Verify that the normalization constant given in Example 8.2is correct for both symmetric and antisymmetric states and is independent of$\mathbf{n}$and ${\mathbf{n}}^{\mathbf{\text{'}}}$?

Symmetric and anti-symmetric state.

The symmetric function is

${\mathbf{\psi }}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{\pi x}}_1}{L}\sin \frac{2{\mathrm{\pi x}}_2}{L}+\sin \frac{2{\mathrm{\pi x}}_1}{L}\sin \frac{{\mathrm{\pi x}}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

role="math" localid="1659952474415" ${\mathbf{\psi }}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\mathbf{(}\mathbf{sin}\frac{{\mathbf{\pi x}}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{\mathbf{2}{\mathbf{\pi x}}_{\mathbf{2}}}{\mathbf{L}}\mathbf{-}\mathbf{sin}\frac{\mathbf{2}{\mathbf{\pi x}}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{{\mathbf{\pi x}}_{\mathbf{2}}}{\mathbf{L}}\mathbf{}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{..}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Equation (1) and (2) can be written as

$\mathbf{\psi }\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{\pi x}}_1}{L}\sin \frac{2{\mathrm{\pi x}}_2}{L}\pm \sin \frac{2{\mathrm{\pi x}}_1}{L}\sin \frac{{\mathrm{\pi x}}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

To verify the normalization condition, evaluate

$\mathrm{N}={\int }_0^{\mathrm{L}}{\int }_0^{\mathrm{L}}{\left|\mathrm{\psi }({\mathrm{x}}_1,{\mathrm{x}}_2)\right|}^2{\mathrm{dx}}_1{\mathrm{dx}}_2$

$=\frac{2}{{\mathrm{L}}^2}{\int }_0^{\mathrm{L}}{\int }_0^{\mathrm{L}}\left(\begin{array}{l}{\sin }^2\frac{4{\mathrm{\pi x}}_1}{\mathrm{L}}{\sin }^2\frac{3{\mathrm{\pi x}}_2}{\mathrm{L}}+{\sin }^2\frac{3{\mathrm{\pi x}}_1}{\mathrm{L}}{\sin }^2\frac{4{\mathrm{\pi x}}_2}{\mathrm{L}}\\ \pm 2\sin \frac{4{\mathrm{\pi x}}_1}{\mathrm{L}}\sin \frac{3{\mathrm{\pi x}}_1}{\mathrm{L}}\sin \frac{4{\mathrm{\pi x}}_2}{\mathrm{L}}\sin \frac{3{\mathrm{\pi x}}_2}{\mathrm{L}}\end{array}\right){\mathrm{dx}}_1{\mathrm{dx}}_2$

$=\left[\begin{array}{l}\frac{2}{{\mathrm{L}}^2}\left({\int }_0^{\mathrm{L}}{\sin }^2\frac{4{\mathrm{\pi x}}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)\left({\int }_0^{\mathrm{L}}{\sin }^2\frac{3{\mathrm{\pi x}}_2}{\mathrm{L}}{\mathrm{dx}}_2\right)\\ +\frac{2}{{\mathrm{L}}^2}\left({\int }_0^{\mathrm{L}}{\sin }^2\frac{3{\mathrm{\pi x}}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)\left({\int }_0^{\mathrm{L}}{\sin }^2\frac{4{\mathrm{\pi x}}_2}{\mathrm{L}}{\mathrm{dx}}_2\right)\\ \pm 2\frac{2}{{\mathrm{L}}^2}{\left({\int }_0^{\mathrm{L}}\sin \frac{4{\mathrm{\pi x}}_1}{\mathrm{L}}\sin \frac{3{\mathrm{\pi x}}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)}^2\end{array}\right].$

$\int {\sin }^2\mathrm{axdx}=\frac{1}{2}\mathrm{x}-\frac{1}{4\mathrm{a}}\sin 2\mathrm{ax}{\int }_0^{\mathrm{L}}{\sin }^2\frac{\mathrm{n\pi x}}{\mathrm{L}}\mathrm{dx}={\left[\frac{1}{2}\mathrm{x}-\frac{\mathrm{L}}{4\mathrm{n\pi }}\sin \frac{2\mathrm{n\pi x}}{\mathrm{L}}\right]}_0^{\mathrm{L}}=\frac{\mathrm{L}}{2}$

The other integral Equation (4) is

$\int \mathrm{sinax}\mathrm{sinbxdx}=\int \left(\frac{1}{2}\cos (\mathrm{a}-\mathrm{b})\mathrm{x}-\frac{1}{2}\cos (\mathrm{a}+\mathrm{b})\mathrm{x}\right)\mathrm{dx}=\frac{\sin (\mathrm{a}-\mathrm{b})\mathrm{x}}{2(\mathrm{a}-\mathrm{b})}-\frac{\sin (\mathrm{a}+\mathrm{b})\mathrm{x}}{2(\mathrm{a}+\mathrm{b})}$

For $\mathrm{a}=\frac{4\mathrm{\pi }}{\mathrm{L}}$and $\mathrm{b}=\frac{3\mathrm{\pi }}{\mathrm{L}}$this is equal to

${\int }_0^{\mathrm{L}}\sin \frac{4\mathrm{\pi x}}{\mathrm{L}}\frac{3\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx}=\frac{1}{2}{\left[\frac{\mathrm{L}}{\mathrm{\pi }}\sin \frac{\mathrm{\pi x}}{\mathrm{L}}-\frac{\mathrm{L}}{7\mathrm{\pi }}\sin \frac{7\mathrm{n\pi x}}{\mathrm{L}}\right]}_{00}^{\mathrm{L}}$

Therefore equation (4) becomes,

$\mathrm{N}=\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2+\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2\pm 2\left({\int }_0^{\mathrm{L}}\sin \frac{4{\mathrm{\pi x}}_1}{\mathrm{L}}\sin \frac{3{\mathrm{\pi x}}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)=\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2+\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2\pm 0=1$