Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Peskin and Schroeder page 38

  1. Oct 30, 2015 #1
    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?
     
  2. jcsd
  3. Oct 30, 2015 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
     
  4. Oct 30, 2015 #3
    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})##.
     
  5. Oct 30, 2015 #4
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Peskin and Schroeder page 38
  1. Peskin, Schroeder (Replies: 8)

Loading...