alternative expression of L^2

**wik_chick88** · September 8th 2011, 06:13 PM

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 $(x_1 x_2 x_3)^2$ ?

**wik_chick88** · September 8th 2011, 06:14 PM

or is $L^2 = L_1^2 + L_2^2 + L_3^2$ ?

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