How is the two-level atom dipole force modified for a real, multi-level atom?

I was trying to compute the dipole force exerted by a laser beam at $$1064$$ nm on a $$^6{\mathrm{Li}}$$ atom (the most important transition being around $$671$$ nm, the laser is greatly red-detuned), when I came upon a conceptual difficulty.

I started from the usual formula for a two-level atom (see Grimm 1999 for instance):

$$U_{\mathrm{dip}}(r) = - \frac{3 \pi c^2}{2 \omega_0^3} \left(\frac{\Gamma}{\omega_0 - \omega} + \frac{\Gamma}{\omega_0 + \omega}\right) I(r),$$

where $$\Gamma$$ is the linewidth of the considered transition, $$\omega_0$$ its frequency and $$I(r)$$ is the local light intensity. Now I have nothing wrong with this formula itself, and I roughly know how to derivate it. However, I was wondering how does this extend to real, multi-level atoms?

I know for instance that this relies partly on the calculation of $$\Gamma$$ for a two-level atom:

$$\Gamma = \frac{\omega_0^3}{3 \pi \varepsilon_0 \hbar c^3} |d|^2,$$ with $$d = \left\langle e \right| q \hat{r} \left| g \right\rangle$$ the dipole matrix element of the transition. But I am pretty sure that this only holds for a two-level system, otherwise you have to sum over the different possible final states.

For a real atom, like $$^6{\mathrm{Li}}$$, I would guess that you would have to sum something over all possible excited states, possibly using a formula that lets explicitly appear $$d$$ instead of $$\Gamma$$. If you consider an atom in its ground state $$2^2 S_{1/2}$$, it is mostly connected to two excited manifolds, which are $$2^2 P_{1/2}$$ (called the $$D_1$$ line), and $$2^2 P_{3/2}$$ ($$D_2$$ line) (see Gehm 2003). This is already complicated, but these two manifolds can be decomposed respectively into $$2$$ and $$3$$ additional manifolds corresponding to different $$F$$ values. Do you have to sum over all of these levels, even though the beam is clearly commpletely off-resonant so I would guess it "doesn't care" about the exact hyperfine structure of $$^6{\mathrm{Li}}$$.

To even add to my confusion, my advisor just told me: "Oh, but look! $$D_1$$ and $$D_2$$ almost have the same $$\Gamma \approx 2\pi \times 6$$ MHz, so just plug $$\Gamma = 2\pi \times 6$$ MHz in the above formula." Aren't you supposed to add them instead (and end up with roughly a factor $$2$$)?

I am as interested in getting an "efficient" way for us, experimentalists, to compute the dipole potential for a given atom, knowing all the transitions linewidth/dipole elements, as I am in getting a better undestanding of the underlying calculations. Can you "forget" about the fine and hyperfine structure when considering very detuned driving? Are there any resources specifically adressing these type of issues (extending two-level atoms results to real atoms)?

Thank you very much for your help.

• Of course you have to sum over all the final states! And of course, contributions from the ones off-resonance will be suppressed by those big denominators. – knzhou Jan 14 at 17:45
• Maybe my question was not clear, but I am already very off-resonant. The light is at $1064$ nm and the considered transition is at $671$ nm. The reason why I consider only the $D_1$ and the $D_2$ lines is that the dipole matrix element from the ground state to the excited state vanishes for other transitions. – QuantumApple Jan 14 at 18:00
• Have you seen that in the Grimm1999 Review that you linked there is a whole section on multi-level atoms after the formula that you gave for the force? Looking at the section it might help. – Wolpertinger Jan 17 at 10:26
• Your intuition that you should sum over states with different values of $d$ is correct. The connection of this to the linewidth is the result of a convenient sum rule that is valid for far-detuned cases (such as the one you are considering), but for conceptual purposes you should understand where this is coming from and how to do the calculation more generally. A nice resource for this is Dan Steck's notes: steck.us/alkalidata/rubidium87numbers.1.6.pdf . They are written specifically for Rb, but can be adapted straightforwardly to other alkali atoms. – Rococo Jan 17 at 15:07
• OP, the relative oscillator strengths of D2 and D1 lines in alkali atoms are usually 2/3 and 1/3 respectively. So if you don't care about getting really accurate results you can weight the D2 and D1 contributions with the relative oscillator strengths I mentioned and sum up the contributions. – wcc Jan 23 at 2:53

