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Does the orbital of an electron change when it experiences a spin-orbit interaction, and if so, how?

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Does the orbital of an electron change when it experiences a spin-orbit interaction, and if so, how?

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blue_leaf77

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The presence of spin-orbit term which is proportional to ##\mathbf{L}\cdot \mathbf{S}## makes the quantum number ##m_l## and ##m_s## no longer good quantum numbers because ##L_z## and ##S_z## do not commute with ##\mathbf{L}\cdot \mathbf{S}##. Instead, it's the total operator ##J_z = L_z + S_z## that commutes with the spin-orbit term. Therefore, the eigenstate of the Hamiltonian must be specified, among others, by ##m_j## associated with the operator ##J_z##. Actually a more complete picture of the effect of spin-orbit coupling is provided by Dirac equation, where the eigenstate has the form of a 4 element column vector, the first two being called "large component" and the last two "small component". But for perturbative effect, such as in low Z hydrogen like ions, the large component dominates the eigenstate and thus we can regard this as the "true eigenstate". As stated before, one of quantum numbers used to specify the state is now ##m_j##, along with its vector's magnitude ##j##, so the spin-angular part of the eigenfucntion looks ##\langle \theta,\phi |l,s,j,m_j \rangle## (as in the usual theory of angular momentum quantum numbers, ##m_j## runs from ##-j## to ##j## with unit increment). This new eigenstate ##\langle \theta,\phi |l,s,j,m_j \rangle## will be a linear combination of the product between spherical harmonics (eigenfunction of ##L^2## and ##L_z##) and spin vector (eigenfunction of ##S^2## and ##S_z##). For example for ##l=1##, ##s=1/2## (obviously for electron), ##j=3/2## and ##m_j=1/2##,

$$ \langle \theta,\phi |1,1/2,3/2,1/2 \rangle = \sqrt{2/3} Y_{10}(\theta,\phi ) \chi_+ + \sqrt{1/3} Y_{11}(\theta,\phi ) \chi_- $$

We see that as a consequence of taking the spin into account, we now have splitted probabilities to find the electron in either spin up or spin down and for our example it's more likely to find the electron in spin up state.

As for the radial part, the exact form of this function for hydrogen like ions looks very complicated despite can be found analytically. But for perturbative effect, the radial part looks just like the usual radial part without spin orbit effect but scaled down toward the nucleus. This shrinkage might be thought of the result of relativistic effect in increasing the mass of the electron.

$$ \langle \theta,\phi |1,1/2,3/2,1/2 \rangle = \sqrt{2/3} Y_{10}(\theta,\phi ) \chi_+ + \sqrt{1/3} Y_{11}(\theta,\phi ) \chi_- $$

We see that as a consequence of taking the spin into account, we now have splitted probabilities to find the electron in either spin up or spin down and for our example it's more likely to find the electron in spin up state.

As for the radial part, the exact form of this function for hydrogen like ions looks very complicated despite can be found analytically. But for perturbative effect, the radial part looks just like the usual radial part without spin orbit effect but scaled down toward the nucleus. This shrinkage might be thought of the result of relativistic effect in increasing the mass of the electron.

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