Nuclear structure III: Nuclear and neutron matter

Stefano Gandolfi

Los Alamos National Laboratory (LANL)

National Nuclear Physics Summer School Massachusetts Institute of Technology (MIT) July 18-29, 2016

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 1 / 29 Fundamental questions in nuclear physics

Physics of nuclei: How do nucleons interact? How are nuclei formed? How can their properties be so different for different A? What’s the nature of closed shell numbers, and what’s their evolution for neutron rich nuclei? What is the equation of state of dense matter? Can we describe simultaneously 2, 3, and many-body nuclei?

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 2 / 29 Uniform nuclear and neutron matter

What is nuclear matter? Easy, an infinite system of nucleons! Infinite systems Symmetric nuclear matter: equal protons and neutrons Pure neutron matter: only neutrons W/o Coulomb: homogeneous Nuclear matter saturates (heavy nuclei, “bulk”) Neutron matter positive pressure Properties of infinite matter important to constrain energy density functionals

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 3 / 29 Neutron Matter

Low-density neutron matter and Cold atoms

“Low-density” means ρ << ρ0

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 4 / 29 Ultracold Fermi atoms

Dilute regime mainly s-wave interaction → T fraction of TF T 0 → ∼ Experimentally tunable interaction Crossover from weakly interacting Fermions (paired) to weakly repulsive Bosons (molecules)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 5 / 29 Ultracold Fermi atoms

Dilute regime mainly s-wave interaction → T fraction of TF T 0 → ∼ Experimentally tunable interaction Crossover from weakly interacting Fermions (paired) to weakly repulsive Bosons (molecules)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 5 / 29 Ultracold Fermi atoms

Dilute regime mainly s-wave interaction → T fraction of TF T 0 → ∼ Experimentally tunable interaction Crossover from weakly interacting Fermions (paired) to weakly repulsive Bosons (molecules)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 5 / 29 Ultracold Fermi atoms

Dilute regime mainly s-wave interaction → T fraction of TF T 0 → ∼ Experimentally tunable interaction Crossover from weakly interacting Fermions (paired) to weakly repulsive Bosons (molecules)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 5 / 29 Ultracold Fermi atoms

Dilute regime mainly s-wave interaction → T fraction of TF T 0 → ∼ Experimentally tunable interaction Crossover from weakly interacting Fermions (paired) to weakly repulsive Bosons (molecules)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 5 / 29 Scattering length

Two-body system with attractive interaction:

a = ∞ a < 0 a > 0

u(r) u(r) u(r)

0 0 0

v(r) v(r) v(r)

0 0 0 r r r 2 ~ No bound states Bound state with E =0 Eb 2 b ∼ 2m a

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 6 / 29 Why neutron matter and cold atoms are so similar?

BCS crossover BEC

1 a k Low density neutron matter F

Very Low Density Neutron Matter: cold atoms

Low density neutron matter unitary limit: → reff r0 a , reff = 0 , a =   | | | | ∞ Only one scale: E = ξEFG →

NN scattering length is large and negative, a = 18.5 fm − NN effective range is small, reff = 2.7 fm

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 7 / 29 BCS crossover BEC

1 a k Low density neutron matter F

Very Low Density Neutron Matter: cold atoms

Low density neutron matter unitary limit: → reff r0 a , reff = 0 , a =   | | | | ∞ Only one scale: E = ξEFG →

Why neutron matter and cold atoms are so similar? NN scattering length is large and negative, a = 18.5 fm − NN effective range is small, reff = 2.7 fm

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 7 / 29 BCS crossover BEC

1 a k Low density neutron matter F

Very Low Density Neutron Matter: cold atoms

Low density neutron matter unitary limit: → reff r0 a , reff = 0 , a =   | | | | ∞ Only one scale: E = ξEFG →

Why neutron matter and cold atoms are so similar? NN scattering length is large and negative, a = 18.5 fm − NN effective range is small, reff = 2.7 fm

