91Ó°ÊÓ

The stationary state of a one-dimensional infinite square wellis:

${\mathbf{Ψ}}_{\mathbf{n}}^{\mathbf{0}}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{a}}}\mathbf{sin}\left(\frac{\mathrm{²ÔÏ€}}{a}x\right)$

a)

For the infinite square well:

$\hat{\mathrm{H}}\text{'}=\mathrm{Î\pm }\mathrm{δ}\left(x-\frac{\mathrm{a}}{2}\right),\mathrm{Î\pm }=\mathrm{con}\mathrm{st}$

Solve the problem by considering the stationary state of a one-dimensional infinite square well, that is:

b)

Use the formula and substitute each value.

For n=1

Note that:$\mathrm{m}≠1,\left(\mathrm{n}=1,\mathrm{m}≠\mathrm{n}\right)$

for$\mathrm{m}=0⇒\sin \left(0\right)=0$

for$\mathrm{m}=0⇒\sin \left(\mathrm{³¾Ï€}\right)=0$

The first three non- zero terms (odd)

$$\begin{gathered} \mathrm{m}=3,5,7;\mathrm{n}=1 \\ {\mathrm{E}}_{\mathrm{n}}^0=\frac{{\mathrm{n}}^2{\mathrm{Ï€}}^2\mathrm{Ä\S }}{2{\mathrm{ma}}^2} \\ ⇒{\mathrm{E}}_1^0=\frac{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}{2{\mathrm{ma}}^2} \\ ⇒{\mathrm{Ψ}}_1^1=\frac{2\mathrm{a}}{\mathrm{a}}\sqrt{\frac{2}{\mathrm{a}}}\left[\frac{\sin \left({\displaystyle \frac{3\mathrm{Ï€}}{2}}\right)}{{\displaystyle \frac{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}{2{\mathrm{ma}}^2}}\left(1-9\right)}\sin \left(\frac{3\mathrm{Ï€³\ae }}{\mathrm{a}}\right)+\frac{\sin \left(\frac{5\mathrm{Ï€}}{2}\right)\sin \left(\frac{5\mathrm{Ï€³\ae }}{\mathrm{a}}\right)}{{\displaystyle \frac{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}{2{\mathrm{ma}}^2}}\left(1-25\right)}+\frac{\mathrm{sinin}\left(\frac{5\mathrm{Ï€}}{2}\right)\sin \left(\frac{5\mathrm{Ï€³\ae }}{\mathrm{a}}\right)}{\frac{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}{2{\mathrm{ma}}^2}\left(1-49\right)}\right] \\ =\frac{2\mathrm{a}}{\mathrm{a}}\sqrt{\frac{2}{\mathrm{a}}}\frac{2{\mathrm{ma}}^2}{{\mathrm{Ï€}}^2}\frac{2{\mathrm{ma}}^2}{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}\left[\frac{1}{8}\sin \left(\frac{3\mathrm{Ï€³\ae }}{\mathrm{a}}\right)-\frac{1}{24}\sin \left(\frac{5\mathrm{Ï€³\ae }}{\mathrm{a}}\right)+\frac{1}{48}\sin \left(\frac{7\mathrm{Ï€³\ae }}{\mathrm{a}}\right)\right] \end{gathered}$$

Proceed further and obtain the result as,

$=\sqrt{\frac{a}{2}}\frac{\mathrm{ma}}{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}\left[\sin \left(\frac{3\mathrm{Ï€³\ae }}{\mathrm{a}}\right)-\frac{1}{3}\sin \left(\frac{5\mathrm{Ï€³\ae }}{\mathrm{a}}\right)+\frac{1}{6}\sin \left(\frac{7\mathrm{Ï€³\ae }}{\mathrm{a}}\right)\right]$