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