[Fradkin - Condenced Matter] 2. The Hubbard model
Vibrations of a One-Dimensional Monatomic Chain
(The Oxford Solid State Basics)
Let us consider a chain of identical atoms of mass $m$ where the equilibrium spacing between atoms is $a$. Let us define the position of the $n$-th atom to be $x_n$ and the equilibrium position of the $n$-th atom to be $x^{eq}_n=na$. We define the small variable $$\delta x_n=x_n-x^{eq}_n.$$
If the system is at low enough temperature we can consider the potential holding the atoms together to be quadratic. With the quadratic interatomic potential, we can write the total potential energy of the chain to be $$V_{tot}=\sum_i V(x_{i+1}-x_i)=\sum_i \frac{\kappa}{2}(x_{i+1}-x_i-a)^2 = \sum_i \frac{\kappa}{2}(\delta x_{i+1}-\delta x_i)^2.$$ The force on the $n$-th mass on the chain is then given by $$F_n=-\frac{\partial V_{tot}}{\partial x_n}=\kappa(\delta x_{n+1}-\delta x_n)+\kappa (\delta x_{n-1}-\delta x_n).$$ Thus we have Newton's equation of motion $$m(\delta \ddot{x}_n)=F_n=\kappa(\delta x_{n+1}+\delta x_{n-1}-2\delta x_n).$$
A normal mode is defined to be a collective oscillation where all particles move at the same frequency. We now attempt a solution to Newton's equations by using an ansatz that describes the normal modes as waves $$\delta x_n=A e^{i\omega t-ik x^{eq}_n}=Ae^{i\omega t-ikna}$$ where $A$ is an amplitude of oscillation, and $k$ and $\omega$ are the wavevector and frequency of the proposed wave. (We also use 'real part' of complex number $\delta x_n$.)
Dispersion relation is periodic in $k\rightarrow k+2\pi/a$ since real space is periodic. To see this, recall $$\delta x_n=A e^{i\omega t-ikna}$$ and put $k\rightarrow k+2\pi/a$ then $$\delta x_n=Ae^{i\omega t-i(k+2\pi/a)na}=Ae^{i\omega t-ikna}e^{-i2\pi n}=Ae^{i\omega t-ikna}$$ where here we have used $$e^{-i2\pi n}=1$$ for any integer $n$. In conclusion, shifting $k\rightarrow k+2\pi/a$ gives us back exactly the same oscillation mode the we had before we shifted $k$. The periodic unit in $k$-space is the Brilloin zone. The "first Brilloiun zone" is a unit cell in $k$-space centered around the point $k=0$.
If a classical harmonic system has a normal oscillation mode at frequency $\omega$ the corresponding quantum system will have eigenstates with energy $$E_n=\hbar \omega(n+\frac{1}{2}).$$
Thus at a given wavevector $k$, there are many possible eigenstates. Each excitation of this "normal mode" by a step up the harmonic oscillator excitation ladder is known as a "phonon". A phonon is a discrete quantum of vibration. In QFT, we understand this as particles of momentum space.
Bloch Theorem
Electron is in periodic potential, then $$\psi(r)=e^{ik\cdot r}u(r)$$
Introduction
The Bloch state of energy $\epsilon_p$, momentum $\vec{p}$, and index $\alpha$ has a wavefunction $\Psi_{\vec{p},\alpha}$, one can construct Wannier states $$\Psi_\alpha(\vec{r}_i)=\frac{1}{\sqrt{N}}\sum_{\vec{p}\in BZ}e^{i\vec{p}\cdot \vec{r}_i}\Psi_{\vec{p},\alpha}(\vec{r}_i)$$ where $\vec{r}_i$ is the location of the $i$th atom and BZ is the Brillouin zone. The Coulomb interaction matrix elements are $$U_{ij,i'j'}=\int d^3r_1d^3r_2\Psi^*_i(\vec{r}_1)\Psi^*_j(\vec{r}_2)\tilde{V}(\vec{r}_1-\vec{r}_2)\Psi_{i'}(\vec{r}_1)\Psi_{j'}(\vec{r}_2)$$ where $\tilde{V}$ is the (screened) Coulomb interaction.
