atom has well defined spin(up and down) and orbital(s,p,d,etc) momentum, but when forming crystals, why the spin degree continues to be good quantum number while orbital momentum is quenched?
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Orbital angular momentum is a good quantum number for the atomic problem because the Coulomb potential between the electron and nucleus is rotationally invariant, but the potential an electron feels in a crystal is not. A non-spherically-symmetric potential can couple states with different $l_z$, and so if $\psi_{l_z}$ were the eigenstate of the spherically symmetric problem with angular momentum $l_z$, then the correct atomic eigenstates in, for example, a cubic or tetragonally symmetric potential separate into linear combinations like $\psi_{l_z} \pm \psi_{-l_z}$ which measures out to total $l_z=0$.
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$\begingroup$ But why does spin continue to be a good quantum number? $\endgroup$– RuslanCommented Dec 28, 2016 at 10:37