(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Derive the following commutation relations from the general commutation relation for the Lorentz generators:

[tex] [J_i,J_j]=i\hbar\epsilon_{ijk}J_k[/tex]

[tex][J_i,K_j]=i\hbar\epsilon_{ijk}K_k[/tex]

[tex][K_i,K_j]=-i\hbar\epsilon_{ijk}J_k[/tex]

2. Relevant equations

The commutator for the Lorentz generators:

[tex][M^{\mu\nu},M^{\rho\sigma}]= i\hbar ((g^{\mu\rho}M^{\nu\sigma} - (\mu \leftrightarrow \nu))-(\rho \leftrightarrow \sigma))[/tex]

[tex]J_i=\frac{1}{2}\epsilon_{ijk}M^{jk}[/tex]

[tex]K_i=M^{i0}[/tex]

3. The attempt at a solution

I've got the first one.

The second two I'm having slight problems and just need help finding my mistake.

For the second commutator, I have an extra factor of 1/2 on the RHS. I start from:

[tex][M^{jk},M^{j0}]= i\hbar ((g^{jj}M^{k0} - (\mu \doublearrow \nu))-(\rho \doublearrow \sigma))[/tex]

Only the first term on the right have side is non zero since all off diagonal g are 0.

Now this implies:

[tex][J_{i},K_j]= \frac{1}{2}\epsilon_{ijk}i\hbar M^{k0}[/tex]

How do I get rid of that pesky 1/2?

Similarly on the third commutator:

I start from the same place and get to the line:

[tex][K_i,K_j]=[M^{i0},M^{k0}]=-i\hbar M^{ij}[/tex]

I can't figure out how to put the RHS in terms of J_k without getting a factor of 2!

Any help will be appreciated. I'm sure its just stupid errors. Thanks!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: QFT: Srendicki Problem 2.4

**Physics Forums | Science Articles, Homework Help, Discussion**