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

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The discussion centers on the commutation relations between the operator $$S_1^x$$ and the total spin operator $$S^2$$. It is established that while $$S_1^x$$ commutes with individual spin squared operators $$S^2_1, S^2_2, S^2_3, S^2_4$$, it does not commute with the total spin operator $$S^2$$. The relationship between $$S^2$$ and $$S_1^x$$ is clarified, leading to the conclusion that $$S^2$$ is equivalent to $$J^2$$. Since $$J^2$$ commutes with the Hamiltonian $$H$$, it is inferred that each component of the total angular momentum operator $$J$$ also commutes with $$H$$. This highlights the intricate relationships within quantum mechanical operators.
Diracobama2181
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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 $$S_1^x$$?
 
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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.
 

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