Quantum Chemistry/Example 1

Find ⟨x⟩, ⟨x²⟩, ⟨p_x⟩ and ⟨p_x²⟩ 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 ${\delta }_x\cdot {\delta }_p\ge \frac{\hslash }{2}\text{where}\hslash =\frac{h}{2\pi }$

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

${\mathrm{\Psi }}_0(x)=(\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}$ Using this wavefunction the average position and the average of the square of the position can be calculated.Template:BookCat

The average position:

$⟨x⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\Psi }}_0(x{)}^{*}\cdot x\cdot {\mathrm{\Psi }}_0(x)dx$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}\cdot x\cdot (\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}dx$

$=(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\cdot e^{\frac{-2\alpha x^2}{2}}dx$

$=(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{-\alpha x^2}dx$

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

$=\left(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\right[\frac{-1}{2\alpha }\cdot e^{-\alpha x^2}{]}_{-\mathrm{\infty }}^{\mathrm{\infty }}$

$=\left(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\right[\frac{-1}{2\alpha }(0)-\frac{1}{2\alpha }(0)]$

$⟨x⟩=0$

The average square of the position:

$⟨x^2⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\Psi }}_0(x{)}^{*}\cdot x^2\cdot {\mathrm{\Psi }}_0(x)dx$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}\cdot x^2\cdot (\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}dx$

$=(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{\frac{-2\alpha x^2}{2}}dx$

$=(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-\alpha x^2}dx$

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

$=\left(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\right[\frac{\sqrt{\pi }\text{erf}(\sqrt{\alpha }x)}{4{\alpha }^{\frac{3}{2}}}-\frac{x{e}^{-\alpha x^2}}{2\alpha }{]}_{-\mathrm{\infty }}^{\mathrm{\infty }}$

$=\left(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\right[\frac{\sqrt{\pi }}{4}\frac{\text{erf}(\sqrt{\alpha }\mathrm{\infty })}{{\alpha }^{\frac{3}{2}}}-\underset{x\to \mathrm{\infty }}{\lim }\frac{x{e}^{-\alpha x^2}}{2\alpha }]-[\frac{\sqrt{\pi }}{4}\frac{\text{erf}(\sqrt{\alpha }-\mathrm{\infty })}{{\alpha }^{\frac{3}{2}}}-\underset{x\to -\mathrm{\infty }}{\lim }\frac{x{e}^{-\alpha x^2}}{2\alpha }]$

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

$=\left(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}\frac{\sqrt{\pi }}{4}\frac{1}{{\alpha }^{\frac{3}{2}}}\right[1-(-1)]$

$=\frac{\sqrt{\alpha }}{\sqrt{\pi }}\frac{\sqrt{\pi }}{4}\frac{1}{{\alpha }^{\frac{3}{2}}}\left[2\right]$

$⟨x^2⟩=\frac{1}{2\alpha }$

The uncertainty on the position:

${\delta }_x=\sqrt{⟨x^2⟩-⟨x{⟩}^2}$

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

${\delta }_x=\sqrt{\frac{1}{2\alpha }}$

The average momentum:

$⟨p_x⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\Psi }}_0(x{)}^{*}\cdot {\hat{p}}_x\cdot {\mathrm{\Psi }}_0(x)dx\text{where}{\hat{p}}_x=-i\hslash (\frac{\partial }{\partial x})$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}\cdot -i\hslash (\frac{\partial }{\partial x})\cdot (\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}dx$

$=(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}-i\hslash \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{\frac{-\alpha x^2}{2}}(\frac{\partial }{\partial x})e^{\frac{-\alpha x^2}{2}}dx$

$=(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}-i\hslash \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{\frac{-\alpha x^2}{2}}(-\alpha x{e}^{\frac{-\alpha x^2}{2}})dx$

$=-\alpha (\frac{\alpha }{\pi }{)}^{\frac{1}{2}}-i\hslash \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{\frac{-\alpha x^2}{2}}\cdot x{e}^{\frac{-\alpha x^2}{2}}dx$

$=-\alpha (\frac{\alpha }{\pi }{)}^{\frac{1}{2}}-i\hslash \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{-\alpha x^2}dx$

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

$=\frac{-{\alpha }^{\frac{3}{2}}}{\sqrt{\pi }}-i\hslash [\frac{1}{2\alpha }{e}^{-2\alpha x^2}{]}_{-\mathrm{\infty }}^{\mathrm{\infty }}$

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

$⟨p_x⟩=0$

The average square of the momentum:

