Spin Matrices for Multiple Particles

[tomdodd4598]

I have two questions, but the second is only worth asking if the answer to the first is yes:
Are the spin matrices for three particles, with the same spin,
σ ⊗ I ⊗ I,
I ⊗ σ ⊗ I and
I ⊗ I ⊗ σ
for particles 1, 2 and 3 respectively, where σ is the spin matrix for a single one of the particles?

I know that the spin matrices for two particles are
σ ⊗ I and
I ⊗ σ
for particles 1 and 2 respectively, so I am guessing that the same is the case for when there are more than two particles.

 

[blue_leaf77]

Yes, it is.

 

[tomdodd4598]

Nice - thanks for the quick answer ;)

What I was wanting to ask was whether the following is a way to generalise the spin matrices for each of N particles:

Where w is the direction, a is the matrix/particle number, N is the number of particles and dim(V) = (2s + 1)^N, where s is the spin of each particle (1/2 for spin 1/2, 1 for spin 1 etc.).

 

[blue_leaf77]

You have to check again what the notations used in that expression mean. For instance, does the first $I_{dim(V)}$ means that it is actually a series of (Kronecker) product of $a-1$ individual identity matrices? If yes, then $dim(V) = 2s+1$, that is, the dimension of the spin space corresponding to a single particle.
I think that is indeed the case there since the dimensions of the first and second $I$ are denoted identically.

 

[tomdodd4598]

Ah yes, I made a mistake there - yeh, the subscript of the identity matrices should be just 2s+1, and the a-1 and N-a under the brackets are the number of Kronecker/tensor products to have - for example, for 8 spin 1 particles, the 3rd matrix would be:

...which is a rather large matrix...

 

[tomdodd4598]

There is a problem though:

If I add together the three spin matrices (z-direction) for 3 spin 1/2 particles, I get:

But surely it should be:

 

[blue_leaf77]

It doesn't really matter as long as you remember, which eigenvector (when you have to calculate it) belongs to which eigenvalue. For example the eigenvector $(0,0,0,1,0,0,0,0)^T$ belongs to eigenvalue $-\hbar$ in the upper matrix, but belongs to $\hbar$ in the lower one.