4.3.4.3 So­lu­tion hydd-c

Ques­tion:

Check that the states

$$\mbox{2p$_x$}= \frac 1{\sqrt 2}\left(-\psi_{211}+\psi_{21-1}... ...$}= \frac{{\rm i}}{\sqrt 2}\left(\psi_{211}+\psi_{21-1}\right)$$

are prop­erly nor­mal­ized.

An­swer:

Find the square norm:

$$\langle\mbox{2p$_x$}\vert\mbox{2p$_x$}\rangle = \frac 12 \le... ...i_{211}+\psi_{21-1}\vert{-}\psi_{211}+\psi_{21-1}\right\rangle$$

or mul­ti­ply­ing out

$$\langle\mbox{2p$_x$}\vert\mbox{2p$_x$}\rangle = \frac 1{2} \... ... + \left\langle\psi_{21-1}\vert\psi_{21-1}\right\rangle\right)$$

or us­ing the or­tho­nor­mal­ity of $\psi_{211}$ and $\psi_{21-1}$.

$$\langle\mbox{2p$_x$}\vert\mbox{2p$_x$}\rangle = \frac 1{2} \left(1 + 0 + 0 + 1\right) = 1.$$

For the state 2p$_y$, re­mem­ber that ${\rm i}$ comes out of the left side of the in­ner prod­uct as $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}$:

$$\langle\mbox{2p$_y$}\vert\mbox{2p$_y$}\rangle = \frac{-{\rm ... ...\psi_{211}+\psi_{21-1}\vert\psi_{211}+\psi_{21-1}\right\rangle$$

The rest goes the same way.