The second quantized Hamiltonian tight binding (in the Wannier-functions basis) is $$H=-\sum_{\vec{r}_i,\vec{r}_j\\ \sigma=\uparrow,\downarrow}\left(c^\dagger_\sigma(\vec{r}_i)t_{ij}c_\sigma(\vec{r}_j)+c^\dagger_\sigma(\vec{r}_j)t_{ij}c_\sigma(\vec{r}_i)\right) \\+ \frac{1}{2}\sum_{i,j,i',j'\\ \sigma,\sigma'=\uparrow,\downarrow}U_{ij,i'j'}c^\dagger_\sigma(\vec{r}_i)c^\dagger_{\sigma'}(\vec{r}_j)c_{\sigma'}(\vec{r}_{j'})c_\sigma(\vec{r}_{i'})$$ where $c^\dagger_\sigma(\vec{r})$ creates an electron at site $\vec{r}$ with spin $\sigma$ and satisfies $$\left\{ c_\sigma(\vec{r}),c^\dagger_{\sigma'}(\vec{r}')\right\}=\delta_{\sigma,\sigma'}\delta_{\vec{r},\vec{r}'}\\ \left\{ c_\sigma(\vec{r}),c_{\sigma'}(\vec{r}')\right\}=0$$
We hope to restrict $t_{ij}$ to nearest neighboring sites: $$t_{ij}=\begin{cases}t& \mbox{if }i,j\mbox{ are nearest neighbors}\\ 0 & \mbox{otherwise}\end{cases}$$ and Coulomb interaction is screened except "on-site" term $$U_{ij,i'j'}=U\delta_{ij}\delta_{i'j'}\delta_{ii'}$$ the resulting model Hamiltonian $$H=-t\sum_{\langle \vec{r},\vec{r}'\rangle\\ \sigma=\uparrow,\downarrow}\left(c^\dagger_\sigma(\vec{r})c_\sigma(\vec{r}')+\mbox{h.c.}\right)+U\sum_{\vec{r}}n_\uparrow(\vec{r})n_\downarrow(\vec{r})$$ is known as the one-band Hubbard model. We have introduced $$n_\sigma(\vec{r})=c^\dagger_\sigma(\vec{r})c_\sigma(\vec{r})$$ from the Pauli principle we get $n_\sigma=0,1$ or $n^2_\sigma=n_\sigma$ at every site.
The Hilbert space of this system is the tensor product of only four state per site, representing $|0\rangle$ as nothing, $|\uparrow \rangle$ as an electron spin up, $|\downarrow \rangle$ as an electron with spin down. and $|\uparrow \downarrow\rangle$ as an up-down pair. The spin operator $\vec{S}(\vec{r})$ is defined by $$\vec{S}(\vec{r})=\frac{\hbar}{2}c^\dagger_\sigma(\vec{r})\vec{\tau}_{\sigma\sigma'}c_{\sigma'}(\vec{r})$$ where $\vec{\tau}$ are the Pauli matrices $$\tau_1=\begin{pmatrix}0&1\\1&0\end{pmatrix},\ \tau_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\ \tau_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ The particle number operator at site $\vec{r}$ (or charge) is $$n(\vec{r})=\sum_\sigma n_\sigma(\vec{r})=\sum_\sigma c^\dagger_\sigma(\vec{r})c_\sigma(\vec{r})\equiv c^\dagger_\sigma(\vec{r})1_{\sigma\sigma'}c_{\sigma'}(\vec{r})$$ and the associated total charge $Q$ is given by $$Q=e\sum_{\vec{r}}n(\vec{r})\equiv eN_e$$
Symmetries of the Hubbard model
$SU(2)$ spin
Suppose we rotate the local spin basis $$c'_\sigma(\vec{r})=U_{\sigma\sigma'}c_{\sigma'}(\vec{r})$$ where $U$ is a $2\times 2$ $SU(2)$ matrix. Namely, given four complex numbers $a,b,c$, and $d$, the matrix $U$ given by $$U=\begin{pmatrix}a&c\\ d&b\end{pmatrix}$$ must satifsy $$U^{-1}=U^\dagger\equiv \left( U^T\right)^*$$ together with the condition $$\det U=1$$ We will parametrize the matrix $U(\vec{\theta})$ as follows: $$U(\vec{\theta})=e^{i\vec{\theta}\cdot\vec{\tau}}=1\cos |\vec{\theta}|+i\sin |\vec{\theta}|\frac{\vec{\theta}\cdot\vec{\tau}}{|\vec{\theta}|}$$ where $1$ represent the $2\times 2$ identity matrix, $\vec{\tau}$ are the Pauli matrices, and the Euler angles $\vec{\theta}=(\theta_1,\theta_2,\theta_3)$ parametrize the $SU(2)$ group.