$⟨p_x^2⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\Psi }}_0(x{)}^{*}\cdot {\hat{p}}_x^2\cdot {\mathrm{\Psi }}_0(x)dx$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}\cdot (-i\hslash \frac{\partial }{\partial x}\left)\right(-i\hslash \frac{\partial }{\partial x})\cdot (\frac{\alpha }{\pi }{)}^{\frac{1}{4}}{e}^{\frac{-\alpha x^2}{2}}dx$

$=\left(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}{i}^2{\hslash }^2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{\frac{-\alpha x^2}{2}}\right(\frac{\partial }{\partial x}\left)\right(\frac{\partial }{\partial x})e^{\frac{-\alpha x^2}{2}}dx$

$=(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}(-1){\hslash }^2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{\frac{-\alpha x^2}{2}}(\frac{\partial }{\partial x}\left)\right(-\alpha x{e}^{\frac{-\alpha x^2}{2}})dx$

$=\alpha \left(\frac{\alpha }{\pi }{)}^{\frac{1}{2}}{\hslash }^2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{\frac{-\alpha x^2}{2}}\right(e^{\frac{-\alpha x^2}{2}}+\alpha x^2{e}^{\frac{-2\alpha x^2}{2}})dx$

$=\frac{{\alpha }^{\frac{3}{2}}}{\sqrt{\pi }}{\hslash }^2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\alpha x^2}dx\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{\frac{-\alpha x^2}{2}}\alpha x^2{e}^{\frac{-2\alpha x^2}{2}}dx$

use $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\alpha x^2}dx=\sqrt{\frac{\pi }{\alpha }}$

$=\frac{{\alpha }^{\frac{3}{2}}}{\sqrt{\pi }}{\hslash }^2(\sqrt{\frac{\pi }{\alpha }}+\alpha \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-\alpha x^2}dx)$

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

$=\frac{{\alpha }^{\frac{3}{2}}}{\sqrt{\pi }}{\hslash }^2(\sqrt{\frac{\pi }{\alpha }}+\alpha [\frac{\sqrt{\pi }\text{erf}(\sqrt{\alpha }x)}{4{\alpha }^{\frac{3}{2}}}-\frac{x{e}^{-\alpha x^2}}{2\alpha }{]}_{-\mathrm{\infty }}^{\mathrm{\infty }})$

$=\frac{{\alpha }^{\frac{3}{2}}}{\sqrt{\pi }}{\hslash }^2(\sqrt{\frac{\pi }{\alpha }}+\alpha [\frac{\sqrt{\pi }}{4{\alpha }^{\frac{3}{2}}}\text{erf}(\sqrt{\alpha }\mathrm{\infty })\underset{x\to \mathrm{\infty }}{\lim }-\frac{x{e}^{-\alpha x^2}}{2\alpha }]-[\frac{\sqrt{\pi }}{4{\alpha }^{\frac{3}{2}}}\text{erf}(\sqrt{\alpha }-\mathrm{\infty })\underset{x\to -\mathrm{\infty }}{\lim }-\frac{x{e}^{-\alpha x^2}}{2\alpha }\left]\right)$

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

$=\frac{{\alpha }^{\frac{3}{2}}}{\sqrt{\pi }}{\hslash }^2(\sqrt{\frac{\pi }{\alpha }}+\alpha [\frac{\sqrt{\pi }}{4{\alpha }^{\frac{3}{2}}}(1)-(0)]-[\frac{\sqrt{\pi }}{4{\alpha }^{\frac{3}{2}}}(-1)-(0)\left]\right)$

$=\frac{{\alpha }^{\frac{3}{2}}}{\sqrt{\pi }}{\hslash }^2[\sqrt{\frac{\pi }{\alpha }}+\alpha (\frac{2\sqrt{\pi }}{4{\alpha }^{\frac{3}{2}}}\left)\right]$

$=\frac{\alpha }{\sqrt{\pi }}{\hslash }^2[\sqrt{\pi }+\frac{1}{2}\sqrt{\pi }]$

$⟨p_x^2⟩=\frac{1}{2}\alpha {\hslash }^2$

The uncertainty on the momentum:

$\delta p_x=\sqrt{⟨p_x^2⟩-⟨p_x{⟩}^2}$

$=\sqrt{\frac{a{\hslash }^2}{2}-0^2}$

$\delta p_x=\sqrt{\frac{a^2}{2}}\hslash$

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

${\delta }_x\cdot \delta p_x=\frac{1}{\sqrt{2\alpha }}\cdot \sqrt{\frac{\alpha }{2}}\hslash$

$=\frac{1}{2}\hslash$

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