Bogoliubov transformation

Remarks

The conditions come from the computation $[b_i, b_k^*] = (UU^* - VV^*)_{ik}$ and $[b_i, b_k] = (UV^T - VU^T)_{ik}$. Example: for $n = 1$, $b = \cosh(r)\, a + \sinh(r)\, a^*$ with $r \in \R$ is a bosonic Bogoliubov transformation, since $\cosh^2 r - \sinh^2 r = 1$; it is the single-mode squeezing transformation. Non-example: $b = a + a^*$ has $U = V = 1$, so $UU^* - VV^* = 0 \ne 1$; indeed $b$ is self-adjoint and $[b, b^*] = 0$. For fermions with $n = 1$, $b = u\, a + v\, a^*$ gives $\set{b, b^*} = \abs{u}^2 + \abs{v}^2$ and $\set{b, b} = 2uv$, so it is a Bogoliubov transformation only if one of $u, v$ is $0$, as for the particle-hole exchange $b = a^*$. A genuine fermionic mixing needs two modes: $b_1 = u\, a_1 + v\, a_2^*$ and $b_2 = u\, a_2 - v\, a_1^*$ with $\abs{u}^2 + \abs{v}^2 = 1$ satisfy the conditions, since here $U = uI$ and $V$ is antisymmetric. If $V = 0$, the condition says that $U$ is a unitary matrix, and the transformation is an ordinary change of one-particle basis.