s-matrix solution of the lippmann-schwinger equation for regular … · 2016. 9. 18. · s-matrix...

S-matrix solution of the Lippmann-Schwinger equation for regular and singular potentials

S-matrix solution of the Lippmann-Schwinger

equation for regular and singular potentials

J. A. Oller

Departamento de FısicaUniversidad de Murcia 1

in collaboration with D. R. Entem

[I] arXiv:1609, next month

[II] Long version, in preparation

TIDS16, Granada, August 23rd, 2016

1Partially funded by MINECO (Spain) and EU, project FPA2013-40483-P


Overview

1 Lippmann-Schwinger equation

2 New exact equation in NR scattering theory

3 N/D method with non-perturbative ∆(A)

4 Regular interactions

5 Singular Interactions

6 T (A) in the complex plane

7 T (A) in the complex plane

8 Conclusions


Lippmann-Schwinger equation

Lippmann-Schwinger equation (LS)

Scattering T -matrix T (z) , Im(z) 6= 0

T (z) =V − VR0(z)T (z)

R0(z) = [H0 − z ]−1

H0 =− 1

2µ∇2

H =H0 + V

• Resolvent of H, R(z):

R(z) =[H − z ]−1

R(z) =R0(z)− R0(z)T (z)R0(z)



• Spectrum of H: H|ψp〉 = Ep|ψp〉Continuous spectrum: Povzner’s result

|ψp〉 =|p〉 − limǫ→0+

R0(Ep + iǫ)T (Ep + iǫ)|p〉

Bound States: Poles in T (p,p′; z) for z ∈ R−

• LS in momentum spaceFor definiteness we consider uncoupled spinless case by now :

T (p′,p, z) = V (p′,p)−∫

d3q

(2π)3V (p′,q)

1q2

2µ − zT (q,p, z)

POTENTIAL

+ . . .

p

p′



• LS in partial waves

Tℓ(p′, p, z) =

1

2

∫ +1

−1dcos θ Pℓ(cos θ)T (p′,p, z)

cos θ =p′ · p

T (p′,p, z) =∞∑

ℓ=0

(2ℓ+ 1)Pℓ(cos θ)Tℓ(p′, p, z)

Tℓ(p′, p, z) =Vℓ(p

′, p) +µ

π2

∫ ∞

0dqq2

Vℓ(p′, q)Tℓ(q, p, z)

q2 − 2µz

Convention: V (p′,p) → −V (p′,p) , T (p′,p, z) → −T (p′,p, z)



• On-shell unitarity (extensively used later)Propagation of real two-body states

p′ =p

Ep =p2

2µ

ImTℓ(p) =µp

2π|Tℓ(p)|2 , p > 0

Im1

Tℓ(p)=− µp

2π

Unitarity cut for p2 > 0



Criterion for Singular Potentials

Vℓ(r) −−−→r→0

αr−γ

α =α+ ℓ(ℓ+ 1)

Potential Ordinary Singular

γ < 2 > 2

γ = 2 1 + 4α > 1 1 + 4α ≤ 1

Ordinary/Regular Potentials:

Standard quantum mechanical treatmentBoundary condition: u(0) = 0 and behavior at ∞No extra free parameters



The One-Pion-Exchange (OPE) potential for the singlet NNinteraction (r > 0):

Yukawa potential V (r) =− τ1 · τ2(

gAmπ

2fπ

)2e−mπr

4πr

Exchange of a pion between two nucleons



In many instances one has singular potentials

Multipole expansion

Φ(x) =

∫

ρ(x′)

|x− x′|d3x ′ = 4π

∑

ℓ,m

qℓm

2ℓ+ 1

Yℓm(θ, φ)

r ℓ+1

qℓm =

∫

Y ∗ℓm(θ

′, φ′)r ′ℓρ(x′)d3x ′



Van der Waals Force (molecular physics)

