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This question is naive but for some reason I'm not getting the expected result.

The generators of $SO(6)$ can be written in this way:

$$(J_{ab})_{cd}=i(\delta_{ac}\delta_{bd}-\delta_{ad}\delta_{bc}),\qquad a,b,c,d=1,...,6\tag{1}$$

The commuting generators are

$$H_k=J_{2k-1\,2k},\qquad k=1,2,3\tag{2}$$

Thus, in this case, $H_1=J_{12},\,H_2=J_{34}$ and $H_3=J_{56}$ are the Cartan generators. These are not diagonal but can be simultaneously diagonalized to have the usual ones

$$H_1^D=E_{11}-E_{22}$$ $$H_2^D=E_{33}-E_{44}\tag{3}$$ $$H_3^D=E_{55}-E_{66}$$

So,

$$H^D_k=SH_kS^{-1}\tag{4}$$

In this particular case, or in general, which matrix should do the work?

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1 Answer 1

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Found the answer in Georgi's Lie algebras in particle physics, page 237:

The Cartan generators can be written in this way

$$H_1=-i J_{12}=\sigma_2\oplus I_2\oplus I_2,$$

$$H_2=-i J_{34}=I_2\oplus\sigma_2\oplus I_2,\tag{1}$$

$$H_3=-i J_{56}=I_2\oplus I_2\oplus\sigma_2,$$

eigenvectors of $\sigma_2$ are

$$\begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix} \begin{pmatrix} 1\\ \pm i \end{pmatrix}=\pm \begin{pmatrix} 1\\ \pm i \end{pmatrix}.\tag{2}$$

So we can construct a simultaneous base for the $H_i$'s:

$$\left\{\begin{pmatrix} 1\\ i\\ 0\\ 0\\ 0\\ 0\\ \end{pmatrix},\begin{pmatrix} 1\\ -i\\ 0\\ 0\\ 0\\ 0\\ \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1\\ i\\ 0\\ 0\\ \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1\\ -i\\ 0\\ 0\\ \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 1\\ i\\ \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 1\\ -i\\ \end{pmatrix}\right\}.\tag{3}$$

As suggested here, we can construct $P$ like this

$$P=\begin{pmatrix} 1 & 1 & & & & \\ i & -1 & & & & \\ & & 1 & 1 & & \\ & & i & -i & & \\ & & & & 1 & 1\\ & & & & i & -i \end{pmatrix},\tag{4}$$

so

$$H_i^d=PH_iP^{-1}.\tag{5}$$

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