Using angular momentum in complex coordinates So given the angular momentum operator:
$$L_{z} = - ih\biggl(x \frac{\mathrm{d}}{\mathrm{d}y} - y \frac{\mathrm{d}}{\mathrm{d}x}\biggr)$$
I know how to write these in terms of polar coordinates (obviously). But say I want to work in complex coordinates, such that:
\begin{align}
z &= x + i y & z^{*} &= x - iy
\end{align}
and
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}z} &= \frac{1}{2} \biggl( \frac{\mathrm{d}}{\mathrm{d}x} - i \frac{\mathrm{d}}{\mathrm{d}y}\biggr) & \frac{\mathrm{d}}{\mathrm{d}z^{*}} = \frac{1}{2} \biggl( \frac{\mathrm{d}}{\mathrm{d}x} + i \frac{\mathrm{d}}{\mathrm{d}y}\biggr)
\end{align}
I know that this should be possible (right?) but no matter how I dot the complex terms, I keep getting $x \frac{\mathrm{d}}{\mathrm{d}x}$ and $y \frac{\mathrm{d}}{\mathrm{d}y}$ terms that don't cancel and shouldn't be there. My textbook (Pringe) uses these substitutions in some problems but doesn't explain very well HOW one goes about performing them.
Is it actually possible to work in these coordinates? Or is this just a notation that no one uses?
 A: If you want to switch from $x,y$ to $z, z^*$, you need to invert your relationships because the opposite direction is needed. The inverse maps are
$$ x = \frac{z+z^*}2 \quad y = \frac{z-z^*}{2i}$$
and 
$$\frac{\partial}{\partial x}\equiv\partial_x = \frac{\partial }{\partial z} + \frac{\partial }{ \partial z^*}, \quad
\frac{\partial }{ \partial y}\equiv \partial_y = i(\frac{\partial }{ \partial_z} -\frac{ \partial }{ \partial z^*}) $$
Note that one has to use the partial derivatives because we're talking about functions of several variables. When you substitute the identities above into the definition of the angular momentum (note that it is $\hbar$, hbar, and not $h$ over there), we get
$$ L_z = -i\hbar (x \partial_y - y\partial _x) =-\frac{i\hbar}{2} (2iz \partial_z-2iz^*\partial_{z^*}) = \hbar(z\partial_z - z^*\partial_{z^*})  $$
The factors of $i$ and $2$ cancel, much like the terms $z\partial_{z^*}$ and its complex conjugate $z^*\partial_z$. The latter have to cancel because $L_z$ is symmetric under rotations around the $z$-axis so the $L_z$ $U(1)$ charges have to cancel in all terms. The "charge" of $z$ exactly cancels with the opposite charge of $\partial / \partial z$ (and similarly for the complex conjugate term) but the mixed terms wouldn't.
Also note that there is no prefactor $i$ in the formula for $L_z$ in terms of $z,z^*$ and their derivatives. That's OK because the $L_z=m$ wave functions behave as $z^m$ or $z^{*-m}$ or some compromise and one may pick an $m$ factor (eigenvalue) by differentiating $z^m$ with respect to $z$, getting $mz^{m-1}$, and multiplying by $z$ again to get $mz^m$. The $i$ is hidden in $z=x+iy$, the relative phase of $x,y$.
