lecture 3: ab initio theory of nuclear forces and nuclear ...lecture 3: ab initio theory of nuclear...

LLNL-PRES-777097This work was performed under the auspices of the U.S. Department of Energy by Lawrence Livermore National Laboratory under contract DE-AC52-07NA27344. Lawrence Livermore National Security, LLC

Lecture 3: Ab Initio Theory of Nuclear Forces andNuclear Structure (continued)

TALENT Course 6: Theory of exploring nuclear reaction experiments

Sofia Quaglioni

LLNL-PRES-7770972

§How to describe the basic interactions of protons and neutrons?

§Can we describe nuclei as many-body systems of interacting protons and neutrons?

Content

Option 2: Solution of Schroedinger equationfor intrinsic + C.M. Hamiltonian

Work with A single particle coordinates and separate intrinsic and center of mass motion

LLNL-PRES-7770974

A-nucleon problem

H(A) = Ti + VNN(|ri-rj|) + V3NA

i=1

A

i<j=1

A

i<j<k=1ijk

s1 s2

r1r2

O

sA…

rA

LLNL-PRES-7770975

1) Solve single-particle problem:

We know how to solve the independent-particle problem

H(A) = Ti + U(ri) = hi

A

i=1

A

i=1

Single-particleHamiltonian

h ϕn(r) = εn ϕn(r)

~

e.g.: U(r) = mΩ2r2−12

LLNL-PRES-7770976

2) The antisymmetric A-nucleon solutions can be build as

We have that:

We know how to solve the independent-particle problem

1ϕk1(r1) ϕk1(r2) … ϕk1(rA)

ϕk2(r1) ϕk2(r2) … ϕk2(rA)

ϕkA(r1) ϕkA(r2) … ϕkA(rA)

… … …A!− det𝛟k =√−

H(A) 𝛟k = Ek 𝛟k with A

i=1Ek = εki

LLNL-PRES-7770977

3) Use the independent-particle model solutions as `basis states’ to build an ansatz for the A-nucleon wave function

What about our original A-nucleon problem?

Ψ (r1,r2,…,rA) = ck 𝛟k(r1,r2,…,rA) k

(H(A)-E) Ψ = 0 è ck (H(A)-E) 𝛟k= 0k

N

N

Maximum number of excitations

LLNL-PRES-7770978

4) Project the equation on the basis states (from the left)


ck ∫𝛟m(r1,…,rA) H(A) 𝛟k(r1,…,rA) dr1…drA*

= E ck ∫𝛟m(r1,…,rA)𝛟k(r1,…,rA) dr1…drAk

*

𝛿mk

k

N

N

Hmk ck = E cmk

N

LLNL-PRES-7770979

The A-nucleon Schrödinger equation becomes a linear algebra eigenvalue problem

§ The elements of the N×N Hamiltonian matrix are

§ And the unknown expansion coefficients ck are the elements of the eigenvector c


H mk = ∫𝛟m(r1,…,rA) H(A) 𝛟k(r1,…,rA) dr1…drA*

H c = E c uknown

LLNL-PRES-77709710

The A-nucleon Schrödinger equation becomes a linear algebra eigenvalue problem

§ The elements of the N×N Hamiltonian matrix are

§ And the unknown expansion coefficients ck are the elements of the eigenvector c


H c = E c uknown

H mk = ⟨𝛟m | H(A) | 𝛟k⟩ Short-hand notation

LLNL-PRES-77709711

§This is an ‘expansion’ technique: uses large (but finite!) expansion on A-body basis states

§Convergence to the exact result is approached (variationally) by increasing N (i.e., basis size)

§Antisymmetrization is trivial

§Did we forget about anything?

Some notes

LLNL-PRES-77709712

§ In the independent-particle problem, in general the c.m. motion is mixed with intrinsic motion, giving rise to spurious effects

§Exception: harmonic oscillator (HO) potential is exactly separable

What about the center of mass motion?

H(HO) = Ti + mΩ2r2A

i=1

~ −12

LLNL-PRES-77709713

§ In the independent-particle problem, in general the c.m. motion is mixed with intrinsic motion, giving rise to spurious effects

§Exception: harmonic oscillator (HO) potential is exactly separable

What about the center of mass motion?

