Heisenberg's Uncertainty Principle
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 }}}
The wavefunction of a quantum harmonic oscillator in the ground state is:
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}}}
The average position:
⟨
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}
use
∫
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}
The average square of the position:
⟨
x
2
⟩
=
∫
−
∞
∞
Ψ
0
(
x
)
∗
⋅
x
2
⋅
Ψ
0
(
x
)
d
x
{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x^{2}\cdot \Psi _{0}(x)\ dx}
=
∫
−
∞
∞
(
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}
use
∫
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 ]}}
use
lim
x
→
∞
erf
(
x
)
=
1
{\displaystyle \lim _{x\to \infty }{\text{erf}}(x)=1}
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 }}}
The uncertainty on the position:
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 }}}}
The average momentum:
⟨
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
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 (}{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx}
=
(
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}
use
∫
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
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}
The average square of the momentum:
⟨
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
4
e
−
α
x
2
2
⋅
(
−
i
ℏ
∂
∂
x
)
(
−
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 {\Biggl (}-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}{\Biggl (}-i\hbar {\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
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}
use
∫
−
∞
∞
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 )}}
use
∫
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
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
]
−
[
π
4
α
3
2
erf
(
α
−
∞
)
lim
x
→
−
∞
−
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 }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}\infty )\lim _{x\to \infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}-\infty )\lim _{x\to -\infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}{\Biggl )}}
use
lim
x
→
∞
erf
(
x
)
=
1
{\displaystyle \lim _{x\to \infty }{\text{erf}}(x)=1}
lim
x
→
−
∞
erf
(
x
)
=
−
1
{\displaystyle \lim _{x\to -\infty }{\text{erf}}(x)=-1}
=
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}}
The uncertainty on the momentum:
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 }
The product of the uncertainty on the position and the uncertainty on the momentum is:
δ
x
⋅
δ
p
x
=
1
2
α
⋅
α
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{\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 }
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{\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ ={\frac {1}{2}}\ \hbar }
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{\displaystyle {\frac {\hbar }{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 }}](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)
![{\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)
![{\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}}}}\ {\text{erf}}({\sqrt {\alpha }}\infty )\lim _{x\to \infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}-\infty )\lim _{x\to -\infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}{\Biggl )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef1c94b9775d2fdabe51f24ddea4aec06cab1f09)
![{\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)
