)}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05fdd5398fc96b43cb68c23b474fc0ac8cdc331f)
![{\displaystyle a[a^{+}(|n>)]{\overset {def\,a,a^{+}}{=}}\left({\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}+i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}\right)\left({\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}-i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c67121e73d11c6514f7d94749d69db068b2028e)
![{\displaystyle ={\frac {m\omega }{2\hbar }}{\hat {x}}^{2}(|n>)-i^{2}{\frac {1}{2\hbar m\omega }}{\hat {p}}^{2}(|n>)-i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {x}}[{\hat {p}}(|n>)]+i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}[{\hat {x}}(|n>)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5238c0a70c8af35f0b61e8d69974271f498c8e20)
![{\displaystyle ={\frac {\hat {H}}{2\hbar \omega }}-{\frac {i}{2\hbar }}[{\hat {x}},{\hat {p}}]|n>={\frac {\hat {H}}{2\hbar \omega }}-{\frac {i}{2\hbar }}(i\hbar {\hat {1}})|n>={\frac {{\hat {H}}|n>}{2\hbar \omega }}+{\frac {{\hat {1}}|n>}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94144cabd6dd71d4a783e9e81b78747d5c145261)
={\hat {1}}(|n>),\qquad [{\hat {H}},{\hat {a}}^{+}](|n>)=\hbar \omega {\hat {a}}^{+}(|n>)\qquad [{\hat {H}},{\hat {a}}](|n>)=-\hbar \omega {\hat {a}}(|n>)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/056e894d4b8086054fe6626349dfc0a5bfda133d)
![{\displaystyle [{\hat {a}},{\hat {a}}^{+}]={\hat {a}}{\hat {a}}^{+}-{\hat {a}}^{+}{\hat {a}}{\overset {theorem}{=}}{\frac {{\hat {H}}|n>}{2\hbar \omega }}+{\frac {{\hat {1}}|n>}{2}}-{\frac {{\hat {H}}|n>}{2\hbar \omega }}-{\frac {{\hat {1}}|n>}{2}}={\hat {1}}|n>}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06626ff8f25b23542e22891977b011f0435e0857)
首先，这可以通过 a、a+ 复合表达式定理再次获得。
