d(α,ɣ)6Li reaction and second lithium puzzle Shubhchintak ECT* Three-body systems in reaction with rare isotopes

Plan of Talk Introduction and Motivation Calculations using Potential model and phase equivalent potential Photon angular distribution and S-factor Conclusion

Some points about Big Bang Nucleosynthesis The primordial nuclei were formed during the first 20 minutes after the Big Bang. This include 2H, 3He, 4He, 7Li and 6Li (very small amount) The primordial reactions start with p + n → d + γ The deuteron’s yield depends on the primordial baryon/photon ratio (ηB). Abundance of heavier elements also depends on ηB. Later on, 7Li also formed by cosmic rays, novae and pulsations in AGB stars. 6Li is formed by cosmic rays.

Observational Evidence in support of BBN model: Expansion of the Universe Cosmic microwave background Abundance of light nuclei like, 2H, 3He, 4He Abundace of 7Li BBN: 7Li/H = (4.56-5.34) x 10-10 J. Cosm. Astrophys. 10, 050 (2014) Observed: 7Li/H = 1.58 −0.28 +0.35 x 10-10 Astron. Astrophys. 522, A26 (2010) First Lithium Puzzle Isotopic ratio of 6Li/ 7Li, BBN: 6Li/ 7Li ~10-5 J. Cosm. Astrophys. 10, 050 (2014). Observed: 6Li/ 7Li ~5 x 10-2 Astrophys. J 644, 229 (2006). Second Lithium Puzzle

Cross section of d(α, γ)6Li Yield of observed and predicted primordial 7Li are quite well established. If 6Li is primordial then its abundance is determined by the d(α, γ)6Li. Therefore, in order to determine the 6Li/7Li, the accurate cross section of d(α, γ)6Li reaction at Big Bang energies (30 ≤ E ≤ 400 keV) are important. Next we will discuss some previous attempts Difficulties at Low energies

Some previous attempts to measure the cross sections of d(α, γ)6Li : First successful attempt was done by Robertson et al., Phys. Rev. Lett. 47, 1867 (1981). Measured cross section around 3+ resonance at 0.712 MeV and at higher energies. 2. P. Mohr et al, also measured cross section at resonance energies. Phys. Rev. C 50, 1543 (1994). 3. Coulomb dissociation of 6Li at 26 MeV/A on Pb. J. Kiener et al., Phys. Rev. C 44, 2195 (1991). Cross section was dominated by E2 transition, their data set only upper limit. 4. 2nd Coulomb dissociation experiment at 150 MeV/A on Pb. F. Hammache et al., Phys. Rev. C 82, 065803 (2010). Breakup was dominated by nuclear breakup hence no information about S-factor . First successful Experiment by LUNA at two Big Bang energies 94 and 134 keV M. Anders, Phys. Rev. Lett. 113, 042501 (2014).

6Li 𝑑(1+) ⨂ 4He (0+) lf =0 binding energy -1.476 MeV Jd = 1 Jα = 0 Jf = 1 For E1 transitions li = 1 For E2 transitions li = 2 3+ resonance at 0.712 MeV

Photon angular distribution and Astrophysical S24 factor using Potential model 𝑑 𝜎 𝜆 𝑑Ω = 𝑘 𝛾 2 − 1 𝑐 𝑑𝒓 𝜙 6 𝐿𝑖 𝑱 𝑟 𝜳 𝑘 + 𝑨 𝝀 𝒌 𝜸 ∗ (𝐫) 2 Aλk(r) is the vector potential of the photon with helicity λ Ψ 𝑘 + 𝜁𝛼,𝜁𝑑;𝒓𝜶𝒅 =𝜙𝛼(𝜁𝛼)𝜙𝑑(𝜁𝑑) 𝜓 + (𝒌,𝒓𝜶𝒅) α-d scattering wavefunction In the long wavelength approximation the charge current density is: 𝐽 𝑟 = 𝑍𝑑𝑒 2𝑚𝑑 [𝛿 𝒓−𝒓𝑑 𝒑 𝑑+ 𝒑 𝑑 𝛿 𝒓−𝒓𝑑 ]+ 𝑍𝛼𝑒 2𝑚𝛼 [𝛿 𝒓−𝒓𝛼 𝒑 𝛼+ 𝒑 𝛼 𝛿 𝒓−𝒓𝛼 ] A. M. Mukhamedzhanov, Shubhchintak, C. A. Bertulani, Phys. Rev. C 93, 045805 (2016)

