Prove Kramers' relation:

$\frac{\mathbf{s}}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{⟨}{\mathit{r}}^{\mathbf{s}}\mathbf{⟩}\mathbf{-}\mathbf{(}\mathbf{2}\mathit{s}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{a}\mathbf{⟨}{\mathit{r}}^{\mathbf{s}\mathbf{-}\mathbf{1}}\mathbf{⟩}\mathbf{+}\frac{\mathbf{s}}{\mathbf{4}}\mathbf{\left[}\mathbf{\right(}\mathbf{2}\mathit{l}\mathbf{+}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{-}{\mathit{s}}^{\mathbf{2}}\mathbf{]}{\mathit{a}}^{\mathbf{2}}\mathbf{⟨}{\mathit{r}}^{\mathbf{s}\mathbf{-}\mathbf{2}}\mathbf{⟩}\mathbf{=}\mathbf{0}$

Which relates the expectation values of rto three different powers $\left(s,s-1,\mathrm{and}s-2\right)$,for an electron in the state ${\mathit{\psi }}_{\mathbf{n}\mathbf{/}\mathbf{m}}$of hydrogen. Hint: Rewrite the radial equation (Equation) in the form

${\mathit{u}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\left[\frac{l(l+1)}{r^2}-\frac{2}{ar}+\frac{1}{n^2{a}^2}\right]\mathit{u}$

And use it to expressrole="math" localid="1658192415441" $\mathbf{\int }\mathbf{\left(}\mathbf{u}{\mathbf{r}}^{\mathbf{s}}{\mathbf{u}}^{\mathbf{\text{'}\text{'}}}\mathbf{\right)}\mathit{d}\mathit{r}$in terms of $\left(r^s\right)\mathbf{,}\left(r^{s-1}\right)\mathbf{and}\mathbf{}\left(r^{s-2}\right)$. Then use integration by parts to reduce the second derivative. Show that $\mathbf{\int }\mathbf{\left(}{\mathbf{ur}}^{\mathbf{s}}{\mathbf{u}}^{\mathbf{\text{'}\text{'}}}\mathbf{\right)}\mathit{d}\mathit{r}\mathbf{=}\mathbf{-}\left(s/2\right)\left(r^{s-1}\right)\mathbf{}\mathbf{and}\mathbf{}\mathbf{\int }\mathbf{\left(}{\mathbf{u}}^{\mathbf{\text{'}}}{\mathbf{r}}^{\mathbf{s}}{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{\right)}\mathbf{dr}\mathbf{=}\mathbf{-}\left[2/\left(s+1\right)\right]\mathbf{\int }\mathbf{\left(}{\mathbf{u}}^{\mathbf{\text{'}\text{'}}}{\mathbf{r}}^{\mathbf{s}\mathbf{+}\mathbf{1}}{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{\right)}\mathbf{dr}$. Take it from there.

The Kramer’s' relation is a relationship between the anticipated values of "nearby" powers of r for the hydrogen atom, named after the Dutch scientist Hans Kramers:

$\frac{s+1}{n^2}⟨r^s⟩-(2s+1)a{r}^{s-1}+\frac{a^{2}s}{4}(2l+1{)}^2-s^2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{r}^{s-2}=0$

Need to establish a relation:

$\frac{s+1}{n^2}⟨r^s⟩-(2s+1)a{r}^{s-1}+\frac{a^{2}s}{4}(2l+1{)}^2-s^2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{r}^{s-2}=0$

Which is equal to $a=\frac{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{ħ}}^2}{m{e}^2}$. Radial part of Schrodinger equation:

$$\begin{gathered} -u^{\text{'}\text{'}}+-\frac{e^2}{4\pi {\epsilon }_0}\frac{2m}{{\hslash }^2}\frac{1}{r}+\frac{l(l+1)}{r^2\hspace{0.33em}}\hspace{0.33em}\hspace{0.33em}u=E\cdot \frac{2m}{{\hslash }^{2}u}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}E=-\frac{m}{2{\hslash }^2}\hspace{0.33em}\hspace{0.33em}\frac{e^2}{4\pi {\epsilon }_0}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{1}{n^2} \\ -u^{\text{'}\text{'}}+-\frac{2}{ar}+\frac{l(l+1)}{r^2\hspace{0.33em}}u=-\frac{1}{a^2{n}^2}u \\ {u}^{\text{'}\text{'}}=\frac{2}{ar}+\frac{l(l+1)}{r^2\hspace{0.33em}}+\frac{1}{a^2{n}^2}u \end{gathered}$$

Calculate the following integral using the given hint:

$$\begin{gathered} \left(u{r}^s{u}^{\text{'}\text{'}}\right)dr=u{r}^s-\frac{2}{ar}+\frac{l(l+1)}{r^2}+\frac{1}{a^2{n}^2}\hspace{0.33em}\hspace{0.33em}udr \\ =-\frac{2}{a}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^{s-1}udr+l(l+1)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^{s-2}udr+\frac{1}{a^2{n}^2}\hspace{0.33em}\hspace{0.33em}u{r}^{s}udr \\ =-\frac{2}{a}{r}^{s-1}+l(l+1)r^{s-2}+\frac{1}{a^2{n}^2}⟨r^s⟩ \end{gathered}$$

But I'm still stumped on the left-hand side of the equation. Use partial integration in this case:

$$\begin{gathered} \left(u{r}^s{u}^{\text{'}\text{'}}\right)dr=u{r}^s{u}^{\text{'}}{|}_0^{\infty }-\frac{d}{dr}\left(u{r}^s\right)u^{\text{'}}dr \\ =-\hspace{0.33em}(u^{\text{'}}{r}^s+su{r}^{s-1}{u}^{\text{'}}dr \\ =-\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{u}^{\text{'}}{r}^s{u}^{\text{'}}dr-s\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^{s-1}{u}^{\text{'}}dr \end{gathered}$$

