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Let $\vec{N}$ and $\vec{S}$ be angular momenta. The quantum number of $\vec{S}$ is $S=1/2$ while for $\vec{N}$ it is all values starting with some integer $\Lambda$ i.e. $N=\Lambda,\Lambda+1,\Lambda+2,...$ and $N_z=\pm\Lambda$ for all $N$. $\vec{N}$ and $\vec{S}$ can couple to $\vec{J}=\vec{N}+\vec{S}$. I want to calculate the matrix element given in the basis $\left|J\Omega;N,S\right\rangle$ where $\Omega=J_z=N_z+S_z$. The element is $$\left\langle J\Omega;N=J+1/2 , S \right| \vec{L} \cdot \vec{S} \left| J\Omega;N=J-1/2,S \right\rangle \, .$$

I understand I have to transform the basis set by Wigner-3j or Clebsch-Gordon from $\left|J\Omega;N,S\right\rangle$ to $\left|N,S; N_z,S_z\right\rangle$ but I'm not sure how to do it in this generality since the only number that is given is $S=1/2$.


For the background here is some link to an old paper: https://journals.aps.org/pr/abstract/10.1103/PhysRev.32.250

I want to understand the off-diagonal terms on page 261 in the determinant above formula (27). I changed the notation to match the modern used term values, namely $\vec{j}_k \rightarrow \vec{N}$.

$\vec{L}$ is the electronic angular momentum, $\vec{N}=\vec{R}+\vec{L}$ the rotational angular momentum including the electronic angular momentum and $\vec{S}$ the spin. $\vec{J}=\vec{N}+\vec{S}$ is the total angular momentum.

Thanks for help

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  • $\begingroup$ This is more or less spin-orbit interaction with $N=L$ so just follow this example. $\endgroup$ May 3, 2020 at 23:41
  • $\begingroup$ Which example? Can you elaborate? $\endgroup$
    – Diger
    May 4, 2020 at 0:15
  • $\begingroup$ This is same as matrix elements of $\vec L\cdot\vec S$. $\endgroup$ May 4, 2020 at 1:41
  • $\begingroup$ Can you mention them? $\endgroup$
    – Diger
    May 4, 2020 at 8:30

1 Answer 1

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You don't need to convert $|J\Omega;N,S\rangle$ to $|N,S;N_z,S_z\rangle$. $$\vec{J}^2=(\vec{N}+\vec{S})^2=\vec{N}^2+\vec{S}^2+2\vec{N}\cdot\vec{S}$$ Notice that $\vec{N}$ and $\vec{S}$ are in different space or different dimensional Hilbert space, so $[\vec{N},\vec{S}]=0$. So, $$\vec{N}\cdot\vec{S}=\frac{\vec{J}^2-\vec{N}^2-\vec{S}^2}{2}$$ Now we know that $\vec{J}^2|J\Omega;N,S\rangle=J(J+1)|J\Omega;N,S\rangle$ and similarly for $\vec{N}$ and $\vec{S}$. Therefore, $$\vec{N}\cdot\vec{S}|J\Omega;N,S\rangle=\frac{J(J+1)-N(N+1)-S(S+1)}{2}|J\Omega;N,S\rangle$$

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  • $\begingroup$ Yes I know that, but apparently this gives $$\langle J\Omega;N=J+1/2,S|J\Omega;N=J-1/2,S\rangle = 0$$ because of the orthogonality. This somehow confuses me, since the result should be $\neq 0$. At first I thought maybe one would have to express it in a different basis, but you are right...It shouldn't depend on that! When you have access to it: What I'm trying to understand is how to obtain the matrix in the work by Hill and van Vleck - On the QM of the Rotational Distortion of Multiplets in Molecular Spectra (Doublet Case Energy formula). Or do you have a better derivation than this old one? $\endgroup$
    – Diger
    May 2, 2020 at 22:41
  • $\begingroup$ @Diger why do you think that this result should be non-zero? $\endgroup$ May 4, 2020 at 0:15
  • $\begingroup$ Sorry...I modified the question. It is not $N$ but $L$. $\endgroup$
    – Diger
    May 4, 2020 at 0:20

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