Introduce the overlap function of bound state of 6Li 𝐼 𝑙𝑓𝑠𝑗𝑓 𝒓𝛼𝑑 = 𝜙𝛼(𝜁𝛼)𝜙𝑑(𝜁𝑑) 𝜙 6𝐿𝑖 (𝜁𝛼,𝜁𝑑;𝒓𝛼𝑑) =𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑝𝑎𝑟𝑡 𝑋 𝐼 𝑙𝑓𝑠𝑗𝑓 (𝑟𝛼𝑑) Radial overlap function 𝐼 𝑙𝑓𝑠𝑗𝑓 𝑟𝛼𝑑 = 𝐶 𝑙𝑓𝑠𝑗𝑓 𝑊 −𝜂𝑓,𝑙𝑓+ 1 2 (2𝜅𝑟𝛼𝑑)/𝑟𝛼𝑑 For rαd > r0 ANC The asymptotic behaviour of radial scattering wavefunction: 𝜓 + (𝑘,𝑟𝛼𝑑)≈ 1 2𝑖 𝑟 𝛼𝑑 𝑒 −𝑖 𝛿 𝑙𝑖𝑠𝑗𝑖 [ 𝐼 𝑙𝑖 𝑘,𝑟𝛼𝑑 − 𝑒 2𝑖 𝛿 𝑙𝑖𝑠𝑗𝑖 𝑂 𝑙𝑖 (𝑘,𝑟𝛼𝑑)] 𝐼 𝑙𝑖 𝑘,𝑟𝛼𝑑 = 𝐺 𝑙𝑖 (𝑘,𝑟𝛼𝑑)−𝑖𝐹 𝑙𝑖 (𝑘,𝑟𝛼𝑑) 𝑂 𝑙𝑖 𝑘,𝑟𝛼𝑑 = 𝐺 𝑙𝑖 𝑘,𝑟𝛼𝑑 +𝑖𝐹 𝑙𝑖 (𝑘,𝑟𝛼𝑑)

Multipole expansion of the Vector potential: 𝑨 𝜆𝑘 𝑟 = 1 2𝜋 ℎ𝑐 𝑘𝛾 𝒆 𝝀 𝑘𝛾 𝑒 𝑖 𝒌 𝜸. 𝒓 = 1 2 2𝜋 𝑘𝛾 𝐿𝑀 2𝐿+1 𝐷 𝑀𝜆 𝐿 (𝜑,𝜃,0)( 𝑨 𝑒𝑘𝛾 𝒓 +𝜆 𝑨 𝑚𝑘𝛾 𝒓 ) 𝐴 𝑒𝑘𝛾𝐿𝑀 𝒓 =2 𝑖𝐿 ℎ𝑐𝑘𝛾 𝐿+1 2𝐿+1 𝑗 𝐿−1 𝑘𝛾𝑟 𝒀 𝐿𝑀 𝐿−1 𝒓 − 𝐿 2𝐿+1 𝑗 𝐿+1 𝑘𝛾𝑟 𝒀 𝐿𝑀 𝐿+1 𝒓 In long wavelength approximation: 𝑗 𝐿 𝑘𝛾𝑟 ≈ 𝑘𝛾𝑟 𝐿 2𝐿+1 ‼ 𝑗 𝐿+1 𝑘𝛾𝑟 ≈ 𝑘𝛾𝑟 𝐿+1 2𝐿+3 ‼ 𝑗 𝐿−1 𝑘𝛾𝑟 ≈ 𝑘𝛾𝑟 𝐿−1 2𝐿−1 ‼ <<

Angular distribution of the photon in the present case is : Effective charge Where we used, s = 1 (channel spin), L, L’ =1, 2

Radiative Capture Cross section Integrate the angular distribution taking L = L’ The astrophysical factor is then calculated as: Most important quantity is the radial integral To Calculate radial overlap function I of final bound state and continuum wave function ψ, we use potential model RADCAP. C.A. Bertulani, Comput. Phys. Commun. 156, 123 (2003)

Potential model In the two body potential model the radial overlap function can be written as: 𝐼 𝑙𝑓𝑠𝑗𝑓 𝑟𝛼𝑑 = 𝑆 𝑙𝑓𝑠𝑗𝑓 1/2 𝜙 𝑙𝑓𝑠𝑗𝑓 𝑟𝛼𝑑 S is the spectroscopic factor of the final bound state. I is not an eigenfunction of Hamiltonian and does not normalized to 1 The tail of the bound state wavefunction is given by: 𝜙 𝑙𝑓𝑠𝑗𝑓 𝑟𝛼𝑑 ≈ 𝑏 𝑙𝑓𝑠𝑗𝑓 𝑊 −𝜂𝑓,𝑙𝑓+ 1 2 (2𝜅𝑟𝛼𝑑)/𝑟𝛼𝑑 For rαd > r0 b is the single particle ANC and it depends upon potential. C = S1/2 b

