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In condensed matter physics, one often encounter a Hamiltonian of the form $$\mathcal{H}=\sum_{\bf{k}} \begin{pmatrix}a_{\bf{k}}^\dagger & a_{-\bf{k}}\end{pmatrix} \begin{pmatrix}A_{\bf{k}} & B_{\bf{k}}\\B_{\bf{k}} & A_{\bf{k}}\end{pmatrix} \begin{pmatrix}a_{\bf{k}} \\ a_{-\bf{k}}^\dagger\end{pmatrix} ,$$ where $a_{\bf{k}}$ is a bosonic operator. A Bogoliubov transformation $$ \begin{pmatrix}a_{\bf{k}} \\ a_{-\bf{k}}^\dagger\end{pmatrix}= \begin{pmatrix}\cosh\theta_{\bf{k}} & -\sinh\theta_{\bf{k}}\\-\sinh\theta_{\bf{k}} & \cosh\theta_{\bf{k}}\end{pmatrix} \begin{pmatrix}\gamma_{\bf{k}} \\ \gamma_{-\bf{k}}^\dagger\end{pmatrix}, $$ with $$\tanh 2\theta_{\bf{k}}=\frac{B_{\bf{k}}}{A_{\bf{k}}}$$ is often used to diagonalized such a Hamiltonian. However, this seems to assume that $|A_{\bf{k}}|> |B_{\bf{k}}|$. Is this true? If so, how else can the Hamiltonian be diagonalized?

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1) Let us replace the momentum indices ${\bf k}$ and $-{\bf k}$ with abstract indices $1$ and $2$, and ignore the momentum summation. The Hamiltonian then reads

$$\tag{1}H ~=~ \begin{pmatrix}a_1^{\dagger} & a_2\end{pmatrix} M\begin{pmatrix}a_1 \\ a_2^{\dagger}\end{pmatrix}, $$

where

$$\tag{2} M~:=\begin{pmatrix}A & B\\B^{*} & A \end{pmatrix}~=~M^{\dagger}, \qquad A~\in~ \mathbb{R},\qquad B~\in~ \mathbb{C}.$$

A Bogoliubov transformation is on the form

$$\tag{3} \begin{pmatrix}a_1 \\ a_2^{\dagger}\end{pmatrix} ~=~U \begin{pmatrix}b_1 \\ b_2^\dagger\end{pmatrix}, \qquad U~:=~\begin{pmatrix}u_{11} & u_{12}\\u_{21} & u_{22}\end{pmatrix}.$$

Because a Bogoliubov transformation should preserve the canonical commutator relations (CCR), it is straightforward to check that the transformation matrix $U$ must belong to the Lie group

$$\tag{4} U(1,1) ~:=~\{U\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid U^{\dagger}\eta U~=~\eta \}, $$

where $$\tag{5} \eta ~:=~\begin{pmatrix}1& 0\\0 & -1\end{pmatrix}$$ is the $1+1$ dimensional Minkowski metric. The Lie group $U(1,1)$ of Bogoliubov transformations is real and non-compact, and it is $4$-dimensional. In fact, one may prove that an element $U\in U(1,1)$ is of the form

$$\tag{6} U~=~\begin{pmatrix}u & v\\wv^* & wu^*\end{pmatrix},$$

where

$$\tag{7}u,v,w~\in~\mathbb{C},\quad |u|^2-|v|^2~=~1 \quad\text{and}\quad |w|~=~1. $$

The Hamiltonian then becomes of the form

$$\tag{8} H ~=~ \begin{pmatrix}b_1^{\dagger} & b_2\end{pmatrix} N\begin{pmatrix}b_1 \\ b_2^{\dagger}\end{pmatrix},$$

where

$$\tag{9} N~:=~U^{\dagger}MU~=~N^{\dagger}.$$

Theorem I: We have the following two invariants under Bogoliubov transformations:

$$\tag{10} \det(N) ~=~\det(M)~=~A^2-|B|^2. $$

$$\tag{11} {\rm tr}(\eta N) ~=~{\rm tr}(\eta M)~=~0.$$

Proof of eq. (11): Use that $\eta N= U^{-1}\eta MU$ and $\eta M$ are connected via a similarity transformation.

Corollary Ia:

The new $2\times 2$ matrix $N$ is of the form

$$\tag{12} N~:=\begin{pmatrix}A^{\prime} & B^{\prime}\\B^{\prime *} & A^{\prime} \end{pmatrix}~=~N^{\dagger}, \qquad A^{\prime}~\in~ \mathbb{R},\qquad B^{\prime}~\in~ \mathbb{C}.$$

It turns out that OP's assertion in the question formulation about diagonalizability is correct, cf. the following Corollary Ib.

Corollary Ib:

(i) The new $2\times 2$ matrix $N$ can only be diagonal if $|A|>|B|$ or $B=0$.

(ii) The new $2\times 2$ matrix $N$ can only be off-diagonal if $|A|<|B|$ or $A=0$.

(iii) If $|A|=|B|\neq 0$, then the new $2\times 2$ matrix $N$ is neither diagonal nor off-diagonal.

Proof of (i): The diagonal matrix $N$ must satisfy

$$ \tag{13} 0~\leq~ A^{\prime 2} ~=~ \det(N) ~=~\det(M)~=~A^2-|B|^2. $$

Hence $|A|>|B|$ or $|A|=|B|$. In the latter case, $A^{\prime}=0$, so $N=0$, and hence $M=0$, and in particular $B=0$.

The proofs of (ii) and (iii) are left as exercises.

3) We will next argue that the case $A<|B|$ is not physically relevant, cf. Theorem II.

Theorem II:

(i) The Hamiltonian $H$ is positive definite if $A >|B|$.

(ii) The spectrum of $H$ is non-negative if $A\geq |B|$.

(iii) The spectrum of $H$ is unbounded from below if $A < |B|$.

A proof of Theorem II is left as an exercise. A proof at the classical level can be established by replacing the operators $a_1$ and $a_2$ with the corresponding classical complex variables, and then investigate the signature of the Hessian for the corresponding classical Hamiltonian function.

4) Finally, the Bogoliubov transformation (3) may be cast in a special relativistic language. One may view the matrix $M$, or equivalently $(A,B)$, as a point in $1+2$ dimensional Minkowski space with time coordinate $A\in \mathbb{R}$ and space coordinates $B\in \mathbb{C}\cong \mathbb{R}^2$. The invariant Minkowski length

$$\tag{14} \det(M)~=~A^2-|B|^2$$

is preserved under the action

$$\tag{15} \rho:U~\mapsto~ (U^{-1})^{\dagger}MU^{-1}$$

of the Lie group $U(1,1)$ of Bogoliubov transformations. Therefore $\rho$ is a Lie group homomorphism

$$\tag{16} \rho: \quad\to\quad O(1,2;\mathbb{R}) $$

into the $3$-dimensional Lorentz group $O(1,2;\mathbb{R})$. The condition $|A| >|B|$ ($|A| < |B|$) is the condition for being a time-like (space-like) vector, respectively. Intuitively, OP's observation concerning diagonalizability may be understood as the fact that one cannot turn a space-like vector into time-like vector by a Lorentz transformation.

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For $A$ and $B$ real, using the parametrization given in the Wikipedia article on BT, we would end up with the condition $B[\cos(\theta_1-\theta_2)\cosh 2r+i\sin(\theta_1-\theta_2)]=A\sinh 2r$. This will lead to the same assumption on $A$ and $B$. –  leongz Feb 7 '13 at 3:07
    
Right. I updated the answer. –  Qmechanic Feb 8 '13 at 2:54
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