91Ó°ÊÓ

The Schrödinger equation for the one-dimensional potential well is given by,

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{ψ}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{8}{\mathbf{π}}^{\mathbf{2}}\mathbf{m}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{[}\mathbf{E}\mathbf{-}\mathbf{U}\left(x\right)\mathbf{]}\mathbf{ψ}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Consider the Schrödinger equation.

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

The unit of the wave function$\mathrm{ψ}$ is${\left(length\right)}^{-1/2}$ , and the unit of$x^2$ is$m^2$ .

From equation (1), the SI units for the first term are,

$$\begin{gathered} \left[\frac{{\mathrm{d}}^2\mathrm{ψ}}{{\mathrm{dx}}^2}\right]=\frac{m^{-1/2}}{m^2} \\ =m^{-2.5} \end{gathered}$$

Similarly, we obtain the SI units for the second term in equation (1).

$$\begin{gathered} \frac{8{\mathrm{π}}^2\mathrm{m}}{h^2}\left[E-U\left(x\right)\right]\mathrm{ψ}=\frac{kg}{{\left[J.s\right]}^2}\left[J\right]\left[m^{-1.5}\right] \\ =\frac{kg}{\left[\left({\displaystyle \frac{kg.m}{s}}.s\right)\right]}\frac{kg.m}{s}\left[m^{-1.5}\right] \\ =m^{-2.5} \end{gathered}$$

Therefore, the terms in the Schrödinger equation have the same dimensions.

The SI unit of the first term in the Schrödinger equation is$m^{-2.5}$ , and the SI unit of the second term in the Schrödinger equation is also$m^{-2.5}$ . So, the common SI unit for each term of theSchrödinger equation is$m^{-2.5}$ .

Therefore, the common SI unit is$m^{-2.5}$ .