Wavefunction and choice of potential First way Adjust the potential parameters to get binding energy. Infinite potential corresponding to three parameters, radius, diffuseness and well depth. Final adjustment can be done using spectroscopic factor Other way is: Adjust WS potential to reproduce the experimental s-wave elastic scattering phase shift and to reproduce the experimental α-d binding energy. Similarly for scattering wave function. F. Hammache et al., Phys. Rev. C 82, 065803 (2010). It include many body effects, so S = 1

Fixing the binding energy -1.476 MeV F. Hammache et al., Phys. Rev. C 82, 065803 (2010). Fixing the binding energy -1.476 MeV For bound state: r0 = 1.20 fm , a = 0.7 fm, V0 = 60.732 MeV For continuum : r0 = 1.25 fm , a = 0.65 fm, V0 = 56.7 MeV However, there are again infinite number of WS potential which differ by ANC Theorem on inverse scattering problem by Gel’fand, Levitan and Marchenko

Phase Equivalent Potential with Correct ANC If V(r) = VN(r) +VC(r) is the α-d potential which reproduce the phase shifts in l = 0, and ϕ(r) is the bound state wave function with ANC b, Phase Equivalent potential is given by, K. Chandan and P. C. Sabatier, Inverse Problems in Quatum Scattering Theory, 𝑉 1 𝑁 𝑟 = 𝑉 𝑁 𝑟 −2𝜆 𝑑2𝐾(𝑟) 𝑑𝑟2 Mukhamedzhanov et al, Phys. Rev. C 83, 055805 (2011). Where, λ = 1 MeV fm2 𝐾 𝑟 =𝑙𝑜𝑔 1+(𝜏−1) 0 𝑟 𝑑𝑥 𝑥2𝜙2(𝑥) The bound state wave function in the potential V(r) = V1N(r) +VC(r) can be expressed in terms of bound state wave function ϕ(r) in the potential V(r) = VN(r) +VC(r) as: 𝜙1(0) 𝜙(0) = 𝜏 1/2 𝜙1 𝑟 = 𝜏 1/2 𝜙(𝑟) [1+(𝜏−1) 0 𝑟 𝑑𝑥 𝑥2𝜙2(𝑥) lim 𝑟→∞ 𝜙1(𝑟) 𝜙(𝑟) = 𝜏 −1/2

τ is the ratio of corresponding ANC’s square Single particle ANC b = 2.69 fm-1/2 Correct ANC C = 2.32 ± 0.12 fm-1/2 Phys. Rev. C 48, 2390 (1993) 𝜏= 𝑏 𝐶 2 =1.34 Difference in the potentials is very small Potentials VN(r) and V1N(r)

Bound state wavefunctions New wavefunction ϕ1(r) has a smaller tail but is higher than old wave function ϕ(r) in the nuclear interior. The total Norm is conserved. Ratio of both wave functions

Why do not multiply ϕ(r) simply by the square root of spectroscopic factor to get correct overlap integral ?? No As V(r) fits the elastic scattering data, therefore it takes into account many-body effects. Any additional factor would be superfluous Incorrect ANC’s should be changed by changing the potentials Exception: Peripheral reaction

Photon’s angular distribution Major inputs: Bound state wavefunction continuum wavefunctions Bound state wavefunctions: Two different potentials are used: Woods-Saxon potential, where to get the correct normalization of the final state overlap function i.e. correct ANC, we introduce spectroscopic factor (S) =(C/b)2 =0.72 Phase equivalent potential described in last section, using S = 1 In both cases overlap function has same asymptotic behaviour being different in the internal region. Continuum wavefunctions: Woods-Saxon potential is used which reproduce the experimental phase shifts for li = 1, 2 For bound state: r0 = 1.20 fm , a = 0.7 fm, V0 = 60.732 MeV For continuum : r0 = 1.25 fm , a = 0.65 fm, V0 = 56.7 MeV

