About the calculation of the Spin-orbit correction for the Hydrogen atom I'm using first order perturbation theory to calculate the energy corrections due to the fine structure of the Hydrogen atom. I'm having some doubts about the calculation of the spin-orbit term. Some people have already asked about this, but not exactly the same question. In particular, I'm interested in using the ${|n,l,s,m_l,m_s\rangle}$ basis, NO the total angular momentum basis,because I want to know how the book (Cohen Tannoudji - Quantum Mechanics Vol2) has obtained a certain result with this basis. I have to calculate the following quantities (here I will be looking just the subspace n=2, l=1):
$$ \left\langle n=2,l=1,s=\frac{1}{2},m_l'',m_s''\left\vert \frac{1}{R^3} L\cdot S\right\vert n=2,l=1,s=\frac{1}{2},m_l,m_s\right\rangle,$$
where R is the operator associated to the radial component of the position, L is the angular momentum operator and S is the Spin operator of the electron.
The literature I have consulted says this term is equal to:
($\int$$\frac{1}{r}$|$\psi$$_{2,1,m_l}$|$^2$ $d\theta d\phi dr$)<n=2,l=1,s=$\frac{1}{2}$,$m_l$'',$m_s$''|L$\cdot$S|n=2,l=1,s=$\frac{1}{2}$,$m_l$,$m_s$>.
I have tried to obtain this result without success, so I would appreciate someone to help me.
This is what I have done so far:
<2,1,$\frac{1}{2}$,$m_l$,$m_s$|$\frac{1}{R^3}$ L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$>
=$\int$<2,1,$\frac{1}{2}$,$m_l$,$m_s$|r><r|$\frac{1}{R^3}$ L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$>r^2 $d\theta d\phi dr$
= $\int$$\psi$*$_{2,1,m_l}$$\frac{1}{r^3}$<r|L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$>r^2 $d\theta d\phi dr$
= $\displaystyle\sum_{m_l''m_s''}$$\int$$\psi$*$_{2,1,m_l}$$\frac{1}{r}$<r|2,1,$\frac{1}{2}$,$m_l$'',$m_s$''><2,1,$\frac{1}{2}$,$m_l$'',$m_s$''|L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$> $d\theta d\phi dr$
=$\displaystyle\sum_{m_l''m_s''}$$\int$$\psi$*$_{2,1,m_l}$$\psi$$_{2,1,m_l''}$$\frac{1}{r}$<2,1,$\frac{1}{2}$,$m_l$'',$m_s$''|L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$> $d\theta d\phi dr$
=$\displaystyle\sum_{m_l''m_s''}$($\int$$\psi$*$_{2,1,m_l}$$\psi$$_{2,1,m_l''}$$d\theta d\phi$ $\int$$\frac{1}{r}$<2,1,$\frac{1}{2}$,$m_l$'',$m_s$''|L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$> $dr$)
Here the sum in $m_l''$ vanishes because of the orthonormality of the wave functions $\psi$$_{2,1,m_l}$
=$\displaystyle\sum_{m_s''}$($\int$|$\psi$$_{2,1,m_l}|^2$$d\theta d\phi$$\int$$\frac{1}{r}$$dr$)<2,1,$\frac{1}{2}$,$m_l$,$m_s$''|L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$>
=($\int$$\frac{1}{r}$|$\psi$$_{2,1,m_l}|^2$$d\theta d\phi$$dr$)$\displaystyle\sum_{m_s''}$<2,1,$\frac{1}{2}$,$m_l$,$m_s$''|L$\cdot$S|2,1,$\frac{1}{2}$,$m_l$,$m_s$>
which is not equal to the expression of the literature, because of the sum in $m_s$, the z-component of the Spin, and also because here I have $m_l$''= $m_l$
What am I missing?
