91Ó°ÊÓ

The ground state of the Hydrogen atom is given by:

$\mathrm{ψ}=\frac{1}{\sqrt{{\mathrm{Ï€²¹}}^3}}{e}^{-r/a}$ ………. (1)

The potential is given by equation4.72as:

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

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:

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

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

localid="1658301584478" $$\begin{gathered} l=\left(-\frac{m}{2\mathrm{Ï€}{\sout{\mathrm{h}}}^2}\right)\left(-\frac{{\sout{h}}^2}{ma}\right)\frac{1}{\sqrt{{\mathrm{Ï€²¹}}^2}}∫\frac{e^{-\left|r-r_0\right|/a}}{\left|r-r_0\right|}\frac{1}{r_0}{e}^{-r_0/a}{d}^3{r}_0 \\ l=\frac{1}{2\mathrm{Ï€²¹}}\frac{1}{\sqrt{{\mathrm{Ï€²¹}}^3}}∫\frac{e^{-\left|r-r_0\right|/a}}{\left|r-r_0\right|}{r}_0^2\sin \left(\mathrm{θ}\right)d{r}_{0}d\mathrm{θ}d\mathrm{Ï•} \end{gathered}$$

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

So, the integral becomes,

$$\begin{gathered} l=\frac{1}{2\mathrm{Ï€²¹}}\frac{1}{\sqrt{{\mathrm{Ï€²¹}}^3}}∫\frac{e^{-\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\mathrm{θ}\right)/a}}{e}^{-r_0/a}}{\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\mathrm{θ}\right)r}}{r}_0^2\sin \left(\mathrm{θ}\right)d{r}_{0}d\mathrm{θ}d\mathrm{Ï•}\frac{1}{a\sqrt{{\mathrm{Ï€²¹}}^3}}{∫}_0^{∞}{r}_0{e}^{-r_0/a} \\ =\left[{∫}_0^{\mathrm{Ï€}}\frac{e^{-\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\mathrm{θ}\right)/a}}}{\sqrt{r^2+r_0^2-2r{r}_0\cos \left(\mathrm{θ}\right)/a}}\sin \left(\mathrm{θ}\right)d\mathrm{θ}\right]d{r}_0 \end{gathered}$$

But,

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

Thus,

$$\begin{gathered} l=-\frac{1}{\sqrt{{\mathrm{Ï€²¹}}^3}}{∫}_0^{∞}{e}^{-r_0/a}\left[e^{-(r_0+r)/a}-e^{-\left|r_0-r\right|/a}\right]d{r}_0 \\ l=-\frac{1}{r\sqrt{{\mathrm{Ï€²¹}}^3}}\left[e^{-r/a}{∫}_0^{∞}{e}^{-2r_0/a}d{r}_0-e^{-r_0/a}{∫}_0^{r}dr-e^{-r/a}{∫}_r^{∞}{e}^{-2r_0/a}d{r}_0\right] \\ l=-\frac{1}{r\sqrt{{\mathrm{Ï€²¹}}^3}}\left[e^{-r/a}\left(\frac{a}{2}\right)-e^{-r/a}\left(r\right)-e^{r/a}{\overline{)\left(-\frac{a}{2}{e}^{-2r_0/a}\right)}}_r^{∞}\right] \end{gathered}$$

On further solving,

$$\begin{gathered} l=-\frac{1}{r\sqrt{{\mathrm{Ï€²¹}}^3}}\left[\frac{a}{2}{e}^{-r/a}-r{e}^{-r/a}-\frac{a}{2}{e}^{r/a}{e}^{-2rla}\right] \\ l=\frac{1}{\sqrt{{\mathrm{Ï€²¹}}^3}}{e}^{-r/a} \\ l=\mathrm{ψ}\left(r\right) \end{gathered}$$

So,

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

Hence, it’s proved.