quantum mechanics - Bogoliubov transformation is not unitary transformation, correct?

To diagonalize quadratic term in the antiferromagnet Heisenberg model, we may introduce the Bogoliubov transformation:$a_k=u_k\alpha_k+v_k\beta_k^\dagger$, $b_k^\dagger=v_k\alpha_k+u_k\beta_k^\dagger$. This transformation can diagonalize the quadratic term in the Hamiltonian:

\begin{align} H &=\sum_k(a^\dagger_ka_k+b^\dagger_kb_k+\gamma_ka^\dagger_kb^\dagger_k+\gamma_ka_kb_k) \\ & =\sum_{\bf{k}} \begin{pmatrix}a_{\bf{k}}^\dagger & b_{\bf{k}}\end{pmatrix} \begin{pmatrix}1 &\gamma_{\bf{k}}\\\gamma_{\bf{k}} & 1\end{pmatrix} \begin{pmatrix}a_{\bf{k}} \\ b_{\bf{k}}^\dagger\end{pmatrix} \\ & =\sum_{\bf{k}} \begin{pmatrix}\alpha_{\bf{k}}^\dagger & \beta_{\bf{k}}\end{pmatrix}\begin{pmatrix}u_k &v_{{k}}\\v_{k} & u_{k}\end{pmatrix} \begin{pmatrix}1 &\gamma_{\bf{k}}\\\gamma_{\bf{k}} & 1_{\bf{k}}\end{pmatrix}\begin{pmatrix}u_k &v_{{k}}\\v_{k} & u_{k}\end{pmatrix} \begin{pmatrix}\alpha_{\bf{k}} \\ \beta_{\bf{k}}^\dagger\end{pmatrix} \\ & =\sum_{\bf{k}} \begin{pmatrix}\alpha_{\bf{k}}^\dagger & \beta_{\bf{k}}\end{pmatrix} \begin{pmatrix}\epsilon_k &0\\0 &\epsilon_k\end{pmatrix} \begin{pmatrix}\alpha_{\bf{k}} \\ \beta_{\bf{k}}^\dagger\end{pmatrix} \end{align}

with $\epsilon_k=\sqrt{1-\gamma_k^2},u_k=\sqrt{\frac{1+\epsilon_k}{2\epsilon_k}},v_k=-\frac{\gamma_k}{\sqrt{2\epsilon_k(1+\epsilon_k)}}$. But the transformation U: $\begin{pmatrix}u_k &v_{{k}}\\v_{k} & u_{k}\end{pmatrix}$ is not unitary, because $u_k,v_k$ are real, $U^\dagger\neq U^{-1}$.

Is the number of bosons not conserved, so the transformation may not be unitary? Are there any restriction on the transformation of boson?