H(HO)~A

i<j=1−12

= Tint + + Tcm + AmΩ2R2mΩ2 (ri-rj)22A

HintHcm

LLNL-PRES-77709714

§ Solutions of independent-particle problem for the HO potential factorize as

§ For each eigenvalue E there will be a set of degenerate eigenstates all sharing the same intrinsic structure but different c.m. excitations

§ This degeneracy can be broken by adding to the Hamiltonian a Lawson term

§ New eigenvalues are 𝐸& = 𝐸 + 𝜆𝑁+,ℏΩ, i.e. 𝑁+, >0 states are pushed to high energy

Elimination of spurious motion of the center of mass

𝐻& →𝐻 + ℒ ℒ = 𝜆 𝐻+,3 − 56ℏ3

LLNL-PRES-77709715

Second Quantization§ One of the most useful representations in many-body theory

• : the state with no particles (the vacuum)

• : creation operator, creates a fermion in the state i

• : annihilation operator, annihilates a fermion in the state i

• Anticommutation relations:

• So that the Slater determinant can be written as:

0

ai

+

ai

+, a

j

+{ }= ai, a

j{ }= 0, ai

+, a

j{ }= ai, a

j

+{ }= δij

Pauli principle in second quantization

ai

ai

+0 = i , a

i

+i = 0

aii = 0 , a

i0 = 0

ai

+aj

+= −a

j

+ai

+

φn

(A)=1

A!

ϕi

r1( ) ϕ

i

r2( ) … ϕ

i

rA( )

ϕj

r1( ) ϕ

j

r2( ) ϕ

j

rA( )

ϕl

r1( ) ϕ

l

r2( ) … ϕ

l

rA( )

= al

+…a

j

+ai

+0 ,

l >> j > i

implicitly assumes we have already chosen the form of the single-particle states, (i = 1,2,3, ... A) as dictated by some mean-field potential

LLNL-PRES-77709716

§ How are Slater determinants (SD) actually represented in a computer program?§ We are dealing with fermions, so a single-particle state is either occupied or empty, which in

computer language translates to either 1’s or 0’s

§ A very useful approach is a bit representation known as M-scheme§ If the mean-field is spherically symmetric, the single-particle states will have good j, mj

§ A single integer represents a complicated slater determinant

§ While the many-body SD states will have goodM, they do not have good J. States of good Jmust be projected and will be a combination of Slater determinants. Same for T and MT

§ Eigenstates obtained by the diagonalization of the Hamiltonian in SD basis will have a good J and (approximately) good T

= 21+ 2

3+ 2

5+ 2

6=106

0p3/2 0s1/20p1/2

1 010 0 1 10

1 113 1 3 11 ----

a1, 32,−12

+a1, 32, 32

+a1, 12, 12

+a0, 12,−12

+0 =

, j, jz

2mj

LLNL-PRES-77709717

§ Many-body HO Slater determinants

— Antisymmetrization is trivial— Good M, MT and parity quantum numbers, but not J and T

• Huge number of basis states

Multi-particle states in the Slater Determinant basis – Summary

r1

σ1

τ1,r2

σ2

τ2,,rA

σ

A

τAa+

la

+

ja+

i 0

=1

A!

ϕi

r1( ) ϕ

i

r2( ) … ϕ

i

rA( )

ϕj

r1( ) ϕ

j

r2( ) ϕ

j

rA( )

ϕl

r1( ) ϕ

l

r2( ) … ϕ

l

rA( )

ϕnljmj

12mt(r,σ ,τ )

= Rnl(r) Y

l(r̂)⊗ χ S

12

(σ )

mj

j

χ T12mt(τ )

LLNL-PRES-77709718

Hard core of nuclear interaction scatters nucleons to high momenta

V(r)Vkk’ = ∫ eik’r V(r)e-ikrdr

= ⟨k| V |k’⟩

Fourier Transform

J = 0 S = 1ℓ = 0 T = 0

LLNL-PRES-77709719

Hard core of nuclear interaction scatters nucleons to high momenta

V(r)Vkk’ = ∫ eik’r V(r)e-ikrdr

= ⟨k| V |k’⟩

Fourier Transform

J = 0 S = 1ℓ = 0 T = 0

Very large N values (basis sizes) are required to reach convergent solution!

