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Consider the following Hamiltonian in $k$-space, quadratic in terms of the $\gamma$ operators: \begin{equation} \hat{H}_2=\frac{1}{2}\sum_k \begin{pmatrix} \gamma_k^\dagger & \gamma_{-k} \end{pmatrix} \begin{pmatrix} A_k & B_k \\ B_k^* & A_k^* \end{pmatrix} \begin{pmatrix} \gamma_k \\ \gamma_{-k}^\dagger \end{pmatrix} + \text{cte}, \end{equation} where $\gamma_{k}=(\gamma_{k,1},\gamma_{k,2},...,\gamma_{k,n_{max}})^T$ is a vector operator.

A Bogoliubov transformation is performed \begin{equation} \begin{pmatrix} \gamma_k\\ \gamma^\dagger_{-k} \end{pmatrix}=T_k \begin{pmatrix} \lambda_k\\ \lambda^\dagger_{-k} \end{pmatrix}, \end{equation} and \begin{equation} T_k^{\dagger} \begin{pmatrix} A_k & B_k \\ B_k^* & A_k^* \end{pmatrix}T_k=\begin{pmatrix} \omega_{k} & 0 \\ 0 & \omega_{k} \end{pmatrix}, \end{equation} where $\omega_{k}=\operatorname{diag}(\omega_{k,1},...,\omega_{k,n_{max}})$, so that the Hamiltonian is diagonalized, and now is given by: \begin{equation} \hat{H}_2=\sum_{k} \sum_{\alpha=1}^{n_{max}} \omega_{k,\alpha}\hat{\lambda}_{k,\alpha}^{\dagger}\hat{\lambda}_{k,\alpha}. \end{equation}

Now, what would the ground state $|\psi_0\rangle$ be? I cannot understand, for example, how could I calculate the expectation value $\langle\gamma_{}i^{\dagger}\gamma_j \rangle$ in terms of the matrices.

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    $\begingroup$ What does "cte" mean? $\endgroup$ Sep 23, 2023 at 3:17

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