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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?

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1 Answer 1

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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 $m > -1$ $$ \langle m'; \frac{1}{2}|L_- S_+| m; -\frac{1}{2}\rangle = \hbar^2 \sqrt{2-m(m-1)} \delta_{m', m-1}; $$

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  • $\begingroup$ You have mixed up the wavefunction and the quantum state: $\langle{r,\theta,\phi}|nlm\rangle=f_{nl}(r)Y_{lm}(\theta,\phi)$. $\endgroup$
    – Christophe
    Mar 15, 2021 at 17:27
  • $\begingroup$ Wave fcuntoin is the Quantum state in space-representation. $\endgroup$
    – ytlu
    Mar 15, 2021 at 18:02
  • $\begingroup$ Thank you for your answer. Unfortunately, what you brote did not clarify my doubt. I think I have not fully understand you answer. Would you explain ir a little more for me? What I used in my calculations was basically three things: 1) <r,$\theta$,$\phi$ | n,l,$m_l$,1/2,$m_s$> = $\psi$$_{nlm}$, where, in this case the wave function $\psi$ is the product of a radial function and a spherical harmonic. 2) $\int$|r><r| r^2 dr^3 = id 3) Summation over all states in the subspace of (n,l) fixed is an identity on that subspace $\endgroup$ Mar 15, 2021 at 18:46
  • $\begingroup$ I added some more. Let me know if you has further question. $\endgroup$
    – ytlu
    Mar 15, 2021 at 18:49
  • $\begingroup$ Thank you, I will read it right now. $\endgroup$ Mar 15, 2021 at 18:50

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