D.66 The tri­an­gle in­equal­ity

The nor­mal tri­an­gle in­equal­ity con­tin­ues to ap­ply for ex­pec­ta­tion val­ues in quan­tum me­chan­ics.

The way to show that is, like other tri­an­gle in­equal­ity proofs, rather cu­ri­ous: ex­am­ine the com­bi­na­tion of ${\skew 6\widehat{\vec J}}_a$, not with ${\skew 6\widehat{\vec J}}_b$, but with an ar­bi­trary mul­ti­ple $\lambda$ of ${\skew 6\widehat{\vec J}}_b$:

\begin{displaymath}
\left\langle\left(\vec J_{a}+\lambda \vec J_{b}\right)^2\ri...
...left\langle\left(J_{z,a}+\lambda J_{z,b}\right)^2\right\rangle
\end{displaymath}

For $\lambda$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 this pro­duces the ex­pec­ta­tion value of $\left(\vec{J}_a+\vec{J}_b\right)^2$, for $\lambda$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1, the one for $\left(\vec{J}_a-\vec{J}_b\right)^2$. In ad­di­tion, it is pos­i­tive for all val­ues of $\lambda$, since it con­sists of ex­pec­ta­tion val­ues of square Her­mit­ian op­er­a­tors. (Just ex­am­ine each term in terms of its own eigen­states.)

If you mul­ti­ply out, you get

\begin{displaymath}
\left\langle\left(\vec J_{a}+\lambda \vec J_{b}\right)^2\right\rangle
=
J^2_a + 2 M \lambda + J^2_b \lambda^2
\end{displaymath}

where $J_a$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $\sqrt{\left\langle{J}_{xa}^2+J_{ya}^2+J_{za}^2\right\rangle}$, $J_b$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $\sqrt{\left\langle{J}_{xb}^2+J_{yb}^2+J_{zb}^2\right\rangle}$, and $M$ rep­re­sents mixed terms that do not need to be writ­ten out. In or­der for this qua­dratic form in $\lambda$ to al­ways be pos­i­tive, the dis­crim­i­nant must be neg­a­tive:

\begin{displaymath}
M^2 - J^2_a J^2_b \mathrel{\raisebox{-.7pt}{$\leqslant$}}0
\end{displaymath}

which means, tak­ing square roots,

\begin{displaymath}
- J_a J_b \mathrel{\raisebox{-.7pt}{$\leqslant$}}M \mathrel{\raisebox{-.7pt}{$\leqslant$}}J_a J_b
\end{displaymath}

and so

\begin{displaymath}
J^2_a - 2 J_a J_b + J^2_b
\mathrel{\raisebox{-.7pt}{$\leqs...
...hrel{\raisebox{-.7pt}{$\leqslant$}}
J^2_a + 2 J_a J_b + J^2_b
\end{displaymath}

or

\begin{displaymath}
\left\vert J_a - J_b\right\vert^2
\mathrel{\raisebox{-.7pt...
...raisebox{-.7pt}{$\leqslant$}}\left\vert J_a + J_b\right\vert^2
\end{displaymath}

and tak­ing square roots gives the tri­an­gle in­equal­ity.

Note that this de­riva­tion does not use any prop­er­ties spe­cific to an­gu­lar mo­men­tum and does not re­quire the si­mul­ta­ne­ous ex­is­tence of the com­po­nents. With a bit of mess­ing around, the az­imuthal quan­tum num­ber re­la­tion $\vert j_a-j_b\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $j_{ab}$ $\raisebox{-.3pt}{$\leqslant$}$ $j_a+j_b$ can be de­rived from it if a unique value for $j_{ab}$ ex­ists; the key is to rec­og­nize that $J$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j+\delta$ where $\delta$ is an in­creas­ing func­tion of $j$ that stays be­low $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$, and the $j$ val­ues must be half in­te­gers. This de­riva­tion is not as el­e­gant as us­ing the lad­der op­er­a­tors, but the re­sult is the same.