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Quick quantum question. I understand the total angular momentum operation [tex]\hat{L}^2 \psi _{nlm} = \hbar\ell(\ell + 1) \psi _{nlm}[/tex] which means the total angular momentum is [tex]L = \sqrt{\hbar\ell(\ell + 1)}[/tex] But how about applying this to an arbitrary superposition of eigenstates such as this [tex]\psi = \frac{1}{\sqrt{2}} \psi _{310} + \frac{1}{\sqrt{2}} \psi _{420}[/tex] and trying to find the expected value of the total angular momentum? Here is my best guess: [tex]\hat{L}^2 \psi = \frac{1}{\sqrt{2}} \hbar \, 1(1+ 1) \psi _{310} + \frac{1}{\sqrt{2}} \hbar \, 2(2 + 1) \psi _{420} = \frac{2}{\sqrt{2}} \hbar \psi _{310} + \frac{6}{\sqrt{2}} \hbar \psi _{420} [/tex] so the expected value is [tex]\langle L ^2 \rangle = \left( \frac{2}{\sqrt{2}} \hbar \right)^2 + \left( \frac{6}{\sqrt{2}} \hbar \right)^2[/tex] which means the expected value of the total angular momentum is maybe [tex]\langle L \rangle = \sqrt{ \left( \frac{2}{\sqrt{2}} \hbar \right)^2 + \left( \frac{6}{\sqrt{2}} \hbar \right)^2}[/tex] Is this correct? Thank you

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# Total angular momentum operator for a superposition

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