Under such a unitary transformation, the spin $\vec{S}$ transforms as follows: $$S'^a(\vec{r})=R^{ab}S^b(\vec{r})=\frac{\hbar}{2}c'^\dagger(\vec{r})\tau^ac'(\vec{r})=\frac{\hbar}{2}c^\dagger(\vec{r})\left( U^{-1}\tau^a U\right) c(\vec{r})$$ where $R^{ab}$ is a rotation matrix induced by the $SU(2)$ transformation of the fermions: $$U^{-1}\tau^a U=R^{ab}\tau^b$$
The axis of quantization can be chosen arbitrarily. Thus, the Hubbard model Hamiltonian shouls not change its from under a rotation of the spin quantization axis. This is not apparent in the standard form of the interaction $$H_1=U\sum_{\vec{r}}n_\uparrow(\vec{r})n_\downarrow(\vec{r})$$ But we can write this in a somewhat different form in which the $SU(2)$ symmetry becomes explicit. Consider the operator $$\sum_{\vec{r}}\left( \vec{S}(\vec{r})\right)^2=\sum_{\vec{r}\\ a=1,2,3} S^a(\vec{r})S^a(\vec{r})$$ By expanding the components and making use of the $SU(2)$ identity $$\sum_{a=1,2,3}\tau^a_{\alpha\beta}\tau^a_{\gamma\delta}=2\delta_{\alpha\delta}\delta_{\beta\gamma}-\delta_{\alpha\beta}\delta_{\gamma\delta}$$ one gets $$\sum_{\vec{r}}\left( \vec{S}(\vec{r})\right)^2=\sum_{\vec{r}}\left(\frac{3}{4}n(\vec{r})-\frac{3}{2}n_\uparrow(\vec{r})n_\downarrow(\vec{r})\right)$$ Thus, we can write $$H_1=U\sum_{\vec{r}}n_\uparrow(\vec{r})n_\downarrow(\vec{r})=-\frac{2U}{3}\vec{S}^2(\vec{r})+\frac{N_eU}{2}$$ The last term is a constant, which can be dropped. The Hamiltonian now has the form $$H=-t\sum_{\langle \vec{r},\vec{r}'\rangle\\ \sigma=\uparrow,\downarrow}\left( c^\dagger_\sigma(\vec{r})c_\sigma(\vec{r}')+\mbox{h.c.}\right)-\frac{2U}{3}\sum_{\vec{r}}\left( \vec{S}(\vec{r})\right)^2+\frac{N_eU}{2}$$ which is manifestly $SU(2)$-invariant.
$U(1)$ charge
We are free to change the phase of the one-particle wavefunction $$c'_\sigma(\vec{r})=e^{i\theta}c_\sigma(\vec{r})$$ Here, $e^{i\theta}$ is an element of the group $U(1)$, and group elements satisfy $$e^{i\theta}e^{i\theta'}=e^{i(\theta+\theta')}$$ The Hamiltonian is invariant under this $U(1)$ transformation. This is nothing but charge conservation. For example, if we had terms that would not conserve charge, like $$c^\dagger_\uparrow (\vec{r})c^\dagger_\downarrow (\vec{r}')\ \rightarrow \ e^{i2\theta}c^\dagger_\uparrow (\vec{r})c^\dagger_\downarrow (\vec{r}')$$ we would not have this invariance.
Suppose not that we couple this system to the electromagnetic field $(A_0,\vec{A})$. We expect theree effects.
1. A Zeeman coupling given by $$H_{Zeemann}=g\sum_{\vec{r}}\vec{S}(\vec{r})\cdot \vec{B}(\vec{r})$$ which couples the spin $\vec{S}(\vec{r})$ with the local magnetic field $\vec{B}(\vec{r})$ so as to alogn it along the $\vec{B}(\vec{r})$ direction.
