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Expert-verified Found in: Page 279 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # For the cubic 3D infinite well wave function${\mathbit{\psi }}\left(x,y,z\right){\mathbf{=}}{\mathbit{A}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\frac{{\mathbf{n}}_{\mathbf{x}}\mathbf{\pi }\mathbf{x}}{\mathbf{L}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\frac{{\mathbf{n}}_{\mathbf{y}}\mathbf{\pi }\mathbf{y}}{\mathbf{L}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}\frac{{\mathbf{n}}_{\mathbf{z}}\mathbf{\pi }\mathbf{z}}{\mathbf{L}}$ show that the correct normalization constant is ${\mathbit{A}}{\mathbf{=}}{\left(2/L\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}$.

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

See the step by step solution

## Step 1:  Given data

Cubic 3D infinite well wave function is given as:

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

## Step 2:  Normalization condition

The normalization condition can be expressed as:

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

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

## Step 3:  To determine the normalization constant

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

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

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

${\int }_{0}^{L}{\int }_{0}^{L}{\int }_{0}^{L}{\int }_{0}^{L}{\left|\psi \left(x,y,z\right)\right|}^{2}dxdydz=1$

$\begin{array}{r}{\int }_{0}^{L} {\int }_{0}^{L} {\int }_{0}^{L} {\left|A\mathrm{sin}\frac{{n}_{x}\pi x}{L}\mathrm{sin}\frac{{n}_{y}\pi y}{L}\mathrm{sin}\frac{{n}_{z}\pi z}{L}\right|}^{2}dxdydz=1\\ {\int }_{0}^{L} {\int }_{0}^{L} {\int }_{0}^{L} {A}^{2}{\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}{\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}{\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dxdydz=1\\ {A}^{2}{\int }_{0}^{L} {\int }_{0}^{L} {\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}{\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}{\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dxdydz=1\\ {A}^{2}{\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}dx{\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}dy{\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dz=1\end{array}$

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,

${\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}dx{\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}dy={\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dz$

Therefore, one segment can be solved as:

${\int }_{0}^{L} {\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}dx=\frac{L}{2}$

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,

${A}^{2}\left(\frac{L}{2}\right)\left(\frac{L}{2}\right)\left(\frac{L}{2}\right)=1\phantom{\rule{0ex}{0ex}}{A}^{2}{\left(\frac{L}{2}\right)}^{3}=1\phantom{\rule{0ex}{0ex}}{A}^{2}={\left(\frac{2}{L}\right)}^{3}\phantom{\rule{0ex}{0ex}}{A}^{2}={\left(\frac{2}{L}\right)}^{3/2}$

## Step 4:  Conclusion

Therefore, the normalization constant is ${\left(\frac{2}{L}\right)}^{3/2}$. ### Want to see more solutions like these? 