of values for the components of $K$, say, to calculate the values of $u\cdot K$.

As I shall show now, one may consider each $u\cdot K$ as a (not necessarily classical) random variable in its own right, and determine their relationship for different $u$ not by ordinary algebra but by the statistics derived from the standard quantum mechanical recipes.

For simplicity, I’ll take for $K$ in place of the angular momentum $J$ the Pauli spin vector $\sigma$ which shows precisely the same problems as $J$. For each unit vector $u\in \Rz^3$, the operator $u\cdot \sigma$ has the simple eigenvalues $1$ and $-1$; thus these are the possible values of $u\cdot \sigma$. The projector to an eigenstate of $u\cdot \sigma$ corresponding to the eigenvalue $1$ is $\Pi_u:= \half(1+u\cdot \sigma)$, with $\tr \Pi_u=1$, and the corresponding projector for the eigenvalue $-1$ is simply $\Pi_{-u}$. For any unit vector $v\in \Rz^3$, the probability for $v\cdot \sigma$ having the value $1$ when $u\cdot \sigma$ has the value $1$ is then, according to ,

[ll]{} P\_[uv]{}&=\_u\_v = (1+u)(1+v)/4\
&=(1+(u+v)+uv+i(u v))/4\
&=(1+uv),

and the probability for $v\cdot \sigma$ having the value $-1$ when $u\cdot\sigma$ has the value $1$ is therefore $P_{u,-v}= \half(1-u\cdot v)$, adding up to 