Phy5670/HubbardModel 2DCalculations

Hubbard Model: 2D Calculations

Expansion of the Hubbard model Hamiltonian into two dimensions allows us to calculate various properties, including the grand canonical potential and corrections to the chemical potential. 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:

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

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

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

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

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

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

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