How Does the Angular Momentum Commutator Derive the Lorentz Algebra?

[spaghetti3451]

In page 39, Peskin and Schroeder write that (3.15) ${\bf{J}}={\bf{x}} \times{\bf{p}}= {\bf{x}}\times(-i \nabla)$ can be used to derive the Lorentz algebra (3.12) for the rotation group: $[J^{i},J^{j}] = i \epsilon^{ijk}J^{k}$.

I am trying to prove it. Here's my attempt. Can you please suggest the next steps?

$[J^{i},J^{j}] = J^{i}J^{j} - J^{j}J^{i} = (\epsilon^{ijk}x^{j}\nabla^{k})(\epsilon^{jki}x^{k}\nabla^{i}) - (\epsilon^{jki}x^{k}\nabla^{i})(\epsilon^{ijk}x^{j}\nabla^{k})$.

Where do I go from here?

 

[vanhees71]

Just put a scalar wave function to the right of the expression and do the differentiation. Then use
$$\epsilon^{jkl} \epsilon^{jmn}=\delta^{km} \delta^{ln}-\delta^{kn} \delta^{lm}$$
and further contractions. It's a bit lengthy but not diffcult.

 

[spaghetti3451]

Thanks for the reply.

I'm just not really sure if my choices of indices in $(\epsilon^{ijk}x^{j}\nabla^{k})(\epsilon^{jki}x^{k}\nabla^{i}) - (\epsilon^{jki}x^{k}\nabla^{i})(\epsilon^{ijk}x^{j}\nabla^{k})$ is sound. I mean, I used the same indices for $J^{i}$ and $J^{j}$'s expanded expressions.

Should I rather do the following?

$[J^{i},J^{j}] = J^{i}J^{j} - J^{j}J^{i} = (\epsilon^{ikl}x^{k}\nabla^{l})(\epsilon^{jmn}x^{m}\nabla^{n}) - (\epsilon^{jmn}x^{m}\nabla^{n})(\epsilon^{ikl}x^{k}\nabla^{l})$.

 

[spaghetti3451]

Thanks! I got the solution in the following link:

http://physics.ucsd.edu/students/courses/fall2009/physics130b/Ang_Mom.pdf