Question

Suppose we put a delta-function bump in the center of the infinite square well:

${\mathbf{H}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{\alpha \delta }\left(x-a/2\right)$

where$\mathit{a}$is a constant.

(a) Find the first-order correction to the allowed energies. Explain why the energies are not perturbed for even$\mathbf{n}$.

(b) Find the first three nonzero terms in the expansion (Equation 6.13) of the correction to the ground state,${\mathit{\Psi }}_{\mathbf{1}}^{\mathbf{1}}$.

Short answer

The first-order correction to the allowed energies$=\frac{2\alpha }{a}{\sin }^2\frac{\mathrm{n\pi }}{2}$

The the first three nonzero terms in the expansion of the correction to the ground state,${\Psi }_1^1\mathrm{is}=\sqrt{\frac{\mathrm{a}}{2}}\frac{\mathrm{m\alpha }}{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}\left[\sin \left(\frac{3\mathrm{\pi x}}{\mathrm{a}}\right)-\frac{1}{3}\sin \left(\frac{5\mathrm{\pi x}}{\mathrm{a}}\right)+\frac{1}{6}\sin \left(\frac{7\mathrm{\pi x}}{\mathrm{a}}\right)\right]$

Step-by-step solution

Stationary state of a one-dimensional infinite square well.

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

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

Step 2: The first-order correction to the allowed energies.

a)

For the infinite square well:

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

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

Step 3: The first three nonzero terms in the expansion.

b)

Use the formula and substitute each value.

For n=1

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

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

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

The first three non- zero terms (odd)

$$\begin{gathered} \mathrm{m}=3,5,7;\mathrm{n}=1 \\ {\mathrm{E}}_{\mathrm{n}}^0=\frac{{\mathrm{n}}^2{\mathrm{\pi }}^2\mathrm{ħ}}{2{\mathrm{ma}}^2} \\ \Rightarrow {\mathrm{E}}_1^0=\frac{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2{\mathrm{ma}}^2} \\ \Rightarrow {\mathrm{\Psi }}_1^1=\frac{2\mathrm{a}}{\mathrm{a}}\sqrt{\frac{2}{\mathrm{a}}}\left[\frac{\sin \left({\displaystyle \frac{3\mathrm{\pi }}{2}}\right)}{{\displaystyle \frac{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2{\mathrm{ma}}^2}}\left(1-9\right)}\sin \left(\frac{3\mathrm{\pi x}}{\mathrm{a}}\right)+\frac{\sin \left(\frac{5\mathrm{\pi }}{2}\right)\sin \left(\frac{5\mathrm{\pi x}}{\mathrm{a}}\right)}{{\displaystyle \frac{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2{\mathrm{ma}}^2}}\left(1-25\right)}+\frac{\mathrm{sinin}\left(\frac{5\mathrm{\pi }}{2}\right)\sin \left(\frac{5\mathrm{\pi x}}{\mathrm{a}}\right)}{\frac{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2{\mathrm{ma}}^2}\left(1-49\right)}\right] \\ =\frac{2\mathrm{a}}{\mathrm{a}}\sqrt{\frac{2}{\mathrm{a}}}\frac{2{\mathrm{ma}}^2}{{\mathrm{\pi }}^2}\frac{2{\mathrm{ma}}^2}{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}\left[\frac{1}{8}\sin \left(\frac{3\mathrm{\pi x}}{\mathrm{a}}\right)-\frac{1}{24}\sin \left(\frac{5\mathrm{\pi x}}{\mathrm{a}}\right)+\frac{1}{48}\sin \left(\frac{7\mathrm{\pi x}}{\mathrm{a}}\right)\right] \end{gathered}$$

Proceed further and obtain the result as,

$=\sqrt{\frac{a}{2}}\frac{\mathrm{ma}}{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}\left[\sin \left(\frac{3\mathrm{\pi x}}{\mathrm{a}}\right)-\frac{1}{3}\sin \left(\frac{5\mathrm{\pi x}}{\mathrm{a}}\right)+\frac{1}{6}\sin \left(\frac{7\mathrm{\pi x}}{\mathrm{a}}\right)\right]$