2. An orbitarl coupling for electrons in a crystal with one-particle Hamiltonian $$H(\vec{p})=\frac{1}{2m_e}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2+V(\vec{r})$$ where $V(\vec{r})$ is the periodic potential imposed by the crystal. In the tight-binding approximation, we must therefore modify the kinetic-energy term according to $$H_0\equiv -t\sum_{\langle \vec{r},\vec{r}\rangle \\ \sigma=\uparrow,\downarrow} \left(c^\dagger_\sigma(\vec{r})c_\sigma(\vec{r}')+\mbox{h.c.}\right)\\ \rightarrow -t\sum_{\langle \vec{r},\vec{r}'\rangle\\ \sigma=\uparrow,\downarrow} \left( c^\dagger_\sigma(\vec{r})e^{\frac{ie}{\hbar c}\int^{\vec{r}'}_{\vec{r}}d\vec{x}\cdot \vec{A}(\vec{x})}c_\sigma(\vec{r}')+c^\dagger_\sigma(\vec{r}')e^{-\frac{ie}{\hbar c}\int^{\vec{r}'}_{\vec{r}}d\vec{x}\cdot \vec{A}(\vec{x})}c_\sigma(\vec{r})\right)$$ We should now check the gauge invariance under the transformation $$\vec{A}'=\vec{A}+\vec{\nabla}\Lambda$$ where $\Lambda$ is an arbitrary function of space and time. We get the change $$A'(\vec{r},\vec{r}')\equiv \int^{\vec{r}'}_{\vec{r}}d\vec{x}\cdot \vec{A}'(\vec{x})=A(\vec{r},\vec{r}')+\Lambda(\vec{r}')-\Lambda(\vec{r})$$ Thus the kinetic-energy temr is gauge-invariant, $$H'_0\equiv -t\sum_{\langle \vec{r},\vec{r}'\rangle\\ \sigma=\uparrow,\downarrow}\left({c'}_\sigma^\dagger(\vec{r})e^{\frac{ie}{\hbar c}A'(\vec{r},\vec{r}')}c'_\sigma(\vec{r}')+\mbox{h.c.}\right)\\ =-t\sum_{\langle \vec{r},\vec{r}'\rangle\\ \sigma=\uparrow,\downarrow}\left( c^\dagger_\sigma(\vec{r})e^{-i\theta(\vec{r})}e^{\frac{ie}{\hbar c}\left(A(\vec{r},\vec{r}')+\Lambda(\vec{r}')-\lambda(\vec{r})\right)}e^{+i\theta(\vec{r}')}c_\sigma(\vec{r}')+\mbox{h.c.}\right)$$ provided that the local change of phase is given by $$\theta(\vec{r})\equiv -\frac{e}{\hbar c}\Lambda(\vec{r})$$
3. An electrostatic coupling given by $$H_{electrostatic}=\sum_{\vec{r},\sigma}eA_0(\vec{r})c^\dagger_\sigma(\vec{r})c_\sigma(\vec{r})$$ which couples the particle density to $A_0(\vec{r})$.
Particle-hole transformation
In the case of a bipartite lattice (i.e. a lattice that is the union of two interpenetrating sublattices $A$ and $B$) we get additional symmetries.
1. First, the sign of $t$ can be changed. Consider the transformation $$c_\sigma(\vec{r})\rightarrow +c_\sigma(\vec{r})\quad \mbox{if}\quad \vec{r}\in A\\ c_\sigma(\vec{r})\rightarrow -c_\sigma(\vec{r})\quad \mbox{if}\quad \vec{r}\in B$$ under which the kinetic energy charges sign: $$tc^\dagger_\sigma(\vec{r})c_\sigma(\vec{r}')\rightarrow -tc^\dagger_\sigma(\vec{r})c_\sigma(\vec{r}'),\quad \vec{r}\in A,\vec{r}'\in B$$ while the potential energy is left unchanged. This transformation leaves the canonical commutation relations unchanged and therefore leaves the spectrum unchanged.
