1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Meng Cheng Sep 25 '15 at 2:23
  • $\begingroup$ I reversed the signs of the inequalities to reflect what I actually want... $\endgroup$ – NSERC Protester Sep 25 '15 at 2:28
  • $\begingroup$ 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). $\endgroup$ – Meng Cheng Sep 25 '15 at 2:51
  • $\begingroup$ 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. $\endgroup$ – Rococo Sep 25 '15 at 4:22
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.