TITLE: $[3]$ representation of group of rotations with basis functions
QUESTION [0 upvotes]: Statement: Show that the $[3]$ representation of the group of rotations about the $z$ axis si completely reducible when referred to $-(x+iy)/\sqrt{2}$, $ z$, $ (x-iy)/ \sqrt{2}$ as basis functions.
To show that the representation is reducible I need to find first the rotational matrix for the new basis functions.
If I were on Cartesian system, the rotational matrix for a certain angle $\theta$ (rotation about $z$ axis) would simply be $$\begin{bmatrix}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
How can I write it in the new basis system?

REPLY [1 votes]: I guess you are referring to $x,y,z$ as the coordinate unit vectors $x=(1,0,0)^t$, $y=(0,1,0)^t$ and $z=(0,0,1)^t$?
In that case, a rotation matrix as you wrote corresponds to the map
\begin{align}
x&\mapsto \cos\theta x + \sin\theta y\\
y&\mapsto -\sin\theta x +\cos\theta y\\
z&\mapsto z
\end{align}
Therefore,
\begin{align}
x+iy&\mapsto (\cos\theta - i\sin\theta)x + (\sin\theta+i\cos\theta)y=(\cos\theta-i\sin\theta)(x+iy)\\
x-iy&\mapsto (\cos\theta +i\sin\theta) x+ (\sin\theta-i\cos\theta)y=(\cos\theta+i\sin\theta)(x-iy)\\
z\mapsto z
\end{align}
It follows that in the basis $\{x+iy,x-iy,z\}$ the matrix for you transformation is
$$
\begin{pmatrix}\cos\theta-i\sin\theta&0&0\\0&\cos\theta+i\sin\theta&0\\0&0&1\end{pmatrix}
$$
Since the rotation $\theta$ is arbitrary, this change of basis simultaneously diagonalizes every rotation about the $z$-axis, and each basis vector spans a 1-dimensional invariant subspace. This shows that the representation is a direct sum of 3 one-dimensional subspaces.
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