2. Now consider the particle-hole transformation $$c_\uparrow(\vec{r})=d_\uparrow (\vec{r})\\ c_\downarrow(\vec{r})=\begin{cases}+d^\dagger_\downarrow (\vec{r}),& \vec{r}\in A\\ -d^\dagger_\downarrow (\vec{r}),& \vec{r}\in B\end{cases}$$ The Hamiltonian $H(t,U)$, changes into $H(t,-U)+UN_\uparrow$, where $N_\uparrow$ is the total number of up spins (which is conserved), since under this transformation we get $$n_\uparrow +n_\downarrow = c^\dagger_\uparrow c_\uparrow + c^\dagger_\downarrow c_\downarrow = d^\dagger_\uparrow d_\uparrow + d_\downarrow d^\dagger_\downarrow = d^\dagger_\uparrow d_\uparrow - d^\dagger_\downarrow d_\downarrow + 1$$ and $$n_\uparrow -n_\downarrow = c^\dagger_\uparrow c_\uparrow - c^\dagger_\downarrow c_\downarrow = d^\dagger_\uparrow d_\uparrow - d_\downarrow d^\dagger_\downarrow = d^\dagger_\uparrow d_\uparrow + d^\dagger_\downarrow d_\downarrow - 1$$ Similarly, the charge $Q$ and the component $S_z$ of the total spin transform as $$Q\rightarrow S_z+1,\quad S_q\rightarrow Q-1$$
The strong-coupling limit
We consider now the strong-coupling limit of the Hubbard model, i.e. $U\rightarrow \infty$.
The half-filled system (U>0)
Recall that the interaction term $$H_{int}\equiv -\frac{2}{3}U\sum_{\vec{r}}\left(\vec{S}(\vec{r})\right)^2$$ forces the spin $\vec{S}$ to be largest if $U$ becomes infinitely large, i.e., doubly occupied sites are forbidden. Only $|\uparrow\rangle$ and $|\downarrow\rangle$ states are kept in this large-$U$ limit at half-filling. (Another explain is $n_\uparrow n_\downarrow$ becomes 0.) This make energy eigenstate degeneracy.
Let $H_0$ and $H_1$ denotethekinetic and interaction terms of the Hubbard Hamiltonian $H$, $$H_0=-t\sum_{\langle \vec{r},\vec{r}'\rangle\\ \sigma=\uparrow,\downarrow}\left(c^\dagger_\sigma(\vec{r})c_\sigma(\vec{r}')+\mbox{h.c.}\right)\\ H_1=U\sum_{\vec{r}}n_\uparrow(\vec{r})n_\downarrow(\vec{r})$$ Let $|\alpha\rangle$ be any of the $2^N$ states with every site occupied by a spin either up or down. Here, $|\alpha\rangle$ is an eigenstate of $H_1$ with eigenvalue $E_1=0$. We will use Brillouin-Wigner perturbation theory. Consider Schrodiner's equation $$H|\Psi\rangle=E|\Psi\rangle$$ where $|\Psi\rangle$ is any eigenstate. We can write $$(E-H_1)|\Psi\rangle=H_0|\Psi\rangle$$ Formally, we get $$|\Psi\rangle=\frac{1}{E-H_1}H_0|\Psi\rangle = \frac{\hat{P}}{E-H_1}H_0|\Psi\rangle+\sum_\alpha |\alpha\rangle \frac{\langle \alpha |H_0|\Psi\rangle}{E-E_1}$$ where $$H_1|\alpha\rangle=E_1|\alpha\rangle$$ and $$\hat{P}=1-\sum_\alpha|\alpha\rangle\langle \alpha|$$ projects out of the unperturbed states. Clearly $\hat{P}$ commutes with $H_1$. Define $|\Psi_\alpha\rangle$ as the soultion of the equation $$|\Psi_\alpha\rangle=|\alpha\rangle+\frac{\hat{P}}{E-H_1}H_0|\Psi_\alpha\rangle$$ Let $a_\alpha$ be given by $$a_\alpha=\frac{\langle \alpha|H_0|\Psi\rangle}{E-E_1}$$ Then we can write $$|\Psi\rangle=\sum_\alpha a_\alpha|\Psi_\alpha\rangle$$ If we iterate solution to first order in powers of $[\hat{P}/(E-H_1)]H_0$, we find $$|\Psi_\alpha\rangle\approx |\alpha\rangle + \frac{\hat{P}}{E-H_1}H_0|\alpha\rangle \approx |\alpha\rangle - \frac{1}{U}H_0|\alpha\rangle$$ since $E-H_1\approx -U$ when $U$ is big. AND $\langle \beta |H_0|\alpha\rangle = 0$ at half-filling, Thus, if we inset $|\Psi_\alpha\rangle$ into $|\Psi\rangle$, and in turn insert $a_\alpha$, we get $$(E-E_1)a_\alpha=\frac{1}{|U|}\sum_{\alpha'}\langle \alpha |H^2_0|\alpha'\rangle a_{\alpha'}$$ This is the same as the Schrodinger equation for the Hamiltonian $H'_0=H^2_0/|U|$, where $H'_0$, at half-filling, is given by $$H'_0=\frac{2t^2}{|U|}\sum_{\langle \vec{r},\vec{r}'\rangle}\vec{S}(\vec{r})\cdot \vec{S}(\vec{r}')$$ In other words, we find the spin one-half quantum Heisenberg antiferromagnet with the exchange coupling $J = 2t^2/|U|$. This result is valid for the half-filled system in any dimension and for any lattice.
