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

Please see this page and give me an advice.

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 $ $.>

### Simultanious eigenstate of Hubbard Hamiltonian and Spin operator in two-site model

Known fact 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 $B|a,b\rangle=b|a,b\rangle$. Hubbard Hamiltonian $H_\te...

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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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