# $SU(2)$ Symmetry of Hubbard Model

I am confused with the $$SU(2)$$ spin rotation symmetry of the fermion Hubbard Hamiltonian. If the Hubbard model has $$SU(2)$$ rotational symmetry, it means that the Hubbard Hamiltonian commutes with the global spin operator in all direction: $$$$[\vec S, H] = 0 ~~,~~ \vec S = \frac{1}{2}\sum_{i} \begin{pmatrix} c^{\dagger}_{i \uparrow} & c^{\dagger}_{i \downarrow} \end{pmatrix} \vec{\sigma} \begin{pmatrix} c_{i \uparrow} \\ c_{i \downarrow} \end{pmatrix}$$$$ Where $$\vec \sigma$$ is the Pauli matrices in vector form. My confusion is that whether $$[\vec S, H] = 0$$ implies $$[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0$$. I have proved that they are equal to zero but I thought that my calculation was wrong. The reason why I think my calculation is wrong since if both $$S_{x}, S_{y} , S_{z}$$ commutes with $$H$$, it means that they simultaneously share the same eigenstates and $$[S_x, S_y] = 0$$. However, we know that from elementary QM course: $$$$[S_{i}, S_{j}] = i \epsilon_{ijk} S_{k} ~~,~~ ijk = xyz$$$$ In my opinion, $$[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0$$ is not true because they cannot satisfy $$SU(2)$$ commutation relation if all commutes with $$H$$. May I know is it true that $$[\vec S, H] = 0$$ implies $$[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0$$?

• In general, it is not true that if $[A,B]=[B, C]=0$, that $[A,C]=0$! For example, take $B=I$ the identity matrix. Everything commutes with the identity but not necessarily with each other. May 17, 2021 at 18:10
• Thank you for your comment . Yes, you are right. I originally thought that there was transitivity in the commutator. I realise that they share the simultaneous eigenstates if all commutators equal zero( e.g. $[H,S_{x,y, z}] = 0$ and $[S_{i} , S_{j}] = 0$. However, due to SU(2) group property, they cannot form the simultaneous eignestates since $[S_{i} , S_{j}] \neq 0$. May 18, 2021 at 4:50

From what I understand of the standard notation, the statement that $$[\vec{S}, H] = 0$$ is exactly the same as saying $$[S_j, H] =0$$ for all $$j$$. Then, to give a proof that a vector operator commutes with some operator simply amounts to proving each component commutes with the given operator.