Friday, 22 June 2018

quantum mechanics - Validity of Bogoliubov transformation


In condensed matter physics, one often encounter a Hamiltonian of the form $$\mathcal{H}=\sum_{\bf{k}} \begin{pmatrix}a_{\bf{k}}^\dagger & a_{-\bf{k}}\end{pmatrix} \begin{pmatrix}A_{\bf{k}} & B_{\bf{k}}\\B_{\bf{k}} & A_{\bf{k}}\end{pmatrix} \begin{pmatrix}a_{\bf{k}} \\ a_{-\bf{k}}^\dagger\end{pmatrix} ,$$ where $a_{\bf{k}}$ is a bosonic operator. A Bogoliubov transformation $$ \begin{pmatrix}a_{\bf{k}} \\ a_{-\bf{k}}^\dagger\end{pmatrix}= \begin{pmatrix}\cosh\theta_{\bf{k}} & -\sinh\theta_{\bf{k}}\\-\sinh\theta_{\bf{k}} & \cosh\theta_{\bf{k}}\end{pmatrix} \begin{pmatrix}\gamma_{\bf{k}} \\ \gamma_{-\bf{k}}^\dagger\end{pmatrix}, $$ with $$\tanh 2\theta_{\bf{k}}=\frac{B_{\bf{k}}}{A_{\bf{k}}}$$ is often used to diagonalized such a Hamiltonian. However, this seems to assume that $|A_{\bf{k}}|> |B_{\bf{k}}|$. Is this true? If so, how else can the Hamiltonian be diagonalized?




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