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 7 / 29 Very Low Density Neutron Matter: cold atoms

Low density neutron matter unitary limit: → reff r0 a , reff = 0 , a =   | | | | ∞ Only one scale: E = ξEFG →

Why neutron matter and cold atoms are so similar? NN scattering length is large and negative, a = 18.5 fm − NN effective range is small, reff = 2.7 fm

BCS crossover BEC

1 a k Low density neutron matter F

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 7 / 29 Unitary Fermi gas

Exact calculation of ξ using AFMC:

0.42

(2) ε 0.40 k

0.38

(4) ξ εk 0.36

N = 38 0.34 N = 48 N = 66 (h) εk

0.32 0 0.1 0.2 0.3 0.4 1/3 1/3 ρ = αN /L

ξ = 0.372(5) Carlson, Gandolfi, Schmidt, Zhang, PRA 84, 061602 (2011) ξ = 0.376(5) Ku, Sommer, Cheuk, Zwierlein, Science 335, 563 (2012) Validation of Quantum Monte Carlo calculations

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 8 / 29 Fermi gas and neutron matter

-1 kF [fm ] 0 0.1 0.2 0.3 0.4 0.5 1 Lee-Yang QMC s-wave 0.9 QMC AV4 Cold Atoms 0.8 FG 0.7 E / 0.6

0.5

0.4 0 2 4 6 8 10 - kF a

Carlson, Gandolfi, Gezerlis, PTEP 01A209 (2012).

Ultracold atoms very useful for nuclear physics!

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 9 / 29 The (condensate) pairs may then form a superfluid:

The pairing gap is basically the energy needed to ”break“ a pair, and then excite the system to its normal energy.

BCS and Fermi superfluidity

BCS (Bardeen-Cooper-Schrieffer) pairs: an arbitrarily small attraction between Fermions can cause a paired state of particles and the system has a lower energy than the normal gas.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 10 / 29 BCS and Fermi superfluidity

BCS (Bardeen-Cooper-Schrieffer) pairs: an arbitrarily small attraction between Fermions can cause a paired state of particles and the system has a lower energy than the normal gas. The (condensate) pairs may then form a superfluid:

The pairing gap is basically the energy needed to ”break“ a pair, and then excite the system to its normal energy.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 10 / 29 D. Page (2012)

Pairing gap and neutron stars

The pairing gap is fundamental for the cooling of neutron stars.

Neutron star crust made of nuclei arranged on a lattice surrounded by a gas of neutrons.

Specific heat suppressed by superfluidity (similarly to the superconducting mechanism).

Cooling dependent to the pairing gap!

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 11 / 29 Pairing gap and neutron stars

The pairing gap is fundamental for the cooling of neutron stars.

Neutron star crust made of nuclei arranged on a lattice surrounded by a gas of neutrons.

Specific heat suppressed by superfluidity (similarly to the superconducting mechanism).

Cooling dependent to the pairing gap!

D. Page (2012)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 11 / 29 Dilute neutron matter

The pairing gap is the energy cost to excite one particle from a BCS (collective) state. Pairing gap of low–density neutron matter vs cold atoms:

-1 kF [fm ] 0 0.1 0.2 0.3 0.4 0.5 0.6

0.6 BCS-atoms QMC Unitarity 0.5

F 0.4

/ E BCS-neutrons ∆ 0.3

0.2 Neutron Matter 0.1 Cold Atoms 0 0 2 4 6 8 10 12 - kF a

Gezerlis, Carlson (2008) Cold atoms results confirmed by experiments!

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 12 / 29 Neutron Matter

Dense neutron matter

“Dense” means ρ (0.5 few times)ρ0 ∼ −

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 13 / 29 For a system made of neutrons and protons, define:

ρn ρp ρ = ρn + ρp , α = − , ρn + ρp Useful relations: 1 α 1 + α ρ = − ρ , ρ = ρ , p 2 n 2 The nuclear matter energy is given by: E N E   Z E   3 2 1 h i (ρ, x) = k(n) + k(p) = ~ k2 (1 + α)5/3 + (1 α)5/3 A A N F A Z F 5 2m F 2 − = f (α) EFG (kF ) .