V (r) = −3

2

αAαB

r6IAIB

IA + IB

In QFT/EFT we treatcomposite objects aspoint like

Physical meaning for r → 0?:

De Broglie length1p>> rA ∼ 2A



Nuclear physicsOPE is singular attractive for the Deuteron in NN scattering

2S+1LJ :3P0 NN partial wave

V (r) =m2

π

12π

(

gA

2fπ

)2

[−4T (r) + Y (r)]

Y (r) =e−mπr

r

T (r) =e−mπr

r

[

1 +3

mπr+

3

(mπr)2

]

Triplet part of the OPE potential V (r) −−−→r→0

g2A

4πf 2πr−3

Quark Models, pNRQCD, pNRQED, QCD’s EFTs, etc



Math: Taking r → 0 for a singular potential

• Full range r ∈]0,∞[: Case, PR60,797(1950)

r

E

• Singular Attractive Potential

Near the origin the solution is thesuperposition of two oscillatorywave functionsOne has to fix a relative phase,ϕ(p) How to do it?

Which are the appropriateboundary conditions?(self-adjoint extensions of theHamiltonian → dϕ(p)/dp = 0)



The potential does not determine uniquely the scattering problemPlesset, PR41,278(1932), Case, PR60,797(1950)

r

E

• Singular Repulsive Potential

There is only one finite(vanishing) reduced wavefunction at r = 0

The solution is fixed

Typically, the phenomenology is not accurate

E.g. this scheme a la Case does not fit well NN phase shifts.



Low-energy EFT paradigm:Contact terms are necessary to reproduce short-distance physics

They are allowed by symmetryThey are required to make loops finite. NonrenormalizableQFT/EFTThey are expected to be relatively important because ofpower-counting


New exact equation in NR scattering theory


Yukaway potential,

V (q) =2g

q2 +m2π

Singularity for q2 = −m2π

1S0 potential: (2S+1LJ)

V (p) =g

2p2log(4p2/m2

π + 1)

Left-hand cut (LHC) discontinuity

p2 <−m2π/4 = L

Born approximation

∆1π(p2) =

V (p2 + i0+)− V (p2 − i0+)

2i= ImV (p2 + i0+) =

gπ

2p2



Full LHC Discontinuity , p2 = −k2 < L

2i∆(p2) =T (p2 + i0+)− T (p2 − i0+)

∆(p2) =ImT (p2 + i0+)

.

.

.

.

The LS generates contributionswith any number of pions to∆(p2), p2 < L

∆nπ(p2) for p2 < −(nm2

π/2)



How to calculate ∆(A)?

G.E. Brown, A.D. Jackson “The Nucleon-Nucleon interaction”,North-Holland, 1976. Page 86: “In practice, of course, we do not

know the exact form of ∆(p2) for a given potential . . . ”

p =ik ± ε , ε = 0+ , p2 = −k2 < L

T (ik ± ε, ik ± ε) =V (ik ± ε, ik ± ε)

+µ

2π2

∫ ∞

0dqq2

V (ik ± ε, q)T (q, ik ± ε)

q2 + k2

T (ik ± ε, ik ± ε), so calculated, IS PURELY REAL!!

You can try to calculate numerically just the first iterated OPE

µ

2π2

∫ ∞

0dqq2

V (ik ± ε, q)V (q, ik ± ε)

q2 + k2∈ R



Notation: A = p2

This is an example of:Not all what you can calculate with a computer is the rightanswer!!• 1st particular method: Algebraic calculationsOPE 1S0 is simple enough

∆1π(p2) =

gπ

2p2θ(L− A)

∆2π(A) =θ(4L− A)

(

g2Am

2π

16f 2π

)2MN

A√−A

log

(

2√−A

mπ

− 1

)

∆3π(A) =θ(9L− A)

(

g2Am

2π

4f 2π

)3 (MN

4π

)2π

4A

∫ 2√−A−mπ

2mπ

dµ11

µ1(2√−A− µ1)

