Prove that $L_i = \epsilon_{ijk} x_ j p_ k $ Prove that $L_i$ ; i = 1, 2, 3 of the ang. momentum operator are related to the Cartesian
components of the position and mometum operators by $L_i = \epsilon_{ijk} x_ j p_ k $ (summation convention: sum over repeated indices).
I got to the part following link how $$L_x=-\iota\hbar(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})$$
$$L_y=-\iota\hbar(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})$$
$$L_z=-\iota\hbar(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})$$
but, the next step is a little too quick
$$L_i=-\iota\hbar\epsilon_{ijk}x_j\frac{\partial}{\partial x_k}  $$
How did we reach this last step ? 
And what trasnformation do I need to do to make the RHS look like 
$$ L_i = \epsilon_{ijk} x_ j p_ k$$
 A: It is simplest to do this by brute force, with $(x,y,z)=(1,2,3)$ and using the explicit values
$\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1$, 
$\epsilon_{213}=\epsilon_{132}=\epsilon_{321}=-1$, with the other combinations $0$.
Then, for instance:
\begin{align}
L_x&=L_1=-i\hbar\epsilon_{1jk}x_j\frac{\partial }{\partial x_k}\, ,\\
&=-i\hbar\left(\epsilon_{123}x_2\frac{\partial }{\partial x_3}+\epsilon_{132}
x_3\frac{\partial}{\partial x_2}\right)\, ,\\
&=-i\hbar\left( x_2\frac{\partial }{\partial x_3}-x_3\frac{\partial }{\partial x_2}\right)\, ,\\
&= -i\hbar\left(y\frac{\partial }{\partial z} - z\frac{\partial }{\partial y}\right).
\end{align}
$L_y$ and $L_z$ are done in the same way as $L_x.$
Finally, since $-i\hbar \partial/\partial x_k=p_k$, the last step follows.
A: This the definition of the cross product: the $i^{th}$ component of the cross product between $x$ and $p$ is $\epsilon_{i\,j\,k}\,x_j\,p_k$. As in ZeroTheHero's answer, you're not really proving anything here, but rather simply expanding the definition out.
The directed area of the parallelogram defined by $x$ and $p$ is the two form $x\wedge p$. Then the Hodge dual, implemented with $\epsilon$ converts this directed plane element into a vector, which is equivalent in three dimensions because the vector fully defines the plane it is normal to. That's how we define the cross product: as a "vectorized" version of the directed area of the parallelogram defined by two vectors.
