A.10 Spin in­ner prod­uct

In quan­tum me­chan­ics, the an­gle be­tween two an­gu­lar mo­men­tum vec­tors is not re­ally de­fined. That is be­cause at least two com­po­nents of a nonzero an­gu­lar mo­men­tum vec­tor are un­cer­tain. How­ever, the in­ner prod­uct of an­gu­lar mo­men­tum vec­tors can be well de­fined. In some sense, that gives an an­gle be­tween the two vec­tors.

An im­por­tant case is the in­ner prod­uct be­tween the spins of two par­ti­cles. It is re­lated to the square net com­bined spin of the par­ti­cles as

$${\widehat S}_{\rm net}^{ 2} = \left({\skew 6\widehat{\vec ... ...{\skew 6\widehat{\vec S}}_1+{\skew 6\widehat{\vec S}}_2\right)$$

If you mul­ti­ply out the right hand side and re­arrange, you find the in­ner prod­uct be­tween the spins as

$$\fbox{$\displaystyle {\skew 6\widehat{\vec S}}_1\cdot{\ske... ...,2} - {\widehat S}_1^{ 2} - {\widehat S}_2^{ 2}\right) $} %$$ (A.29)

Now an el­e­men­tary par­ti­cle has a def­i­nite square spin an­gu­lar mo­men­tum

$${\widehat S}^{ 2} = s(s+1)\hbar^{ 2}$$

where $s$ is the spin quan­tum num­ber. If the square com­bined spin also has a def­i­nite value, then so does the dot prod­uct be­tween the spins as given above.

As an im­por­tant ex­am­ple, con­sider two fermi­ons with spin $s_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $s_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$. These fermi­ons may be in a sin­glet state with com­bined spin $s_{\rm {net}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Or they may be in a triplet state with com­bined spin $s_{\rm {net}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. If that is plugged into the for­mu­lae above, the in­ner prod­uct be­tween the spins is found to be

$$\fbox{$\displaystyle \mbox{singlet:}\quad {\skew 6\widehat... ...widehat{\vec S}}_2 = {\textstyle\frac{1}{4}} \hbar^{ 2} $} %$$ (A.30)

Based on that, you could ar­gue that in the sin­glet state the an­gle be­tween the spin vec­tors is 180$\POW9,{\circ}$. In the triplet state the an­gle is not zero, but about 70$\POW9,{\circ}$.