Prove that $L_i = \epsilon_{ijk} x_ j p_ k $

Question

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 $$ L_i = \epsilon_{ijk} x_ j p_ k$$

Answer 1

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.

Answer 2

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.