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The Hubbard model reads $$H = -t \sum_{\langle ij \rangle, \sigma} c_{j\sigma}^\dagger c_{i\sigma} + U\sum_i n_{i\uparrow}n_{i\downarrow} $$

In the large $U$ limit and at half-filling, the Hubbard model tranforms to $t$-$J$ model $$H = -t \sum_{\langle ij \rangle, \sigma} c_{j\sigma}^\dagger c_{i\sigma} + J\sum_{\langle ij \rangle}S_i\cdot S_j$$ where $J=4t^2/U$. Obviously, ground state favors anti-parallel spin configurations but it is not obvious to see whether it is the Neel state $(\uparrow\downarrow\uparrow\downarrow...) - (\downarrow\uparrow\downarrow\uparrow...)$ or the dimerized state $(\uparrow\downarrow)(\uparrow\downarrow)...$ where $(\uparrow\downarrow) = \uparrow\downarrow - \downarrow\uparrow$. Calculation shows, in 1-D, dimerized state has lower energy (I'm not very clear about the calculation.)

It is also known that holon and doublon are not bounded in 1D (absence of Mott transition, Lieb-Wu's analytic result + VMC simulations). While in higher dimensions, Mott transitions do occur and dimerization is suppressed.

It seems that the occurance of dimerization is correlated with the absence of Mott transition. So are there any pictures that show why Neel states lead to bounded doublon-holon?

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At half-filling, the Hubbard model is reduced to the Heisenberg model without hopping, not the $t$-$J$ model. In the large $U$ limit, holons and doublons are simply gapped from low-energy. The dimerization/Neel order and the Mott transition are not related. Because there are dimerized Mott insulators and spin-liquid (Mott insulator without Neel order) in 2D, which are conter examples to your statement. The 1D system is just a special case. –  Everett You May 12 '13 at 14:53
    
I calculated two-site Hubbard model and found the double occupation is $\sim (t/U)^2$. As long as $U/t$ is finite, double occupation is finte, there is no Mott transition in this sense. I argue that the dimerized ground state can be viewed as decoupled two-site Hubbard models and so there is no Mott transition. My point is that dimerization -> no Mott transition. If my arguement is false, then what's the physical mechanism that bound holons and doublons? –  ChenChao May 13 '13 at 2:42
    
In your opinion, if the wave function contains finite weight of the double occupation component, then it is not Mott. Then there would be no Mott insulator in any case, because as long as $U/t$ is finite, the double occupancy weight will not vanish. 1. Mott transition is a phase transition of many-body system and can not be understood locally. 2. The character of Mott is the Mott gap, not the vanishing weight of double occupancy. Even if the wave function contains double occupancy, its excitation can still be gapped, and hence in a Mott phase. –  Everett You May 13 '13 at 5:43
    
I see, thanks! Suppose I could show that in the variational ansatz, the doublon and holon were attractive to each other, will it be convincing to predict the Mott transition even though I do not scan the filling? –  ChenChao May 13 '13 at 15:14
    
I think attraction is not important. One should show that the doublon and holon are gapped. A metal gets Mott simply because of the development of the charge gap, but not the confinement of charge. –  Everett You May 14 '13 at 2:47

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