(a) Use the recursion formula (Equation 4.76) to confirm that when$\mathit{I}\mathbf{=}\mathit{n}\mathbf{-}\mathbf{1}$ the radial wave function takes the form

${\mathit{R}}_{\mathbf{n}\left(n-1\right)}\mathbf{=}{\mathit{N}}_{\mathbf{n}}{\mathit{r}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{n}\mathbf{a}}$ and determine the normalization constant by direct integration.

(b) Calculate 200a and $<r^2>$ for states of the form ${\mathit{\psi }}_{\mathbf{n}\left(n-1\right)\mathbf{m}}\mathbf{·}$

(c) Show that the "uncertainty" in $\mathit{r}\left({\delta }_r\right)$ is$<r>\mathbf{/}\sqrt{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}$for such states. Note that the fractional spread in decreases, with increasing (in this sense the system "begins to look classical," with identifiable circular "orbits," for large ). Sketch the radial wave functions for several values of, to illustrate this point.

A recursive formula is one that defines each term in a series in terms of the term before it (s). As an example: An arithmetic sequence's recursive formula is $a_n=a_{n-1}+d$geometric sequence's recursive formula is $a_n=a_{n-1}r$.

(a)

The radial equation is given by:

$N_n={\left(\frac{2}{na}\right)}^{n+\frac{1}{2}}\frac{1}{\sqrt{\left(2n\right)!}}$

where:

$V\left(\rho \right)=\sum _{j‐0}^{\infty }{c}_j{\rho }^j\rho =\kappa r\kappa =\frac{\sqrt{-2mE}}{\sout{h}}$

and the coefficients turned out to satisfy the recursion formula:

$c_{j+1}=\frac{2\left(j+I+1\right)-2n}{\left(j+1\right)\left(j+2\left(I+1\right)\right)}{c}_j$

let $I=n-1$thus:

$R_{n\left(n-1\right)}=\frac{1}{r}{\rho }^n{e}^{-p}v\left(\rho \right)$

but:

$$\begin{gathered} c_1=\frac{2\left(n-n\right)}{\left(1\right)\left(2n\right)}{c}_0=0 \\ \rho =\frac{r}{na} \\ \Rightarrow v\left(\rho \right)=c_0 \end{gathered}$$

so:

$R_{n\left(n-1\right)}\left(r\right)=\frac{c_0}{{\left(an\right)}^n}{r}^{n-1}{e}^{-r/an}$

combine the constant out front into a single constant, $N_n=c_0/{\left(an\right)}^n$, so:

$R_{n\left(n-1\right)}=N_n{r}^{n-1}{e}^{-r/na}$

Normalize the wave function, that is:

role="math" localid="1656315281010" ${\int }_0^{\infty }{\left|R\right|}^2{r}^{2}dr=1$

to do the integral I used the integral-calculator.com. Thus:

$$\begin{gathered} {\int }_0^{\infty }{\left|R\right|}^2{r}^{2}dr={\left(N_n\right)}^2{\int }_0^{\infty }{r}^{2n}{e}^{-2r/na}dr \\ ={\left(N_n\right)}^2\left(2n\right)!{\left(\frac{na}{2}\right)}^{2n+1} \\ =1 \end{gathered}$$

$N_n={\left(\frac{2}{na}\right)}^{n+\frac{1}{2}}\frac{1}{\sqrt{\left(2n\right)!}}$

Therefore, the normalization constant $N_n$by direct integration is$N_n={\left(\frac{2}{na}\right)}^{n+\frac{1}{2}}\frac{1}{\sqrt{\left(2n\right)!}}$

(b)

Now find the expectation value of r and r² , first we find it for $r^I$then we let $I=1$and $I=2$as:

for $I=1$, we have:

For $I=2$:

role="math" localid="1656316300775"

Therefore, the value of and is and

(c)

The uncertainty in $r$is given by:

localid="1656316682565"

substitute from part $\left(b\right)$to get:

localid="1656316692244"

Thus :

Now we need to sketch few wave functions with different $n$values, combine $N_n$with $R_{n\left(n-1\right)}$in part $\left(a\right)$, so we get:

$R_{n\left(n-1\right)}={\left(\frac{2}{na}\right)}^{n+\frac{1}{2}}\frac{1}{\sqrt{\left(2n\right)!}}{r}^{n-1}{e}^{-r/na}$to plot this function /set $a=1$(for simplicity) and plot $r$from 0 to 200 (actully from 0 to 200a ), I used python to plot it and the code is shown in the following picture:

Therefore, the uncertainty is