LLNL-PRES-77709720

Basis dimension grows rapidly with A!

Convergence can be a challenge!

LLNL-PRES-77709721

§ Introduce unitary transformation: 𝒰 (𝒰⨥𝒰 = 𝟙)

Effective interactions from unitary transformations of bare Hamiltonian

E = ⟨Ψ| H(A) |Ψ⟩

E = ⟨Ψ| 𝒰⨥𝒰 H(A) 𝒰⨥𝒰 |Ψ⟩

= (⟨Ψ|𝒰⨥) 𝒰H(A)𝒰⨥ (𝒰|Ψ⟩)

= ⟨Ψ| H(A) |Ψ⟩~ ~ ~

Bare Hamiltonian,

wave function

EffectiveHamiltonian,

wave function

LLNL-PRES-77709722

Example: Similarity renormalization group (SRG) transformation

⟨H𝛌 = 𝒰𝛌H 𝒰⨥𝛌 k|

H(2) |k’⟩~ dH𝛌 = -4 [𝜼(𝛌),H𝛌 ]

d𝛌 𝛌5~ ~

Flow parameter

ʹk

k⟨k| H(2) |k’⟩𝛌0 > 𝛌1> 𝛌2 …

Plane wave

𝛌~

Two-body Hamiltonian in momentum space

LLNL-PRES-77709723

§ Start from:

§ Compute derivative with respect to 𝛼 = =>

§ Setting with hermitean

§ Customary choice in nuclear physics …kinetic energy operator

Similarity renormalization group (SRG) transformation

Hα =Uα HUα+

UαUα+ =Uα

+Uα =1

dHα

dα=dUα

dαHUα

+ +UαHdUα

+

dα=dUα

dαUα

+UαHUα+ +UαHUα

+Uα

dUα+

dα

=dUα

dαUα

+Hα +HαUα

dUα+

dα= ηα,Hα[ ]

anti-Hermitiangenerator

ηα ≡dUα

dαUα

+ = −ηα+

ηα = Gα,Hα[ ] Gα

Gα = T

LLNL-PRES-77709724


dH𝛌 = -4 [𝜼(𝛌),H𝛌 ]d𝛌 𝛌5~ ~

𝛌 = 20 fm-1

⟨k| H(2) |k’⟩𝛌~

Low and high momentum components coupled


H(2) |k’⟩~

Initial condition

Hα=0 = Hλ=∞ = H λ 2 =1/ α

LLNL-PRES-77709725


dH𝛌 = -4 [𝜼(𝛌),H𝛌 ]d𝛌 𝛌5~ ~

𝛌 = 2 fm-1

⟨k| H(2) |k’⟩𝛌~

Low and high momentum components de-coupled


H(2) |k’⟩~

Hα=0 = Hλ=∞ = H λ 2 =1/ α

Initial condition

LLNL-PRES-77709726


dH𝛌 = -4 [𝜼(𝛌),H𝛌 ]d𝛌 𝛌5~ ~

𝛌 = 2 fm-1

⟨k| H(2) |k’⟩𝛌~

Can work with smaller N values (basis sizes)!


H(2) |k’⟩~

Hα=0 = Hλ=∞ = H λ 2 =1/ α

Initial condition

LLNL-PRES-77709727


dH𝛌 = -4 [𝜼(𝛌),H𝛌 ]d𝛌 𝛌5~ ~

𝛌 = 2 fm-1

⟨k| H(2) |k’⟩𝛌~


H(2) |k’⟩~

See: Bogner, Furnstahl, Schwenk, Prog. Part. Nucl. Phys. 65 (2010)

Hα=0 = Hλ=∞ = H λ 2 =1/ α

Initial condition

LLNL-PRES-77709728

§This sounds to good to be true …

§What’s the catch?

Question

LLNL-PRES-77709729

§The transformation (e.g., SRG) generates a ‘new’, softer NN interaction

§Unitarily equivalent to the bare NN potential in the two-nucleon sector only!

