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- Homework Statement
- I am trying to compute the Fourier transform of the 2D ##t-V## model for the case ##t=0##.

- Relevant Equations
- $$\hat H = -t \displaystyle \sum_{\langle i,j\rangle} ( \hat c_i^{\dagger} \hat c_j + \hat c_j^{\dagger} \hat c_i) + V \sum_{\langle i, j \rangle} \hat n_i \hat n_j$$

To compute the Fourier transform of the ##t-V## model for the case where ##t = 0##, we start by expressing the Hamiltonian in momentum space. Given that the hopping term ##t## vanishes, we only need to consider the potential term:

$$\hat{H} = V \sum_{\langle i, j \rangle} \hat{n}_i \hat{n}_j$$

Let's express this in terms of creation and annihilation operators in momentum space. The Fourier transform of the electron annihilation operator ##\hat{c}_i## is given by:

$$\hat{c}_i = \frac{1}{\sqrt{N}} \sum_k e^{-ikr_i} \hat{c}_k$$

where ##N## is the total number of lattice sites, ##k## is the wavevector, and ##r_i## is the position of lattice site ##i##.

Therefore, the electron number operator ##\hat{n}_i## can be expressed as:

$$\hat{n}_i = \hat{c}_i^\dagger \hat{c}_i = \frac{1}{N} \sum_{k,k'} e^{i(k'-k)r_i} \hat{c}_{k'}^\dagger \hat{c}_k$$

Hence, the Hamiltonian in momentum space becomes:

$$\hat{H} = \frac{V}{N^2} \sum_{\langle i, j \rangle} \sum_{k,k'} e^{i(k'-k)(r_j-r_i)} \hat{c}_{k'}^\dagger \hat{c}_k \hat{c}_{k}^\dagger \hat{c}_{k'}$$

I wonder if we can further simplify, is my attempt correct? Is it possible to compute the energy eigenvalues like in the case of ##V=0## where the solution corresponds ##e_k = -2t(cos(kx)+cos(ky))##.

$$\hat{H} = V \sum_{\langle i, j \rangle} \hat{n}_i \hat{n}_j$$

Let's express this in terms of creation and annihilation operators in momentum space. The Fourier transform of the electron annihilation operator ##\hat{c}_i## is given by:

$$\hat{c}_i = \frac{1}{\sqrt{N}} \sum_k e^{-ikr_i} \hat{c}_k$$

where ##N## is the total number of lattice sites, ##k## is the wavevector, and ##r_i## is the position of lattice site ##i##.

Therefore, the electron number operator ##\hat{n}_i## can be expressed as:

$$\hat{n}_i = \hat{c}_i^\dagger \hat{c}_i = \frac{1}{N} \sum_{k,k'} e^{i(k'-k)r_i} \hat{c}_{k'}^\dagger \hat{c}_k$$

Hence, the Hamiltonian in momentum space becomes:

$$\hat{H} = \frac{V}{N^2} \sum_{\langle i, j \rangle} \sum_{k,k'} e^{i(k'-k)(r_j-r_i)} \hat{c}_{k'}^\dagger \hat{c}_k \hat{c}_{k}^\dagger \hat{c}_{k'}$$

I wonder if we can further simplify, is my attempt correct? Is it possible to compute the energy eigenvalues like in the case of ##V=0## where the solution corresponds ##e_k = -2t(cos(kx)+cos(ky))##.

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