Fermi gas (1/2)

Non interacting two-components Fermi gas (non-relativistic): E 3 2 (k ) = ~ k2 = E (k ) , N F 5 2m F FG F 2 1/3 where kF = (3π ρ) .

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 14 / 29 The nuclear matter energy is given by: E N E   Z E   3 2 1 h i (ρ, x) = k(n) + k(p) = ~ k2 (1 + α)5/3 + (1 α)5/3 A A N F A Z F 5 2m F 2 − = f (α) EFG (kF ) .

Fermi gas (1/2)

Non interacting two-components Fermi gas (non-relativistic): E 3 2 (k ) = ~ k2 = E (k ) , N F 5 2m F FG F 2 1/3 where kF = (3π ρ) . For a system made of neutrons and protons, define:

ρn ρp ρ = ρn + ρp , α = − , ρn + ρp Useful relations: 1 α 1 + α ρ = − ρ , ρ = ρ , p 2 n 2

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 14 / 29 Fermi gas (1/2)

Non interacting two-components Fermi gas (non-relativistic): E 3 2 (k ) = ~ k2 = E (k ) , N F 5 2m F FG F 2 1/3 where kF = (3π ρ) . For a system made of neutrons and protons, define:

ρn ρp ρ = ρn + ρp , α = − , ρn + ρp Useful relations: 1 α 1 + α ρ = − ρ , ρ = ρ , p 2 n 2 The nuclear matter energy is given by: E N E   Z E   3 2 1 h i (ρ, x) = k(n) + k(p) = ~ k2 (1 + α)5/3 + (1 α)5/3 A A N F A Z F 5 2m F 2 − = f (α) EFG (kF ) .

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 14 / 29 Fermi gas (2/2)

For small asymmetries, α 0, the function f (α) can be expanded ≈ 5 5 f (α) = 1 + α2 + α4 + ..., 9 243 And thus the equation of state is given by:

E 3 5 2 2 4 (ρ, x) = EFG (ρ)+ EFG (ρ)α + = ESNM +α S(ρ)+α S4(ρ)+..., A 5 9 ···

where ESNM is the energy of symmetric nuclear matter (α = 0) and S(ρ) is the symmetry energy given by

1 ∂2 S(ρ) = [E(ρ, x)] EPNM (ρ) ESNM (ρ) , 2 ∂α2 α=0 ' −

and EPNM is the energy of pure neutron matter (α = 1).

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 15 / 29 Then, around ρ0 we can expand:  2 L ρ ρ0 Ksym ρ ρ0 Esym = S0 + − + − + ..., 3 ρ0 18 ρ0

where L is the slope of the symmetry energy, and Ksym is the symmetry compressibility.

Symmetry energy

Around density ρ0 nuclear matter saturates, thus

∂ESNM (ρ) = 0 ∂ρ ρ=ρ0 and we can expand as

 2  3 ρ ρ0 ρ ρ0 ESNM = E0 + α − + β − + ..., ρ0 ρ0 Pure neutron matter instead does not saturate, thus also linear power in ρ is fine.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 16 / 29 Symmetry energy

Around density ρ0 nuclear matter saturates, thus

∂ESNM (ρ) = 0 ∂ρ ρ=ρ0 and we can expand as

 2  3 ρ ρ0 ρ ρ0 ESNM = E0 + α − + β − + ..., ρ0 ρ0 Pure neutron matter instead does not saturate, thus also linear power in ρ is fine.

Then, around ρ0 we can expand:  2 L ρ ρ0 Ksym ρ ρ0 Esym = S0 + − + − + ..., 3 ρ0 18 ρ0

where L is the slope of the symmetry energy, and Ksym is the symmetry compressibility.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 16 / 29 Pure neutron systems are the probe! Why to study symmetry energy?

