Does $$S_1^x$$ commute with $$S^2$$?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
Diracobama2181
Messages
70
Reaction score
3
Homework Statement
Consider a tetrahedron with four spin (1/2) particles, one at each of the vertices. Suppose the Hamiltonian is given by $$H=\sum_{i\neq q}S_iS_j$$. Show that all three components of the total spin $$J =\sum_{i}S_i$$ of the system commutes with $$H$$.
Relevant Equations
$$S^2=S_{1}^2+S_{2}^2+S_{3}^2+S_{4}^2+2\sum_{i\neq q}S_iS_j$$
$$H$$ can be rewritten as $$H=\frac{1}{2}(S^2-S_{1}^2-S_{2}^2-S_{3}^2-S_{4}^2)$$. Let's focus on the x component, $$J^x=\sum_{i}S_i^x$$. Now $$S_1^x$$ commutes with $$S^2_1, S^2_2, S^2_3, S^2_4$$, but does it commute with $$S^2$$? If not, what is the exact relation between $$S^2$$ and $$S_1^x$$?
 
Physics news on Phys.org
Nevermind, figured it out. $$S_1^x$$ does not commute with $$S^2$$. However, $$S^2=J^2$$, so $$J^2$$ commutes with $$H$$,which I believe implies each component of $$J$$ also commutes.