Question

(a) Plugs=0,s=2, and s=3into Kramers' relation (Equation 6.104) to obtain formulas for $\left(r^{-1}\right)\mathbf{,}\left(r\right)\mathbf{,}\left(r^{-2}\right)\mathbf{,}\mathbf{and}\mathbf{}\left(r^3\right)$. Note that you could continue indefinitely, to find any positive power.

(b) In the other direction, however, you hit a snag. Put in $\mathit{s}\mathbf{=}\mathbf{-}\mathbf{1}$, and show that all you get is a relation between role="math" localid="1658216018740" $\mathbf{\left(}{\mathbf{r}}^{\mathbf{-}\mathbf{2}}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left(}{\mathbf{r}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}$.

(c) But if you can get $\mathbf{\left(}{\mathbf{r}}^{\mathbf{-}\mathbf{2}}\mathbf{\right)}$by some other means, you can apply the Kramers' relation to obtain the rest of the negative powers. Use Equation 6.56(which is derived in Problem 6.33) to determine $\mathbf{\left(}{\mathbf{r}}^{\mathbf{-}\mathbf{3}}\mathbf{\right)}$ , and check your answer against Equation 6.64.

Short answer

(a)The Kramer's relation value of

$$\begin{gathered} s=1\Rightarrow ⟨r⟩=\frac{a}{2}[3n^2-l(l+1\left)\right] \\ s=2\Rightarrow r^2=\frac{n^2{a}^2}{2}[5n^2-3l(l+1)+1] \\ s=3\Rightarrow r^3=\frac{n^2{a}^3}{8}[35n^4+25n^2-30n^{2}l(l+1)+3l^2(l+1{)}^2-6l(l+1)] \end{gathered}$$

(b) The value of $s=-1$is $r^{-3}=\frac{1}{al(l+1)}{r}^2$

(c) The value of $\left(r^{-3}\right)$is $\frac{1}{a^3{n}^{3}I\left(I+1\right)\left(I+{\displaystyle \frac{1}{2}}\right)}$

Step-by-step solution

Given information

The values of to be substituted in the Kramer’s relation are 0,1,2,3.

$⟨\frac{1}{r^2}⟩=\frac{1}{(l+1/2)n^3{a}^2}.$ . . . . . . 6.56

Determine the formula for Kramer's relation

When computing, in particular, perturbative corrections to the hydrogen spectrum, the Kramer’s relationship is crucial since those calculations call for expectation values of the radial Hamiltonian.

It gives the expectation values of close powers of r for the hydrogen atom.

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

Solve the s=0,  s=1,  s=2 and s=3 into Kramer’s' relation

a)

Several values for s must be inserted into Kramer's relation.

s=0

$$\begin{gathered} \frac{1}{n^2}-a{r}^{-1}=0 \\ {r}^{-1}=\frac{1}{a{n}^2} \end{gathered}$$

For s=1,

$$\begin{gathered} \frac{2}{n^2}\left(r\right)-3a+\frac{1}{4}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(2l+1{)}^2-1a^2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{r}^{-1}=0 \\ ⟨r⟩=\frac{n^2}{2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{a^2}{4}\hspace{0.33em}\hspace{0.33em}{r}^{-1}\hspace{0.33em}\hspace{0.33em}(-4l^2-4l)+3a \\ =\frac{n^2}{2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}-a^2\frac{1}{a{n}^2}l(l+1)+3a \\ \Rightarrow ⟨r⟩=\frac{a}{2}[3n^2-l(l+1)] \end{gathered}$$

Also for s=2 ,

$\frac{3}{n^2}{r}^2-5a\left(r\right)+\frac{1}{2}(2l+1{)}^2-4a^2=0$

$$\begin{gathered} r^2=\frac{n^2}{3}5a\left(r\right)+\frac{a^2}{2}\left(4l^2+4l-3\right) \\ =\frac{n^2}{3}\frac{5a^2}{2}\left(3n^2-l^2-l\right)-\frac{a^2}{2}\left(4l^2+4l-3\right) \\ =\frac{n^2{a}^2}{6}\left[15n^2-9l^2-9l+3\right] \\ \Rightarrow r^2=\frac{n^2{a}^2}{6}\left[5n^2-3l\left(l+1\right)+1\right] \end{gathered}$$

Lastly, for s=3,

$\frac{4}{n^2}{r}^3-7a{r}^2+\frac{3}{4}(2l+1{)}^2-9a^2\left(r\right)=0$

$$\begin{gathered} r^3=\frac{n^2}{4}7a{r}^2-\frac{3a^2}{4}\left(4l^2+4l-8\right)\left(r\right) \\ =\frac{n^2}{4}7a\frac{n^2{a}^2}{2}5n^2-3l\left(i+1\right)+1-3a^2\left(l\left(l+1\right)-2\right)\frac{a}{2}3n^2-l\left(l+1\right) \\ =\frac{n^2{a}^2}{8}\left[35n^4-21n^{2}l\left(l+1\right)+7n^2-9n^{2}l\left(l+1\right)+3l^2{\left(l+1\right)}^2+18n^2-6l\left(l+1\right)n\right] \\ \Rightarrow r^3=\frac{n^2{a}^2}{8}\left[35n^4+25n^2-30n^{2}l\left(l+1\right)+3l^2{\left(l+1\right)}^2-6l\left(l+1\right)\right] \end{gathered}$$

Step 4: Show that for s=-1 relation between (r-2) and (r-3)

(b)

In Kramer's relation, substitute $s=-1$,

$$\begin{gathered} a{r}^2-\frac{1}{4}\left[\left(2l+1\right)-1\right]a^2{r}^{-3}=0 \\ {r}^{-3}=\frac{1}{al\left(l+1\right)}{r}^{-2} \end{gathered}$$

Thus, for the value of s= -1 , the obtained result is a relation between $r^{-3}$and $r^{-2}$.

Step 5: Determine (r-3)

c)

Evaluate from $r^{-2}$from the previous step,

$$\begin{gathered} r^{-2}=\frac{1}{a^2{n}^{2}l+{\displaystyle \frac{1}{2}}} \\ {r}^{-3}=\frac{1}{al\left(l+1\right)}\frac{1}{a^2{n}^{2}l+{\displaystyle \frac{1}{2}}} \\ =\frac{1}{a^3{n}^{3}l\left(l+1\right)l+{\displaystyle \frac{1}{2}}} \end{gathered}$$

Therefore, the value of $\left(r^{-3}\right)$is $\frac{1}{a^3{n}^{3}l\left(l+1\right)l+{\displaystyle \frac{1}{2}}}$.