QFT: Srendicki Problem 2.4

[G01]

Homework Statement

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

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

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

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

Homework Equations

The commutator for the Lorentz generators:

$$[M^{\mu\nu},M^{\rho\sigma}]= i\hbar ((g^{\mu\rho}M^{\nu\sigma} - (\mu \leftrightarrow \nu))-(\rho \leftrightarrow \sigma))$$$$J_i=\frac{1}{2}\epsilon_{ijk}M^{jk}$$

$$K_i=M^{i0}$$

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:

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

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

Now this implies:

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

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:

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

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!

 

[diazona]

For part b, I don't think you should be getting to $[M^{jk},M^{j0}$ at all. It looks like you may be mixing up two different indices: the j in
$$[J_i,K_j] = i\hbar\epsilon_{ijk}K_{k}$$
is not the same as the j in
$$J_i = \frac{1}{2}\epsilon_{ijk}M^{jk}$$
It might help to rewrite the definitions of J and K as
$$J_a = \frac{1}{2}\epsilon_{abc}M^{bc}$$
$$K_d = M^{d0}$$
to keep the different indices straight.

Then again, I tried to do the problem myself and I'm getting the same extra factor of 1/2 that you are...