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There is an elegant formulation of the Bogolyubov transformation in terms of Clifford algebras. Note that a quadratic Hamiltonian (noted by a hat), is a hermitian element of the representation of a Clifford algebra on the Hilbert space, $$\mathbb{C} l_{2k}\xrightarrow{~~Q~~} \text{End}(\Lambda(\mathbb{C}^{k})),~~~~\hat H=Q(H)$$ In which case, noting that quadratic polynomials are anti-hermitian, $$Q(e_ie_j)^\dagger=Q(e_je_i)=-Q(e_ie_j)$$ we can write the Hamiltonian as a purely imaginary element of degree two (a quadratic polynomial in the elements of the Clifford algebra): $$H\in Cl_{2k}^{(2)}\otimes i\subset \mathbb{C} l_{2k}.\tag{$k=2n+1$}$$ In other words, the space of quadratic field theories is a real vector space (of dimension $2k$ choose $2$), and has the basis $\{e_ie_j\}_{i<j}$. Since via identifying $Cl_{2k}^{(2)}\simeq \mathfrak{so}(2k)$ we get $$Cl_{2k}^{(2)}\otimes i\simeq \Lambda^2(\mathbb{R}^{2k})\otimes i,$$ Therefore we can naturally identify the space of quadratic field theories with the space of real antisymmetric matrices, $$ Cl^{(2)}_{2k}\otimes i=\{\text{space of quadratic QFT's}\} \xrightarrow{\simeq}\{\text{antisymmetric matrices on classical phase space}\}$$ via the following linear isomorphism: $$\sum_{i<j}\lambda_{ij}e_ie_j\mapsto \omega,~~~~~~~~\omega_{ij}=\lambda_{ij}/2,~~~\omega_{ji} = -\lambda_{ij}/2$$ In which case the diagonal elements $D$ are precisely the antisymmetric matrices which happen to be in canonical form (i.e. Darboux coordinates): $$H\in D=\sum_{j}\lambda_jz_j\bar z_j=2i\sum_{j}\lambda_je_{2j-1}e_{2j}\mapsto 2i\sum_j e_{2j-1}\wedge e_{2j}\in \Lambda^2(\mathbb{R}^{2k})\otimes i$$ But Wikipedia states that every antisymmetric matrix may be put in canonical form by a special orthogonal transformation. So the diagonalization will be given by a complexified special orthogonal transformation: $$S\otimes 1\in SO(2k)\otimes \mathbb{C}$$ But we already established that every algebra automorphism is a complexified orthogonal transformation: $$R\otimes 1\in O(2k)\otimes \mathbb{C}$$ So the algorithm works.