Recent content by Ben4000

I am not sure how you can infer that \hat{L}_{z}^{2} = (\hbar m)^{2} from L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle

Yes \left\langle \hat{L}^{2} \right\rangle = \left\langle l,m\right| L^{2} \left|l,m\right\rangle \left\langle l,m\right| L^{2} \left|l,m\right\rangle = \left\langle l,m\right| L_{x}^{2}+L_{y}^{2}+L_{z}^{2} \left|l,m\right\rangle \left\langle l,m\right|...

I can show that <Lx>=0 using the ladder opertators, but i don't think this is what is wanted from this question... how do i use [Ly,Lz]=i(hbar)Lx to prove <Lx> = 0?

Homework Statement Using the fact that ,\left\langle \hat{L}_{x}^{2} \right\rangle = \left\langle \hat{L}_{y}^{2} \right\rangle show that \left\langle \hat{L}_{x}^{2} \right\rangle = 1/2 \hbar^{2}(l(l+1)-m^{2}. The Attempt at a Solution L^{2} \left|l,m\right\rangle = \hbar^{2}l(l+1)...

Homework Statement Express Lx in terms of the commutator of Ly and Lz and, using this result, show that <Lx>=0 for this particle. The Attempt at a Solution [Ly,Lz]=i(hbar)Lx <Lx>=< l,m l Lx l l,m> then what?