θ(µ1 − 2mπ)

∫ µ1−mπ

mπ

dµ21

µ2(2√−A− µ2)



∆4π(A) =θ(16L− A)

(

g2Am

2π

4f 2π

)4 (MN

4π

)3π

4A

∫ 2√−A−mπ

3mπ

dµ11

µ1(2√−A− µ1)

×θ(µ1 − 3mπ)

∫ µ1−mπ

2mπ

dµ21

µ2(2√−A− µ2)

×θ(µ2 − 2mπ)

∫ µ2−mπ

mπ

dµ31

µ3(2√−A− µ3)

.

This can be generalize for a diagram with n pions to

∆nπ(A) =θ(n2L− A)

(

g2Am

2π

4f 2π

)n (MN

4π

)n−1 π

4A

×n−1∏

j=1

θ(µj−1 − (n + 1− j)mπ)

∫ µj−1−mπ

(n−j)mπ

dµj1

µj(2√−A− µj)

with µ0 = 2√−A



Formal solution of the integral equation (IE):

∆(A, µ) =∆1π(A) +

(

MNA

π2

)

θ(µ− 2mπ)

∫ µ−mπ

mπ

dµ∆1π(A)∆(A, µ)

µ(2√−A− µ)

∆(A) =∆(A, 2√−A)

• Asymptotic solution for√−A ≫ mπ

∆(A) =λπ2

MNAe

2λ√−A

arctanh

(

1− mπ√−A

)

λ =gMN

2π



• 2nd GENERAL method:Analytic extrapolation of the LS from its integral expression

f(ν) = ∆v(ν, k)− θ(k − 2mπ − ν)m

2π2

∫ k−mπ

mπ+ν

dν1k2 − ν21

∆v(ν, ν1)f(ν1)

∆(A) = − f(−k)

2A, k =

√−A , v(p1, p2) = p1p2V (p1, p2)

IE : − k +mπ < ν < k −mπ

The limits in the IE ARE FINITE

The denominator never vanishes , |ν1| < k in the IE

NO FREE PARAMETERS

Reason: Contact interactions (monomials) do not contribute tothe discontinuity of T (A)Short-distance physics is not resolved→ Contact interactions



f(ν) = ∆v(ν, k)− θ(k − 2mπ − ν)m

2π2

∫ k−mπ

mπ+ν

dν1k2 − ν21

∆v(ν, ν1)f(ν1)

∆(A) = − f(−k)

2A, k =

√−A , v(p1, p2) = p1p2V (p1, p2)

It can be applied to:

Any local potential (spectral decomposition:)

V (p′,p) =2

π

∫ ∞

m0

dmmη(m2)

q2 +m2, q = p′ − p

Higher partial waves, ℓ ≥ 0

Coupled Channels

Nonlocal potentials due to relativistic corrections



log-log plot for 1S0 ∆(A); gA = 6.80

1e-05

0.0001

0.001

0.01

0.1

1

10

100

1000

10000

1 10 100 1000 10000 100000 1e+06

|∆(A

)| (|L

|-1)

|A| (|L|)

∆1π, ∆2π, ∆3π, ∆4π, Asymptotic sol. (dots) |A| ≫ m2π

Full solution ∆(A)


N/D method with non-perturbative ∆(A)


Once we now the exact ∆(A) for a given potential we can useS-matrix theory to solve the LS: N/D method with the full ∆(A)

TJℓS(A) =NJℓS(A)

DJℓS(A)

NJℓS(A) has Only LHC

DJℓS(A) has Only RHC

RHC

ǫ → 0

R → ∞

CI

ǫ → 0

R → ∞CII

−m2π4

LHC



Uncoupled Partial Waves

Exact knowledge of discontinuities

Tℓ(A) =Nℓ(A)

Dℓ(A)

