4.4.3.1 So­lu­tion esdb2-a

Ques­tion:

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

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

where both $\psi_{211}$ and $\psi_{21-1}$ are eigen­func­tions of the to­tal en­ergy Hamil­ton­ian $H$ with eigen­value $E_2$ and of square an­gu­lar mo­men­tum $\L ^2$ with eigen­value $2\hbar^2$; how­ever, $\psi_{211}$ is an eigen­func­tion of $z$ an­gu­lar mo­men­tum $\L _z$ with eigen­value $\hbar$, while $\psi_{21-1}$ is one with eigen­value $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$. Eval­u­ate the ex­pec­ta­tion val­ues of en­ergy, square an­gu­lar mo­men­tum, and $z$ an­gu­lar mo­men­tum in the 2p$_x$ state us­ing in­ner prod­ucts. (Of course, since 2p$_x$ is al­ready writ­ten out in terms of the eigen­func­tions, there is no sim­pli­fi­ca­tion in this case.)

An­swer:

For en­ergy you have,

$$\left\langle{E}\right\rangle =\frac 12 \langle -\psi_{211}+\psi_{21-1}\vert H\vert{-}\psi_{211}+\psi_{21-1}\rangle .$$

By the de­f­i­n­i­tion of eigen­func­tion, the prod­ucts with $H$ sim­plify:

$$\left\langle{E}\right\rangle =\frac 12 \langle -\psi_{211}+\psi_{21-1}\vert{-}E_2\psi_{211}+E_2\psi_{21-1}\rangle .$$

Mul­ti­ply­ing out fur­ther, while not­ing that on ac­count of or­tho­nor­mal­ity of the eigen­states,

$$\langle\psi_{211}\vert\psi_{211}\rangle = \langle\psi_{21-1}... ...i_{21-1}\rangle = \langle\psi_{21-1}\vert\psi_{211}\rangle =0,$$

you get $\left\langle{E}\right\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_2$.

Sim­i­larly, for the square an­gu­lar mo­men­tum,

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

or mul­ti­ply­ing out

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

mul­ti­ply­ing out fur­ther to $\left\langle{L^2}\right\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\hbar^2$.

For the $z$ an­gu­lar mo­men­tum,

$$\langle L_z\rangle =\frac 12 \langle -\psi_{211}+\psi_{21-1}\vert\L _z\vert{-}\psi_{211}+\psi_{21-1}\rangle .$$

or mul­ti­ply­ing out

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

mul­ti­ply­ing out fur­ther to $\left\langle{L_z}\right\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.