Do Problem 6.6 in 3-dimensional rectangular coordinates. That is, solve the “particle in a box” problem for a cube.

The time independent Schrodinger equation with U = 0 or a particle in a cube of dimensional l.

$\frac{-h^2}{2m}[\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}]+U(x,y,z)\psi (x,y,z)=E\psi (x,y,z)$.

The coordinates of a point in a three-dimensional coordinate system are its distances from a set of perpendicular lines intersecting at an origin, such as two on a plane or three in space.

Write the time independent Schrodinger equation with U = 0 for a particle in a cube of dimensional l.

$\frac{-h^2}{2m}[\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}]+U(x,y,z)\psi (x,y,z)=E\psi (x,y,z)$

Consider the potential $U(x,y,z)=0$inside the cube with side l.

$U(x,y,z)=\left\{\begin{array}{l}\begin{array}{cc}0& 0<x,y,z<l\end{array}\\ \begin{array}{cc}\infty & \text{Otherwise}\end{array}\end{array}\right.$

Put $U(x,y,z)=U_1\left(x\right)+U_2\left(y\right)+U_3\left(z\right)$into Schrodinger equation.

$\frac{-h^2}{2m}\frac{{\partial }^2\psi \left(r\right)}{\partial x^2}-\frac{h^2}{2m}\frac{{\partial }^2\psi \left(r\right)}{\partial y^2}-\frac{h^2}{2m}\frac{{\partial }^2\psi \left(r\right)}{\partial z^2}+U_1\left(x\right)\psi \left(r\right)++U_2\left(y\right)\psi \left(r\right)++U_3\left(z\right)\psi \left(r\right)=0$ ….. (1)

The above solution is of the form$\psi \left(r\right)={\psi }_1\left(x\right){\psi }_2\left(y\right){\psi }_3\left(z\right)$.

Put $\psi \left(r\right)={\psi }_1\left(x\right){\psi }_2\left(y\right){\psi }_3\left(z\right)$in equation (1).

$\begin{array}{l}[\frac{-h^2}{2m}\frac{{\partial }^2{\psi }_1\left(x\right)}{\partial x^2}+U_1\left(x\right){\psi }_1\left(x\right)]+{\psi }_2\left(y\right){\psi }_3\left(z\right)\\ [\frac{-h^2}{2m}\frac{{\partial }^2{\psi }_2\left(y\right)}{\partial y^2}+U_2\left(y\right){\psi }_2\left(y\right)]+{\psi }_1\left(x\right){\psi }_3\left(z\right)\\ [\frac{-h^2}{2m}\frac{{\partial }^2{\psi }_3\left(z\right)}{\partial z^2}+U_3\left(z\right){\psi }_3\left(z\right)]+{\psi }_2\left(y\right){\psi }_1\left(x\right)\end{array}\}=(E_1+E_2+E_3){\psi }_1\left(x\right){\psi }_2\left(y\right){\psi }_3\left(z\right)$

Compare the left-hand side and the right-hand side of the above equation.

$\begin{array}{c}[\frac{-h^2}{2m}\frac{{\partial }^2{\psi }_1\left(x\right)}{\partial x^2}+U_1\left(x\right){\psi }_1\left(x\right)]+{\psi }_2\left(y\right){\psi }_3\left(z\right)=E_1{\psi }_1\left(x\right)\\ [\frac{-h^2}{2m}\frac{{\partial }^2{\psi }_2\left(y\right)}{\partial y^2}+U_2\left(y\right){\psi }_2\left(y\right)]+{\psi }_1\left(x\right){\psi }_3\left(z\right)=E_2{\psi }_2\left(y\right)\\ [\frac{-h^2}{2m}\frac{{\partial }^2{\psi }_3\left(z\right)}{\partial z^2}+U_3\left(z\right){\psi }_3\left(z\right)]+{\psi }_2\left(y\right){\psi }_1\left(x\right)=E_3{\psi }_3\left(z\right)\end{array}$

Multiplying the three one-dimensional Schrodinger equation yields the solution to three-dimensional Schrodinger equations.

$\begin{array}{c}{E}_1=\frac{{\pi }^2{h}^2}{2m{l}^2}{n}_1\\ E_2=\frac{{\pi }^2{h}^2}{2m{l}^2}{n}_2\\ E_3=\frac{{\pi }^2{h}^2}{2m{l}^2}{n}_3\end{array}$

And,

$\begin{array}{c}{\psi }_1\left(x\right)=\sqrt{\frac{2}{l}}\sin \left(\frac{\pi n_1}{l}\right)\\ {\psi }_2\left(y\right)=\sqrt{\frac{2}{l}}\sin \left(\frac{\pi n_2}{l}\right)\\ {\psi }_3\left(z\right)=\sqrt{\frac{2}{l}}\sin \left(\frac{\pi n_3}{l}\right)\end{array}$

Sum all the energy.

$\begin{array}{c}E=E_1+E_2+E_3\\ =\frac{{\pi }^2{h}^2}{2m}\left(\frac{n_x^2+{n_y}^2+{n_z}^2}{l^2}\right)\end{array}$

Write the solution to the Schrodinger wave equation.

$\begin{array}{c}\psi \left(r\right)={\psi }_1\left(x\right){\psi }_2\left(y\right){\psi }_3\left(z\right)\\ =\sqrt{\frac{8}{V}}\sin \left(\frac{\pi n_x}{l}x\right)\sin \left(\frac{\pi n_y}{l}y\right)\sin \left(\frac{\pi n_z}{l}z\right)\end{array}$

Here,$V=l^3$.