Show that the ground state of hydrogen (Equation 4.80) satisfies the integral form of the Schrödinger equation, for the appropriateV and E(note that Eis negative, so $\mathit{k}\mathbf{=}\mathit{i}\mathit{k}$, where $\mathit{k}\mathbf{=}\sqrt{\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{E}\mathbf{/}\mathbf{h}}$).

The ground state of the Hydrogen atom is given by:

$\psi =\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}{e}^{-r/a}$ ………. (1)

The potential is given by equation4.72as:

$$\begin{gathered} V=\frac{e^2}{4\mathrm{\pi }{\tilde{\mathrm{o}}}_0\mathrm{r}} \\ V=\frac{{\sout{h}}^2}{ma}\frac{1}{r} \end{gathered}$$

and the energy is negative, so we can write as:

$$\begin{gathered} k=\frac{\sqrt{2mE}}{\sout{h}} \\ k=\frac{i\sqrt{-2mE}}{\sout{h}} \\ k=\frac{i}{a} \end{gathered}$$

We need to show that equation (1) satisfies the integral form of the Schrodinger equation, that is:

$-\frac{\mathrm{m}}{2\mathrm{\pi }\sout{{\mathrm{h}}^2}}{\int }_{}^{}\frac{{\mathrm{e}}^{\mathrm{ik}\left|\mathrm{r}-{\mathrm{r}}_0\right|}}{\left|\mathrm{r}-{\mathrm{r}}_0\right|}\mathrm{V}\left({\mathrm{r}}_0\right)\mathrm{\psi }\left({\mathrm{r}}_0\right){\mathrm{d}}^3{\mathrm{r}}_0=\mathrm{\psi }\left(\mathrm{r}\right)$ ……. (2)

We need to do the integral on the LHS the wave function in equation (1), so we get:

localid="1658301584478" $$\begin{gathered} l=\left(-\frac{m}{2\mathrm{\pi }{\sout{\mathrm{h}}}^2}\right)\left(-\frac{{\sout{h}}^2}{ma}\right)\frac{1}{\sqrt{{\mathrm{\pi a}}^2}}\int \frac{e^{-\left|r-r_0\right|/a}}{\left|r-r_0\right|}\frac{1}{r_0}{e}^{-r_0/a}{d}^3{r}_0 \\ l=\frac{1}{2\mathrm{\pi a}}\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}\int \frac{e^{-\left|r-r_0\right|/a}}{\left|r-r_0\right|}{r}_0^2\sin \left(\theta \right)d{r}_{0}d\theta d\varphi \end{gathered}$$

Where we've taken the z axis to be parallel to r, since for the purposes of the integral, is constant. We have,

localid="1658301661656" $\left|r-r_0\right|=\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)}$

So, the integral becomes,

$$\begin{gathered} l=\frac{1}{2\mathrm{\pi a}}\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}\int \frac{e^{-\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)/a}}{e}^{-r_0/a}}{\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)r}}{r}_0^2\sin \left(\theta \right)d{r}_{0}d\theta d\varphi \frac{1}{a\sqrt{{\mathrm{\pi a}}^3}}{\int }_0^{\infty }{r}_0{e}^{-r_0/a} \\ =\left[{\int }_0^{\mathrm{\pi }}\frac{e^{-\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)/a}}}{\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)/a}}\sin \left(\theta \right)d\theta \right]d{r}_0 \end{gathered}$$

But,

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}\frac{e^{-\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)/a}}}{\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)r}}\sin \left(\theta \right)d\theta =-\frac{a}{r{r}_0}{\overline{)e^{-\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\theta \right)/a}}}}_0^{\mathrm{\pi }} \\ =-\frac{a}{r{r}_0}\left[e^{-(r-r_0)/a}-e^{-\left|r-r_0\right|/a}\right] \end{gathered}$$

Thus,

$$\begin{gathered} l=-\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}{\int }_0^{\infty }{e}^{-r_0/a}\left[e^{-(r_0+r)/a}-e^{-\left|r_0-r\right|/a}\right]d{r}_0 \\ l=-\frac{1}{r\sqrt{{\mathrm{\pi a}}^3}}\left[e^{-r/a}{\int }_0^{\infty }{e}^{-2r_0/a}d{r}_0-e^{-r_0/a}{\int }_0^{r}dr-e^{-r/a}{\int }_r^{\infty }{e}^{-2r_0/a}d{r}_0\right] \\ l=-\frac{1}{r\sqrt{{\mathrm{\pi a}}^3}}\left[e^{-r/a}\left(\frac{a}{2}\right)-e^{-r/a}\left(r\right)-e^{r/a}{\overline{)\left(-\frac{a}{2}{e}^{-2r_0/a}\right)}}_r^{\infty }\right] \end{gathered}$$

On further solving,

$$\begin{gathered} l=-\frac{1}{r\sqrt{{\mathrm{\pi a}}^3}}\left[\frac{a}{2}{e}^{-r/a}-r{e}^{-r/a}-\frac{a}{2}{e}^{r/a}{e}^{-2rla}\right] \\ l=\frac{1}{\sqrt{{\mathrm{\pi a}}^3}}{e}^{-r/a} \\ l=\psi \left(r\right) \end{gathered}$$

So,

$-\frac{\mathrm{m}}{2\mathrm{\pi }\sout{{\mathrm{h}}^2}}{\int }_{}^{}\frac{{\mathrm{e}}^{\mathrm{ik}\left|\mathrm{r}-{\mathrm{r}}_0\right|}}{\left|\mathrm{r}-{\mathrm{r}}_0\right|}\mathrm{V}\left({\mathrm{r}}_0\right)\mathrm{\psi }\left({\mathrm{r}}_0\right){\mathrm{d}}^3{\mathrm{r}}_0=\mathrm{\psi }\left(\mathrm{r}\right)$

Hence, it’s proved.