91影视

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

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

When computing, in particular, perturbative corrections to the hydrogen spectrum, the Kramer鈥檚 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.

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

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,

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

Also for s=2 ,

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

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

Lastly, for s=3,

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

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

(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 $r^{-2}$.

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 $\frac{1}{a^3{n}^{3}l\left(l+1\right)l+{\displaystyle \frac{1}{2}}}$.