Solution for Sakurai problem 3.12: What must we know in addition to [(Subscript[S, x])], [(Subscript[S, y])], [(Subscript[S, z])] to completely characterize the density matrix of a spin 1 system?  I guessed that we also needed [((Subscript[S, x])^2)], [((Subscript[S, y])^2)], [((Subscript[S, z])^2)], but that and tr(ρ) = 1, was not enough to find ρ, since those relations do not provide nine equations for the nine unknowns.  Omitting the trace requirement, but also introducing variables for the other second order products [(Subscript[S, x])(Subscript[S, y])], [(Subscript[S, y])(Subscript[S, z])], [(Subscript[S, z])(Subscript[S, x])] was enough to solve the problem.  Printing out all those products is helpful to show why this is the case: you can see visually that these products appear to span the space of 3x3 matrices.

isq = 1/Sqrt[2] ;
j = {
isq {{0, 1,0}, {1,0,1}, {0, 1,0}},
I isq {{0, -1,0}, {1,0,-1}, {0, 1,0}},
{{1,0,0}, {0,0,0}, {0, 0,-1}}
} ;
eig = Eigensystem[#] &/@ j ;
(*eig
#[[2]] &/@ eig*)
rho = {{a,b,c},{d,e,f},{g, h,i}} ;

sys = {
sx == Tr[rho.j[[1]]] &&
sy == Tr[rho.j[[2]]] &&
sz == Tr[rho.j[[3]]] &&
sxx == Tr[rho.j[[1]].j[[1]]] &&
syy== Tr[rho.j[[2]].j[[2]]] &&
szz == Tr[rho.j[[3]].j[[3]]] &&
sxy== Tr[rho.j[[1]].j[[2]]]&&
syz == Tr[rho.j[[2]].j[[3]]]  &&
szx == Tr[rho.j[[3]].j[[1]]] 
} ;

coeff = Solve[sys,{a,b,c,d,e,f,g,h,i}] // Flatten ;

(* http://mathematica.stackexchange.com/a/100294/10 *)
MakeBoxes[sx,StandardForm]:=RowBox[{"[",SubscriptBox["S","x"],"]"}] ;
MakeBoxes[sy,StandardForm]:=RowBox[{"[",SubscriptBox["S","y"],"]"}] ;
MakeBoxes[sz,StandardForm]:=RowBox[{"[",SubscriptBox["S","z"],"]"}] ;
MakeBoxes[sxx,StandardForm]:=RowBox[{"[",SubscriptBox["S","x"], SubscriptBox["S","x"],"]"}] ;
MakeBoxes[syy,StandardForm]:=RowBox[{"[",SubscriptBox["S","y"], SubscriptBox["S","y"],"]"}] ;
MakeBoxes[szz,StandardForm]:=RowBox[{"[",SubscriptBox["S","z"], SubscriptBox["S","z"],"]"}] ;
MakeBoxes[sxy,StandardForm]:=RowBox[{"[",SubscriptBox["S","x"], SubscriptBox["S","y"],"]"}] ;
MakeBoxes[syz,StandardForm]:=RowBox[{"[",SubscriptBox["S","y"], SubscriptBox["S","z"],"]"}] ;
MakeBoxes[szx,StandardForm]:=RowBox[{"[",SubscriptBox["S","z"], SubscriptBox["S","x"],"]"}] ;

(*cdisp = coeff /. {
sx->Row[{"[",Subscript[S,x],"]"}]/ℏ,
sy->Row[{"[",Subscript[S,y],"]"}]/ℏ,
sz->Row[{"[",Subscript[S,z],"]"}]/ℏ,
sxx->Row[{"[",Subscript[S,x]^2,"]"}]/(ℏ^2),
syy->Row[{"[",Subscript[S,y]^2,"]"}]/(ℏ^2),
szz->Row[{"[",Subscript[S,z]^2,"]"}]/(ℏ^2),
sxy->Row[{"[",Subscript[S,x],Subscript[S,y],"]"}]/(ℏ^2),
syz->Row[{"[",Subscript[S,y],Subscript[S,z],"]"}]/(ℏ^2),
szx->Row[{"[",Subscript[S,z],Subscript[S,x],"]"}]/(ℏ^2)
} ;
Row[{"ρ = ",1/(2ℏ^2), 2 ℏ^2 rho /. cdisp // Simplify// MatrixForm }]*)
Row[{"ρ = ",1/(2ℏ^2), 2 ℏ^2 rho /. coeff // Simplify// MatrixForm }]

(*Works because these products appear to span the space of the 3x 3 matrices:*)
jj[m_] := Row[ {(Subscript[S, m])/ℏ," = ", j[[m]] //  MatrixForm }];
jj[m_, n_] := Row[ {(Subscript[S, m])(Subscript[S, n])/ℏ^2," = ", j[[m]].j[[n]] //  MatrixForm }];
p = {{1,2},{2,3},{3,1},{1,1},{2,2},{3,3}} ;
{jj[#] &/@ Range[3],jj @@@ p} // Flatten// TableForm (*@@@ == Apply*)

(*Note for future (above): a pure function that takes two args can be done like: *)
(*Apply[#1^#2&,{2,3}]
#1^#2 & @@@ {{2,3}}*)

(*{sx^2,sxy} /. {sx -> (Subscript[J, x])/ℏ, sxy ->  (Subscript[J, x]) (Subscript[J, y])/(ℏ^2)}
{sx^2,sxy}/.{sx->Subscript[J,x],sxy->Subscript[J,x] Subscript[J,y]}*)
(*{sx^2,sxy}/.{sx->Row[{"[",Subscript[J,x],"]"}]}*)
(*coeff /. sx->Row[{"[",Subscript[J,x],"]"}]*)