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Q18E

Expert-verifiedFound in: Page 279

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**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 ${\mathit{A}}{\mathbf{=}}{{\left(2/L\right)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}$****.**

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

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

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

$\begin{array}{r}{\int}_{0}^{L}\u200a{\int}_{0}^{L}\u200a{\int}_{0}^{L}\u200a{\left|A\mathrm{sin}\frac{{n}_{x}\pi x}{L}\mathrm{sin}\frac{{n}_{y}\pi y}{L}\mathrm{sin}\frac{{n}_{z}\pi z}{L}\right|}^{2}dxdydz=1\\ {\int}_{0}^{L}\u200a{\int}_{0}^{L}\u200a{\int}_{0}^{L}\u200a{A}^{2}{\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}{\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}{\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dxdydz=1\\ {A}^{2}{\int}_{0}^{L}\u200a{\int}_{0}^{L}\u200a{\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}{\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}{\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dxdydz=1\\ {A}^{2}{\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}dx{\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}dy{\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dz=1\end{array}$

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,

${\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}dx{\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{y}\pi y}{L}dy={\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{z}\pi z}{L}dz$

Therefore, one segment can be solved as:

${\int}_{0}^{L}\u200a{\mathrm{sin}}^{2}\frac{{n}_{x}\pi x}{L}dx=\frac{L}{2}$

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,

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

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

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