求基态量子谐振子的⟨x⟩、⟨x2⟩、⟨px⟩和⟨px2⟩，然后确定位置和动量的测不准量。位置和动量的测不准量之积是否与海森堡测不准原理一致？
δ x ⋅ δ p ≥ ℏ 2 where ℏ = h 2 π {\displaystyle \qquad \delta _{x}\cdot \delta _{p}\geq {\frac {\hbar }{2}}\qquad {\text{where}}\ \hbar ={\frac {h}{2\pi }}}
基态量子谐振子的波函数为
Ψ 0 ( x ) = ( α π ) 1 4 e − α x 2 2 {\displaystyle \qquad \qquad \Psi _{0}(x)={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}} 利用此波函数，可以计算平均位置和位置平方的平均值。
平均位置
⟨ x ⟩ = ∫ − ∞ ∞ Ψ 0 ( x ) ∗ ⋅ x ⋅ Ψ 0 ( x ) d x {\displaystyle \qquad \qquad \qquad \qquad \langle x\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x\cdot \Psi _{0}(x)\ dx}
= ∫ − ∞ ∞ ( α π ) 1 4 e − α x 2 2 ⋅ x ⋅ ( α π ) 1 4 e − α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \quad \ =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx}
= ( α π ) 1 2 ∫ − ∞ ∞ x ⋅ e − 2 α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x\cdot e^{\frac {-2\alpha x^{2}}{2}}\ dx}
= ( α π ) 1 2 ∫ − ∞ ∞ x e − α x 2 d x {\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }xe^{-\alpha x^{2}}\ dx}
使用 ∫ x e − α x 2 d x = − 1 2 α e − α x 2 + C {\displaystyle \int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C}
= ( α π ) 1 2 [ − 1 2 α ⋅ e − α x 2 ] − ∞ ∞ {\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}\cdot e^{-\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }}
= ( α π ) 1 2 [ − 1 2 α ( 0 ) − 1 2 α ( 0 ) ] {\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}}
⟨ x ⟩ = 0 {\displaystyle \qquad \qquad \qquad \qquad \langle x\rangle =0}
位置的平均平方
⟨ x 2 ⟩ = ∫ − ∞ ∞ Ψ 0 ( x ) ∗ ⋅ x 2 ⋅ Ψ 0 ( x
= ∫ − ∞ ∞ ( α π ) 1 4 e − α x 2 2 ⋅ x 2 ⋅ ( α π ) 1 4 e − α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x^{2}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx}
= ( α π ) 1 2 ∫ − ∞ ∞ x 2 e − 2 α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx}
= ( α π ) 1 2 ∫ − ∞ ∞ x 2 e − α x 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx}
使用 ∫ x 2 e − α x 2 d x = π erf ( α x ) 4 α 3 2 − x e − α x 2 2 α + C {\displaystyle \int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C}
= ( α π ) 1 2 [ π erf ( α x ) 4 α 3 2 − x e − α x 2 2 α ] − ∞ ∞ {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}_{-\infty }^{\infty }}
= ( α π ) 1 2 [ π 4 erf ( α ∞ ) α 3 2 − lim x → ∞ x e − α x 2 2 α ] − [ π 4 erf ( α − ∞ ) α 3 2 − lim x → − ∞ x e − α x 2 2 α ] {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}\ \infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to \infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}-\infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to -\infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}}
使用 lim x → ∞ erf ( x ) = 1 {\displaystyle \lim _{x\to \infty }{\text{erf}}(x)=1} 以及 lim x → − ∞ erf ( x ) = − 1 {\displaystyle \lim _{x\to -\infty }{\text{erf}}(x)=-1}
= ( α π ) 1 2 π 4 1 α 3 2 [ 1 − ( − 1 ) ] {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\biggl [}1-(-1){\biggl ]}}
= α π π 4 1 α 3 2 [ 2 ] {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\sqrt {\alpha }}{\sqrt {\pi }}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\bigl [}2{\bigl ]}}
⟨ x 2 ⟩ = 1 2 α {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle ={\frac {1}{2\alpha }}}
位置的不确定性
δ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\langle x^{2}\rangle -\langle x\rangle ^{2}}}} = 1 2 α − 0 {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ ={\sqrt {{\frac {1}{2\alpha }}-0}}}
δ x = 1 2 α {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\frac {1}{2\alpha }}}}
动量的平均值
