Question

Question: The function $\mathbf{\phi }\left(x\right)$displayed in Eq. 38-27 can describe a free particle, for which the potential energy is $\mathbf{U}\left(x\right)\mathbf{=}\mathbf{0}$ in Schrodinger’s equation (Eq. 38-19). Assume now that $\mathbf{U}\left(x\right)\mathbf{=}{\mathbf{U}}_{\mathbf{0}}\mathbf{=}\mathbf{a}$ constant in that equation. Show that Eq. 38-27 is a solution of Schrodinger’s equation, with $\mathbf{k}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{h}}\sqrt{\mathbf{2}\mathbf{m}\left(E-U_0\right)}$giving the angular wave number k of the particle.

Short answer

It has been proved that Eq. 38-27 is a solution of Schrodinger’s equation, with

$\mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{h}}\sqrt{2\mathrm{m}(\mathrm{E}-{\mathrm{U}}_0)}$giving the angular wave number k of the particle

Step-by-step solution

The given data

The function $\psi \left(x\right)$displayed in Eq. 38-27 can describe a free particle, for which the potential energy $\mathrm{U}\left(x\right)=0$is in Schrodinger’s equation (Eq. 38-19).

Assume now that $\mathrm{U}\left(x\right)={\mathrm{U}}_0=\mathrm{a}$constant in that equation.

Concept and Formula used

Schrodinger’s equation is

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{8}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{m}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{[}\mathbf{E}\mathbf{-}\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{]}\mathbf{\psi }\mathbf{=}\mathbf{0}$

The function
$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is

$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{ikx}}$

Here, we need to solve a special case of Schrodinger’s equation where the potential energy is $\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{U}}_{\mathbf{0}}\mathbf{=}\mathbf{constant}$.

Solve Schrodinger’s equation

For $\mathrm{U}={\mathrm{U}}_0$, Schrodinger’s equation becomes

$\frac{{\mathrm{d}}^2\mathrm{\psi }}{{\mathrm{dx}}^2}+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{{\mathrm{h}}^2}\left[\mathrm{E}-{\mathrm{U}}_0\right]\mathrm{\psi }=0$

Substitute $\psi ={\psi }_0{e}^{ikx}$

Solve for k

The second derivative is

$\begin{array}{rcl}\frac{d^2\psi }{d{x}^2}& =& -k^2{\psi }_0{e}^{ikx}\\ & =& -k^2\psi \end{array}$

The result is

$-{\mathrm{k}}^2\mathrm{\psi }+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{{\mathrm{h}}^2}\left[\mathrm{E}-{\mathrm{U}}_{\mathrm{o}}\right]\mathrm{\psi }=0$.

Solving for k,

$\begin{array}{rcl}k& =& \sqrt{\frac{8{\mathrm{\pi }}^2\mathrm{m}}{h^2}}\left[E-U_0\right]\\ & =& \frac{2\mathrm{\pi }}{h}\sqrt{2m\left[E-U_0\right]}\end{array}$

Hence, $\psi \left(x\right)=A{e}^{ikx}$is a solution of Schrodinger’s equation, with $k=\frac{2\mathrm{\pi }}{h}\sqrt{2m(E-U_0)}$giving the angular wave number k of the particle.