4.4.2.1 So­lu­tion esdb-a

Ques­tion:

The 2p$_x$ pointer state of the hy­dro­gen atom was de­fined as

$$\frac 1{\sqrt 2}\left(-\psi_{211}+\psi_{21-1}\right).$$

What are the ex­pec­ta­tion val­ues of en­ergy, square an­gu­lar mo­men­tum, and $z$ an­gu­lar mo­men­tum for this state?

An­swer:

Note that the square co­ef­fi­cients of the eigen­func­tions $\psi_{211}$ and $\psi_{21-1}$ are each $\frac 12$, so each has a prob­a­bil­ity $\frac 12$ in the 2p$_x$ state.

Eigen­func­tion $\psi_{211}$ has an en­ergy eigen­value $E_2$, and so does $\psi_{21-1}$, so the ex­pec­ta­tion value of en­ergy in the 2p$_x$ state is

$$\left\langle{E}\right\rangle =\frac 12 E_2 + \frac 12 E_2 = E_2 = - 3.4 \mbox{ eV.}$$

This is as ex­pected since the only value that can be mea­sured in this state is $E_2$.

Sim­i­larly, eigen­func­tion $\psi_{211}$ has a square an­gu­lar mo­men­tum eigen­value $2\hbar^2$, and so does $\psi_{21-1}$, so the ex­pec­ta­tion value of square an­gu­lar mo­men­tum in the 2p$_x$ state is that value,

$$\langle L^2\rangle =\frac 12 2\hbar^2 + \frac 12 2\hbar^2 = 2\hbar^2.$$

Eigen­func­tion $\psi_{211}$ has a $z$ an­gu­lar mo­men­tum eigen­value $\hbar$, and $\psi_{21-1}$ has $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$, so the ex­pec­ta­tion value of $z$ an­gu­lar mo­men­tum in the 2p$_x$ state is

$$\langle L_z\rangle =\frac 12 \hbar - \frac 12 \hbar = 0$$

Mea­sure­ments in which the $z$ an­gu­lar mo­men­tum is found to be $\hbar$ av­er­age out against those where it is found to be $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$.