Question

Show that the normalization constant${\mathrm{A}}_{\mathrm{n}}$ for the wave functions of a particle in a rigid box has the value given in Equation 40.26.

Short answer

${\mathrm{A}}_{\mathrm{n}}=\sqrt{\frac{2}{\mathrm{L}}}$

Step-by-step solution

Given information.

We have given the one dimensional potential box.

We have to find the normalization constant of the wave function.

Simplify

The wave function of the particle in one dimensional box is given by,

${\mathrm{\psi }}_{\mathrm{n}}={\mathrm{A}}_{\mathrm{n}}\sin \frac{\mathrm{n\pi x}}{\mathrm{L}}$

then suing the normalization condition,

$$\begin{gathered} \int {\psi }^{*}\psi dx=1 \\ \int A_n^2{\sin }^2\frac{n\pi x}{L}=1 \\ \int A_n^2\mathit{(}\mathit{1}\mathit{-}\mathit{cos}\frac{\mathit{2}n\pi x}{L}\mathit{)}=1 \\ {\mathrm{A}}_{\mathrm{n}}^2{\left[\mathrm{x}-\sin \frac{\mathit{2}n\pi x}{L}\left(\frac{\mathrm{L}}{\mathrm{2}\mathrm{n}\mathrm{\pi }}\right)\right]}_0^{\mathrm{L}}=1 \\ {\mathrm{A}}_{\mathrm{n}}=\sqrt{\frac{1}{\mathrm{L}}} \end{gathered}$$