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While reading about $LS$ coupling, and the fine structure of atomic energy levels, of various electron-electron configurations, I came across two different representations.

For example, if we consider a 2 electron system - $4p4d$ for example, the various states can be represented in the coupled basis and the uncoupled basis.

In the coupled basis, the representation is of the form $|L,S,J,m_j\rangle$, with $(2j+1)$ microstates.

In the uncoupled basis, the representation is of the form $|L,S,m_L,m_S\rangle$, with $(2l+1)(2s+1)$ microstates.

The coupled basis representation is usually denoted as term symbols while determining the energy levels. The total no. of microstates in both the representations is obviously the same, however, there is no one-to-one correspondence between the representations in the coupled and un-coupled basis. We can use Clebsch Gordon coefficients to depict, coupled basis eigenstates as a superposition of uncoupled basis eigenstates. So, the following equation should be true

$|L,S,J,m_j\rangle = \sum C_i |L_i,S_i,m_{L_i},m_{S_i}\rangle$

My question is, why do we need the coupled basis, to determine energy levels? Shouldn't we be able to determine the energy levels using the uncoupled basis eigenstates? For example, there are 12 energy levels associated with $4p4d$ corresponding to total of 60 microstates. However, if we write them in the uncoupled basis, we are still able to obtain 60 microstates, but how are we going to arrange them in energy levels ?

Till, spin-orbit interaction term, the no. of energy levels in both basis, is same, and arranged using Hunds first and second rules. Using Hund's third rule, we can determine which one of the following has a higher energy $^3D_2$ or $^3D_1$. But if we are provided with $|1,1,-1,0\rangle$ and $|1,1,1,1\rangle$ in the uncoupled basis ($|L,S,m_l,m_s\rangle$ ), how can we find out which one has a higher energy ? Here both $L$ and $S$ are the same, so we have terms, and so we are not able to compare their levels.

Is this because $m_l$ and $m_s$ are bad quantum numbers, that we need $J$ to determine the energy levels? Or is it something else that I'm missing ?

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Is this because $m_l$ and $m_s$ are bad quantum numbers, that we need $J$ to determine the energy levels?

Basically, yes, just that. In the presence of spin-orbit coupling, neither $\vec L$ nor $\vec S$ is conserved (only $\vec J$ is), so the uncoupled basis is not an eigenbasis of the hamiltonian.

If you discard the spin-orbit coupling, of course, then all of those states will be degenerate, and you can use whichever basis is most convenient.

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  • $\begingroup$ Thank you so much, but I have a question. If we represent the states in the coupled basis, aren't we losing information about the orbital angular momentum and spin angular momentum orientations? Like say an electron state is $^3P_1 (m_j=0)$, and we act the $L_z$ operator on this. What will we get, will this even be an eigenket of this operator? We only have information about the total angular momentum in that case. Is this 'sacrifice' absolutely necessary ? $\endgroup$ Aug 16 at 8:59
  • $\begingroup$ Seems to me, that if we represent this in an uncoupled basis, we retain information about z components of orbital angular momentum and spin. However, we can't find individual energy levels. But if we represent in coupled basis, we are able to find the energy levels, but we lose information about the z component of the angular momentum and spin separately. Is there any way, we can have all the information ? That is, any way, we can know all $L,S,m_l,m_s,J,m_j$ and the energy levels at the same time ? $\endgroup$ Aug 16 at 9:02
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    $\begingroup$ @NakshatraGangopadhay That's the whole point of the coupling. Since $[H,L_z]\neq 0$, it makes no sense to worry about whether the $H$ eigenstates have well-defined $L_z$ (and, in general, they won't be). $H$ and $L_z$ are incompatible. $\endgroup$ Aug 16 at 9:03
  • $\begingroup$ Thank you so much, I think it is more clear to me now. $\endgroup$ Aug 16 at 9:04

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