Question

Check the uncertainty principle for the wave function in the equation? Equation 2.129.

Short answer

Uncertainty of the wave equation $p^2={\left(\frac{m\alpha }{h}\right)}^2$

Step-by-step solution

Define the Schrodinger equation

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

Determine the uncertainty principle

$\mathrm{\Psi }\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{m\alpha }}{\mathrm{h}}}{\mathrm{e}}^{-\frac{\mathrm{m\alpha x}}{{\mathrm{h}}^2}}$when $\mathrm{x}\ge 0$

$\mathrm{\Psi }\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{m\alpha }}{\mathrm{h}}}{\mathrm{e}}^{\frac{\mathrm{m\alpha x}}{{\mathrm{h}}^2}}$ when$\mathrm{x}\le 0$

$\mathrm{x}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x\Psi }}^2\mathrm{dx}=0$

${\mathrm{x}}^2=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x}}^2{\mathrm{\Psi }}^2\mathrm{dx}=\frac{2\mathrm{m\alpha }}{{\mathrm{h}}^2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{x}}^2{\mathrm{e}}^{-\frac{2\mathrm{m\alpha }}{{\mathrm{h}}^2}}\mathrm{dx}$

$\mathrm{x}=\frac{{\mathrm{h}}^4}{2{\mathrm{m}}^2{\mathrm{\alpha }}^2}$

${\mathrm{\sigma }}_{\mathrm{x}}=\frac{{\mathrm{h}}^2}{\sqrt{2}\mathrm{m\alpha }}$

$\frac{\mathrm{d\Psi }}{\mathrm{dx}}=-\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2}{\mathrm{e}}^{\frac{-\mathrm{m\alpha x}}{{\mathrm{h}}^2}}$ for $x\ge 0$

$\frac{\mathrm{d\Psi }}{\mathrm{dx}}=-\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2}{\mathrm{e}}^{\frac{\mathrm{m\alpha x}}{{\mathrm{h}}^2}}$ for $\mathrm{x}\le 0$

$\frac{\mathrm{d\Psi }}{\mathrm{dx}}={\left(\frac{\sqrt{\mathrm{m\alpha }}}{\mathrm{h}}\right)}^3\left[-\mathrm{\theta }\left(\mathrm{x}\right){\mathrm{e}}^{\frac{-\mathrm{m\alpha x}}{{\mathrm{h}}^2}}+\mathrm{\theta }(-\mathrm{x}){\mathrm{e}}^{\frac{\mathrm{m\alpha x}}{{\mathrm{h}}^2}}\right]$

$\frac{{\mathrm{d}}^2\mathrm{\Psi }}{{\mathrm{dx}}^2}={\left(\frac{\sqrt{\mathrm{m\alpha }}}{\mathrm{h}}\right)}^3\left[-\mathrm{\delta }\left(\mathrm{x}\right){\mathrm{e}}^{\frac{-\mathrm{m\alpha x}}{{\mathrm{h}}^2}}+\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2}\mathrm{\theta }\left(\mathrm{x}\right){\mathrm{e}}^{\frac{-\mathrm{m\alpha x}}{{\mathrm{h}}^2}}-\mathrm{\delta }(-\mathrm{x}){\mathrm{e}}^{\frac{\mathrm{m\alpha x}}{{\mathrm{h}}^2}}+\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2}\mathrm{\theta }(-\mathrm{x}){\mathrm{e}}^{\frac{\mathrm{m\alpha x}}{{\mathrm{h}}^2}}\right]$

Here,

$\mathrm{\delta }(-\mathrm{x})=\mathrm{\delta }\left(\mathrm{x}\right)$,

$\mathrm{f}\left(\mathrm{x}\right)\mathrm{\delta }\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)\mathrm{\delta }\left(\mathrm{x}\right)$

$\mathrm{\theta }(-\mathrm{x})+\mathrm{\theta }\left(\mathrm{x}\right)=1$

$\frac{{\mathrm{d}}^2\mathrm{\Psi }}{{\mathrm{dx}}^2}={\left(\frac{\sqrt{\mathrm{m\alpha }}}{\mathrm{h}}\right)}^3\left[-2\mathrm{\delta }\left(\mathrm{x}\right)+\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2}{\mathrm{e}}^{\frac{-\mathrm{m\alpha x}}{{\mathrm{h}}^2}}\right]$

$\mathrm{p}=0$

$p^2=-h^2\underset{-\infty }{\overset{\infty }{\int }}\Psi \frac{d^2\Psi }{d{x}^2}dx$

$p^2=-h^2{\left(\frac{\sqrt{m\alpha }}{h}\right)}^3\underset{-\infty }{\overset{\infty }{\int }}{e}^{-\frac{m\alpha x}{h^2}}\left[-2\delta \left(x\right)+\frac{m\alpha }{h}{e}^{-\frac{m\alpha x}{h^2}}\right]dx$

$p^2={\left(\frac{m\alpha }{h}\right)}^2\left[2-2\frac{m\alpha }{h^2}\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{2m\alpha }{h^2}}dx\right]$

$\begin{array}{c}{p}^2={\left(\frac{m\alpha }{h}\right)}^2\left[1-\frac{m\alpha }{h^2}\frac{h^2}{2m\alpha }\right]\\ ={\left(\frac{m\alpha }{h}\right)}^2\end{array}$

$p^2={\left(\frac{m\alpha }{h}\right)}^2$

${\sigma }_p=\frac{m\alpha }{h}$

$\begin{array}{c}{\sigma }_x{\sigma }_p=\frac{h^2}{\sqrt{2}m\alpha }\frac{m\alpha }{h}\\ =\sqrt{2}\frac{h}{2}>\frac{h}{2}\end{array}$