Theory

Esym, L

Neutron stars Experiments

Neutron matter equation of state

Why do we care? EOS of neutron matter gives the symmetry energy and its slope. Assume that NN is very good - fit scattering data with very high precision. Three-neutron force (T = 3/2) very weak in light nuclei, while T = 1/2 is the dominant part. No direct T = 3/2 experiments.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 17 / 29 Pure neutron systems are the probe! Why to study symmetry energy?

Theory

Esym, L

Neutron stars Experiments

Neutron matter equation of state

Why do we care? EOS of neutron matter gives the symmetry energy and its slope. Assume that NN is very good - fit scattering data with very high precision. Three-neutron force (T = 3/2) very weak in light nuclei, while T = 1/2 is the dominant part. No direct T = 3/2 experiments.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 17 / 29 Pure neutron systems are the probe! Why to study symmetry energy?

Theory

Esym, L

Neutron stars Experiments

Neutron matter equation of state

Why do we care? EOS of neutron matter gives the symmetry energy and its slope. Assume that NN is very good - fit scattering data with very high precision. Three-neutron force (T = 3/2) very weak in light nuclei, while T = 1/2 is the dominant part. No direct T = 3/2 experiments.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 17 / 29 Why to study symmetry energy?

Theory

Esym, L

Neutron stars Experiments

Neutron matter equation of state

Why do we care? EOS of neutron matter gives the symmetry energy and its slope. Assume that NN is very good - fit scattering data with very high precision. Three-neutron force (T = 3/2) very weak in light nuclei, while T = 1/2 is the dominant part. No direct T = 3/2 experiments. Pure neutron systems are the probe!

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 17 / 29 Theory

Esym, L

Neutron stars Experiments

Neutron matter equation of state

Why do we care? EOS of neutron matter gives the symmetry energy and its slope. Assume that NN is very good - fit scattering data with very high precision. Three-neutron force (T = 3/2) very weak in light nuclei, while T = 1/2 is the dominant part. No direct T = 3/2 experiments. Pure neutron systems are the probe! Why to study symmetry energy?

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 17 / 29 Neutron matter equation of state

Why do we care? EOS of neutron matter gives the symmetry energy and its slope. Assume that NN is very good - fit scattering data with very high precision. Three-neutron force (T = 3/2) very weak in light nuclei, while T = 1/2 is the dominant part. No direct T = 3/2 experiments. Pure neutron systems are the probe! Why to study symmetry energy?

Theory

Esym, L

Neutron stars Experiments

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 17 / 29 Let’s start from Hamiltonians used for nuclei: -20 1 + + − + 2 2 + 7/2 + 4 + 0 + + − 2 + -30 0 3 5/2 0 + 0 + 0 4 6 1 + 5/2 − 2 + 1 + He He 7/2 − 8 2 + 4 + 6 1/2 − He + 3 + 7/2 + + -40 Li − 1 + + 4 3/2 3 + 1 5/2 − 2 + + 2 + 7/2 + 7 1 + − 3 Li 2 + 4 7/2 − 2 + 3/2 + -50 8 + 3 3 + Li + 3/2 1 + + 2 5/2 + + 4 0 + − 1 2 + Argonne v 18 1/2 + -60 8 5/2 − 3 Be 1/2 + 2 + with UIX or Illinois-7 − 1 + 3/2 1 + Energy (MeV) GFMC Calculations -70 9 Be 3 + 1 June 2011 10 B -80 AV18 -90 AV18 AV18 +UIX +IL7 Expt. 0 + 12 -100 C

Carlson, et al., RMP (2015)

Neutron matter at nuclear densities

At nuclear densities neutron matter cannot be modeled as in the dilute regime. Nucleon-nucleon (and three-nucleon) interaction become very important.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 18 / 29 Neutron matter at nuclear densities