Im1

Tℓ(A)= −ρ(A) ≡ µ

√A

2πA > 0 (RHC)

ImDℓ(A) = −Nℓ(A)ρ(A) A > 0 (RHC)

ImNℓ(A) = Dℓ(A)∆(A) A < L (LHC)



(m1,m2) N/D equations for D(A) and N(A)

N/Dm1 m2

N(A) =

m1∑

i=1

νi (A− C )m1−i +(A− C )m1

π

∫ L

−∞

dk2 ∆(k2)D(k2)

(k2 − A)(k2 − C )m1

D(A) =

m2∑

i=1

δi (A− C )m2−i − (A− C )m2

π

∫ ∞

0

dq2ρ(q2)N(q2)

(q2 − A)(q2 − C )m2

N(A) is substituted in D(A)

Linear IE for D(A) arises

D(0) = 1. To fix a floating constant in the ratioT (A) = N(A)/D(A)


Regular interactions


• N/D01: Regular solution for an ordinary potential

Scattering is completely fixed by the potential

N(A) =1

π

∫ L

−∞

dωLD(ωL)∆(ωL)

(ωL − A)

D(A) = 1− A

π

∫ ∞

0dωR

ρ(ωR)N(ωR)

(ωR − A)ωR

= 1− iµ√A

2π2

∫ L

−∞

dωL∆(ωL)D(ωL)

√ωL

(√ωL +

√A)



• N/D11: Additional subtraction in N(A) is fixed in terms ofscattering length

D(A) = 1 + ia√A+ i

MN

4π2

∫ L

−∞

dωLD(ωL)∆(ωL)

ωL

A√A+

√ωL

N(A) = −4πa

MN

+A

π

∫ L

−∞

dωLD(ωL)∆(ωL)

(ωL − A)ωL

Effective Range Expansion (ERE)

kcotδ(k) = −1

a+

1

2rk2 +

∑

i=2

vik2i



• N/D12: Additional subtraction in D(A), r is fixed

D(A) = 1 + ia√A− ar

2A− i

MNA

4π2

∫ L

−∞

dωLD(ωL)∆(ωL)

ωL

×[

√A

(√ωL +

√A)

√ωL

− i

aωL

]

N(A) = −4πa

MN

+A

π

∫ L

−∞

dωLD(ωL)∆(ωL)

(ωL − A)ωL

The results are just dependent on ∆(A) (input potential) andexperimental ERE parameters



• N/D22: Additional subtraction in N(A), v2 is fixed

D(A) = (1− 2v2

rA)(1 + ia

√A)− ar

2A

+iMN

4π2A

∫ L

−∞

dωLD(ωL)∆(ωL)

ω2L

×[

A√A+

√ωL

+ i2

ra2ωL

(1 + ia√ωL)(1 + ia

√A)

]

N(A) = −4πa

MN

+ A8πav2

MN r+

A

π

∫ L

−∞

dωLD(ωL)∆(ωL)

ω2L

×[

A

(ωL − A)+

2

raωL

(1 + ia√ωL)

]

The more subtractions are included the more perturbative N/D iswith respect to ∆(A). ∆nπ(A) contributes for A < n2L



Example: 1S0 projected Yukawa potential

gA = 1.26

0

2

4

6

8

10

12

14

0 50 100 150 200 250 300 350 400

δ1S 0

k (MeV)

(a)

gA = 6.80 → a = −23.75 fm

50

100

150

200

250

300

350

400

450

0 50 100 150 200 250 300 350 400δ1

S 0

k (MeV)

(c)

LS (black dots), N/D01 (red line)1π, 2π, 3π,4π

Non-perturbative problemOnly Full ∆(A) gives agreementwith LS



ERE parameters from N/D01 and N/D11

gA = 6.80

as r v2 v3 v4 v5 v6

Regular (fm) (fm) (fm3) (fm5) (fm7) (fm9) (fm11)N/D01

∆1π 1.66 0.714 -0.168 0.847 -5.35 35.5 -241∆2π 3.53 2.03 -5.70 10−2 3.38 -27.4 234 -1.99 103