 A: You are mixed up the wave funciton, the radial part and the angular part, A wave fcuntoin spacified by indexes $n$, $l$, $m$:
$$ \tag{1}
 \Psi_{nlm}(r, \theta, \phi) u_s(\theta, \phi) = f_{nl}(r) Y_{lm}(\theta,\phi) u_s(\theta, \phi) \to f_{nl}(r) | l, m\rangle \otimes |S, s\rangle \\
 \equiv f_{nl}(r) | l, m ; S s\rangle  .
$$
Givs the spin-orbital interation $H_{so} = \frac{L\cdot S}{r^3}$,  the matrix element
$$ \tag{2}
 \langle 2, 1, m'; \frac{1}{2} s'|H_{so}| 2, 1, m; \frac{1}{2}, s\rangle \\
=  \left\{\iiint r^2dr \frac{1}{r^3} f_{2,1}^2(r) \right\} \int_0^\pi \sin\theta d\theta \int_0^{2\pi}d\phi Y_{1m'}^*(\theta,\phi)Y_{\frac{1}{2},s'}^*(\theta,\phi) \{\mathbf{L}\cdot\mathbf{S}\} Y_{1m}(\theta,\phi)Y_{\frac{1}{2},s}(\theta,\phi) \\
= \left\{\iiint r^2dr \frac{1}{r^3} f_{2,1}^2(r) \right\} \langle  1, m'; \frac{1}{2} s'|\mathbf{L}\cdot \mathbf{S}| 1, m; \frac{1}{2}, s\rangle
$$
The intergral inside the curry braket is a constant, $V_{so} = \{ \iiint r^2dr \frac{1}{r^3} f_{2,1}^2(r) \}$ depends on $n=2$ and $l=1$ (all fiexed). We may now omit the notation $l=1$ and $S=\frac{1}{2}$ for clarification.
$$
 \langle 2, 1, m'; \frac{1}{2} s'|H_{so}| 2, 1, m; \frac{1}{2}, s\rangle = V_{so} \langle  m'; s'|\mathbf{L}\cdot \mathbf{S}| m; s\rangle
$$
Then expand $\mathbf{L}\cdot \mathbf{S}$ in terms of $L_+ S_-$, $L_- S_+$, and $L_z S_z$ combinations, if you don't want to use the simple relation  $\mathbf{L}\cdot \mathbf{S} = \frac{J^2-L^2-S^2}{2} = \frac{J^2-2-\frac{3}{4}}{2}$, where $L^2 = l(l+1) = 1 \times 2= 2$ and $S^2 = \frac{1}{2} \frac{3}{2} = \frac{3}{4}$ (all in unit of $\hbar^2$), and the value depends solely on $J=\frac{1}{2}$ or $\frac{3}{2}$.
In the hard way, We expand $\mathbf{L}\cdot \mathbf{S}$
$$ \tag{3}
\mathbf{L}\cdot \mathbf{S} = L_x S_x + L_y S_y + L_z s_z = L_z S_z - L_+ S_- - L_- S_+
$$
Substitute Eq.(3) into Eq.(2):
$$ \tag{4}
 \langle 2, 1, m'; \frac{1}{2} s'|H_{so}| 2, 1, m; \frac{1}{2}, s\rangle \\
= V_{so} \langle  m'; s'| L_z S_z - L_+ S_- - L_- S_+| m; s\rangle
$$
The three angular terms in Eq.(4).
The first term $L_z S_z$:
$$
 \langle  m'; s'|L_z S_z| m; s\rangle = m s \hbar^2 \delta_{m', m} \delta_{s',s}
$$
For $L_+ S_-$, the non-vanished term are $s'=-\frac{1}{2}$ and  $s=\frac{1}{2}$, and $m<1$
$$
 \langle  m'; -\frac{1}{2}|L_+ S_-| m; \frac{1}{2}\rangle = \hbar^2 \sqrt{2-m(m+1)} \delta_{m', m+1}; 
$$
For $L_- S_+$, the non-vanished term are $s'=\frac{1}{2}$ and  $s=-\frac{1}{2}$, and $m > -1$
$$
 \langle  m'; \frac{1}{2}|L_- S_+| m; -\frac{1}{2}\rangle = \hbar^2 \sqrt{2-m(m-1)} \delta_{m', m-1}; 
$$
