A Simultanious eigenstate of Hubbard Hamiltonian and Spin operator in tw

schwarzg

Summary
Hubbard Hamiltonian seems not to commute to S^2 operator. help

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?

<Moderator's note: use  to enclose inlined equations, not .>

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DrClaude

Mentor
$\Rightarrow$ Thus, eigenstates of $H_\text{hub}$ are also eigenstates of $\vec{S}^2$.
That is incorrect. Commutating operators have a common basis of eigenstates, but that does not mean that any eigenstate of one operator is also an eigenstate of the other. In the case of degenerate eigenstates of one operator, there might be only certain linear combinations of these eigenstates that will result in eigenstates of the other operator. This appears to be what you have here.

schwarzg

In the case of degenerate eigenstates of one operator, there might be only certain linear combinations of these eigenstates that will result in eigenstates of the other operator.
Thank you for your explanation. However, the ground state of the Hubbard model is non-degenerated. Thus, it is not the case you mentioned. Thus I still guess that $|g\rangle$ must be an eigenstate of $\vec{S}^2$ still. Is there are another error in my logic?

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