# Bosonic operators

[EDITED 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:

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

[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?

[EDIT. Answer: Now that the matrix is unitary, the linear combination of the bosonics operators are also bosonic operators.]

• You wrote down the conditions for operators to be bosonic. Did you try to straightforwardly check if your new operators fulfil those conditions?
– noah
Sep 10 at 13:45
• What is the connection between the ${\boldsymbol \phi}$ and the $A$'s? And why write the inverse of the orthogonal matrix $Ri$ n such a strange way? It's just the transpose of $R$ surely? Sep 10 at 18:56
• The vector $\phi$ is formed by three bosonic operators $a$, $b$ and $c$. They follow these three conditions, where $Aℓ=a,b,c$. The matrix $R$ was given like that, so the operation $RR^{\dagger}=2\mathbb{I}$, that might be the source of the issue here. Thank you.
– koy
Sep 10 at 20:21

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.