∆3π 1.80 1.15 -8.71 10−2 0.924 -5.69 36.9 -247∆4π -6.89 13.7 47.5 356 2.63 103 1.97 104 1.46 105

Full -23.75 8.90 18.7 89.8 411 2.00 103 8.98 103


Singular Interactions

Analytical properties determine the solutions for singularpotentials

Singular Interactions. E.g. NLO and NNLO 1S0 potentials

Potentials are calculated within Chiral Perturbation TheoryLow-energy EFT of QCD

∆(A) is now divergent

limA→−∞∆(A)NLOBorn/k2 < 0 , ∆(A)NNLOBorn /k3 < 0

Attractive singular interaction

N/D11 and N/D22 are convergent

N/D01 (at least one parameter is needed) and N/D12 are notconvergent



We compare with

LS renormalized with one contact term:

V (p1, p2) → V (p1, p2) + C0

V (p1, p2) → V (p1, p2) + C0 + C1(p21 + p22) + . . .

LS is insensitive to C1, etc.

Repulsive Singular Potential: LS is insensitive to all Ci



NLO NNLO

-10

0

10

20

30

40

50

60

70

50 100 150 200 250 300 350 400

δ1S 0

k (MeV)

-30

-20

-10

0

10

20

30

40

50

60

70

50 100 150 200 250 300 350 400

δ1S 0

k (MeV)

N/D11 , N/D22

Subtractive-renormalized LS: pointsOne counterterm It cannot reproduce r

Yang,Elster,Phillips, PRC80,044002(’09)

a = −23.75 , r = 2.655 fm , v2 = −0.6265 fm3


T (A) in the complex plane


• As a bonus the non-perturbative-∆ N/D method allows tocalculate T (A) for A ∈ C in the 1st/2nd Riemann sheet

This is not trivial with LS

Look for bound states, resonances and virtual states

Bound State A = (ik)2

Binding energy of near threshold bound state, gA = 7.45One does not need to solve Schrodinger equation

Poles of T (A) ↔ zeros of D(A)



• As a bonus the non-perturbative-∆ N/D method allows tocalculate T (A) for A ∈ C in the 1st/2nd Riemann sheet

This is not trivial with LS

Binding energy of near threshold bound state, gA = 7.45One does not need to solve Schrodinger equation

Poles of T (A) ↔ zeros of D(A)

A = (ik)2 N/D01 N/D11 Schrodinger

∆1π 2.02∆2π 2.18∆3π 2.21∆4π 0.89 2.22

Non-perturbative 2.22 2.22 2.22



Anti-bound (virtual) state for 1S0

T−1II (A) = T−1

I (A) + 2iρ(A)

=DI + NI 2iρ(A)

NI

, Im√A ≥ 0

Look for zero of DII (A) . E = A/MN =

N/D11:−0.070 (LO) , −0.067 (NLO,NNLO) MeV

For the other N/Dm1m2 : −0.066 MeV always



Back to Yukawa 1S0gA = 1.26

0

2

4

6

8

10

12

14

0 50 100 150 200 250 300 350 400

δ1S 0

k (MeV)

(b)

gA = 6.80 → a = −23.75 fm

-140

-120

-100

-80

-60

-40

-20

0

20

40

60

0 50 100 150 200 250 300 350 400

δ1S 0

k (MeV)

(d)

LS (black dots), N/D11 (red line) 1π, 2π, 3π,4π

• Perturbative-∆ N/D + subtractions JAO et al PRC93,024002(2016);

PRC89,014002(’14); PRC86,034005(’12);PRC84,054009(’11)



Two-nucleon reducible diagrams [II]; Guo,Rıos,JAO,PRC89,014002(’14);

Similar size to the other NLOirreducible diagrams

.

