# Quantum Chemistry/Example 1

Find ⟨x⟩, ⟨x$^2$⟩, ⟨p$_x$⟩ and ⟨p$_x$$^2$⟩ 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 $\qquad \delta _{x}\cdot \delta _{p}\geq {\frac {\hbar }{2}}\qquad {\text{where}}\ \hbar ={\frac {h}{2\pi }}$
The wavefunction of a quantum harmonic oscillator in the ground state is:
$\qquad \qquad \Psi _{0}(x)={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}$ Using this wavefunction the average position and the average of the square of the position can be calculated.
The average position:
$\qquad \qquad \qquad \qquad \langle x\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x\cdot \Psi _{0}(x)\ dx$
$\qquad \qquad \qquad \qquad \quad \ =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x\cdot e^{\frac {-2\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }xe^{-\alpha x^{2}}\ dx$
                   use $\int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl{\frac {-1}{2\alpha }}\cdot e^{-\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }$
$\qquad \qquad \qquad \qquad \quad \ ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\biggl{\frac {-1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}$
$\qquad \qquad \qquad \qquad \langle x\rangle =0$
The average square of the position:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot x^{2}\cdot \Psi _{0}(x)\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot x^{2}\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx$
                               use $\int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl{\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}_{-\infty }^{\infty }$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}{\Biggl{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}\ \infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to \infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl{\frac {\sqrt {\pi }}{4}}{\frac {{\text{erf}}({\sqrt {\alpha }}-\infty )}{\alpha ^{\frac {3}{2}}}}\ -\ \lim _{x\to -\infty }{\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}$
                               use $\lim _{x\to \infty }{\text{erf}}(x)=1$ and $\lim _{x\to -\infty }{\text{erf}}(x)=-1$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\biggl1-(-1){\biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\sqrt {\alpha }}{\sqrt {\pi }}}\ {\frac {\sqrt {\pi }}{4}}\ {\frac {1}{\alpha ^{\frac {3}{2}}}}\ {\bigl2{\bigl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \langle x^{2}\rangle ={\frac {1}{2\alpha }}$
The uncertainty on the position:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\langle x^{2}\rangle -\langle x\rangle ^{2}}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ ={\sqrt {{\frac {1}{2\alpha }}-0}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta _{x}={\sqrt {\frac {1}{2\alpha }}}$
The average momentum:
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}\cdot \Psi _{0}(x)\ dx\qquad \qquad \qquad \qquad \qquad \qquad {\text{where}}\ {\widehat {p}}_{x}=-i\hbar {\Bigl{\frac {\partial }{\partial x}}{\Bigr )}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot -i\hbar {\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggl-\alpha x\ e^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\cdot x\ e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =-\alpha {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}-i\hbar \int _{-\infty }^{\infty }x\ e^{-\alpha x^{2}}\ dx$
                            use $\int xe^{-\alpha x^{2}}\ dx={\frac {-1}{2\alpha }}e^{-\alpha x^{2}}+C$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar \ {\biggl{\frac {1}{2\alpha }}\ e^{-2\alpha x^{2}}{\biggl ]}_{-\infty }^{\infty }$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {-\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}-i\hbar {\biggl{\frac {1}{2\alpha }}(0)-{\frac {1}{2\alpha }}(0){\biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}\rangle =0$
The average square of the momentum:
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle =\int _{-\infty }^{\infty }\Psi _{0}(x)^{*}\cdot {\widehat {p}}_{x}^{2}\cdot \Psi _{0}(x)\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int _{-\infty }^{\infty }{\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\cdot {\Biggl-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}{\Biggl-i\hbar {\frac {\partial }{\partial x}}{\Biggr )}\cdot {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{4}}e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}i^{2}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\ e^{\frac {-\alpha x^{2}}{2}}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}(-1)\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\Biggl{\frac {\partial }{\partial x}}{\Biggr )}\ {\biggl-\alpha xe^{\frac {-\alpha x^{2}}{2}}{\biggl )}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\alpha {\Bigl{\frac {\alpha }{\pi }}{\Bigr )}^{\frac {1}{2}}\hbar ^{2}\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}{\biggle^{\frac {-\alpha x^{2}}{2}}+\alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}{\biggl )}\ dx$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx\int _{-\infty }^{\infty }e^{\frac {-\alpha x^{2}}{2}}\ \alpha x^{2}e^{\frac {-2\alpha x^{2}}{2}}\ dx$
                                        use $\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\ dx={\sqrt {\frac {\pi }{\alpha }}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha \int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}\ dx{\biggl )}$
                               use $\int x^{2}e^{-\alpha x^{2}}\ dx={\frac {{\sqrt {\pi }}{\text{erf}}({\sqrt {\alpha }}x)}{4\alpha ^{\frac {3}{2}}}}\ -\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}+C$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl{\frac {{\sqrt {\pi }}\ {\text{erf}}({\sqrt {\alpha }}\ x)}{4\alpha ^{\frac {3}{2}}}}-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\biggl ]}_{-\infty }^{\infty }{\Biggl )}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}\infty )\lim _{x\to \infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}\ -\ {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ {\text{erf}}({\sqrt {\alpha }}-\infty )\lim _{x\to -\infty }-\ {\frac {xe^{-\alpha x^{2}}}{2\alpha }}{\Biggl ]}{\Biggl )}$
                               use $\lim _{x\to \infty }{\text{erf}}(x)=1$ and $\lim _{x\to -\infty }{\text{erf}}(x)=-1$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (1)-\ (0){\Biggl ]}\ -\ {\Biggl{\frac {\sqrt {\pi }}{4\alpha ^{\frac {3}{2}}}}\ (-1)-\ (0){\Biggl ]}{\Biggl )}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha ^{\frac {3}{2}}}{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\frac {\pi }{\alpha }}}+\alpha {\biggl{\frac {2{\sqrt {\pi }}}{4\alpha ^{\frac {3}{2}}}}\ {\biggl )}{\Biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad ={\frac {\alpha }{\sqrt {\pi }}}\ \hbar ^{2}{\Biggl{\sqrt {\pi }}+{\frac {1}{2}}{\sqrt {\pi }}{\Biggl ]}$
$\qquad \qquad \qquad \qquad \qquad \qquad \langle p_{x}^{2}\rangle ={\frac {1}{2}}\alpha \hbar ^{2}$
The uncertainty on the momentum:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\langle p_{x}^{2}\rangle -\langle p_{x}\rangle ^{2}}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ ={\sqrt {{\frac {a\hbar ^{2}}{2}}-0^{2}}}$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_{x}={\sqrt {\frac {a^{2}}{2}}}\ \hbar$
The product of the uncertainty on the position and the uncertainty on the momentum is:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta _{x}\cdot \delta p_{x}={\frac {1}{\sqrt {2\alpha }}}\cdot {\sqrt {\frac {\alpha }{2}}}\ \hbar$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ ={\frac {1}{2}}\ \hbar$
This is equal to ${\frac {\hbar }{2}}$, therefore, a quantum harmonic oscillator in the ground state is consistent with the Heisenberg Uncertainty Principle.