Rotation operator in the matrix representation in the standard basis I recently been introduced to rotation operators in Quantum-Mechanics.
Can someone please provide an explanation on what is means to represent a operator for the rotation of a quantum-mechanical system around the z-axis in the matrix representation in the standard basis.
Any help would be greatly appreciated. 
Thanks
 A: (Disclaimer: I'm confused by the phrase 'standard basis' and think that is non-standard terminology. However, there's only one thing I think it could be.)
From the general theory of angular momentum we know that a spin $j$ object has a Hilbert space $\mathcal{H}$ of dimension $2j+1$. This Hilbert space is spanned by kets $|j,m\rangle$ where $-j \leq m \leq +j$ and the $m$'s differ by integers. The $m$'s correspond to the eigenvalues of the operator $J_z$ whilst the $j$'s are the related to the total angular momentum $J^2$. For this discussion, the $j$'s simply label which representation (Hilbert space) you are working in.
Take for example, $l=1$. Then $L_z$ is an operator whose eigenkets are precisely the $|1,m\rangle$s with eigenvalues $m$. That is, $L_z$ is diagonal in this basis with matrix representation:
$$ L_z = \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$
It is important to note that the meaning of this matrix is the representation of an infinitesimal rotation. More precisely, for a finite rotation $\theta$ we can break it up into $N$ infinitesimal rotations $\theta/N$. This rotation then has representation:
$$U(\theta/N) = I - i\theta L_z/N$$
Multiplying all of these out and using the identity $\lim_{N\to\infty} (1+x/N)^N=\exp(x)$ gives:
$$U(\theta) = \exp(-i \theta L_z)$$
Exponentiating a diagonal matrix is easy:
$$U(\theta) = \begin{pmatrix}e^{-i\theta} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{i\theta} \end{pmatrix} $$
Notice that this is not the usual rotation in matrix in $R^3$. However, it is related to it by a change of basis, just one that uses complex numbers.
