How to Derive Specific Commutation Relations for Lorentz Generators?

[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...