4.3.4.2 So­lu­tion hydd-b

Ques­tion:

Check from the con­di­tions

$$n > l \mathrel{\raisebox{-1pt}{$\geqslant$}}\vert m\vert$$

that $\psi_{200}$, $\psi_{211}$, $\psi_{210}$, and $\psi_{21-1}$ are the only states of the form $\psi_{nlm}$ that have en­ergy $E_2$. (Of course, all their com­bi­na­tions, like 2p$_x$ and 2p$_y$, have en­ergy $E_2$ too, but they are not sim­ply of the form $\psi_{nlm}$, but com­bi­na­tions of the ba­sic so­lu­tions $\psi_{200}$, $\psi_{211}$, $\psi_{210}$, and $\psi_{21-1}$.)

An­swer:

Since the en­ergy is given to be $E_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_2$, you have $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2. The az­imuthal quan­tum num­ber $l$ must be a smaller non­neg­a­tive in­te­ger, so it can only be 0 or 1. In case $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the ab­solute value of the mag­netic quan­tum num­ber $m$ can­not be more than zero, al­low­ing only $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is the $\psi_{200}$ state. In the case that $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, the ab­solute value of $m$ can be up to one, al­low­ing $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 0, and $\vphantom{0}\raisebox{1.5pt}{$-$}$1.