# Hubbard model in the t>>U limit

I know one can obtain the t-J model from the Hubbard one by taking the limit $t\ll U$ in the following Hamiltonian:

$H= -t\sum_{i\neq j}a_{i\sigma}^\dagger a_{j\sigma}+U\sum_i n_{i\uparrow}n_{i\downarrow}$

But how would you go on about solving for the "reciprocal" model, that is, when you take the limit $t\gg U$, using perturbation theory, knowing that you treat the second guy as a perturbation Hamiltonian? I was suggested to go read a chapter in a condensed matter textbook treating the nearly-free electron in 1D but was that a valid reading to do?

• Your first statement is simply wrong: $t-J$ model is NOT obtained in the $t\gg U$ limit, rather in the $t\ll U$ limit. – Meng Cheng Sep 25 '15 at 2:23
• I reversed the signs of the inequalities to reflect what I actually want... – NSERC Protester Sep 25 '15 at 2:28
• For $U\ll t$, we perfectly understand the situations: this is where perturbative expansion really works. The answer is basically a Fermi liquid, if the filling is not at one half (so the Fermi surface is not nesting). – Meng Cheng Sep 25 '15 at 2:51
• The standard way to include weak interactions is the Hartree-Fock approximation. I would try googling for things like "Hartree-Fock condensed matter" or looking for this subject in any condensed matter textbook. – Rococo Sep 25 '15 at 4:22

## 1 Answer

The nearly free electron model is a good thing to read about, but it's slightly different then what you want. The nearly-free electron model starts with non-interacting electrons that are not bound by any periodic potential, and treats the potential as a perturbation. You want to START with non-interacting electrons in a periodic potential, and treat the interactions as a perturbation. Let's just think about what that would entail.

Assuming 1D for the moment (feel free to generalize to higher dimensions), our base Hamiltonian is $H=t\sum_i (c_i^\dagger c_{i+1}+h.c.)$ This Hamiltonian has eigenvectors which are created by the Fourier transform of the creation operators, $c_k^\dagger = \frac{1}{\sqrt{N}}\sum_j e^{ikj}c_j^\dagger$, with energy $E_k=-2t\cos(k)$. To find the first-order energy shift in a state $|\Psi_k>=c^\dagger_k|0>$, we just braket it with our perturbation. The perturbation is $H^1=\sum_i Un_{i\uparrow}n_{i\downarrow}$. If we take $<\Psi_k|H^1|\Psi_k>$, we get zero. You'd expect this, since in the one-electron sector, we expect turning on interactions will do nothing.

If you consider wavefunctions involving two particles, things get more interesting. If the two particles are both spin-up or both spin-down, nothing happens, since $U$ is only an interaction BETWEEN spin up and spin down electrons. So let's consider a state $c_{k\uparrow}^\dagger c_{k'\downarrow}^\dagger|0>$. Then, considering the braket gives

$$\begin{array}{rcl} <0|c_{k\uparrow} c_{k'\downarrow}H^1c_{k\uparrow}^\dagger c_{k'\downarrow}^\dagger|0> & = & \frac{1}{N}\sum_{jll'} U e^{i(kl+k'l')}<0|c_{k\uparrow} c_{k'\downarrow}n_{j\uparrow}n_{j\downarrow}c_{l\uparrow}^\dagger c_{l'\downarrow}^\dagger|0>\\ &=&\frac{1}{N}\sum_{j} U e^{i(k+k')j}<0|c_{k\uparrow} c_{k'\downarrow}n_{j\uparrow}n_{j\downarrow}c_{j\uparrow}^\dagger c_{j\downarrow}^\dagger|0>\\ &=&\frac{1}{N}\sum_{j} U e^{i(k+k')j}<0|c_{k\uparrow} c_{k'\downarrow}c_{j\uparrow}^\dagger c_{j\downarrow}^\dagger|0>\\ &=&\frac{1}{N^2}\sum_{jll'} U e^{i(k+k')j}e^{-i(kl+k'l')}<0|c_{l\uparrow} c_{l'\downarrow}c_{j\uparrow}^\dagger c_{j\downarrow}^\dagger|0>\\ &=&\frac{1}{N^2}\sum_{j} U e^{i(k+k')j}e^{-i(k+k')j}<0|c_{j\uparrow} c_{j\downarrow}c_{j\uparrow}^\dagger c_{j\downarrow}^\dagger|0>\\ &=&\frac{U}{N} \end{array}$$

Where $N$ is the number of sites.

I may have done some of the math wrong, but that's the idea. You can generalize to 3+ electron states if you really feel like it.

Let me know/feel free to edit if you see any algebra mistakes!