Mean of powers of r for hydrogen wave functions (Pt. 1) | Quantum Science Philippines

## Mean of powers of r for hydrogen wave functions (Pt. 1)

Problems:

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

2.) Show that $\big \langle r^{2} \big \rangle=\frac{a_{0}^{2}n^{2}}{2}\big[5n^{2}+1-3l(l+1)\big]$

Solutions:

To show that $\big \langle r \big \rangle$ and $\big \langle r^{2} \big \rangle$, we can use Kramer’s relation which is given as $\frac{k+1}{n^{2}}\big \langle r^{k} \big \rangle-\big(2k+1\big)a_{0}\big \langle r^{k-1} \big \rangle+\frac{ka_{0}^{2}}{4}\Big[\big(2l+1\big)^{2}-k^{2}\Big]\big \langle r^{k-2} \big \rangle=0$

where $\big \langle r^{k} \big \rangle=\int_{0}^{\infty}r^{k+2}\big[R_{nl}\big(r\big)\big]^{2}dr$

(1) For $k=1$: $\frac{2}{n^{2}}\big \langle r \big \rangle-\big(2+1\big)a_{0}\big \langle r^{0} \big \rangle+\frac{a_{0}^{2}}{4}\Big[\big(2l+1\big)^{2}-1\Big]\big \langle r^{1-2} \big \rangle=0$ $\frac{2}{n^{2}}\big \langle r \big \rangle-3a_{0}+\frac{a_{0}^{2}}{4}\Big[\big(2l+1\big)^{2}-1\Big]\big \langle r^{-1} \big \rangle=0$

But note that: $\big \langle r^{0} \big \rangle=1$ and $\big \langle r^{-1} \big \rangle=\big \langle \frac{1}{r} \big \rangle=\frac{1}{a_{0}n^{2}}$.

So, $\frac{2}{n^{2}}\big \langle r \big \rangle-\big(2+1\big)a_{0}\big \langle r^{0} \big \rangle+\frac{a_{0}^{2}}{4}\Big[\big(2l+1\big)^{2}-1\Big]\big \langle r^{1-2} \big \rangle=0$ $\frac{2}{n^{2}}\big \langle r \big \rangle-3a_{0}+\frac{a_{0}^{2}}{4}\Big[\big(2l+1\big)^{2}-1\Big]\frac{1}{a_{0}n^{2}}=0$ $\frac{2}{n^{2}}\big \langle r \big \rangle-\big(2+1\big)a_{0}\big \langle r^{0} \big \rangle+\frac{a_{0}^{2}}{4}\Big[\big(2l+1\big)^{2}-1\Big]\big \langle r^{1-2} \big \rangle=0$ $\frac{2}{n^{2}}\big \langle r \big \rangle-3a_{0}+\frac{4a_{0}^{2}}{4}\Big[l^{2}+l\Big]\frac{1}{a_{0}n^{2}}=0$

Therefore, $\big \langle r \big \rangle=\frac{a_{0}}{2}\big[3n^{2}-l\big(l+1\big)\big]$.

(2) For $k=2$: $\frac{3}{n^{2}}\big \langle r^{2} \big \rangle-5a_{0}\big \langle r \big \rangle+\frac{2a_{0}^{2}}{4}\Big[\big(2l+1\big)^{2}-4\Big]\big \langle r^{0} \big \rangle=0$ $\frac{3}{n^{2}}\big \langle r^{2} \big \rangle-\frac{5a_{0}^{2}}{2}\big[3n^{2}-l\big(l+1\big)\big]+\frac{a_{0}^{2}}{2}\Big[4l^{2}-4l-3\Big]=0$

since $\big \langle r \big \rangle=\frac{a_{0}}{2}\big[3n^{2}-l\big(l+1\big)\big].$

Therefore, $\big \langle r^{2} \big \rangle=\frac{a_{0}^{2}n^{2}}{2}\big[5n^{2}+1-3l\big(l+1\big)\big]$.