§ Induces 3-body and, in general, up to A-body forces even starting from an NN potential

Notes on effective interactions

LLNL-PRES-77709730

§ Evolution induces many-nucleon terms (up to A)

§ determined in A=2 system; determined in A=3 system, etc.

§ In actual calculations so far only terms up to

§ Three types of SRG-evolved Hamiltonians used — NN only: Start with initial T+VNN and keep — NN+3N-induced: Start with initial T+VNN and keep— NN+3N-full: Start with initial T+VNN+VNNN and keep

SRG evolution for A-nucleon system

Hα = Hα[1] + Hα

[2] + Hα[3] + Hα

[4] +...+ Hα[A]

Hα[3]

Hα[1] + Hα

[2]

Hα[1] + Hα

[2] + Hα[3]

Hα[1] + Hα

[2] + Hα[3]

Varying 𝜆 provides diagnostic tool to asses contribution of omitted many-body terms, tests unitarity

Hα[3]

LLNL-PRES-77709731

Example: convergence of 4He ground-state energy with chiral NN+3N forces

Jurgenson et al., PRL 103, 082501

SRG NN+3N

Bare NN+3N

LLNL-PRES-77709732

Example: convergence of 4He ground-state energy with chiral NN+3N forces

1 2 3 4 5 10 20λ [fm

−1]

−29

−28

−27

−26

−25

−24

Gro

und-S

tate

Ener

gy [

MeV

]

NN-onlyNN+NNN-induced+NNN-initial

4He N

3LO (500 MeV)

Expt.

Figure 29: Alpha particle binding energy during SRG evolution. Thecurves correspond to those in Fig. 28.

7.6 7.8 8 8.2 8.4 8.6 8.8

Eb(3H) [MeV]

24

25

26

27

28

29

30

Eb(4

He)

[M

eV]

Tjon line for NN-only potentialsSRG NN-only

SRG NN+NNN (λ >1.7 fm−1

)

8.45 8.528.2

28.3

28.4

N3LO

λ=3.0λ=1.2

λ=2.5

λ=1.5

λ=2.0

λ=1.8

Expt.

(500 MeV)

Figure 30: Correlation plot of the binding energies of the alpha parti-cle and the triton. The dotted line connects (approximately) the locusof points found for phenomenological potentials, and is known as theTjon line [40].

potential. The change in energy with λ reflects the vi-olation of unitarity by omission of the induced three-body force. When this induced 3NF is included (“NN+ NNN–induced”), the energy is independent of λ forA = 3. If we now turn to the alpha particle (4He) inFig. 29, we see similar behavior, except now the inclu-sion of the induced 3NF does not lead to a completelyflat curve at the lowest λ values. If there is sufficientconvergence, this is a signal of missing induced 4NF.

In both cases, it is evident that starting with an initialNN-only interaction (in this case, an N3LO(500 MeV)

1 2 3 4 5 10λ

−40

−30

−20

−10

0

10

20

30

40

g.s

. E

xp

ecta

tio

n V

alu

e

<Trel><V

NN>

<V3N

>

1 2 3 4 5 10λ

−1−0.5

0

<Trel><V

NN>

<V3N

>

3H

h- ω = 28

Nmax

= 18

Nmax

-A3 = 32 NN+NNN

Figure 31: Contributions to the triton binding energy during SRG evo-lution. Plotted are the expectation values of the kinetic energy, thetwo-body potential, and the three-body potential [17].

interaction [29]), does not reproduce experiment. Thethird line in each plot of Figs. 28 and 29 shows that aninitial 3NF (labeled NNN) contribution leads to a goodreproduction of experiment. The triton energy is part ofthe fit of this initial force, but the alpha particle energyis a prediction. Note that the magnitude of the NN-onlyvariation is comparable to the initial 3NF needed. Thisis an example of the natural size of the 3NF being mani-fested by the running of the potential (which is, in effect,the beta function).

