# For which irreps can normalized spin vectors be chosen continuously, for all possible spins?

Let $(u_x,u_y,u_z)$ be a unit vector. Then a normalized vector for spin-1/2 in this direction is given by normalizing a column from the pure density matrix: $\frac{1}{\sqrt{2+2u_z}}\left(\begin{array}{c}1+u_z\\u_x+iu_y\end{array}\right)$. This solution works except when $u_z=-1$. So this gives a normalization choice of spin vector that is continuous over the possible eigenstates of spin with the exception of spin down = $\left(\begin{array}{c}0\\1\end{array}\right)$.

It is impossible to choose normalized spin-1/2 vectors for all possible eigenstates of spin-1/2. This is a fact that is related to Berry-Pancharatnam phase. The best you can do is to leave a single point, as in the above, where there is an essential discontinuity.

What I'd like to know is what other j have spin-j vectors unable to be normalized. That is, given all possible eigenstates of spin-j, is it possible to choose one from each ray, such that the choice is continuous?

Partial Solution: Per Dr. Motl's answer below, I'll down the situation for a natural representation of spin-1. Choose $S_x = i\left(\begin{array}{ccc}0&0&0\\0&0&+1\\0&-1&0\end{array}\right)$ $S_y = i\left(\begin{array}{ccc}0&0&-1\\0&0&0\\+1&0&0\end{array}\right)$ $S_z = i\left(\begin{array}{ccc}0&+1&0\\-1&0&0\\0&0&0\end{array}\right)$.

It's easy to verify that these satisfy the $SU(2)$ rules $[S_x,S_y] = iS_z$, etc. Then the spin operator for spin in the $u = (u_x,u_y,u_z)$ direction is given by: $S_u = i\left(\begin{array}{ccc}0&u_z&-u_y\\-u_z&0&u_x\\u_y&-u_x&0\end{array}\right)$

and the solution with spin=0 in the $u$ direction is given by the vector $u$. And sure enough, it's impossible to choose the vectors for spin=1 as a continuous function of $u$.

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For $j=1$, you can find a normalized continuous vector for each direction, namely the $u$ vector itself - properly transformed in the right basis.
However, I suppose you want the spin-$j$ vectors to be those of the maximum component $j_z=j$; strictly speaking, you haven't written it but it's the only interpretation I can imagine - that generalizes the $j=1/2$ case you described. That kills the previous paragraph because the vector $u$ itself as a function on the sphere is the solution for $j_z=0$. In particular, for $u=(0,0,1)$, the vector corresponding to the function $z$ is nothing else than the $m=0$ component of the triplet. Recall that $m=\pm 1$ correspond to the functions $x\pm iy$ on the sphere.
If you were trying to solve the problem for $j=1$ and $j_z=\pm 1$, you would fail to find a solution again. In that case, the problem would become totally equivalent to the combing of the sphere - because you would be looking for tangential vectors to the sphere's surface - but it is impossible to comb the sphere.
Because of these two no-go cases for nonzero $j_z$, I suppose that the problem can only be solved for $j$ being integer and $j_z=0$ in which case the solution is given by the symmetric tensor power of the solution for $j=1$, $j_z=0$, with the trace removed. I believe that all cases when you want a continuous $(2j+1)$-tuplet with $j_z$ along the variable axis being nonzero - which includes all half-integral $j$ cases as well as most integral $j$ cases, except those with $j_z=0$ - will suffer from the "combing the sphere" problem.