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

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?

• You should distinguish between as algebra and as vector space when doing the gradings. – AHusain Jun 21 '16 at 22:12

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.
• Be more careful about your $\otimes$. – AHusain Jun 21 '16 at 22:15