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 ${\skew 6\widehat{\vec J}}_b$:

$$\left\langle\left(\vec J_{a}+\lambda \vec J_{b}\right)^2\r... ...eft\langle\left(J_{z,a}+\lambda J_{z,b}\right)^2\right\rangle$$

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

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

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:

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

which means, taking square roots,

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

and so

$$J^2_a - 2 J_a J_b + J^2_b \mathrel{\raisebox{-.7pt}{$\le... ...el{\raisebox{-.7pt}{$\leqslant$}} J^2_a + 2 J_a J_b + J^2_b$$

or

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

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.