Photon’s angular distribution Red lines : Method 1 Green lines : Method 2 70 keV 100 keV Interference of E1 and E2 200 keV 400 keV Quadrupole (E2) Dipole (E1) A. M. Mukhamedzhanov, Shubhchintak, C. A. Bertulani, Phys. Rev. C 93, 045805 (2016)

What we get??? Angular distribution gives peak around 50 deg. Peak position slightly depends upon energy. Both methods give same angular distribution, confirming that at low energies d(α,γ)6Li is completely peripheral. In the direct measurement by LUNA, the detector was placed at a 90 deg angle. M. Anders et al., Phys. Rev. Lett. 113, 042501 (2014). So new measurements with a better geometry can significantly improve the accuracy of the data

Astrophysical Factor At low energies (<100 keV), E1 cross section dominates and at higher energies E2 dominates. A. M. Mukhamedzhanov, Shubhchintak, C. A. Bertulani, Phys. Rev. C 93, 045805 (2016) Data: Square boxes: M. Anders, Phys. Rev. Lett. 113, 042501 (2014). Solid dots : J Kiner et al. Phys. Rev. c 44, 2195 (1991). Cross: P. Mohr et al., Phys. Rev. C 50, 1543 (1994). Triangles : R. G. H. Robertson et al., Phys. Rev. Lett. 47, 1867 (1981). Long Dashed-dotted line: K. M. Nollett et al., Phys. Rev. C 63, 024003 (2001). Dashed-dotted line: F. Hammache et al., Phys. Rev. C 82, 065803 (2010). E2 E1 E1+E2

Very good agreement with data from LUNA . Our reaction rates should also agree with LUNA and are significantly lower than those adopted by NACRE. C. Angulo et al., Nucl phys A 656, 3 (1999). Our S24 factor is lower than the other previous calculations at Big Bang energies. Example: at 70 keV , S24 = 2.58 MeVnb (Present) S24 = 4.0 MeVnb K. M. Nollett et al., Phys. Rev. C 63, 024003 (2001). S24 = 3.16 MeVnb F. Hammache et al., Phys. Rev. C 82, 065803 (2010). ANC used was 2.7 fm-1/2

Present work and LUNA data give same abundance of 6Li Second Lithium Puzzle Present work and LUNA data give same abundance of 6Li 6Li/H = (0.74 ± 0.16) x 10-14 Using baryon to photon ratio 6.047 x10-10 Plank Collaboration arXiv: 1502.01589 Previous Estimates: 6Li/H = (0.90 - 1.77) x 10-14 J. Cosm. Astrophys. Phys. 10, 050 (2014). 6Li/H = (1.23 - 1.32) x 10-14 Astrophys. J. 744, 158 (2012). Latest estimate of 7Li abundance 7Li/H = (5.1 ± 0.4) x 10-10 Phys. Rev. Lett. 113, 042501 (2014) 6Li/7Li = (1.5 ± 0.3) x 10-5 Previous Estimates: 6Li/7Li = (2 – 3.3) x 10-5 J. Cosm. Astrophys. Phys. 10, 050 (2014). 6Li/7Li = (2.3) x 10-5 Astrophys. J. 744, 158 (2012).

Observed 6Li/7Li ratio in nine stars is ~5 x10-2 LUNA Experiment and present calculations give a strong limit of 6Li/7Li ratio i.e. (1.5 ± 0.3) x 10-5. Observed 6Li/7Li ratio in nine stars is ~5 x10-2 Astrophys. J 644, 229 (2006). Second Lithium problem is getting more difficult to resolve. There is need to reconcile both the Big Bang model predictions and the observational data. Hopes: Recent observations using 3D non-local thermodynamic equilibrium modelling technique results in decreased 6Li/7Li ratio in 4 stars. K. Lind et al., Astron. Astrophys. 554, A96 (2013)

Conclusions Thanks for your attention 1. First and general expression of the angular distribution of photon for (α,γ) reaction is derived. 2. Using proper bound state wavefunction with correct ANC, we calculate the photon angular distribution for the reaction d(α, γ)6Li . 2. The angular distribution resulted from the interference of E1 and E2 transition give peak at 50 deg. 3.Our finding can significantly improve the accuracy of the data for new measurements. 4. Our calculated astrophysical factor matches with the experimental data from LUNA. This proves the power of ANC method. 5. The isotopic ratio 6Li/7Li = (1.5 ± 0.3) x 10-5 calculated by LUNA and also here gives the strong limit. 6. To resolve second lithium puzzle, there is need to reconcile both the Big Bang model predictions and the observational data or to explain the three-orders of magnitude difference. Thanks for your attention