Bosonic operators

Question

We have a vector of bosonic operators, such that: \begin{equation*} \vec{\phi} = (a, b, c)^{\text{T}} \; , \end{equation*} and the following commutation rules, \begin{equation*} \begin{split} [A_{\ell}, A_{\ell^{'}}] &=0 \\ [A^{\dagger}_{\ell}, A^{\dagger}_{\ell^{'}}] &=0 \\ [A_{\ell}, A^{\dagger}_{\ell^{'}}] &=\delta_{\ell,\ell^{'}} \; , \end{split} \end{equation*} for each operator, where $A_{\ell} = a,b,c$.

If now we applied a rotation to the vector, such that: \begin{equation*} \vec{\phi^{'}} = R^{-1}\vec{\phi} \; , \end{equation*} where:

$R = \frac{1}{2} \begin{bmatrix} -1 & 1 & \sqrt{2} \\ \sqrt{2} & \sqrt{2} & 0 \\ -1 & 1 & -\sqrt{2} \end{bmatrix} \; ,$

[EDIT: the matrix $R$ had a typo and wasn't unitary. Where it reads $1/2$, it was $1/\sqrt{2}$.]

are the new operators also bosonic? Is this type of linear combination of bosonic operators also a bosonic operator?

Answer 1

As you have written it, $R$ is not unitary nor orthogonal, as $$RR^\dagger=R^\dagger R=2\mathbb{I}.$$ Using the proper normalization for $R$ (with $1/2$ out front instead of $1/\sqrt{2}$) will ensure that it is unitary and thus that the resulting operators will also satisfy bosonic commutation relations.