Solving Spin-Spin Interaction for Total Spin Commutation

[Diracobama2181]

TL;DR

How do I go about showing that the total Spin component commutes with the hamiltonian

Let there be four spin (1/2) particles at the corners of a tetrahedron, coupled such that the total hamiltonian is given by $$H=\sum_{i\neq j} S_{i} \cdot S_{J}$$.
How would I go about showing that each component of the total spin $$\sum_{i} S_{i}$$ commutes with the hamiltonian.

Work so far:
I know
$$\sum_{i} S_{i}^{x}=S_{1}^{x}+S_{2}^{x}+S_{3}^{x}+S_{4}^{x}$$
so $$[\sum_{i} S_{i}^{x},H]=[S_{1}^{x},H]+[S_{2}^{x},H]+[S_{3}^{x},H]+[S_{4}^{x},H]$$
From there, I can use $$H=S_{1}S_{2}+S_{1}S_{3}+S_{1}S_{4}+S_{2}S_{3}+S_{2}S_{4}+S_{3}S_{4}$$
I can also use $$S_{i}\cdot S_{j}=S_{i}^{x}S_{j}^{x}+S_{i}^{y}S_{j}^{y}+S_{i}^{z}S_{j}^{z}$$

As you can see, it gets rather messy. I just want to know if my reasoning is correct thus far and or if I am over complicating my solution.
Thank you.

 

[azntoon]

A:Yes, you are on the right track. What you need to do is expand out the commutator:$$[\sum_i S_i^x,H]=\sum_{i \neq j}[S_i^x,S_i\cdot S_j] + [S_j^x,S_i\cdot S_j].$$Using the fact that $[S_i^x,S_i\cdot S_j] = 0$ (this follows from the definition of the dot product and the fact that $S^x$ is linear in $S$), we see that the above equals$$[\sum_i S_i^x,H] = \sum_{i\neq j}[S_j^x,S_i\cdot S_j].$$Now using the fact that for any two $2\times 2$ matrices $A,B$ we have $[A,B]= AB - BA$, we can expand out the commutator on the right hand side to get$$[\sum_i S_i^x,H] = \sum_{i\neq j}(S_i\cdot S_j S_j^x - S_j^x S_i\cdot S_j).$$Finally, using the fact that $S^x$ is hermitian, we see that the above equals$$[\sum_i S_i^x,H] = \sum_{i\neq j}(S_i\cdot S_j S_j^x - S_i\cdot S_j S_j^x) = 0.$$The same argument can be applied to show that $[\sum_i S_i^y,H] = 0$ and $[\sum_i S_i^z,H] = 0$.