Mean of powers of r for hydrogen wave functions (Pt. 2) | Quantum Science Philippines

## Mean of powers of r for hydrogen wave functions (Pt. 2)

Problems:

1.) Show that $\big \langle \frac{1}{r} \big \rangle=\frac{1}{a_{0}}\frac{1}{n^{2}}$

2.) Show that $\big \langle \frac{1}{r^{2}} \big \rangle=\frac{1}{a_{0}^{2}\big(l+\frac{1}{2}\big)}\frac{1}{n^{3}}$

Solutions:

(1) Given that $\big \langle r^{k} \big \rangle=\int_{0}^{\infty}r^{k+2}\big[R_{nl}\big(r\big)\big]^{2}dr$, $\big \langle \frac{1}{r} \big \rangle=\big \langle r^{-1} \big \rangle=\int_{0}^{\infty}r\big[R_{nl}\big(r\big)\big]^{2}dr$.

But note that $R_{nl}\big(r\big)^{2}=\frac{u^{2}\big(r\big)}{r^{2}}$,

so $\big \langle \frac{1}{r} \big \rangle=\int_{0}^{\infty}\frac{u^{2}\big(r\big)}{r}dr$. $\frac{-\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}u_{nl}\big(r\big)+\Bigg[\frac{l\big(l+1\big)\hbar^{2}}{2\mu r^{2}}-\frac{e^{2}}{r}\Bigg]u_{nl}\big(r\big)=E_{n}u_{nl}\big(r\big)$.

But $E_{n}=-\frac{\mu e^{4}}{2\hbar^{2}n^{2}}$. Substitution and rearranging gives: $\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}=\frac{l\big(l+1\big)}{r^{2}}-\frac{2\mu e^{2}}{\hbar^{2}}\frac{1}{r}+\frac{\mu^{2}e^{4}}{\hbar^{4}n^{2}}$.

To find $\big \langle \frac{1}{r} \big \rangle$, treat the electron charges as continuous variable $\frac{\partial}{\partial e}\Bigg[\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}\Bigg]=-\frac{4\mu e}{\hbar^{2}}\frac{1}{r}+\frac{4\mu^{2}e^{3}}{\hbar^{4}n^{2}}$

Multiplying both sides by $u_{nl}^{2}(r)$ and integrating over r, $\int_{0}^{\infty}u_{nl}^{2}(r)\frac{\partial}{\partial e}\Bigg[\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}\Bigg]dr=-\frac{4\mu e}{\hbar^{2}}\int_{0}^{\infty}\frac{u_{nl}^{2}(r)}{r}dr+\frac{4\mu^{2}e^{3}}{\hbar^{4}n^{2}}\int_{0}^{\infty}u_{nl}^{2}(r)dr$

Note that: $\int_{0}^{\infty}u_{nl}^{2}(r)\frac{\partial}{\partial e}\Bigg[\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}\Bigg]dr=\int_{0}^{\infty}u_{nl}(r)\frac{\partial u_{nl}''(r)}{\partial e}dr-\int_{0}^{\infty}u_{nl}''(r)\frac{\partial u_{nl}(r)}{\partial e}dr=0$

and $\int_{0}^{\infty}\frac{u_{nl}^{2}(r)}{r}dr=\big \langle \frac{1}{r} \big \rangle$

So, $\frac{4\mu e}{\hbar^{2}}\big \langle \frac{1}{r} \big \rangle =\frac{4\mu^{2}e^{3}}{\hbar^{4}n^{2}}$ $\big \langle \frac{1}{r} \big \rangle=\frac{\mu e^{2}}{\hbar^{2}n^{2}}$.

Recall that the Bohr radius, $a_{0}=\frac{\hbar_{2}}{\mu e^{2}}$,

hence, $\big \langle \frac{1}{r} \big \rangle$ can be written as: $\big \langle \frac{1}{r} \big \rangle=\frac{1}{a_{0}}\frac{1}{n^{2}}$.

(2) Similarly, $\big \langle \frac{1}{r^{2}} \big \rangle=\big \langle r^{-2} \big \rangle=\int_{0}^{\infty}\big[R_{nl}\big(r\big)\big]^{2}dr.$

But note that $R_{nl}(r)^{2}=\frac{u_{nl}^{2}\big(r\big)}{r^{2}}$,

so $\big \langle \frac{1}{r^{2}} \big \rangle=\int_{0}^{\infty}\frac{u_{nl}^{2}\big(r\big)}{r^{2}}dr.$ $\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}=\frac{l\big(l+1\big)}{r^{2}}-\frac{2\mu e^{2}}{\hbar^{2}}\frac{1}{r}+\frac{\mu^{2}e^{4}}{\hbar^{4}n^{2}}.$

To find $\big \langle r^{-2} \big \rangle$, treat the orbital quantum number l as continuous variable.
So, $\frac{\partial}{\partial l}\Bigg[\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}\Bigg]=\frac{2l +1}{r^{2}}-0+-\frac{2\mu^{2}e^{4}}{\hbar^{4}n^{3}}$

Multiplying both sides by $u_{nl}^{2}(r)$ and integrating over r, $\int_{0}^{\infty}u_{nl}^{2}(r)\frac{\partial}{\partial l}\Bigg[\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}\Bigg]dr=(2l+1)\int_{0}^{\infty}u_{nl}^{2}(r)\frac{1}{r^{2}}dr-\frac{2\mu^{2}e^{4}}{\hbar^{4}n^{3}}\int_{0}^{\infty}u_{nl}^{2}(r)dr$

Note that: $\int_{0}^{\infty}u_{nl}^{2}(r)\frac{\partial}{\partial l}\Bigg[\frac{u_{nl}''\big(r\big)}{u_{nl}\big(r\big)}\Bigg]dr=\int_{0}^{\infty}u_{nl}(r)\frac{\partial u_{nl}''(r)}{\partial l}dr-\int_{0}^{\infty}u_{nl}''(r)\frac{\partial u_{nl}(r)}{\partial l}dr=0$

and $\int_{0}^{\infty}u_{nl}^{2}(r)dr=1$

Hence, $(2l+1)\big \langle \frac{1}{r^{2}} \big \rangle = \frac{2\mu^{2}e^{4}}{\hbar^{4}n^{3}}$ $\big \langle \frac{1}{r^{2}} \big \rangle =\frac{2\mu^{2}e^{4}}{\hbar^{4}n^{3}} \bigg(\frac{1}{2l+1}\bigg)$ $\big \langle \frac{1}{r^{2}} \big \rangle =\frac{1}{a_{0}^{2}\bigg(l+\frac{1}{2}\bigg)}\frac{1}{n^{3}}$