Away from half-filling
??
The weak-coupling limit
Similar to weakly coupled electron gas, and it called Fermi liquid.
For free fermions, the Hamiltonian reduces to the kinetic-energy term. For the Hubbard model we have $$H_0=-\sum_{\vec{r}\vec{r}'\\ \sigma=\uparrow,\downarrow}\left(c^\dagger_\sigma(\vec{r})t(\vec{r}-\vec{r}')c_\sigma(\vec{r}')+c^\dagger_\sigma(\vec{r}')t(\vec{r}-\vec{r}')c_\sigma(\vec{r})\right)$$ It is convenient to go to Fourier space (momentum). Assume that we are in $d$ space dimensions and that the lattice has $N^d$ sites with $N$ even (for simplicity). With $V\equiv N^d$, we define $$c_\sigma(\vec{r})=\frac{1}{V}\sum_{\vec{k}}e^{i\vec{k}\cdot\vec{r}}c_\sigma(\vec{k})$$ where $$\vec{k}=\frac{2\pi}{N}(n_1,\cdots,n_d)-(\pi,\cdots,\pi)$$ and $1\le n_i\le N$. Thus the momenta $k_i$ vary over the range $-\pi + 2\pi/N\le k_i \le \pi$. In the termodynatic limit $N\rightarrow \infty$, $2\pi/N\rightarrow 0$ and the $k$s becomes uniformly distributed in the interval $-\pi \le k_i\le \pi$, the Brillouin zone. The canonical (anti)commutation relations $$\{c^\dagger_\sigma(\vec{r}),c_{\sigma'}(\vec{r}')\}=\delta_{\sigma,\sigma'}\delta_{\vec{r},\vec{r}'}$$
Then $$c_\sigma(\vec{r})=\int^\pi_{-\pi}\frac{d^dk}{(2\pi)^d}e^{i\vec{k}\cdot\vec{r}}c_{\sigma(\vec{k})}$$ and $$\{c^\dagger_\sigma(\vec{k}),c_{\sigma'}(\vec{k}')\}=(2\pi)^d\delta_{\sigma,\sigma'}\delta^{(d)}(\vec{k}-\vec{k}')$$ The kinetic energy then takes the form $$H_0=-\sum_{\vec{r},\vec{r}'\\ \sigma=\uparrow,\downarrow} \left(c^\dagger_\sigma(\vec{r})t(\vec{r}-\vec{r}')c_\sigma(\vec{r}')+c^\dagger_\sigma(\vec{r}')t(\vec{r}-\vec{r}')c_\sigma(\vec{r})\right)\\ =-\sum_{\vec{r},\vec{r}'\\ \sigma=\uparrow ,\downarrow} t(\vec{r}-\vec{r}')\int \frac{d^dk}{(2\pi)^d}\int \frac{d^dk'}{(2\pi)^d}\left(e^{-i\vec{k}\cdot\vec{r}+i\vec{k}'\cdot\vec{r}'}c_\sigma^\dagger(\vec{k})c_\sigma(\vec{k}')+\mbox{h.c.}\right)$$ If by $t(\vec{k})$ we denote the Fourier transform of $t(\vec{l})$, $$t(\vec{k})=\sum_{\vec{l}}t(\vec{l})e^{-i\vec{k}\cdot\vec{l}}$$ we can write $$\sum_{\vec{r},\vec{r}'}t(\vec{r}-\vec{r}')e^{-i\vec{k}\cdot\vec{r}+i\vec{k}'\cdot\vec{r}'}=t(\vec{k})(2\pi)^d\delta^{(d)}\left(\vec{k}-\vec{k}'\right)$$ For this case $$t(\vec{r}-\vec{r}')\equiv t(\vec{l})=\begin{cases}t& \mbox{for nearest neighbors}\\ 0& \mbox{otherwise}\end{cases}$$ we get $$t(\vec{k})=2t\sum^d_{j=1}\cos k_j$$ and a free Hamiltonian of the form $$H_0=\sum_{\sigma=\uparrow,\downarrow}\int \frac{d^dk}{(2\pi)^d}\epsilon(\vec{k})c^\dagger_\sigma(\vec{k})c_\sigma(\vec{k})$$ with $$\epsilon(\vec{k})=-t(\vec{k})=-2t\sum^d_{j=1}\cos k_j$$
If $\mathcal{N}$ is the number of particles and $N$ the number of sites, we get $$\mathcal{N}=2N\int^{k_F}_{-k_F}\frac{dk}{2\pi}=\frac{2Nk_F}{\pi}$$ and $$k_F\equiv = \frac{\pi\mathcal{N}}{2N}\equiv \frac{\pi}{2}\rho$$ where $\rho$ is the linear density.
