# alternative expression of L^2

• Sep 8th 2011, 07:13 PM
wik_chick88
alternative expression of L^2
show that $L^2$ may be alternative expressed as
$L^2 = L_{0}^{2} + L_0 + L_{-}L_{+} = L_{0}^{2} - L_0 + L_{+}L_{-}$
and that $[L^2, L_{\alpha}] = 0$ for $\alpha = 1, 2, 3$

i know that the generators are $L_{\alpha} = ix_{\alpha}$
and $L_+ = L_1 + iL_2, L_- = L_1 - iL_2$
but i can't get $L^2 = L_{0}^{2} = L_0 + L_{-}L_{+} = L_{0}^{2} - L_0 + L_{+}L_{-}$ no matter which way i try it!

what is $L^2$ anyway? is it $(x_1 x_2 x_3)^2$?
• Sep 8th 2011, 07:14 PM
wik_chick88
Re: alternative expression of L^2
or is $L^2 = L_1^2 + L_2^2 + L_3^2$?