# Problem with Bogoliubov transformations of an operator

I have a set of Bogoliubov transformation as follows: $$a(\beta) = a \cos \theta (\beta) - \tilde{a}^\dagger \sin \theta (\beta)\\ \tilde{a}(\beta) = \tilde{a} \cos \theta (\beta) + a^\dagger \sin \theta (\beta)$$ And their corresponding equations obtained by taking the adjoint of both equations. Now, we define the following: $$A = \begin{pmatrix}a\\\tilde{a}^\dagger \end{pmatrix}$$ And say that $A$ transforms according to the following unitary transformations: $$U(\theta)AU^\dagger(\theta)=\bar{U}(\theta)A\\\\ \text{where}$$

\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)

• If the vector notation is confusing just write $U A_i U^\dagger = \sum_j \bar{U}_{ij} A_j$. It is then clear that both sides have the same dimensionality. Oct 16, 2016 at 12:04
• Writing it like this becomes a simple matter of multiplication of U to $A_i$, i.e. $a$. Sure. But I'm actually trying to obtain $\bar{U}$ using the above transformation. Obtaining it will be much harder (explicitly trying to solve the linear equations) than just looking at how A transforms when left and right multiplied by U and $U^\dagger$ respectively. Oct 16, 2016 at 12:09
• The solution is most easily found by taking $\theta$ to be infinitesimal and integrating. Or alternatively wholesale via the Baker-Campbell-Hausdorff formula. Neither of these is particularly laborious as the algebra of creation and annihilation operators is very simple. I cannot see a way of finding $\bar{U}$ without considering the action of the rotation $U$ on $A_i$ Oct 16, 2016 at 12:18
• Just a side-note, @pyroscepter; you don't have to use  to render equations here; instead use as it renders the same result here and saves time too.
– user36790
Oct 16, 2016 at 12:21

Based on some of the expressions that are provided, it seems the issue is one of notation. So, I'm going to be ratherpedantic. Let's define $$\{a, \tilde{a}\} \equiv \left\{N^{(1)}_{ij}, N^{(2)}_{ij}\right\}$$ so that $$A \equiv N^p_{ij}$$ where $p$ is an index that selects either $a$ or $\tilde{a}$ and $ij$ are indices that run from 1 to $n$ and represent the indices of the creation and annihilation operators. Now we can say that $U$ operates on the $ij$-indices, $$U\equiv U_{ij}$$ whereas $\bar{U}$ operates on the $p$ index $$\bar{U}\equiv \bar{U}_{pq}$$
So then the expression $$U(\theta)AU^\dagger(\theta)=\bar{U}(\theta)A$$ in my pedantic notation becomes (dropping the $\theta$) $$\sum_{ij} U_{ki}N^p_{ij} U^{\dagger}_{jl}= \sum_q \bar{U}_{pq}N^q_{kl} .$$