To determine the energy quantization condition

For the infinite and finite square well

$\mathrm{U}(\mathrm{x}<0)=\mathrm{\infty }$

$\mathrm{y}(\mathrm{x}<0)=0$

Inside of the well, where$0<\mathrm{x}<\mathrm{L}$ and the potential is zero

$\mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{Asin}\left(\mathrm{kx}\right)+\mathrm{Bcos}\left(\mathrm{kx}\right)$

Here, $\mathrm{k}=\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}$, $\mathrm{E}$is the energy, $\mathrm{m}$is the mass, and$\mathrm{\hslash }$ is the reduced Planck's constant.

For outside of the well where there is potential energy${\mathrm{U}}_0$ and where$\mathrm{x}>\mathrm{L}$

$\mathrm{\psi }\left(\mathrm{x}\right)={\mathrm{Ce}}^{-\mathrm{\alpha x}}$

Here $\mathrm{\alpha }=\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{h}}$

Ignore the ${\mathrm{Ce}}^{\mathrm{\alpha x}}$component since it diverges as$\mathrm{x}$ goes to infinity.

The wave function$\mathrm{\psi }$must be continuous at the edges of the well at$\mathrm{X}=0$and $\mathrm{L}$.

For the case of $\mathrm{x}=0$, find that$\mathrm{B}$must be zero for the function to be continuous.

$0=\mathrm{Asin}\left(0\right)+\mathrm{Bcos}\left(0\right)$

$\mathrm{B}=0$

In the case of$\mathrm{x}=\mathrm{L}$

$\mathrm{Asin}\left(\mathrm{kL}\right)={\mathrm{Ce}}^{-\mathrm{\alpha L}}$ …… (1)

Here, $\frac{\mathrm{d\psi }}{\mathrm{dx}}$must also be continuous at $\mathrm{x}=\mathrm{L}$

$\frac{\mathrm{d\psi }}{\mathrm{dx}}\left(\mathrm{Asin}\right(\mathrm{kL}\left)\right)=\frac{\mathrm{d\psi }}{\mathrm{dx}}\left({\mathrm{Ce}}^{-\mathrm{\alpha L}}\right)$

$\mathrm{kAcos}\left(\mathrm{kL}\right)=-{\mathrm{aCe}}^{-\mathrm{\alpha L}}$ …… (2)

Divide the equation (2) with the equation (1) and simplify it

$\frac{\mathrm{kAcos}\left(\mathrm{kL}\right)}{\mathrm{Asin}\left(\mathrm{kL}\right)}=\frac{-{\mathrm{\alpha Ce}}^{-\mathrm{aL}}}{{\mathrm{Ce}}^{-\mathrm{aL}}}$

$\mathrm{kcot}\left(\mathrm{kL}\right)=-\mathrm{\alpha }$

Substitute $\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{h}}$ for $\mathrm{\alpha }$ and $\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}$for $\mathrm{k}$in the above equation and $\mathrm{si}$.

$\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\cot \left(\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}\right)=-\frac{\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}}{\mathrm{h}}$

$\sqrt{\mathrm{E}}\cot \left(\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}\right)=-\sqrt{{\mathrm{U}}_0-\mathrm{E}}$.

Hence, the energy quantization condition is $\sqrt{\mathrm{E}}\cot \left(\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\mathrm{L}\right)=-\sqrt{{\mathrm{U}}_0-\mathrm{E}}$.