⟨ p x ⟩ = ∫ − ∞ ∞ Ψ 0 ( x ) ∗ ⋅ p ^ x ⋅ Ψ 0 ( x ) d x where p ^ x = − i ℏ ( ∂ ∂ x ) {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}\cdot \Psi _{0}(x)\ dx\qquad \qquad \qquad \qquad \qquad \qquad {\text{where}}\ {\widehat {p}}_{x}=-i\hbar {\Bigl (}{\frac {\partial }{\partial x}}{\Bigr )}}
= ∫ − ∞ ∞ ( α π ) 1 4 e − α x 2 2 ⋅ − i ℏ ( ∂ ∂ x ) ⋅ ( α π ) 1 4 e − α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot -i\hbar {\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx}
= ( α π ) 1 2 − i ℏ ∫ − ∞ ∞ e − α x 2 2 ( ∂ ∂ x ) e − α x 2
= ( α π ) 1 2 − i ℏ ∫ − ∞ ∞ e − α x 2 2 ( − α x e − α x 2 2 ) d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl (}-\alpha x\ e^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx}
= − α ( α π ) 1 2 − i ℏ ∫ − ∞ ∞ e − α x 2 2 ⋅ x e − α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\cdot x\ e^{\frac {-\alpha x^{2}}{2}}\ dx}
= − α ( α π ) 1 2 − i ℏ ∫ − ∞ ∞ x e − α x 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }x\ e^{-\alpha x^{2}}\ dx}
= − α 3 2 π − i ℏ [ 1 2 α e − 2 α x 2 ] − ∞ ∞ {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar \ {\biggl [}{\frac {1}{2\alpha }}\ e^{-2\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }}
= − α 3 2 π − i ℏ [ 1 2 α ( 0 ) − 1 2 α ( 0 ) ] {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar {\biggl [}{\frac {1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}}
⟨ p x ⟩ = 0 {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =0}
动量的平均平方
⟨ p x 2 ⟩ = ∫ − ∞ ∞ Ψ 0 ( x ) ∗ ⋅ p ^ x 2 ⋅ Ψ 0 ( x) d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}^{2}\cdot \Psi _{0}(x)\ dx}
= ( α π ) 1 2 i 2 ℏ 2 ∫ − ∞ ∞ e − α x 2 2 ( ∂ ∂ x ) ( ∂ ∂ x ) e − α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}i^{2}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx}
= ( α π ) 1 2 ( − 1 ) ℏ 2 ∫ − ∞ ∞ e − α x 2 2 ( ∂ ∂ x ) ( − α x e − α x 2 2 ) d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}(-1)\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ {\biggl (}-\alpha xe^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx}
= α ( α π ) 1 2 ℏ 2 ∫ − ∞ ∞ e − α x 2 2 ( e − α x 2 2 + α x 2 e − 2 α x 2 2 ) d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl (}e^{\frac {-\alpha x^{2}}{2}}+\alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}{\biggl )}\ dx}
= α 3 2 π ℏ 2 ∫ − ∞ ∞ e − α x 2 d x ∫ − ∞ ∞ e − α x 2 2 α x 2 e − 2 α x 2 2 d x {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\ \alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx}
使用 ∫ − ∞ ∞ e − α x 2 d x = π α {\displaystyle \int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx={\sqrt {\frac {\pi }{\alpha }}}}
= α 3 2 π ℏ 2 ( π α + α ∫ − ∞ ∞ x 2 e − α x 2 d x ) {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha \int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx{\biggl )}}
= α 3 2 π ℏ 2 ( π α + α [ π erf ( α x ) 4 α 3 2 − x e − α x 2 2 α ] − ∞ ∞ ) {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl [}{\frac {{\sqrt {\pi }}\ {\text{erf}}({\sqrt {\alpha }}\ x)}{4\alpha ^{\frac {3}{2}}}}-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\biggl ]}_{-\infty }^{\infty }{\Biggl )}}
= α 3 2 π ℏ 2 ( π α + α [ π 4 α 3 2 erf ( α ∞ ) lim x → ∞ − x e − α x 2 2 α <
= α 3 2 π ℏ 2 ( π α + α [ π 4 α 3 2 ( 1 ) − ( 0 ) ] − [ π 4 α 3 2 ( − 1 ) − ( 0 ) ] ) {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (1)-\ (0){\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (-1)-\ (0){\Biggl ]}{\Biggl )}}
= α 3 2 π ℏ 2 [ π α + α ( 2 π 4 α 3 2 ) ] {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl (}{\frac {2{\sqrt {\pi }}}{4\alpha ^{\frac {3}{2}}}}\ {\biggl )}{\Biggl ]}}
