Physics Asked by ersbygre1 on June 6, 2021
If we have two angular momenta, $j_1$ and $j_2$, which couple to the total angular momentum $J$, we can choose between two sets of basis systems,
$$ (j_1,m_1,j_2,m_2);text{ vs. };(j_1,j_2,J,M). $$
The second one is useful, because $J$ being the total angular momentum of the system corresponds to a conserved quantity and plays the role of the generator of rotations.
This leads to my question: when is it useful to instead consider the first basis set, i.e. when is it useful to consider decoupled angular momenta?
(I know one example is the Paschen–Back effect, but are there other cases as well..?)
Whenever we have multiple basis it’s generally a good idea to look at the energy scales of different parts of the Hamiltonian. For example, consider the hydrogen atom Hamiltonian with the spin-orbit coupling: $$H=H_{coul} + H_{SO}$$ where $H_{coul}$ is the coulomb term and $H_{SO}$ is the spin-orbit coupling term. The energy scale of the coulomb term is of the order of $10$ eV and that of the spin-orbit coupling is of the order of $10^{-5}$ eV. What this means is that we can treat the spin-orbit as a perturbation to the coulomb term. So in this case, as the coulomb term commutes with individual z-component of angular momentum operators the $j_1,m_1;j_2,m_2$ basis is easier to work with.
With multielectron systems a third term comes into picture. The electron-electron correlation. And this term doesn’t commute with the individual $L_z$. So for these systems, if the correlation energy is much lower than spin-orbit (both are generally much lower than coulomb), then we can use the decoupled basis generally known as $ls$ basis. This is heavily used in spectroscopy, known as Russel-Saunders coupling. On the other hand if the correlation energy is higher than the spin-orbit coupling (as is the case in some actinides) we use the coupled basis generally known as $jj$ basis.
Answered by Superfast Jellyfish on June 6, 2021
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