I have a set of Bogoliubov transformation as follows: \begin{equation} a(\beta) = a \cos \theta (\beta) - \tilde{a}^\dagger \sin \theta (\beta)\\ \tilde{a}(\beta) = \tilde{a} \cos \theta (\beta) + a^\dagger \sin \theta (\beta) \end{equation} And their corresponding equations obtained by taking the adjoint of both equations. Now, we define the following: \begin{equation} A = \begin{pmatrix}a\\\tilde{a}^\dagger \end{pmatrix} \end{equation} And say that $A$ transforms according to the following unitary transformations: \begin{equation} U(\theta)AU^\dagger(\theta)=\bar{U}(\theta)A\\\\ \text{where}\end{equation}
\begin{align}U(\theta) &= \exp{\left(-\theta\left(\tilde{a} a - \tilde{a}^\dagger a^\dagger\right) \right)}\\ \bar{U}(\theta) &=\begin{pmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{align} $a,\tilde{a} $ are the annihilation operators of the particle and antiparticle respectively. Problem here is, I'm not sure of the correctness of the transformation equation. If the creation and annihilation operators are $n\times n$ matrices, then the dimension of $U$ is $n\times n,$ and the left hand side of the transformation equation isn't valid since an $n\times n$ matrix can never left multiply with another $2n\times n$ matrix.
I've been trying to resolve this dilemma but very little progress. The same sort of unitary operator transformations are given everywhere, but are valid only if the operator is a matrix of the same dimensionality as the unitary matrix. These sort of equations are also given in group theory in order to establish the lie algebra of any group. But the same problems are encountered over there as well. Where am I going wrong?
Reference: Ashok Das: "Finite Temperature Field Theory", chapter 3, equations (3.29)-(3.32)
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