Spin operator for an ensemble Given that there are N number of particles and for each one of them $\hat{\sigma}_x$ is the usual Pauli matrix. How can we show that the Pauli matrix for the entire ensemble has the form $\frac{1}{N}\Sigma_{n} \hat{\sigma}_x^n$ ?
 A: If I understnand your question correctly, I'm not sure why you want the $1/N$ factor but that's just a normalization.  Anyways, it's enough to do it explicitly for $N=2$ and you can fill in the rest for arbitrary $N$.
Your Hilbert space is of dimension $2^N=4$ for $N=2$, and the $4$ states of the form $\vert \pm \rangle_1\vert\pm \rangle_2$ are a basis for your space.
Because they are projections and infinitesimal generators, the spin operators are additive in the sense that the total spin along $\hat x$ is
$$
\hat \Sigma_x= \hat \sigma^1_x\otimes \hat 1 + \hat 1\otimes \hat \sigma^2_x := \hat \sigma^1_x+\hat \sigma^2_x\tag{1}
$$
where 
$$
\hat \sigma^1_x\otimes 1 \vert \pm \rangle_1\vert \pm \rangle_2 := 
\left[\hat \sigma_x\vert\pm\rangle_1\right]\left[\hat 1\vert\pm\rangle_2\right]
=\left[\hat \sigma_x\vert\pm\rangle_1\right]\vert\pm\rangle_2
$$
and similarly for $\hat \Sigma_y$ and $\hat\Sigma_z$.  Note that I've interpreted my (formal) expression for $\hat \Sigma_x$ somewhat loosely on the right hand side of (1) as this is the usual interpretation of the sum you give in your question
In this way, the action of $\hat \sigma^1_i \otimes \hat 1$ commutes with the action of $\hat 1\otimes \hat\sigma^2_j$ for $i,j=x,y,z$ so that $\hat\Sigma_x,\hat \Sigma_y$ and $\hat \Sigma_z$ commute on the $su(2)$ algebra.  The key point is that $\sigma^k_i$ acts only on the $k$'th factor in the product state.
Up to notation and $1/N$, this is the result you want.  For $N=3$ the operators have the form 
$$
\hat\Sigma_k=\hat\sigma^1_k\otimes \hat 1\otimes \hat 1
+\hat 1\otimes \hat \sigma^2_k\otimes \hat 1+ \hat 1\otimes \hat 1\otimes 
\sigma^3_k
$$
acting in the obvious way on product states of the type $\vert \pm\rangle_1\vert\pm\rangle_2\vert\pm\rangle_3$.
