Use the <canonical commutation relation> for the <creation and annihilation operators> and the transformation derived above. Directly,
$$
[a'_i,a_k'{}^\dagger]=\sum_j(\overline A_{ij}A_{kj}-\overline B_{ij}B_{kj}),\qquad
[a'_i,a'_k]=\sum_j(-\overline A_{ij}\overline B_{kj}+\overline B_{ij}\overline A_{kj}).
$$
Requiring these to equal $\delta_{ik}$ and zero, respectively, and conjugating the equations yields the <canonical identities for a bosonic Bogoliubov transformation>:
$$
\boxed{AA^\dagger-BB^\dagger=I,\qquad AB^T=BA^T.}
$$
The commutator of the primed creation operators then vanishes by taking adjoints. These are exactly the corresponding positive-mode norm and positive-negative orthogonality conditions for the <Bogoliubov transformation>. They are sufficient for these algebraic commutators whenever the sums are well defined. Infinitely many modes introduce further questions of convergence and whether the two choices admit a common <bosonic Fock space> representation; the matrix identities alone do not settle unitary implementability.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 311 / 4 / iii / b / Solution

{title2=$AA^\dagger-BB^\dagger=I,\quad AB^T=BA^T$}

For mode mixing $\psi'_i=\sum_j(A_{ij}\psi_j+B_{ij}\overline\psi_j)$, the <Klein-Gordon inner product> extracts $a'_i=\sum_j(\overline A_{ij}a_j-\overline B_{ij}a_j^\dagger)$. The <canonical commutation relation> is preserved exactly when $AA^\dagger-BB^\dagger=I$ and $AB^T=BA^T$. These follow by computing $[a'_i,a_k'{}^\dagger]$ and $[a'_i,a'_k]$; they are also the positive- and negative-frequency mode orthonormality identities. In infinitely many modes, convergence and the existence of a common <bosonic Fock space> representation require additional analysis; the expected total particle number is finite only if $B$ is a <Hilbert-Schmidt operator>.

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 Incoming links: Canonical identities for a bosonic Bogoliubov transformation