D.67 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\r...
...eft\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}{$\le...
...el{\raisebox{-.7pt}{$\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{-.7...
...aisebox{-.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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, 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.