View source for Phy5670/HubbardModel 2DCalculations

=Hubbard Model: 2D Calculations=

Expansion of the Hubbard model Hamiltonian into two dimensions allows us to calculate various properties. In 2D, the Hamiltonian can be written as:

<math> H = -t \sum_{r=1}^{n_x} \sum_{s=1}^{n_y} \sum_{\sigma = \uparrow,\downarrow}  
(c_{r,s,\sigma}^{\dagger} c_{r+1,s,\sigma} + c_{r,s,\sigma}^{\dagger} c_{r,s+1,\sigma} + h.c. )
+ U \sum_{r=1}^{n_x} \sum_{s=1}^{n_y} c_{r,s,\uparrow}^{\dagger} c_{r,s,\downarrow}^{\dagger}c_{r,s,\downarrow}c_{r,s,\uparrow} </math>

The grand canonical potential, Omega, is best calculated by using coherent state path integral. The grand partition function is defined as:

<math> Z = Tr \big[e^{-\beta (H - \mu N)} \big] = e^{-\beta \Omega} </math>

which can be expanded as:

<math> Z = Z_{0} \big\langle e^{-S_{int}} \big\rangle = Z_{0} e^{-\langle S_{int} \rangle} e^{\frac{1}{2}( \langle S_{int}^2 \rangle - \langle S_{int} \rangle^2)} = e^{-\beta \Omega_0}e^{-\beta \Omega_1}e^{-\beta \Omega_2} </math>

<math> S_{int} = \int_{0}^{\beta} d\tau H_{int} (\tau) </math>

which utilizes cumulant expansion. We begin to calculate the grand canonical potential by analyzing the contribution from <math> Z_0 </math>:

<math> Z_{0} = \prod_k (1+e^{-\beta(E_k - \mu)})^2 = e^{2 \sum_k ln(1+e^{-\beta(E_k - \mu)})} = e^{-\beta \Omega_0} </math>

<math> \Omega_0 = -\frac{2}{\beta} \sum_k ln(1+e^{-\beta(E_k - \mu)}) </math>

Now we look at the contribution from the first order cumulant expansion. First we'll need to convert Hint to momentum space:

<math> c_{r,s,\sigma} = \frac{1}{\sqrt{M}} \sum_{k_x,k_y} e^{i k_x \cdot r} e^{i k_y \cdot s} c_{k_x,k_y,\sigma} </math>

<math> H_{int} = U \sum_{r=1}^{n_x} \sum_{s=1}^{n_y} c_{r,s,\uparrow}^{\dagger} c_{r,s,\downarrow}^{\dagger}c_{r,s,\downarrow}c_{r,s,\uparrow} </math>

<math> = \frac{U}{M^2} \sum_{k_{x_1}...k_{x_4}} \sum_{k_{y_1}...k_{y_4}} e^{-ik_{x_1} r}e^{-ik_{x_2} r}e^{ik_{x_3} r}e^{ik_{x_4} r}e^{-ik_{y_1} s}e^{-ik_{y_2} s}e^{ik_{y_3} s}e^{ik_{y_4} s} c_{k_{x_1},k_{y_1},\uparrow}^{\dagger} c_{k_{x_2},k_{y_2},\downarrow}^{\dagger}c_{k_{x_3},k_{y_3},\downarrow}c_{k_{x_4},k_{y_4},\uparrow} </math>

<math> = \frac{U}{M} \sum_{k_{x_1}...k_{x_4}} \sum_{k_{y_1}...k_{y_4}} \delta_{k_{x_1}+k_{x_2},k_{x_3}+k_{x_4}} \delta_{k_{y_1}+k_{y_2},k_{y_3}+k_{y_4}} c_{k_{x_1},k_{y_1},\uparrow}^{\dagger} c_{k_{x_2},k_{y_2},\downarrow}^{\dagger}c_{k_{x_3},k_{y_3},\downarrow}c_{k_{x_4},k_{y_4},\uparrow} </math>

For simplicity, we will combine the <math> k_x </math> and <math> k_y </math> into a single index as <math> k </math>. Evaluating the Kronecker deltas yields: 

<math> H_{int} = \frac{U}{M} \sum_{k,k',q} c_{k,\uparrow}^{\dagger} c_{k',\downarrow}^{\dagger} c_{k'+q,\downarrow} c_{k-q,\uparrow} </math>

The only contraction combination possible, due to orthogonal spins, results in the following set of Green's functions:

<math> \langle S_{int} \rangle = \bigg \langle \int_{0}^{\beta} d\tau \frac{U}{M} \sum_{k,k',q} c_{k,\uparrow}^{\dagger}(\tau) c_{k',\downarrow}^{\dagger}(\tau) c_{k'+q,\downarrow}(\tau) c_{k-q,\uparrow}(\tau) \bigg \rangle </math>

<math> = \int_{0}^{\beta} d\tau \frac{U}{M} \sum_{k,k'} \big \langle \mathcal{G}_{0}(k\tau,k\tau)\mathcal{G}_{0}(k'\tau,k'\tau) \big \rangle </math>

<math> = \int_{0}^{\beta} d\tau \frac{U}{M} \sum_{k,k'} n_{F}(\epsilon_{k}) n_{F}(\epsilon_{k'}) </math>

<math> = {\beta} \frac{U}{M} \sum_{k,k'} n_{F}(\epsilon_{k}) n_{F}(\epsilon_{k'}) = \beta \Omega_1 </math>

<math> \Omega_1 = \frac{U}{M} \sum_{k,k'} n_{F}^2(\epsilon_k) </math>

Combining both terms, the grand canonical potential to first order is:

<math> \Omega = \Omega_0 + \Omega_1 = -\frac{2}{\beta} \sum_k ln(1+e^{-\beta(E_k - \mu)}) + \frac{U}{M} \sum_{k} n_{F}^2(\epsilon_k) </math>