# Calculating off-diagonal Matrix Element

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

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

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, $$\vec{N}\cdot\vec{S}|J\Omega;N,S\rangle=\frac{J(J+1)-N(N+1)-S(S+1)}{2}|J\Omega;N,S\rangle$$
• 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? May 2, 2020 at 22:41
• Sorry...I modified the question. It is not $N$ but $L$. May 4, 2020 at 0:20