Spin and isotope shift

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  • Thread starter kelly0303
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Hello! The isotope shift for an atomic transition is usually parameterized as:

$$\delta\nu = K\frac{m_1m_2}{m_1-m_2}+F\delta<r^2>$$

where ##m_{1,2}## are the masses of the 2 isotopes, ##\delta<r^2>## is the change in the mean square charge radius between the 2 isotopes and K and F are some parameters having to do with the electronic transition that is considered. I am a bit confused about how the spin-orbit coupling comes into play. For example, assume that we ignore the spin orbit coupling for now, and we have a transition from an S to a P state. The parameters of this transition are ##\delta\nu_{S-P}## (which we measure) and ##F_{S-P}## and ##K_{S-P}## (which are usually calculated numerically). If we account for the spin orbit coupling (assume it is of the form ##A S\cdot L##), the P state will get split, say, into ##P_{1/2}## and ##P_{3/2}##. Now we have 2 isotope shifts: ##\delta\nu_{S-P_{1/2}}## and ##\delta\nu_{S-P_{3/2}}## and a value of K and F for each of the 2. Is there any relationship between the ##\delta\nu_{S-P}## and ##\delta\nu_{S-P_{1/2}}## and ##\delta\nu_{S-P_{3/2}}##? Or between ##K_{S-P}## and ##K_{S-P_{1/2}}## and ##K_{S-P_{3/2}}## and same for F? Thank you!
 

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Twigg
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I'm really not too sure because I'm not totally confident I know where the constants K and F come from. But my hunch says that these two terms correspond very roughly to the two terms in equation 6.87 of Griffiths. I think F is proportional to the coefficient that for hydrogen is ##\frac{\mu_0 g_p e^2}{3m_p m_e} \langle\mathbf{S}_p \cdot \mathbf{S}_e\rangle## where ##\mathbf{S}_p## is the nuclear spin and ##\mathbf{S}_e## is the electron spin. It's definitely not the same, because the term in Griffiths neglects the nuclear charge radius ##\langle r^2 \rangle## and uses the form for a contact interaction, which assumes ##\langle r^2 \rangle \rightarrow 0##.

If my hunch is correct, then you can see how you would calculate K for the D1 and D2 lines by evaluating the expectation value $$K \propto \langle \frac{3(\mathbf{S}_p \cdot \hat{r})(\mathbf{S}_e \cdot \hat{r}) - \mathbf{S}_p \cdot \mathbf{S}_e}{r^3} \rangle$$ where little r now means the electron radius.

For example, to get K in the ##P_{3/2}## state, you'd need to expand ##|J=\frac{3}{2},m_J \rangle## into spherical harmonics using the Clebsch-Gordon coefficients. Then you could perform the angular integral (the expectation value I put on its own line above), for the ##P_{3/2}## state. You could repeat the process for the ##P_{1/2}## state, and take the difference. That would be related to the difference in K between D1 and D2 lines. Does that help?

I also couldn't tell you how good is the dipole approximation that Griffiths makes (first term in eqn 6.86). It's possible some theory folks who were cited in your previous threads have considered magnetic quadrupole and higher order interactions.
 
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