At nuclear densities neutron matter cannot be modeled as in the dilute regime. Nucleon-nucleon (and three-nucleon) interaction become very important. Let’s start from Hamiltonians used for nuclei: -20 1 + + − + 2 2 + 7/2 + 4 + 0 + + − 2 + -30 0 3 5/2 0 + 0 + 0 4 6 1 + 5/2 − 2 + 1 + He He 7/2 − 8 2 + 4 + 6 1/2 − He + 3 + 7/2 + + -40 Li − 1 + + 4 3/2 3 + 1 5/2 − 2 + + 2 + 7/2 + 7 1 + − 3 Li 2 + 4 7/2 − 2 + 3/2 + -50 8 + 3 3 + Li + 3/2 1 + + 2 5/2 + + 4 0 + − 1 2 + Argonne v 18 1/2 + -60 8 5/2 − 3 Be 1/2 + 2 + with UIX or Illinois-7 − 1 + 3/2 1 + Energy (MeV) GFMC Calculations -70 9 Be 3 + 1 June 2011 10 B -80 AV18 -90 AV18 AV18 +UIX +IL7 Expt. 0 + 12 -100 C

Carlson, et al., RMP (2015)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 18 / 29 Scattering data and neutron matter

How much can we trust the nucleon-nucleon interactions?

In a scattering event with energy Elab two nucleons have p k Elab m/2 , kF ≈ → that correspond to 3/2 (Elab m/2) kF ρ . → ≈ 2π2

−3 Elab=150 MeV corresponds to about 0.12 fm . −3 Elab=350 MeV to 0.44 fm .

−3 Argonne potentials useful to study dense matter above ρ0=0.16 fm , other (soft) interactions not clear

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 19 / 29 What is the Symmetry energy?

symmetric nuclear matter pure neutron matter

Symmetry energy

E = -16 MeV 0 Nuclear saturation 0 -3 ρ0 = 0.16 fm

Assumption from experiments:

−3 ESNM (ρ0) = 16MeV , ρ0 = 0.16fm , Esym = EPNM (ρ0) + 16 −

At ρ0 we access Esym by studying PNM

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 20 / 29 Neutron matter

Equation of state of neutron matter using the AV80+UIX Hamiltonian.

120

100 AV8’+UIX AV8’ 80

60

exp. Esym 40

20 Energy per Neutron (MeV)

0 0 0.1 0.2 0.3 0.4 0.5 -3 Neutron Density (fm )

Incidentally these can be considered as ”extremes” with respect to the measured Esym.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 21 / 29 Neutron matter

Three-neutron interaction uncertainty:

120

100 different 3N

80

60

Esym = 33.7 MeV 40

20 Energy per Neutron (MeV)

0 0 0.1 0.2 0.3 0.4 0.5 -3 Neutron Density (fm )

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 22 / 29 Which one dominates?

Neutron matter

Main sources of uncertainties:

Experimental: Esym Theoretical: form of three-neutron interaction not totally understood

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 23 / 29 Which one dominates?

Neutron matter

Main sources of uncertainties:

Experimental: Esym Theoretical: form of three-neutron interaction not totally understood

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 23 / 29 Which one dominates?

Neutron matter

Main sources of uncertainties:

Experimental: Esym Theoretical: form of three-neutron interaction not totally understood

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 23 / 29 Neutron matter

Main sources of uncertainties:

Experimental: Esym Theoretical: form of three-neutron interaction not totally understood

Which one dominates?

