91Ó°ÊÓ

The function $\mathrm{ψ}\left(x\right)$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.

Schrodinger’s equation is

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

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

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

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

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

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

Substitute $\mathrm{ψ}={\mathrm{ψ}}_0{e}^{ikx}$

The second derivative is

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

The result is

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

Solving for k,

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

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