For the cubic 3D infinite well wave function

$\mathit{\psi }\left(x,y,z\right)\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{x}}\mathbf{\pi }\mathbf{x}}{\mathbf{L}}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{y}}\mathbf{\pi }\mathbf{y}}{\mathbf{L}}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{z}}\mathbf{\pi }\mathbf{z}}{\mathbf{L}}$show that the correct normalization constant is$\mathit{A}\mathbf{=}{\left(2/L\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}$.

Cubic 3D infinite well wave function is given as:

$\psi \mathit{(}\mathit{x}\mathit{,}\mathit{y}\mathit{,}\mathit{z}\mathit{)}\mathit{=}Asin\frac{{\mathit{n}}_{\mathit{x}}\mathit{\pi }\mathit{x}}{\mathit{L}}sin\frac{{\mathit{n}}_{\mathit{y}}\mathit{\pi }\mathit{y}}{\mathit{L}}sin\frac{{\mathit{n}}_{\mathit{z}}\mathit{\pi }\mathit{z}}{\mathit{L}}$

The normalization condition can be expressed as:

$\mathbf{\oint }\left|\psi \left(x,y,z\right)\right|\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{x}}\mathbf{\pi }\mathbf{x}}{\mathbf{L}}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{y}}\mathbf{\pi }\mathbf{y}}{\mathbf{L}}\mathit{s}\mathit{i}\mathit{n}\frac{{\mathbf{n}}_{\mathbf{z}}\mathbf{\pi }\mathbf{z}}{\mathbf{L}}$

We know that the probability of finding the particle will always be 1.

Evaluate the normalization condition for a given 3D infinite well solution to determine the value of the normalization constant A as:

For a 3D infinite well, its wave function is given by

$\psi \mathit{(}x\mathit{,}y\mathit{,}z\mathit{)}\mathit{=}Asin\frac{n_x\pi x}{L}sin\frac{n_y\pi y}{L}sin\frac{n_z\pi z}{L}$

Here,

L is the edge length of the square well.

We know that the probability density ${\left|\psi \left(x,y,z\right)\right|}^2$integrated over the volume of the 3D box must equal 1. Thus, we have

${\int }_0^L{\int }_0^L{\int }_0^L{\int }_0^L{\left|\psi \left(x,y,z\right)\right|}^{2}dxdydz=1$

$\begin{array}{r}{\int }_0^L {\int }_0^L {\int }_0^L {\left|A\sin \frac{n_x\pi x}{L}\sin \frac{n_y\pi y}{L}\sin \frac{n_z\pi z}{L}\right|}^{2}dxdydz=1\\ {\int }_0^L {\int }_0^L {\int }_0^L A^2{\sin }^2\frac{n_x\pi x}{L}{\sin }^2\frac{n_y\pi y}{L}{\sin }^2\frac{n_z\pi z}{L}dxdydz=1\\ A^2{\int }_0^L {\int }_0^L {\int }_0^L {\sin }^2\frac{n_x\pi x}{L}{\sin }^2\frac{n_y\pi y}{L}{\sin }^2\frac{n_z\pi z}{L}dxdydz=1\\ A^2{\int }_0^L {\sin }^2\frac{n_x\pi x}{L}dx{\int }_0^L {\sin }^2\frac{n_y\pi y}{L}dy{\int }_0^L {\sin }^2\frac{n_z\pi z}{L}dz=1\end{array}$

1)

All three integrals multiplying each other are of the same form, for different values of $\left(n_x,n_y,n_z\right)$ so next evaluation just one of them for an arbitrary n.

Visualizing the dependence of the trigonometric functions graphically leads to the conclusion that,

${\int }_0^L {\sin }^2\frac{n_x\pi x}{L}dx{\int }_0^L {\sin }^2\frac{n_y\pi y}{L}dy={\int }_0^L {\sin }^2\frac{n_z\pi z}{L}dz$

Therefore, one segment can be solved as:

${\int }_0^L {\sin }^2\frac{n_x\pi x}{L}dx=\frac{L}{2}$

Thus, all three integrals multiplying each other in equation (1) are equal to each other and to $\frac{L}{2}$.

Therefore, the equation (1) becomes,

$$\begin{gathered} A^2\left(\frac{L}{2}\right)\left(\frac{L}{2}\right)\left(\frac{L}{2}\right)=1 \\ {A}^2{\left(\frac{L}{2}\right)}^3=1 \\ {A}^2={\left(\frac{2}{L}\right)}^3 \\ {A}^2={\left(\frac{2}{L}\right)}^{3/2} \end{gathered}$$

Therefore, the normalization constant is ${\left(\frac{2}{L}\right)}^{3/2}$.