# A simple equality of Generalized Lorentz Operators

1. Nov 30, 2013

### tamiry

1. The problem statement, all variables and given/known data
Hi

we have Lorentz operators
$J^{\mu\nu} = i(x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu})$

and these have
$[J^{\mu\nu}, J^{\rho\sigma}] = i(\eta^{\nu\rho}J^{\mu\sigma} + \eta^{\mu\sigma}J^{\nu\rho} - \eta^{\mu\rho}J^{\nu\sigma} - \eta^{\nu\sigma}J^{\mu\rho})$

Now define generalized rotation operators, for i, j, k space coordinates
$M^{i} = \epsilon_{ijk}J^{jk}$

Show that $M^{i}$ have the SU(2) algebra. i.e.
$[M^{i}, M^{j}] = i\epsilon_{ijk}M^{k}$

2. Relevant equations
(all is above)

3. The attempt at a solution
I've done a few attempts and failed. so I tried taking an example
$M^{1}= \epsilon_{123}J^{23}+\epsilon_{132}J^{32} = J^{23} - J^{32}$
$M^{2}= \dots = J^{31} - J^{13}$

and now
$[M^{1},M^{2}] = [J^{23}, J^{31}] - [J^{32}, J^{31}] + [J^{23}, J^{13}] - [J^{32}, J^{13}]$

well $J^{ij} = -J^{ji}$ so the second term negates the first one ($J^{32} for J^{23}$) and like wise the fourth and third term. So all in all I get zero. and that's no SU(2) :(

where did I go wrong?

thanks a lot for reading this
Tamir

2. Nov 30, 2013

### TSny

Hello, tamiry.

Did you leave out a factor of 1/2 here?

Those terms don't cancel. Watch the signs carefully.

3. Nov 30, 2013

### tamiry

but of course...
thanks!