Chapter 12, Problem 12.44E

Show that a variation theory treatment of $\text{H}$ using $\varphi =e^{-kr}$ as an unnormalized trial function yields the correct minimum-energy solution for the hydrogen atom when the specific expression for $k$ is determined.

Interpretation Introduction

Interpretation:

The variation theory treatment of $\text{H}$ using $\varphi =e^{-kr}$ as an unnormalized trial function yields the correct minimum-energy solution for the hydrogen atom is to be shown.

Concept introduction:

According to the variation theory, the lower the energy of a system the better is the approximation. Variation theory is based on the fact that any test system has average energy equal to or greater than the ground state energy of that system. The advantage of the variation theory is that any wavefunction can be taken for any test system.

Answer to Problem 12.44E

The variation theory treatment of $\text{H}$ using $\varphi =e^{-kr}$ as an unnormalized trial function yield the correct minimum-energy solution for the hydrogen atom has been shown.

Explanation of Solution

The expression for trial function can be written as given below.

$\langle E_{\text{trial}}\rangle =\frac{{\displaystyle \int {\varphi }^{*}\widehat{H}\varphi d\tau }}{{\displaystyle \int {\varphi }^{*}\varphi d\tau }}$

Substitute the values in the above expression as follows.

$\begin{array}{rll}\langle E_{\text{trial}}\rangle & =& \frac{{\displaystyle \int {\varphi }^{*}\widehat{H}\varphi d\tau }}{{\displaystyle \int {\varphi }^{*}\varphi d\tau }}\\ & =& \frac{{\displaystyle \int e^{-kr}\left(-\frac{{\hslash }^2}{2\mu }\left(\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)\right)-\frac{e^2}{4\pi {\epsilon }_{0}r}\right)e^{-2kr}d\tau }}{{\displaystyle \int e^{-2kr}d\tau }}\\ & =& \frac{-\frac{{\hslash }^2}{2\mu }{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-kr}\left(\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial e^{-kr}}{\partial r}\right)\right)dr}-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{e^2}{4\pi {\epsilon }_{0}r}{e}^{-2kr}dr}}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}dr}}\\ & =& \frac{-\frac{{\hslash }^2}{2\mu }{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-kr}\left(\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\left(-k\right)e^{-kr}\right)\right)dr}-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{e^2}{4\pi {\epsilon }_{0}r}{e}^{-2kr}dr}}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}dr}}\end{array}$

The above equation can be simplified as given below.

$\begin{array}{l}=\frac{-\frac{{\hslash }^2}{2\mu }{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-kr}\left(\frac{-k}{r^2}\left(2r{e}^{-kr}-k{r}^2{e}^{-kr}\right)\right)dr}-{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{e^2}{4\pi {\epsilon }_{0}r}{e}^{-2kr}dr}}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}dr}}\\ =\frac{\frac{{\hslash }^{2}k}{2\mu }{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}\left(\frac{2}{r}-k\right)dr}-\frac{e^2}{4\pi {\epsilon }_0}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{e^{-2kr}}{r}dr}}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}dr}}\\ =\frac{\left(\frac{{\hslash }^{2}k}{\mu }-\frac{e^2}{4\pi {\epsilon }_0}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}\left(\frac{1}{r}\right)dr}-\frac{{\hslash }^2{k}^2}{2\mu }{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}dr}}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}dr}}\\ =\frac{\left(\frac{{\hslash }^{2}k}{\mu }-\frac{e^2}{4\pi {\epsilon }_0}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}\left(\frac{1}{r}\right)dr}}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}dr}}-\frac{{\hslash }^2{k}^2}{2\mu }\end{array}$

Solve the integral as follows.

$\begin{array}{rll}& =& \frac{\left(\frac{{\hslash }^{2}k}{\mu }-\frac{e^2}{4\pi {\epsilon }_0}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}\left(\frac{1}{r}\right)4\pi r^{2}dr}}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}4\pi r^{2}dr}}-\frac{{\hslash }^2{k}^2}{2\mu }\\ & =& \frac{\left(\frac{{\hslash }^{2}k}{\mu }-\frac{e^2}{4\pi {\epsilon }_0}\right)4\pi {\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}rdr}}{4\pi {\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2kr}{r}^{2}dr}}-\frac{{\hslash }^2{k}^2}{2\mu }\\ & =& \frac{\left(\frac{{\hslash }^{2}k}{\mu }-\frac{e^2}{4\pi {\epsilon }_0}\right)\frac{1}{{\left(2k\right)}^2}}{\frac{2}{{\left(2k\right)}^3}}-\frac{{\hslash }^2{k}^2}{2\mu }\\ & =& \left(\frac{{\hslash }^{2}k}{\mu }-\frac{e^2}{4\pi {\epsilon }_0}\right)\frac{2k}{2}-\frac{{\hslash }^2{k}^2}{2\mu }\end{array}$

$\begin{array}{rll}& =& \left(\frac{{\hslash }^2{k}^2}{\mu }-\frac{e^{2}k}{4\pi {\epsilon }_0}\right)-\frac{{\hslash }^2{k}^2}{2\mu }\\ & =& \frac{{\hslash }^2{k}^2}{2\mu }-\frac{e^{2}k}{4\pi {\epsilon }_0}\end{array}$

The value of $k$ can be calculated by differentiating the above expression with respect to $k$ as follows.

$\begin{array}{rll}\langle E_{\text{trial}}\rangle & =& \frac{{\hslash }^2{k}^2}{2\mu }-\frac{e^{2}k}{4\pi {\epsilon }_0}\\ \frac{\partial \langle E_{\text{trial}}\rangle }{\partial k}& =& \frac{2{\hslash }^{2}k}{2\mu }-\frac{e^2}{4\pi {\epsilon }_0}& =& 0\\ k& =& \frac{e^2}{4\pi {\epsilon }_0}\times \frac{4{\pi }^2\mu }{h^2}\\ & =& \frac{\pi \mu e^2}{h^2{\epsilon }_0}\end{array}$

Substitute this value in the expression for trial energy as given below.

$\begin{array}{rll}\langle E_{\text{trial}}\rangle & =& \frac{{\hslash }^2{k}^2}{2\mu }-\frac{e^{2}k}{4\pi {\epsilon }_0}\\ & =& \frac{h^2}{8\mu }{\left(\frac{\pi \mu e^2}{h^2{\epsilon }_0}\right)}^2-\frac{e^2}{4\pi {\epsilon }_0}\left(\frac{\pi \mu e^2}{h^2{\epsilon }_0}\right)\\ & =& \frac{e^4\mu }{8h^2{\epsilon }_0^2}-\frac{e^4\mu }{4h^2{\epsilon }_0^2}\\ & =& -\frac{e^4\mu }{8h^2{\epsilon }_0^2}\end{array}$

The exact energy of hydrogen atom in ground state is given by the expression written below.

$\begin{array}{rll}\langle E\rangle & =& -\frac{e^4\mu }{8h^2{\epsilon }_0^2{n}^2}\\ & =& -\frac{e^4\mu }{8h^2{\epsilon }_0^2}\end{array}$

From the above equation it is clear that the exact energy of hydrogen atom in ground state is same as the minimized trial energy.

Conclusion

The variation theory treatment of $\text{H}$ using $\varphi =e^{-kr}$ as an unnormalized trial function yields the correct minimum-energy solution for the hydrogen atom has been shown.