The nature of the evolution is illustrated in Fig. 30,which is a correlation plot of the binding energies ineach nucleus. The dotted line is known as the Tjonline for NN-only phenomenological potentials. It wasfound that different potentials that fit NN scattering datagave different binding energies, but that they clusteredaround this line. With the SRG evolution starting withjust an NN potential, the path follows the line, passingfairly close to the experimental point. With an initialNNN force and keeping the induced 3-body part, thetrajectory is greatly reduced (see inset), at least until λis small.

Figures 31 and 32 show individual contributions tothe energy in the form of ground-state matrix elementsof the kinetic energy, two-body, three-body, and (im-plied) four-body potentials. The hierarchy of contri-butions is quite clear but the graphs also manifest thestrong cancellations between the NN and kinetic energycontributions. These cancellations magnify the impactof higher-body forces. Even so, it appears that a trunca-tion including the NNN but omitting higher-body forces

R.J. Furnstahl / Nuclear Physics B (Proc. Suppl.) 228 (2012) 139–175154

π π π

c1, c3, c4 cD cE

Figure 27: Leading three-body forces from chiral EFT. These contri-butions represent three different ranges: long-range 2-pion exchange,short-range contact with one-pion exchange, and pure contact interac-tion.

energy correct. In fact, low-energy effective theory tellus generalized diagrams such as those in Fig. 26 withfour or more legs imply that there are A-body forces(and operators) initially!

However, there is a natural hierarchy predicted fromchiral EFT, whose leading contributions are given inFig. 27 (we’ll return to this in Section 4.1 and supplyadditional details). If we stop the flow equations be-fore induced A-body forces are unnaturally large or ifwe can tailor the SRG Gs to suppress their growth, wewill be ok. (Another option is to choose a non-vacuumreference state, which is what is done with in-mediumSRG, to be discussed later.) Note that analytic boundson A-body growth have not been derived, so we need toexplicitly monitor the contribution in different systems.But the bottom line that makes the SRG attractive asa method to soften nuclear Hamlitonians is that it is atractable method to evolve many-body operators.

To include the 3NF using SRG with normal-orderingin the vacuum, we start with the SRG flow equationdHs/ds = [[Gs,Hs],Hs] (e.g., with Gs = Trel). Theright side is evaluated without solving bound-state orscattering equations, unlike the situation with Vlow k, sothe SRG can be applied directly in the three-particlespace. The key observation is that for normal-orderingin the vacuum, A-body operators are completely fixed inthe A-particle subspace. Thus we can first solve for theevolution of the two-body potential in the A = 2 space,with no mention of the 3NF (either initial or induced),and then use this NN potential in the equations appliedto A = 3.

What about spectator nucleons? There is a decou-pling of the 3NF part. We can see this from the first-quantized version of the SRG flow equation,

dVs

ds=

dV12

ds+

dV13

ds+

dV23

ds+

dV123

ds= [[Trel,Vs],Hs] , (42)

where we isolate the contributions from each pair andthe 3NF. Using each SRG equation for the two-bodyderivatives, we can cancel them against terms on the

1 2 3 4 5 6 7 10 20λ [fm−1 ]

−8.6

−8.4

−8.2

−8.0

−7.8

−7.6

−7.4G

roun

d-St

ate

Ene

rgy

[MeV

]

NN-onlyNN + NNN-inducedNN + NNN

3 H

Expt.

Figure 28: Triton binding energy during SRG evolution. The threecurves are for an initial potential with only NN components wherethe induced 3NF is not kept (“NN-only”), for the same initial NNpotential but keeping the induced 3NF (“NN + NNN-induced”), andwith an initial NNN included as well (“NN + NNN”).

right side. The result is [12]:

dV123

ds= [[T12,V12], (T3 + V13 + V23 + V123)]

+ {123→ 132} + {123→ 231}+[[Trel,V123],Hs] . (43)

The key is that there are no “multi-valued” two-bodyinteractions remaining (i.e., dependence on the excita-tion energy of unlinked spectators); all the terms areconnected. An implementation of these equations in amomentum basis would be very useful and has very re-cently been achieved by Hebeler [37]. But an alternativeapproach has also succeeded: a direct solution in a dis-crete basis [38, 16, 17].