.

.

.

All pion lines must be put on-shell −→ A ≤ −n2M2π/4.

As n increases their physical contribution fades away.

This only occurs for the imaginary part!



ERE parameters from N/D01 and N/D11

gA = 6.80

as r v2 v3 v4 v5 v6

Regular (fm) (fm) (fm3) (fm5) (fm7) (fm9) (fm11)

∆1π 1.66 0.714 -0.168 0.847 -5.35 35.5 -241∆2π 3.53 2.03 -5.70 10−2 3.38 -27.4 234 -1.99 103

∆3π 1.80 1.15 -8.71 10−2 0.924 -5.69 36.9 -247∆4π -6.89 13.7 47.5 356 2.63 103 1.97 104 1.46 105

Full -23.75 8.90 18.7 89.8 411 2.00 103 8.98 103

N/D11

∆1π -23.75 8.56 15.3 60.5 221 906 3.08 103

∆2π -23.75 8.80 17.7 80.4 346 1.60 103 6.71 103

∆3π -23.75 8.88 18.4 87.5 395 1.90 103 8.40 103

∆4π -23.75 8.90 18.6 89.3 408 1.98 103 8.87 103

Full -23.75 8.90 18.7 89.8 411 2.00 103 8.98 103

The more subtractions are included the more perturbative N/D iswith respect to ∆(A).



LS renormalized with a counterterm

We add to the Yukawa potential Local Terms:

Monopole Form Factor:

V1(p′, p) = C 1

0

Λ2

2p′pQ0(zΛ)

Q0(zΛ) =1

2log

(

zΛ + 1

zΛ − 1

)

, zΛ =p′2 + p2 + Λ2

2p′p

Gaussian type form factors

V2(p′, p) = C 2

0 e−

(

p′

Λ

)4−( p

Λ)4

V3(p′, p) = C 3

0 e−

(

p′

Λ

)6−( p

Λ)6



gA = 1.26 , a = −23.70 fm , Λ = 100 GeV

0

10

20

30

40

50

60

70

80

0 50 100 150 200 250 300 350 400

δ1S 0

k (MeV)

LS (black dots; Λ = 100 GeV)N/D11 (red line) , 1π, 2π, 3π,4π



75.0

75.5

76.0

76.5

77.0

77.5

78.0

78.5

0 100 2 10-5 4 10-5 6 10-5 8 10-5 1 10-4

δ1S 0

(k=

400

MeV

)

1/Λ (MeV-1)

V1, V2, V3. Λ → ∞ by a linear fit.Subtractive renormalized LS (solid black line)N/D11 by dashed lines, 1π, 2π, 3π,4π, Full ∆(A)



Yukawa 1S0

-10

0

10

20

30

40

50

60

70

80

50 100 150 200 250 300 350 400

δ1S 0

k (MeV)

N/D11, N/D12, N/D33

Phase shifts: Granada analysis Navarro et al. PRC88,064002(’13)



G.E. Brown, A.D. Jackson “The Nucleon-Nucleon interaction”,North-Holland, 1976. Page 86: “In practice, of course, we do not

know the exact form of ∆(p2) for a given potential and the N/Dequations do not represent a practical alternative to the exact

solution of the LS equation for potential scattering. . . ”

Now (2016), this statement is superseded


Conclusions

Conclusions

A new non-singular IE allows to calculate the exact ∆(A) inpotential scattering for a given potential

One can calculate the scattering amplitude for regular/singularpotentials from its analytical/unitarity properties

Any proper solution for singular potentials can be found withthis method

We reproduce the LS outcome with/without one counterterm


Conclusions

It can be straightforwardly used in the whole complex plane(bound states, resonances, virtual states)

It can reproduce accurately 1S0 phase shifts

s-matrix solution of the lippmann-schwinger equation for regular … · 2016. 9. 18. · s-matrix...

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