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Expectation value of angular momentum
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[QUOTE="fluidistic, post: 4789205, member: 122352"] [h2]Homework Statement [/h2] A particle is under a central potential. Initially its wave function is an eigenfunction ##\psi## such that ##\hat {\vec L ^2} \psi = 2 \hbar ^2## , ##\hat L_3 \psi =0##. Calculate the expectation value of ##\hat {\vec L}## for all times. [h2]Homework Equations[/h2] ##\frac{d}{dt}<\hat {\vec L}>=\frac{i}{\hbar } [\hat H , \hat {\vec L}]##. I get that the quantum numbers of the eigenfunction are ##l=1##, ##m=0##. Not sure if that helps. [h2]The Attempt at a Solution[/h2] Using the relevant equation, I get that ##\frac{d}{dt}<\hat {\vec L}>=0## because H commutes with L because it's a central potential problem. This means that the expectation value of the operator L is constant in time. So I can calculate it at t=0, and use the information that at ##t=0##, ##\Psi (x,t=0)## is an eigenfunction (of the Hamiltonian but also of L^2 and L_z apparently). So here's my problem: ##<\hat {\vec L}>(0)=<\psi , \hat {\vec L} \psi>##. I don't know how to calculate this. I've checked in Nouredine Zetilli's book because its said to have many solved problems but I couldn't find a single problem where one have to calculate the expected value of a vector operator such as ##\hat {\vec L}##. Using intuition I guess I should rewrite that expectation value in terms of the expectation values of L^2 and L_z (=L_3) but I don't know how I could do this. Here's an attempt: ##\hat {\vec L^2}=L_x^2+L_y^2+L_z^2##. ##\Rightarrow (L_x^2+L_y^2) \psi =2\hbar ^2 \psi##. Stuck here. Here's another attempt: I want the expectation value of ##<L_x \hat i + L_y \hat j + L_z \hat k>##. I know I can't write this as ##<L_x>+<L_y>+<L_z>##. So I'm stuck here. Any help is appreciated. [/QUOTE]
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Expectation value of angular momentum
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