Differentiating an arbitrary vector function with angular momentum operator in quantum mechanics This is part of a quantum mechanics problem, where the eigenvalues of angular momentum operator in 3D are supposed to be calculated. The 2 particle wave function is given as:
$$\Psi(\textbf{r}_\textbf{1},\textbf{r}_\textbf{2})=f(\textbf{r}_\textbf{1}^2)g(\textbf{r}_\textbf{2}^2)h(\textbf{r}_\textbf{1},\textbf{r}_\textbf{2})$$
where, f and g are arbitrary functions and h is given. Now the orbital angular momentum is given by:
$$\textbf{L}=\textbf{r}\times\textbf{p}=-i\hbar\textbf{r}\times\nabla$$
It is clear to me, that when operating with $\textbf{L}_1$ for example, the $\textbf{r}_\textbf{2}$ will be a constant. What confuses me though, is how to deal with the arbitrary functions. This is my thinking (for example for the $\textbf{r}_\textbf{1}$ case):
$$\textbf{L}_1\Psi=-i\hbar g(\textbf{r}_2^2)h(\textbf{r}_\textbf{1},\textbf{r}_\textbf{2})(\textbf{r}_1\times\nabla_1)f(\textbf{r}_1^2)-i\hbar f(\textbf{r}_1^2)g(\textbf{r}_1^2)(\textbf{r}_1\times\nabla_1)h(\textbf{r}_\textbf{1},\textbf{r}_\textbf{2})$$
The operation of $h$ is easy, since it is given, but I would like to check, if my thinking is correct for the arbitrary function $f$:
$$(\textbf{r}_1\times\nabla_1)f(\textbf{r}_1^2)=\textbf{r}_1\times f(\nabla \textbf{r}_1^2)\propto \textbf{r}_1\times\textbf{r}_1=0$$
and we may thus concern ourselves only with differentiating $h$. I've deduced this for a couple of simple examples, such as $f(\textbf{r}_1^2)=\alpha\textbf{r}_1^2$. I apologize if I've written something incredibly stupid, but I am relatively new to vector analysis.
Thank you for your help!
 A: Yes, you are correct! However, you should replace $f(\nabla ||\mathbf{r_1}^2||)$ with $\nabla f( ||\mathbf{r_1}^2||)$ in your last line of workings. 
Assume cartesian coordinates, and let $\mathbf{r_1} = (r_1,r_2,r_3)$. We have \begin{align} [(\mathbf{r_1} \times \nabla_1)f]_i &= \epsilon_{ijk} r_j \partial_k f(||\mathbf{r_1}||^2) \\ &= \epsilon_{ijk} r_j \frac{\partial f}{||\mathbf{r_1}||^2} \partial_k ||\mathbf{r_1}||^2 \\&=2 \frac{\partial f}{||\mathbf{r_1}||^2} \epsilon_{ijk}r_j r_k \\ &= 2\frac{\partial f}{||\mathbf{r_1}||^2} (\mathbf{r_1} \times \mathbf{r_1})_i \\ &= 0. \end{align}
Summation over $j$ and $k$ are implied. This is a short hand way of showing $(\mathbf{r}_1 \times \nabla_1) f = \mathbf{0} \iff \mathbf{L}_1 f = \mathbf{0}$ using the Levi-Civita symbol, see the section on cross products. 
You can use also use this method (or expand the equation into components) to explicitly show $\mathbf{L}_1(fh) = \mathbf{L}_1(f)h + f \mathbf{L}_1h$. We then deduce $\mathbf{L}_1(fgh) = g \mathbf{L}_1(fh) = fg\mathbf{L}_1(h)$.
Identically, we have $\mathbf{L_2}{\Psi} = fg \mathbf{L_2}{h}$. Thus, if you define $\mathbf{L} = \mathbf{L}_1 + \mathbf{L}_2$, which seems sensible from my limited knowledge of QM, you only need to worry about differentiating h when applying this operator to $\Psi$ :-).
