4.4.3.2 So­lu­tion esdb2-b

Ques­tion:

Con­tin­u­ing the pre­vi­ous ques­tion, eval­u­ate the stan­dard de­vi­a­tions in en­ergy, square an­gu­lar mo­men­tum, and $z$ an­gu­lar mo­men­tum in the 2p$_x$ state us­ing in­ner prod­ucts.

An­swer:

For en­ergy you have,

$$\sigma_E^2 = \langle(H-E_2)^2\rangle =\frac 12 \langle -\psi... ...si_{21-1}\vert(H-E_2)^2\vert{-}\psi_{211}+\psi_{21-1}\rangle .$$

or mul­ti­ply­ing out, not­ing that $H\psi_{21\pm 1}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_2\psi_{21\pm 1}$, so that $(H-E-2)\psi_{21\pm 1}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0,

$$\sigma_E^2 = \frac 12 \langle -\psi_{211}+\psi_{21-1}\vert{-}0\psi_{211}+0\psi_{21-1}\rangle .$$

which is zero. The same way, $\sigma_{L^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

For $z$ an­gu­lar mo­men­tum, you have, since the ex­pec­ta­tion value is zero,

$$\sigma_{L_z}^2=\langle(\L _z-0)^2\rangle =\frac 12 \langle -... ..._{21-1}\vert(\L _z-0)^2\vert{-}\psi_{211}+\psi_{21-1}\rangle .$$

or mul­ti­ply­ing out,

$$\sigma_{L_z}^2=\frac 12 \langle -\psi_{211}+\psi_{21-1}\vert -\hbar^2\psi_{211}+\hbar^2\psi_{21-1}\rangle$$

which mul­ti­plies out to $\hbar^2$, so $\sigma_{L_z}$ it­self is $\hbar$.