The idea is that the SRG flow equation is an opera-tor equation, and thus we can choose to evolve in anybasis. If one chooses a discrete basis, than a separateevolution of the three-body part is not needed. This wasfirst done for nuclei by Jurgenson and collaborators in2009 using an anti-symmetrized Jacobi harmonic oscil-lator (HO) basis [16]. The technology for working withsuch a basis had already been well established for appli-cations to the no-core shell model (NCSM) [39]. Thisapproach leads to SRG-evolved matrix elements of thepotential directly in the HO basis, which is just whatis needed for many-body applications such as NCFC orcoupled cluster.

In Fig. 28, the comparison of two-body-only to fulltwo-plus-three-body evolution is shown for the triton(3H). The NN-only curve uses the evolved two-body

R.J. Furnstahl / Nuclear Physics B (Proc. Suppl.) 228 (2012) 139–175 153 omitted induced 4N

omitted induced 3N

Jurgenson et al., PRL 103, 082501

LLNL-PRES-77709733

Ab initio no-core shell model (NCSM)

§Bare/effective (e.g., SRG-evolved) NN+3N forces

§ Superposition of HO wave functions

§ Jacobi/single-particle coordinates

§Preserves translational invariance of intrinsic wave function

§ ‘Diagonalizes’ Hamiltonian matrix

§A ≲ 16

1max += NN

Nmax … maximal allowed HO excitation above the lowest

possible A-nucleon configuration

Full Nmax space: All basis states with N ≤ Nmax

LLNL-PRES-77709734


ΨSDA = cSDNjΦSDNj

HO (!r 1,!r 2 , ... ,

!r A )j∑

N=0

Nmax

∑ =ΨA ϕ000 (!RCM )

ΨSDA = cSDNjΦSDNj

HO (!r 1,!r 2 , ... ,

!r A )j∑

N=0

Nmax

∑ =ΨA ϕ000 (!RCM )

ΨA = cNiΦNiHO ( !η 1,

!η 2 ,...,

!η A−1)

i∑

N=0

Nmax

∑






§A ≲ 16

LLNL-PRES-77709735

§ Jacobi-coordinate HO basis

Small dimensions (no center of mass d.o.f; Coupled to J, T and parity)

NN (and NNN) interaction depends on Jacobi (relative) coordinates and/or momenta

Complicated antisymmetrization (involved, specialized algebra)

−Formalism depends on number of nucleons A

—Most efficient for A = 3,4 systems

Jacobi-coordinate vs. Slater Determinant HO basis

LLNL-PRES-77709736

§ Slater Determinant HO basis

Antisymmetrization is trivial; powerful second quantization representation

Center of mass d.o.f included, but can be removed exactly

−Physical eigenstates contain 0hW center of mass motion

Huge dimensions (M-scheme basis, no good J, T, only MJ, MT , and parity)

3N interaction needs to be transformed to Slater determinant basis

−Huge number of input matrix elements

—Most efficient for A>4 systems

Jacobi-coordinate vs. Slater Determinant HO basis

LLNL-PRES-77709737







§A ≲ 16

Works well if wave function is localized(well-bound states)

LLNL-PRES-77709738


Example: energy spectrum of nuclear states of the 10B nucleus

Helped to point out the fundamental importance of 3N forces in structure calculations

PRL 99, 042501 (2007)

Ener

gy sp

ectr

um o

f nuc

lear

stat

es (M

eV)

NN+3N

LLNL-PRES-77709739







§A ≲ 16

Does not works as well for nuclei with exoticdensities (halo nuclei)

LLNL-PRES-77709740







§A ≲ 16

Definitively not adapted to the description of

scattering wave functions!

LLNL-PRES-77709741

§Green’s function Monte Carlo

§Nuclear Lattice Effective Field Theory

§Coupled Cluster theory

§ In-Medium SRG

§Gorkov-Green function theory

§Many-Body Perturbation Theory

§Ab initio valence-space shell model

Ab initio community extremely successful in describing the static properties of nuclei

Explosion of ab initio methods pushing to medium-mass nuclei

What about the dynamics between nuclei (scattering States with E>0)?

lecture 3: ab initio theory of nuclear forces and nuclear ...lecture 3: ab initio theory of nuclear...

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