Total spin in multiparticle system

[hideelo]

If I have a single spin 1/2 particle, I know that it's total spin in any direction with unit vector can be computed by using the operator σ•n where σ_i are the pauli matrices.

Suppose however I had a multiparticle system, is there a generalization of the pauli matrices (which let's call ρ_i) so that the the expectation value of ρ•n is the total spin in the n direction?

 

[jambaugh]

Yes... but it depends somewhat on the statistics of the particles in that multi-particle system. But we can start with the assumption of distinguishable particles with Maxwell-Boltzmann statistics in which case the system Hilbert space is the tensor product of the Hilbert spaces of the individual particles and the spin operators are elements of the corresponding operator algebra.

Now the operators in the algebra have two roles, finite elements and "infinitesimal" generators. That is to say we have both Lie group and Lie algebra represented in the same operator algebra. This distinction is important when you consider how one maps to the tensor product algebra.

For example the unitary operator for a finite rotation, say $R$ for a single particle would, in the composite, correspond to a tensor product of rotations rotating each factor system (each particle) the same. The resulting composite operator would be: $\tilde{R} = R_1\otimes R_2\otimes \ldots$.

For infinitesimal generators, which are what correspond to observables like spin we would however map to the composite algebra differentially. For $R=e^{i\theta J}$ we would have the composite operator: $\tilde{R}= e^{i\theta\tilde{J}}$
and thus (with a little algebra you can show...):
$$\tilde{J} = J_1\otimes \mathbb{I}\otimes\mathbb{I}\otimes\cdots + \mathbb{I}\otimes J_2 \otimes \mathbb{I}\cdots + \ldots$$

(Notice that the indexing subscript I am using here merely indicates that each particle in the ensemble may have a distinct spin representation. If you assume each particle is isomorphic to the others you can drop the subscripts.)

Very often the expression above for the infinitesimal generator is abbreviated, by identifying say $J_2 \equiv \mathbb{I}\otimes J_2 \otimes \mathbb{I}\cdots$ and thus writing $$\tilde{J} = J_1 + J_2 + \ldots$$. But that's skipping the explicit extension. You should be aware of it before you start dropping it's explicit presentation.Now from here, if there are statistical considerations you would extract out the specific action on the symmetrized or anti-symmetrized subspaces according to whether your particles are Bosons or Fermions.