# What is the ground state of a Hamiltonian in $k$-space after Bogoliubov transformation? [duplicate]

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

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

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.

• What does "cte" mean? Sep 23, 2023 at 3:17