Letting the standard Clifford algebra of dimension $2k$ be denoted by $Cl\_{2k}$, let's denote the corresponding complex Clifford algebra via $$\mathbb{C}l\_{2k}\equiv Cl\_{2k}\otimes\_{\mathbb{R}}\mathbb{C}.$$ 
This has both a $\mathbb Z\_2$-grading and a $\mathbb Z$-grading, which we distinguish by the formulae
$$\mathbb{C}l\_{2k}\simeq \mathbb{C}l\_{2k}^{(0)}\oplus \cdots \oplus \mathbb{C}l\_{2k}^{(2k)},$$
$$\mathbb{C}l\_{2k}\simeq \mathbb{C}l\_{2k}^0\oplus \mathbb{C}l\_{2k}^1.~~~~~~~~~$$
Every orthogonal transformation $R:\mathbb{R}^{2k}\to \mathbb{R}^{2k}$ induces an algebra automorphism $\mathbb{C}l\_{2k}(R):\mathbb{C}l\_{2k}\to \mathbb{C}l\_{2k}$. I now want to consider the orbits of elements of 
$$Cl^{(2)}\_{2k}\subset \mathbb Cl^{(2)}\_{2k}$$
under this action, namely, the real elements of degree two. Specifically, fix an orthonormal basis $\{e\_1,\cdots e\_{2k}\}$ of $\mathbb{R}^{2k}$. Then, there is this special subspace $D\subset Cl^{(2)}\_{2k}$, which I'll call the **diagonal elements**, which can be written as
$$v=\sum\_{j=1}^{k}\lambda\_j\bar z\_j z\_j,~~~~z\_j = e\_{2j-1}+ie\_{2j},~~~\bar z\_j = e\_{2j-1}-ie\_{2j},~~~\lambda\_j\in\mathbb{R}$$
Note that $D\simeq \mathbb{R}^k$ as a real vector space. No matter what basis we choose, for every element $v\in Cl^{(2)}\_{2k}$, there exists an algebra automorphism known as a *Bogoliubov transformation*, originating from an orthogonal transformation $R$ which diagonalizes $v$, that is,
$$\forall v\in Cl^{(2)}\_{2k},~~\exists R:~~~\mathbb{C}l(R)v\in D.$$
Is there a nice proof of this statement using this language of Clifford algebras?



Why does the Bogolyubov transformation work? - In language of Clifford Algebras?
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.


