Graduate Diagonalization of Hubbard Hamiltonian

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To diagonalize the Hubbard Hamiltonian in the context of DFT, one can apply a linear (Bogoliubov) transformation to the creation and annihilation operators. This transformation must maintain the commutation relations while simplifying the Hamiltonian to a diagonal form. The resulting problem can be approached using standard linear algebra techniques. Understanding these transformations is crucial for effectively solving the Hubbard model. Mastery of this process is essential for applying the Hubbard model in DFT studies.
Guilherme
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Hi guys! I am starting to study Hubbard model with application in DFT and I have some doubts how to solve the Hubbard Hamiltonian. I have the DFT modeled to Hubbard, where the homogeneous Hamiltonian is

$$ H = -t\sum_{\langle i,j \rangle}\sigma (\hat{c}_{i\sigma}^{\dagger}\hat{c}_{j\sigma} + H.c.) + \sum_i v_i^{eff} \hat{c}_{i\sigma}^{\dagger}\hat{c}_{i\sigma} $$

How do I diagonalize it?

Thanks in advance.
 
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Guilherme said:
Hi guys! I am starting to study Hubbard model with application in DFT and I have some doubts how to solve the Hubbard Hamiltonian. I have the DFT modeled to Hubbard, where the homogeneous Hamiltonian is

$$ H = -t\sum_{\langle i,j \rangle}\sigma (\hat{c}_{i\sigma}^{\dagger}\hat{c}_{j\sigma} + H.c.) + \sum_i v_i^{eff} \hat{c}_{i\sigma}^{\dagger}\hat{c}_{i\sigma} $$

How do I diagonalize it?
.
Make an arbitrary linear (Bogoliubov) transformation of creation and annihilation operators, work out the conditions that preserve the commutation rules and the conditions that make the resulting Hamiltonian diagonal, and you get a standard problem from linear algebra.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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