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

Now, recall the radial equation

\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.

Now recall the radial equation

\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}}

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