[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.]

  • 2
    $\begingroup$ You wrote down the conditions for operators to be bosonic. Did you try to straightforwardly check if your new operators fulfil those conditions? $\endgroup$
    – noah
    Sep 10 at 13:45
  • $\begingroup$ 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? $\endgroup$
    – mike stone
    Sep 10 at 18:56
  • $\begingroup$ 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. $\endgroup$
    – 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.

  • $\begingroup$ Apparently the question provides an incorrect rotation matrix. I was thinking that something else would be needed if that wasn't the issue. Thank you. $\endgroup$
    – koy
    Sep 10 at 20:09
  • $\begingroup$ @yko awesome, glad to help! $\endgroup$ Sep 10 at 20:10

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