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:

$H=-t\sum _{r=1}^{n_{x}}\sum _{s=1}^{n_{y}}\sum _{\sigma =\uparrow ,\downarrow }(c_{r,s,\sigma }^{\dagger }c_{r+1,s,\sigma }+c_{r,s,\sigma }^{\dagger }c_{r,s+1,\sigma }+h.c.)+U\sum _{r=1}^{n_{x}}\sum _{s=1}^{n_{y}}c_{r,s,\uparrow }^{\dagger }c_{r,s,\downarrow }^{\dagger }c_{r,s,\downarrow }c_{r,s,\uparrow }$

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

$Z=Tr{\big [}e^{-\beta (H-\mu N)}{\big ]}=e^{-\beta \Omega }$

which can be expanded as:

$Z=Z_{0}{\big \langle }e^{-S_{int}}{\big \rangle }=Z_{0}e^{-\langle S_{int}\rangle }e^{{\frac {1}{2}}(\langle S_{int}^{2}\rangle -\langle S_{int}\rangle ^{2})}=e^{-\beta \Omega _{0}}e^{-\beta \Omega _{1}}e^{-\beta \Omega _{2}}$

$S_{int}=\int _{0}^{\beta }d\tau H_{int}(\tau )$

which utilizes cumulant expansion. We begin to calculate the grand canonical potential by analyzing the contribution from $Z_{0}$:

$Z_{0}=\prod _{k}(1+e^{-\beta (E_{k}-\mu )})^{2}=e^{2\sum _{k}ln(1+e^{-\beta (E_{k}-\mu )})}=e^{-\beta \Omega _{0}}$

$\Omega _{0}=-{\frac {2}{\beta }}\sum _{k}ln(1+e^{-\beta (E_{k}-\mu )})$

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

$c_{r,s,\sigma }={\frac {1}{\sqrt {M}}}\sum _{k_{x},k_{y}}e^{ik_{x}\cdot r}e^{ik_{y}\cdot s}c_{k_{x},k_{y},\sigma }$

$H_{int}=U\sum _{r=1}^{n_{x}}\sum _{s=1}^{n_{y}}c_{r,s,\uparrow }^{\dagger }c_{r,s,\downarrow }^{\dagger }c_{r,s,\downarrow }c_{r,s,\uparrow }$

$={\frac {U}{M^{2}}}\sum _{k_{x_{1}}...k_{x_{4}}}\sum _{k_{y_{1}}...k_{y_{4}}}e^{-ik_{x_{1}}r}e^{-ik_{x_{2}}r}e^{ik_{x_{3}}r}e^{ik_{x_{4}}r}e^{-ik_{y_{1}}s}e^{-ik_{y_{2}}s}e^{ik_{y_{3}}s}e^{ik_{y_{4}}s}c_{k_{x_{1}},k_{y_{1}},\uparrow }^{\dagger }c_{k_{x_{2}},k_{y_{2}},\downarrow }^{\dagger }c_{k_{x_{3}},k_{y_{3}},\downarrow }c_{k_{x_{4}},k_{y_{4}},\uparrow }$

$={\frac {U}{M}}\sum _{k_{x_{1}}...k_{x_{4}}}\sum _{k_{y_{1}}...k_{y_{4}}}\delta _{k_{x_{1}}+k_{x_{2}},k_{x_{3}}+k_{x_{4}}}\delta _{k_{y_{1}}+k_{y_{2}},k_{y_{3}}+k_{y_{4}}}c_{k_{x_{1}},k_{y_{1}},\uparrow }^{\dagger }c_{k_{x_{2}},k_{y_{2}},\downarrow }^{\dagger }c_{k_{x_{3}},k_{y_{3}},\downarrow }c_{k_{x_{4}},k_{y_{4}},\uparrow }$

For simplicity, we will combine the $k_{x}$ and $k_{y}$ into a single index as $k$. Evaluating the Kronecker deltas yields:

$H_{int}={\frac {U}{M}}\sum _{k,k',q}c_{k,\uparrow }^{\dagger }c_{k',\downarrow }^{\dagger }c_{k'+q,\downarrow }c_{k-q,\uparrow }$

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

$\langle S_{int}\rangle ={\bigg \langle }\int _{0}^{\beta }d\tau {\frac {U}{M}}\sum _{k,k',q}c_{k,\uparrow }^{\dagger }(\tau )c_{k',\downarrow }^{\dagger }(\tau )c_{k'+q,\downarrow }(\tau )c_{k-q,\uparrow }(\tau ){\bigg \rangle }$

$=\int _{0}^{\beta }d\tau {\frac {U}{M}}\sum _{k,k'}{\big \langle }{\mathcal {G}}_{0}(k\tau ,k\tau ){\mathcal {G}}_{0}(k'\tau ,k'\tau ){\big \rangle }$

$=\int _{0}^{\beta }d\tau {\frac {U}{M}}\sum _{k,k'}n_{F}(\epsilon _{k})n_{F}(\epsilon _{k'})$

$={\beta }{\frac {U}{M}}\sum _{k,k'}n_{F}(\epsilon _{k})n_{F}(\epsilon _{k'})=\beta \Omega _{1}$

$\Omega _{1}={\frac {U}{M}}\sum _{k,k'}n_{F}^{2}(\epsilon _{k})$

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

$\Omega =\Omega _{0}+\Omega _{1}=-{\frac {2}{\beta }}\sum _{k}ln(1+e^{-\beta (E_{k}-\mu )})+{\frac {U}{M}}\sum _{k}n_{F}^{2}(\epsilon _{k})$