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 23 / 29 Neutron matter

Equation of state of neutron matter using Argonne forces:

100 Esym = 35.1 MeV (AV8’+UIX)

Esym = 33.7 MeV 80 Esym = 32 MeV

Esym = 30.5 MeV (AV8’) 60

40

20 Energy per Neutron (MeV)

0 0 0.1 0.2 0.3 0.4 0.5 -3 Neutron Density (fm ) Gandolfi, Carlson, Reddy, PRC (2012)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 24 / 29 Symmetry energy

Many experimental efforts to measure Esym (or S0) and its slope L:

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 25 / 29 Neutron matter and symmetry energy

From the EOS, we can fit the symmetry energy around ρ0 using L ρ 0.16 Esym(ρ) = Esym + − + 3 0.16 ···

70

65 AV8’+UIX

60

55 Esym=33.7 MeV 50

L (MeV) 45 Esym=32.0 MeV 40

35 AV8’ 30 30 31 32 33 34 35 36 Esym (MeV) Gandolfi et al., EPJ (2014) Tsang et al., PRC (2012)

Very weak dependence to the model of 3N force for a given Esym. Knowing Esym or L useful to constrain 3N! (within this model...)

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 26 / 29 Neutron matter and the ”puzzle” of the three-body force

20 -20 AV8’+UIX 1+ + + − AV8’+IL7 + 2 2 7/2 + 4+ 0 + + − 2 + 18 -30 0 3 5/2 + 0+ 0 AV8’ − 0 + 4 1+ 5/2 2+ 1 He 6 − + He 7/2 8 2+ 4 6 − He + + Li 1/2 1+ 3 7/2 4+ 16 -40 − + 1+ 5/2+ + 3/2 3 − 2 + 2+ 7/2 + 1 − 3 7 + 4+ 7/2 + Li 2 − 2 3/2 3+ + 14 -50 8 + 3 + 3/2 + + Li 2 + 1 4 + 5/2 1+ + Argonne v 0 1/2− 2 18 − 3+ -60 8 5/2 + 12 Be + 2 with UIX or Illinois-7 1/2 + 3/2− 1 1+ Energy (MeV) GFMC Calculations -70 9 3+ 1 June 2011 Be 10 10B -80

Energy per Neutron [MeV] 8 AV18

-90 AV18 AV18 + +UIX +IL7 Expt. 0 6 12 -100 C 0.04 0.06 0.08 0.1 0.12 0.14 0.16 -3 ρ [fm ]

Note: AV8’+UIX and (almost) AV8’ are stiff enough to support observed neutron stars, but AV8’+IL7 too soft. How to reconcile with nuclei??? →

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 27 / 29 Neutron matter at N2LO

EOS of pure neutron matter at N2LO, R0=1.0 fm.

18 N2LO (D2,E1) 18 N2LO (D2,E ) 16 P 16 N2LO (D2, Eτ) AV8’+UIX AV8’ 14 14 12 12 10

(MeV) 10 8 E/A 8 6 6 4 4 2 Energy per Neutron (MeV) 2 0 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0 3 n (fm− ) 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 -3 Neutron Density (fm )

Lynn, et al., PRL (2016). Note: the above (but not all) chiral Hamiltonian able to describe A=3,4,5 nuclei and neutron matter reasonably.

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 28 / 29 End for today...

Summary of this lecture: Low-density neutron matter and cold atoms Superfluidity Dense matter: free Fermi gas Dense matter with interaction

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 29 / 29 End for today...

Summary of this lecture: Low-density neutron matter and cold atoms Superfluidity Dense matter: free Fermi gas Dense matter with interaction

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 29 / 29 End for today...

Summary of this lecture: Low-density neutron matter and cold atoms Superfluidity Dense matter: free Fermi gas Dense matter with interaction

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 29 / 29 End for today...

Summary of this lecture: Low-density neutron matter and cold atoms Superfluidity Dense matter: free Fermi gas Dense matter with interaction

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 29 / 29 End for today...

Summary of this lecture: Low-density neutron matter and cold atoms Superfluidity Dense matter: free Fermi gas Dense matter with interaction

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 29 / 29 Summary of this lecture: Low-density neutron matter and cold atoms Superfluidity Dense matter: free Fermi gas Dense matter with interaction

End for today...

Stefano Gandolfi (LANL) - [email protected] Nuclear and neutron matter 29 / 29