Angular momentum quantum mechanics operator In classical mechanics the angular momentum is given as $\bf L = r \times p$ and when going  to quantum mechanics you replace $\bf r$ and $\bf p$ by their respective quantum operators, namely $\bf \hat{r} = r$ and ${\bf \hat{p}} = -i \hbar \bf{\nabla}$, that is $${\bf \hat{L}}= -i \hbar {\bf r} \times \nabla.\tag{1}$$ Shouldn't one use a symmetrised form for $\bf\hat{L}$, i.e. $${ {\bf \hat{L}} = -i \hbar \frac{1}{2} (\ \bf r \times \nabla  - \nabla \times r )}?\tag{2}$$
 A: No, because by definition of the vector product, it is antisymmetric and thus your definition of L would be 2 times the "normal" L. (check it by computing each component!)
L is defined by the correspondance principle (+ its commutation relations).
A: To give an explicit answer with a calculation, consider the product: $$\vec{r}\times\nabla=\hat{i}(y\partial_z-z\partial_y)+\hat{j}(z\partial_x-x\partial_z)+\hat{k}(x\partial_y-y\partial_x)$$
Notice that since $[r_i,\partial_j]=0$ for $i\neq j$, you can rewrite this: $$\vec{r}\times\nabla=\hat{i}(\partial_zy-\partial_yz)+\hat{j}(\partial_xz-\partial_zx)+\hat{k}(\partial_yx-\partial_xy) = -\nabla\times\vec{r}$$
So that finally: $$-i\hbar\vec{r}\times\nabla = -i\hbar\frac{1}{2}(\vec{r}\times\nabla-\nabla\times\vec{r})$$
So the two operators are actually the same.  
A: No, because when expanding $L_k$ you see that only components $X_k$ and $P_h$ with $h \neq k$ enter the formula, and they commute. So you can safely use the classical definition also in QM and the symmetrised formula would produce the same result.
A: The canonical commutation relations (CCR) are $$[X_i, P_j] = i\hbar \delta_{ij}$$ so for example $X$ commutes with $P_Y$ and $P_Z$.
When expanding the cross product, its components are $$ L_k = \epsilon_{ijk} X_i P_j$$ but notice the Levi-Civita symbol vanishes when two indices are equal. Hence you never have terms $X_i P_i$, so you are never actually multiplying observables that do not commute.
Hence no need to symmetrize.