= α π ℏ 2 [ π + 1 2 π ] {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha }{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\pi }}+{\frac {1}{2}}{\sqrt {\pi }}{\Biggl ]}}
⟨ p x 2 ⟩ = 1 2 α ℏ 2 {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle ={\frac {1}{2}}\alpha \hbar ^{2}}
动量的误差
δ p x = ⟨ p x 2 ⟩ − ⟨ p x ⟩ 2 {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\langle p_{x}^{2}\rangle -\langle p_{x}\rangle ^{2}}}}
= a ℏ 2 2 − 0 2 {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ ={\sqrt {{\frac {a\hbar ^{2}}{2}}-0^{2}}}}
δ p x = a 2 2 ℏ {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\frac {a^{2}}{2}}}\ \hbar }
位置误差和动量误差的乘积为
δ x ⋅ δ p x = 1 2 α ⋅ α 2 ℏ {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta _{x}\cdot \delta p_{x}={\frac {1}{\sqrt {2\alpha }}}\cdot {\sqrt {\frac {\alpha }{2}}}\ \hbar }
= 1 2 ℏ {\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ ={\frac {1}{2}}\ \hbar }
这等于 ℏ 2 {\displaystyle {\frac {\hbar }{2}}} ，因此，处于基态的量子谐振子与海森堡测不准原理一致。
利用此波函数，可以计算平均位置和位置平方的平均值。
![{\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}\cdot e^{-\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac9fb1c65a597f89866342c2e242e7711e1d9a29)
![{\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4623aa0982f3eb60f69e1479d2c8c77c3eed8123)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}_{-\infty }^{\infty }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb23fc1dd04638c74e2de7541a12aece057d2d4d)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}\ \infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to \infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}-\infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to -\infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04e997cc1fe15cf3b2cfc94d2736f1de8080f542)
以及 
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\biggl [}1-(-1){\biggl ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/161d24f11477ac620d924af9c081675de7ac873f)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\sqrt {\alpha }}{\sqrt {\pi }}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\bigl [}2{\bigl ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/833fad4fa713f013750145ea0e55fe097c3e2d67)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar \ {\biggl [}{\frac {1}{2\alpha }}\ e^{-2\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53b31b9136a4ed2fd06e9416ecadf14fc3b519a5)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar {\biggl [}{\frac {1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea3d6d135db173403156ee6e3bbad8ffc2f4da58)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl [}{\frac {{\sqrt {\pi }}\ {\text{erf}}({\sqrt {\alpha }}\ x)}{4\alpha ^{\frac {3}{2}}}}-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\biggl ]}_{-\infty }^{\infty }{\Biggl )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bea1950eca25f0f4a96c89dbe270e9ca1073bf7)
以及 ![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (1)-\ (0){\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (-1)-\ (0){\Biggl ]}{\Biggl )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d97ef74620922703644cbaf46dda6cf159b98ea1)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl (}{\frac {2{\sqrt {\pi }}}{4\alpha ^{\frac {3}{2}}}}\ {\biggl )}{\Biggl ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9f74dcc8fcb2866bac5fbd88c9454e0d78e88e7)
![{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha }{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\pi }}+{\frac {1}{2}}{\sqrt {\pi }}{\Biggl ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bddbe327a1a3da86c9bbfa4ffa542d64e97a5f84)
，因此，处于基态的量子谐振子与海森堡测不准原理一致。
