I'm trying to understand why the eigen value of the total angular momentum [itex]L^{2}[/itex] is [itex]\hbar ^2 l(l+1)[/itex]. The proofs I have seen go like this. Using the ladder operators [itex]L_{\pm} = L_x \pm iL_y[/itex] we can see and the [itex]|l, m \rangle[/itex] state with maximum value of m (eigen value of [itex]L_z[/itex] )(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\langle l,m_{max} | L^2 | l, m_{max} \rangle = \langle l, m_{max} | L_{-}L_+ + L_z^2 + \hbar L^z |l,m_{max}\rangle[/itex]

= [itex]l^2 \hbar ^2 + \hbar ^2 l = \hbar ^2 l(l + 1)[/itex]

Because [itex]L_+[/itex] acting on the state with highest [itex]m[/itex] value annihilates it. So far so good, as long as I can prove that [itex]L_+ |l , m_{max} \rangle = 0 [/itex] , I'll be happy.

Now all proof's I have seen for the eigen values of [itex]L_+[/itex] use the same identity again but now say

[itex]\langle l, m | L_-L_+| l , m \rangle = \langle l, m | L^2 - L_z^2 - \hbar L_z | l , m \rangle = \hbar ^2 \left( l(l+1) - m(m+1) \right) [/itex]

and claim that when [itex]l = m[/itex] this is zero. So you are now using the fact that [itex]L^2 |l , m \rangle = \hbar ^2 l(l+1)[/itex] which was what I wanted to prove in the first place. Is there an independent proof that avoids this cycle?

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# Eigen value of total angular momentum

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