I have done it for an atom which possesses only a $$(LS)J$$ coupling. In your case, you will have to adapt the methods, which will be painful -- I know what I'm talking about, because I once had do exactly the same calculate for Lithium as well. Unfortunately, I don't have these calculations anymore.
Using the polarisation $$\vec \epsilon_{q}$$, the ac-Stark shift of the state $$|\alpha\rangle$$ is given by $$\Delta E_{\alpha}^{(q)} = % \frac{E_{0}^2}{4 \pi \hbar c} \sum_{\alpha^{\prime}} \;% sign_{\alpha^{\prime}, \alpha} \;% % \left| \; % |\mu_{\alpha^{\prime}, q, \alpha}|^2 % \frac{\lambda_{\alpha^{\prime}, \alpha} \, % \lambda_{\textrm{laser}}^2}{% \lambda_{\textrm{laser}}^2% - \lambda_{\alpha^{\prime}, \alpha}^2} % % \; \right|%$$ with the atomic transition wavelength $$\lambda_{\alpha^{\prime}, \alpha}$$ and the (electric dipole) transition matrix element $$\mu_{\alpha^{\prime}, \vec q, \vec \alpha} =% e \, \langle\alpha^{\prime}| % r \cdot \epsilon_{q}|\alpha\rangle %%% = e\, \langle \gamma^{\prime} n^{\prime} (S L^{\prime}) J^{\prime} m_{J^\prime} | \vec r \cdot \vec \epsilon_{q} |\gamma n (S L) J m_{\textrm{J}} \rangle$$ Hint: The sign of the ac-Stark shift should be put explicitly. $$sign_{\alpha^{\prime}, \alpha}=-1$$ for the energetically lower state, while it is $$+1$$ for the upper state.
Be aware that NIST defines $$|\mu_{\alpha^{\prime}, q, \alpha}|^2 = \frac{3 \, \epsilon_{0}\, h \, \lambda_{\alpha^\prime, \alpha}^3}{16 \pi^3} \, % A^{(\textrm{NIST})}_{\vec \alpha^\prime, \alpha} % \times % \langle{\gamma n J \, 1\, m_{\textrm{J}} \, q|}{\gamma^{\prime} n^{\prime} (J 1)J^{\prime} \, m_{\textrm{J}^{\prime}}}\rangle^{2}$$ where the last term is the square of the Clebsch-Gordan coefficient. Since Lithium possesses a nuclear spin $$I_{\textrm{spin}}$$ the formula is given by (please cross-check) \begin{align} \label{eq:ACStarkShift} E_{j} = \frac{3\pi}{2}\, c^2 \, I \sum_{k} \;% &\frac{A_{k,j}}{\omega_{k,j}^3} % \left(% \frac{1}{\omega_{k,j}+\omega_{\textrm{laser}}} + % NO MINUS !!! \frac{1}{\omega_{k,j}-\omega_{\textrm{laser}}}% \right) \\% &\times % (2\,J_k +1) \, % (2\,F_j+1) \, (2\,F_k+1) % \nonumber \\% &\times % \left|% \begin{pmatrix} % J_k & 1 & J_j \\ % -M_k & q & M_j % \end{pmatrix} % \left\{ % \begin{matrix} % J_k & I_{\textrm{spin}} & F_k \\ % F_j & 1 & J_j % \end{matrix} % \right\} % \right|^2 % \nonumber % \end{align} Notation: The indices $$j$$ and $$k$$ are for the energetically lower and upper state, %are understood to be multi-indices, $$q$$ specifies the polarisation of the light ($$q=0, \pm1$$ for $$\pi$$, $$\sigma^{(\pm)}$$ light), and the round $$(\ldots)$$ and curly $$\{\ldots\}$$ brackets denote the Wigner-3j and Wigner-6j symbols, respectively.