4.3.2.2 So­lu­tion hydb-b

Ques­tion:

Use the generic ex­pres­sion

$$\psi_{nlm} = -\frac{2}{n^2} \sqrt{\frac{(n-l-1)!}{[(n+l)!a_0... ...+1}\left(\frac{2\rho}n\right) e^{-\rho /n} Y_l^m(\theta ,\phi)$$

with $\rho$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r$$\raisebox{.5pt}{$/$}$​$a_0$ and $Y_l^m$ from the spher­i­cal har­mon­ics ta­ble to find the ground state wave func­tion $\psi_{100}$. Note: the La­guerre poly­no­mial $L_1(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1-x$ and for any $p$, $L_1^p$ is just its $p$-​th de­riv­a­tive.

An­swer:

You get, sub­sti­tut­ing $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0:

$$\psi_{100} = -\frac{2}{1^2} \sqrt{\frac{0!}{[1!a_0]^3}} \lef... ...1^1\left(\frac{2\rho}1\right) e^{-\rho /1} Y_0^0(\theta ,\phi)$$

where 0! = 1! = 1, $L_1^1(x)$ is the first de­riv­a­tive of $L_1(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1-x$ with re­spect to $x$, which is $\vphantom{0}\raisebox{1.5pt}{$-$}$1, and $Y_0^0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1$\raisebox{.5pt}{$/$}$​$\sqrt{4\pi}$ ac­cord­ing to the ta­ble. So you get

$$\psi_{100} = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}.$$

as in the pre­vi­ous ques­tion.