Fermi surface??
The expectation value of the occupation number $$n_{\vec{k}}=\sum_{\sigma=\uparrow,\downarrow}c^\dagger_\sigma(\vec{k})c_\sigma(\vec{k})$$ has a jump at the Fermi surface both in the free case with interaction (see Fig. 2.5).
Correlation functions
The fermion Green function plays an important role in the theory. We can define it in terms of field operators in the Heisenberg representation $$c_\sigma(\vec{r},t)=e^{iHt}c_\sigma(\vec{r})e^{-iHt}$$ The fermion propagator is defined by $$G_{\sigma\sigma'}(\vec{r},t;\vec{r}',t')=-i\langle \mbox{Gnd}|Tc_\sigma(\vec{r},t)c^\dagger_{\sigma'}(\vec{r}',t')|\mbox{Gnd}\rangle$$ where $|\mbox{Gnd}\rangle$ stands for the ground state of the system and $T$ means a time-ordered product of operators, $$TA(t)B(t')=A(t)B(t')\theta(t-t')\pm B(t')A(t)\theta(t'-t)$$ with a $+(-)$ sign for bosons (fermions) and $$\theta(t)=\begin{cases}1& \mbox{if }t>0\\ 0 & \mbox{if }t<0\end{cases}$$ For a translationally invariant and time-independent system, we can write $G_{\sigma\sigma'}(\vec{r},t;\vec{r}',t')$ in terms of its Fourier transform $$G_{\sigma\sigma'}(\vec{r},t;\vec{r}',t')=\int \frac{d^dk}{(2\pi)^d}\int\frac{d\omega}{2\pi}e^{i\left( \vec{k}\cdot(\vec{r}-\vec{r}')-\omega(t-t')\right)}G_{\sigma\sigma'}(\vec{k},\omega)$$ In principle $G_{\sigma\sigma'}(\vec{k},\omega)$ is a $2\times 2$ spin matrix. In this case of non-interacting system, $G^{(0)}_{\sigma\sigma'}(\vec{k},\omega)$ is very simple to compute. The result is $$G_{\sigma\sigma'}(\vec{k},\omega)=\delta_{\sigma\sigma'}\lim_{\nu\rightarrow 0+}\left(\frac{\theta\left(\epsilon(\vec{k})-\epsilon_F\right)}{\omega-\epsilon(\vec{k})+i\nu}+\frac{\theta\left(\epsilon_F-\epsilon(\vec{k})\right)}{\omega-\epsilon(\vec{k})-i\nu}\right)$$ The poles of $G^0_{\alpha\beta}(\vec{k},\omega)$ exhibit the physical one-particle excitation spectrum $$\omega=\epsilon(\vec{k})$$
Weakly interacting system (a Fermi liquid): propagator look like?? Fermi surface~~
(1) The density correlation function
The density correlation function measures the strength of the density fluctuations in a physical system.
(2) The current correlation function
Hubbard model in the absence of external electromagnetic fields
(3) The spin correlation function
represents the amplitude of the ferromagnetic order parameter $M$
(4) The Cooper-pair correlation function
Bardeen–Cooper–Schrieffer (BCS) theory of superconductivity