And you'll have to do it twice more for these new unknown integrals:

$$\begin{gathered} u{r}^s{u}^{\text{'}}dr=-\hspace{0.33em}\frac{d}{dr}\left(u{r}^s\right)udr \\ =-\hspace{0.33em}{u}^{\text{'}}{r}^{s}udr-s{r}^{s-1} \end{gathered}$$

$u^{\text{'}}{r}^{s}udr=\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^s{u}^{\text{'}}dr$Because radial functions are real, not complex.

$$\begin{gathered} \Rightarrow 2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{u}^{\text{'}}{r}^{s}udr=-s\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{r}^{s-1} \\ \Rightarrow \hspace{0.33em}u{r}^s{u}^{\text{'}}dr=-\frac{s}{2}{r}^{s-1} \end{gathered}$$

In the equation, the first integral must be calculated. This is how we go about it:

$$\begin{gathered} (u^{\text{'}}\text{'}{r}^{s+1}{u}^{\text{'}})dr=-\hspace{0.33em}{u}^{\text{'}}\frac{d}{dr}\left(r^{s+1}{u}^{\text{'}}\right)dr \\ =-(s+1)\hspace{0.33em}\hspace{0.33em}{u}^{\text{'}}{r}^s{u}^{\text{'}}dr-\hspace{0.33em}\hspace{0.33em}{u}^{\text{'}}{r}^{s+1}{u}^{\text{'}\text{'}}dr2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left(u^{\text{'}\text{'}}{r}^{s+1}{u}^{\text{'}}\right)dr \\ =-(s+1)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{u}^{\text{'}}{r}^s{u}^{\text{'}}dr \end{gathered}$$

$\Rightarrow \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{u}^{\text{'}}{r}^s{u}^{\text{'}}dr=-\frac{2}{s+1}\hspace{0.33em}\hspace{0.33em}\left(u^{\text{'}\text{'}}{r}^{s+1}{u}^{\text{'}}\right)dr$

$$\begin{gathered} u^{\text{'}}\text{'}=-\frac{2}{ar}+\frac{l(l+1)}{r^2}+\frac{1}{a^2{n}^2} \\ u{u}^{\text{'}}{r}^s{u}^{\text{'}}dr=-\frac{2}{s+1}\hspace{0.33em}-\frac{2}{ar}+\frac{l(l+1)}{r^2}+\frac{1}{a^2{n}^2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^{s+1}{u}^{\text{'}}dr \\ =-\frac{2}{s+1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}-\frac{2}{a}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^s{u}^{\text{'}}dr+l(l+1)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^{s-1}{u}^{\text{'}}dr+\frac{1}{a^2{n}^2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^{s+1}{u}^{\text{'}}dr \end{gathered}$$

Using equation get:

$$\begin{gathered} u^{\text{'}}{r}^s{u}^{\text{'}}dr=-\frac{2/}{s+1})\hspace{0.33em}\hspace{0.33em}\frac{\hspace{0.33em}s}{a}{r}^{s-1}+l(l+1)-\frac{s-1}{2}\hspace{0.33em}\hspace{0.33em}{r}^{s-2}-\frac{1}{a^2{n}^2}\frac{s+1}{2}⟨r^s⟩ \\ =-\frac{2}{as}\frac{s}{+1}{r}^{s-1}+\frac{s-1}{s+1}l(l+1)r^{s-2}+\frac{1}{n^2{a}^2}⟨r^s⟩ \end{gathered}$$

Equation on second page is:

$(u{r}^s{u}^{\text{'}}\text{'})dr=-\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{u}^{\text{'}}{r}^s{u}^{\text{'}}dr-s\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}u{r}^{s-1}{u}^{\text{'}}dr$

On the left hand side, there is an equation from the second page, and on the right hand side, there are equations from the third page and an equation from the second page. All of those equations are now inserted into the previous expression:

$$\begin{gathered} -\frac{2}{a}{r}^{s-1}+l(l+1)r^{s-2}+\frac{1}{a^2{n}^2}⟨r^s⟩=\frac{2}{a}\frac{s}{s+1}{r}^{s-1}-\frac{s-1}{s+1}l(l+1)r^{s-2}-\frac{1}{n^2{a}^2}⟨r^s⟩+\frac{s(s-1)}{2}{r}^{s-2} \\ {r}^{s-2}l(l+1)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}1+\frac{s-1}{s+1}-\frac{s\left(s-1\right)}{2}{r}^{s-1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{s}{s+1}+1+\frac{2}{a^2{n}^2}⟨r^s⟩=0 \\ {r}^{s-2}l(l+1)\frac{2s}{s+1}-\frac{s\left(s-1\right)}{2}\frac{2}{a} \\ r1\frac{2s+1}{s+1}+\frac{2}{a^2{n}^2}⟨r^s⟩=0\hspace{0.33em}\frac{a^2(s+1)}{2} \end{gathered}$$

$$\begin{gathered} r^{s-2}{a}^{2}l(l+1)s-\frac{s(s^2-1)}{4}-(2s+1)a{r}^{s-1}+\frac{s+1}{n^2}⟨r^s⟩=0 \\ {r}^{s-2}\frac{a^{2}s}{4}(2l+1{)}^2-s^2-(2s+1)a{r}^{s-1}+\frac{s+1}{n^2}⟨r^s⟩=0 \end{gathered}$$

The proved Kramer’s' relation of equation

$r^{s-2}\frac{a^{2}s}{4}(2l+1{)}^2-s^2-(2s+1)a{r}^{s-1}+\frac{s+1}{n^2}⟨r^s⟩=0$