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: \begin{equation} [\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} \end{equation} 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: \begin{equation} [S_{i}, S_{j}] = i \epsilon_{ijk} S_{k} ~~,~~ ijk = xyz \end{equation} 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 $?

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    $\begingroup$ 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. $\endgroup$
    – jsborne
    May 17, 2021 at 18:10
  • $\begingroup$ 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 $. $\endgroup$
    – Ricky Pang
    May 18, 2021 at 4:50

1 Answer 1


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.


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