A.8 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

$$\fbox{$\displaystyle {\skew 6\widehat{\vec S}}_1\cdot{\s... ... - {\widehat S}_1^{\,2} - {\widehat S}_2^{\,2}\right) $} %$$ (A.28)

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

$$\fbox{$\displaystyle \mbox{singlet:}\quad {\skew 6\wideh... ...ehat{\vec S}}_2 = {\textstyle\frac{1}{4}} \hbar^{\,2} $} %$$ (A.29)

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$\POW9,{\circ}$.