# Quantum Chemistry/Example 1

Find ⟨x⟩, ⟨x2⟩, ⟨px⟩ and ⟨px2⟩ for a quantum harmonic oscillator in the ground state, then determine the uncertainty on the position and momentum. Is the product of the uncertainty on position and momentum consistent with the Heisenberg's Uncertainty Principle?

 Heisenberg's Uncertainty Principle ${\displaystyle \qquad \delta _{x}\cdot \delta _{p}\geq {\frac {\hbar }{2}}\qquad {\text{where}}\ \hbar ={\frac {h}{2\pi }}}$

The wavefunction of a quantum harmonic oscillator in the ground state is:

${\displaystyle \qquad \qquad \Psi _{0}(x)={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}}$ Using this wavefunction the average position and the average of the square of the position can be calculated.

The average position:

${\displaystyle \qquad \qquad \qquad \qquad \langle x\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x\cdot \Psi _{0}(x)\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \quad \ =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x\cdot e^{\frac {-2\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }xe^{-\alpha x^{2}}\ dx}$

 use ${\displaystyle \int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C}$

${\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}\cdot e^{-\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }}$

${\displaystyle \qquad \qquad \qquad \qquad \quad \ ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl [}{\frac {-1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}}$

${\displaystyle \qquad \qquad \qquad \qquad \langle x\rangle =0}$

The average square of the position:

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x^{2}\cdot \Psi _{0}(x)\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x^{2}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx}$

 use ${\displaystyle \int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}_{-\infty }^{\infty }}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}\ \infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to \infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}-\infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to -\infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}}$

 use ${\displaystyle \lim _{x\to \infty }{\text{erf}}(x)=1}$ and ${\displaystyle \lim _{x\to -\infty }{\text{erf}}(x)=-1}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\biggl [}1-(-1){\biggl ]}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\sqrt {\alpha }}{\sqrt {\pi }}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\bigl [}2{\bigl ]}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle ={\frac {1}{2\alpha }}}$

The uncertainty on the position:

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\langle x^{2}\rangle -\langle x\rangle ^{2}}}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ ={\sqrt {{\frac {1}{2\alpha }}-0}}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\frac {1}{2\alpha }}}}$

The average momentum:

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}\cdot \Psi _{0}(x)\ dx\qquad \qquad \qquad \qquad \qquad \qquad {\text{where}}\ {\widehat {p}}_{x}=-i\hbar {\Bigl (}{\frac {\partial }{\partial x}}{\Bigr )}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot -i\hbar {\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl (}-\alpha x\ e^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\cdot x\ e^{\frac {-\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }x\ e^{-\alpha x^{2}}\ dx}$

 use ${\displaystyle \int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar \ {\biggl [}{\frac {1}{2\alpha }}\ e^{-2\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar {\biggl [}{\frac {1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =0}$

The average square of the momentum:

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}^{2}\cdot \Psi _{0}(x)\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot {\Biggl (}-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}{\Biggl (}-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}i^{2}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}(-1)\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl (}{\frac {\partial }{\partial x}}{\Biggr )}\ {\biggl (}-\alpha xe^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\alpha {\Bigl (}{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl (}e^{\frac {-\alpha x^{2}}{2}}+\alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}{\biggl )}\ dx}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\ \alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx}$

 use ${\displaystyle \int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx={\sqrt {\frac {\pi }{\alpha }}}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha \int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx{\biggl )}}$

 use ${\displaystyle \int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl [}{\frac {{\sqrt {\pi }}\ {\text{erf}}({\sqrt {\alpha }}\ x)}{4\alpha ^{\frac {3}{2}}}}-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\biggl ]}_{-\infty }^{\infty }{\Biggl )}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}\infty )\lim _{x\to \infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}-\infty )\lim _{x\to -\infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}{\Biggl )}}$

 use ${\displaystyle \lim _{x\to \infty }{\text{erf}}(x)=1}$ and ${\displaystyle \lim _{x\to -\infty }{\text{erf}}(x)=-1}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl (}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (1)-\ (0){\Biggl ]}\ -\ {\Biggl [}{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (-1)-\ (0){\Biggl ]}{\Biggl )}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl (}{\frac {2{\sqrt {\pi }}}{4\alpha ^{\frac {3}{2}}}}\ {\biggl )}{\Biggl ]}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha }{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl [}{\sqrt {\pi }}+{\frac {1}{2}}{\sqrt {\pi }}{\Biggl ]}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle ={\frac {1}{2}}\alpha \hbar ^{2}}$

The uncertainty on the momentum:

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\langle p_{x}^{2}\rangle -\langle p_{x}\rangle ^{2}}}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ ={\sqrt {{\frac {a\hbar ^{2}}{2}}-0^{2}}}}$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\frac {a^{2}}{2}}}\ \hbar }$

The product of the uncertainty on the position and the uncertainty on the momentum is:

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta _{x}\cdot \delta p_{x}={\frac {1}{\sqrt {2\alpha }}}\cdot {\sqrt {\frac {\alpha }{2}}}\ \hbar }$

${\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ ={\frac {1}{2}}\ \hbar }$

This is equal to ${\displaystyle {\frac {\hbar }{2}}}$, therefore, a quantum harmonic oscillator in the ground state is consistent with the Heisenberg Uncertainty Principle.