Known fact

  1. If two operators $A$ and $B$ commute, $[A,B]=0$, they have simultaneous eigenstates. That means $A|a,b\rangle=a|a,b\rangle$ and $B|a,b\rangle=b|a,b\rangle$.
  2. Hubbard Hamiltonian $H_\text{hub}$ is symmetric w.r.t. SU(2) spin operators. Thus $[H,S^z]=[H,\vec{S}^2]=0$.

$\Rightarrow$ Thus, eigenstates of $H_\text{hub}$ are also eigenstates of $\vec{S}^2$.

To verify the above conclusion, I set a two-site Hubbard model with one $\uparrow$ and one $\downarrow$ fermions.
$$H=-t\sum_{\sigma=\uparrow,\downarrow}(c^\dagger_{1\sigma}c_{2\sigma}+h.c.)+U\sum_{i=1,2}(n_{i\uparrow} n_{i,\downarrow})$$ Without interaction, the ground state of the Hamiltonian is given by $$|g\rangle=\frac{1}{2}(|\uparrow\downarrow ,0\rangle+|\uparrow,\downarrow\rangle+|\downarrow,\uparrow\rangle+|0,\uparrow\downarrow\rangle)$$
The total spin operator $\vec{S}^2$ is written in fermionic operator as $$\vec{S}^2=\frac{1}{2}(S^+ S^- + S^- S^+)+(S^z)^2\\=\sum_{i=1,2}\sum_{j=1,2}[\frac{1}{2}(S_i^+ S_j^- + S_i^- S_j^+)+S_i^z S_j^z]$$ where each local operators are $$ S^+_i=c^\dagger_{i,\uparrow} c_{i,\downarrow}\\ S^-_i=c^\dagger_{i,\downarrow} c_{i,\uparrow}\\ S^z_i=\frac{1}{2}(c^\dagger_{i,\uparrow} c_{i,\uparrow}-c^\dagger_{i,\downarrow} c_{i,\downarrow}) $$ Since $|g\rangle$ is an eigenstate of $H$, so it must be an eigenstate of $\vec{S}^2$. However, for each basis, we have $$\vec{S}^2|\uparrow\downarrow,0\rangle=0\\ \vec{S}^2|\uparrow,\downarrow\rangle=|\uparrow,\downarrow\rangle+|\downarrow,\uparrow\rangle\\ \vec{S}^2|\downarrow,\uparrow\rangle=|\uparrow,\downarrow\rangle+|\downarrow,\uparrow\rangle\\ \vec{S}^2|0,\uparrow\downarrow\rangle=0$$ Thus we have $$\vec{S}^2|g\rangle=|\uparrow,\downarrow\rangle+|\downarrow,\uparrow\rangle\\ \neq j(j+1)|g\rangle $$

Where does the contradiction come from?

  • $\begingroup$ While I don't think this is the source of your problem, it is worth noting that it is not true that if 2 operators commute then all eigenstates of one are eigenstates of the other. It is instead true that there exists a basis of mutual eignestates for the 2 operators, but this is a weaker statement. For example if I have 2 states $|a,b\rangle$ and $|a,b'\rangle$ with $b\ne b'$ then $|a,b\rangle + |a,b'\rangle$ is an eigenstate of $A$ but not $B$. This is possible due to the degeneracy of the eigenvalue $a$. $\endgroup$ – By Symmetry Aug 29 at 11:31
  • $\begingroup$ You are right for the degenerated case. However, the ground state of the two-site Hubbard model is non-degenerated. Such a non-degenerate state can be also the case that you mensioned? $\endgroup$ – Juhee Lee Aug 30 at 2:37

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