Skip to main content
removed a typo, fixed grammar
Source Link
wsc
  • 5.5k
  • 33
  • 31

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 $l_z$, I seem to recall thatthen the correct atomic eigenstates in, for example, a cubic or tetragonally symmetric potential potential for example separate into linear combinations like $\psi_{l_z} \pm \psi_{-l_z}$ which measures out to total $l_z=0$.

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 $l_z$, I seem to recall that the correct atomic eigenstates in a cubic or tetragonally symmetric potential potential for example separate into linear combinations like $\psi_{l_z} \pm \psi_{-l_z}$ which measures out to total $l_z=0$.

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 $l_z=0$.

Source Link
wsc
  • 5.5k
  • 33
  • 31

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 $l_z$, I seem to recall that the correct atomic eigenstates in a cubic or tetragonally symmetric potential potential for example separate into linear combinations like $\psi_{l_z} \pm \psi_{-l_z}$ which measures out to total $l_z=0$.