Question

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 $\mathit{k}\mathbf{=}\sqrt{\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{E}\mathbf{/}\mathbf{h}}$).

Short answer

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

Hence, it’s proved.

Step-by-step solution

Given.

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:

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

To show that the ground state of hydrogen satisfies the integral form of the Schrödinger equation. 

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

The integral values.

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.