91Ó°ÊÓ

Symmetric and anti-symmetric state.

The symmetric function is

${\mathbf{ψ}}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{Ï€³\ae }}_1}{L}\sin \frac{2{\mathrm{Ï€³\ae }}_2}{L}+\sin \frac{2{\mathrm{Ï€³\ae }}_1}{L}\sin \frac{{\mathrm{Ï€³\ae }}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{…}\mathbf{…}\mathbf{…}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

role="math" localid="1659952474415" ${\mathbf{ψ}}_{\mathbf{s}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\mathbf{(}\mathbf{sin}\frac{{\mathbf{Ï€³\ae }}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{\mathbf{2}{\mathbf{Ï€³\ae }}_{\mathbf{2}}}{\mathbf{L}}\mathbf{−}\mathbf{sin}\frac{\mathbf{2}{\mathbf{Ï€³\ae }}_{\mathbf{1}}}{\mathbf{L}}\mathbf{sin}\frac{{\mathbf{Ï€³\ae }}_{\mathbf{2}}}{\mathbf{L}}\mathbf{}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{…}\mathbf{…}\mathbf{…}\mathbf{..}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

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

$\mathbf{ψ}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\frac{\sqrt{\mathbf{2}}}{\mathbf{L}}\left(\sin \frac{{\mathrm{Ï€³\ae }}_1}{L}\sin \frac{2{\mathrm{Ï€³\ae }}_2}{L}Â\pm \sin \frac{2{\mathrm{Ï€³\ae }}_1}{L}\sin \frac{{\mathrm{Ï€³\ae }}_2}{L}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{…}\mathbf{…}\mathbf{…}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

To verify the normalization condition, evaluate

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

$=\frac{2}{{\mathrm{L}}^2}{∫}_0^{\mathrm{L}}{∫}_0^{\mathrm{L}}\left(\begin{array}{l}{\sin }^2\frac{4{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}{\sin }^2\frac{3{\mathrm{Ï€³\ae }}_2}{\mathrm{L}}+{\sin }^2\frac{3{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}{\sin }^2\frac{4{\mathrm{Ï€³\ae }}_2}{\mathrm{L}}\\ Â\pm 2\sin \frac{4{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}\sin \frac{3{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}\sin \frac{4{\mathrm{Ï€³\ae }}_2}{\mathrm{L}}\sin \frac{3{\mathrm{Ï€³\ae }}_2}{\mathrm{L}}\end{array}\right){\mathrm{dx}}_1{\mathrm{dx}}_2$

$=\left[\begin{array}{l}\frac{2}{{\mathrm{L}}^2}\left({∫}_0^{\mathrm{L}}{\sin }^2\frac{4{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)\left({∫}_0^{\mathrm{L}}{\sin }^2\frac{3{\mathrm{Ï€³\ae }}_2}{\mathrm{L}}{\mathrm{dx}}_2\right)\\ +\frac{2}{{\mathrm{L}}^2}\left({∫}_0^{\mathrm{L}}{\sin }^2\frac{3{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)\left({∫}_0^{\mathrm{L}}{\sin }^2\frac{4{\mathrm{Ï€³\ae }}_2}{\mathrm{L}}{\mathrm{dx}}_2\right)\\ Â\pm 2\frac{2}{{\mathrm{L}}^2}{\left({∫}_0^{\mathrm{L}}\sin \frac{4{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}\sin \frac{3{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)}^2\end{array}\right].$

$∫{\sin }^2\mathrm{axdx}=\frac{1}{2}\mathrm{x}−\frac{1}{4\mathrm{a}}\sin 2\mathrm{ax}{∫}_0^{\mathrm{L}}{\sin }^2\frac{\mathrm{²ÔÏ€³\ae }}{\mathrm{L}}\mathrm{dx}={\left[\frac{1}{2}\mathrm{x}−\frac{\mathrm{L}}{4\mathrm{²ÔÏ€}}\sin \frac{2\mathrm{²ÔÏ€³\ae }}{\mathrm{L}}\right]}_0^{\mathrm{L}}=\frac{\mathrm{L}}{2}$

The other integral Equation (4) is

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

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

${∫}_0^{\mathrm{L}}\sin \frac{4\mathrm{Ï€³\ae }}{\mathrm{L}}\frac{3\mathrm{Ï€³\ae }}{\mathrm{L}}\mathrm{dx}=\frac{1}{2}{\left[\frac{\mathrm{L}}{\mathrm{Ï€}}\sin \frac{\mathrm{Ï€³\ae }}{\mathrm{L}}−\frac{\mathrm{L}}{7\mathrm{Ï€}}\sin \frac{7\mathrm{²ÔÏ€³\ae }}{\mathrm{L}}\right]}_{00}^{\mathrm{L}}$

Therefore equation (4) becomes,

$\mathrm{N}=\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2+\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2Â\pm 2\left({∫}_0^{\mathrm{L}}\sin \frac{4{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}\sin \frac{3{\mathrm{Ï€³\ae }}_1}{\mathrm{L}}{\mathrm{dx}}_1\right)=\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2+\frac{2}{{\mathrm{L}}^2}{\left(\frac{\mathrm{L}}{2}\right)}^2Â\pm 0=1$