Hadrons and Nuclei : Single Hadrons

Lattice Summer SchoolLattice Summer School

Martin Savage

Summer 2007

University of Washington

Mass Spectrum of Mesons| - M esons S=C=B=T =0, qq

¼§ m¼§ = 139:57MeV ¿ = 2:6£ 10¡ 8 s J ¼= 0¡

¼0 m¼0 = 134:96MeV ¿ = 0:83£ 10¡ 16 s J ¼= 0¡

´0 m´ = 548:6MeV ¿¡ 1 = 0:9keV J ¼= 0¡

´00 m´0 = 957:6MeV ¿¡ 1 = 0:3MeV J ¼= 0¡

½0;§ m½= 770MeV ¿¡ 1 = 154MeV J ¼= 1¡

! 0 m! = 783MeV ¿¡ 1 = 9:9MeV J ¼= 1¡

Á0 mÁ = 1020MeV ¿¡ 1 = 4:2MeV J ¼= 1¡

A1 mA 1 = 1275MeV ¿¡ 1 » 300MeV J ¼= 1+

J =Ã mJ =Ã = 3:1GeV ¿¡ 1 = 88keV J ¼= 1¡

¨ m¨ = 9:5GeV ¿¡ 1 = 52keV J ¼= 1¡

Mass Spectrum of Light Baryons| - Baryons S = C = B = T = 0, qqq

p mp = 938:28MeV ¿ > 1033yrs J ¼ = 12

+

n mn = 939:57MeV ¿ = 898§ 16 s J ¼ = 12

+

¢ m¢ » 1230MeV ¿¡ 1 » 120MeV J ¼ = 32

+

| - Baryons S = 1 C = B = T = 0, sqq

¤ m¤ = 1115:6MeV ¿ = 2:6£ 10¡ 10s J ¼ = 12

+

§ § m§ § = 1197:3MeV ¿ = 1:5£ 10¡ 10s J ¼ = 12

+

§ 0 m§ 0 = 1192MeV ¿ = 6£ 10¡ 20s J ¼ = 12

+

SU(2) Flavor Symmetry : Isospin

u u u

d d d

SU(3)C

SU(2)

Local color transformations

Global flavor transformationsIsospin

If mu = md then SU(2) would be an exact symmetry of QCD

mu - md << so SU(2) is an approximate symmetry of QCD

SU(3) Flavor Symmetry

u u u

d d d

SU(3)C

SU(3)

Local color transformations

Global flavor transformationss s s

If mu = md = ms then SU(3) would be an exact symmetry of QCD

mi - mj << so SU(3) is an approximate symmetry of QCD

Mesons : SU(2)

qa = ud( ) q V q

Vector symmetry : L = R = V

M ba = qb°5qa ¡ 1

2±b

a qc°5qc

M =

µ¼0=

p2 ¼+

¼¡ ¡ ¼0=p

2

¶

= 1p2¼a¿a

M V M V

Mesons : SU(3)

qa = uds( ) q V q

M V M V

M =

0

@¼0=

p2+´=

p6 ¼+ K +

¼¡ ¡ ¼0=p

2+´=p

6 K 0

K ¡ K 0 ¡ 2=6

1

A

M ba = qb°5qa ¡ 1

3±b

a qc°5qc

( Lectures by Claude Bernard )

Meson Correlation Functions and Interpolating Fields

d-quark propagator u-quark propagator

hd(x)°5u(x)¡d(y)°5u(y)

¢yi = hTr [ D(x ! y) °5 U(y ! x) °5 ]i

= hTr£

D(x ! y) Uy(x ! y)¤i

hO(x;y)i »

ZDq Dq DA¹ O(x;y) ei

Rd4z L (q;q;A ¹ )

e.g. ¼+

Zd3xhd(x)°5u(x)

¡d(y)°5u(y)

¢yi ! Z¼

e¡ m¼tE

2m¼+ ::

Baryons : SU(2)

B°i j k »

hq®;a

i q¯ ;bj q° ;c

k ¡ q®;ai q° ;c

j q¯ ;bk

i²abc (C°5)®

spin

flavor

Babc =1

p6

( ²ab Nc + ²acNb )

N =

µpn

¶

B ! VVVB

Va®Vb

¯ ²ab ! ²®

Baryons : SU(3)

B°i j k »

hq®;a

i q¯ ;bj q° ;c

k ¡ q®;ai q° ;c

j q¯ ;bk

i²abc (C°5)®

spin

flavor

Babc = 1p6

¡²abd Bd

c + ²acdBdb

¢

B =

0

@§ 0=

p2+¤=

p6 § + p

§ ¡ ¡ § 0=p

2+¤=p

6 n¥¡ ¥0 ¡ 2=6¤

1

A

B ! VVVB

Va®Vb

¯ ²abc ! ²® ° Vy°c

Light Baryons : SU(3)

S=0, I=1/2

S=-1, I=0,1

S=-2, I=1/2

S=0, I=3/2

S=-1, I=1

S=-2, I=1/2

S=-3, I=0

J ¼ = 32

+

J ¼ = 12

+

Charmed (or Bottom) Baryons :SU(3)

Mixing / ms ¡ mmQ

Sl=1, I=1

Sl=1, I=1/2

Sl=1, I=0

Sl=0, I=0

Sl=0, I=1/2

Light Quark Masses and Spurions

L = q i°¹ @¹ q ¡ qmqqSU(3) invariant breaks SU(3)

mq =

0

@mu 0 00 md 00 0 ms

1

A

BUT : let mq ! VmqV y and then both terms are SU(3) invariant

Then we can simply use the Wigner-Eckhart Theorem to constructinvariant matrix elements.

, K, Masses : Gell-Mann—Okubo Mass Relation

Construct all possible group invariants that can contributeMore insertions of M and mq

m2K = ®(m+ms) + 2 (2m+ms)

m2´ = 2

3®(m+ 2ms) + 2 (2m+ ms)

m2¼ = 2®m+ 2 (2m+ ms)

L = ®Tr [ M M mq ] + ¯ Tr [ M M ]Tr [ mq ] + ::

= ¡ m2¼ ¼+¼¡ + ::

4m2K ¡ m2

¼ = 3m2´

p, n, Masses : Gell-Mann—Okubo Mass Relation

L = ¡ M0 Tr£

BB¤

¡ ®Tr£

Bmq B¤

¡ ¯ Tr£

BB mq

¤¡ ° Tr

£BB

¤Tr [ mq ] + ::

MN = M0 + (2m+ms)° +m®+ms¯

2MN + 2M¥ = M§ + 3M¤

Quark Masses from Lattice( Claudes lectures )

mu=md = 0:43§ 0§ 0:01§ 0:08

M s=m = 27:4§ 0:1§ 0:4§ 0:0§ 0:1

MILC collaboration , ¹ = 2 GeV

Homework 1 : Check the validity of GMO mass relation

amongst the pseudo-Goldstone bosons using Particle Data group compliations.

What is the violation as a percent of the pion mass?

Derive the masses of the at one-insertion of the light quark mass matrix

Latest Lattice results : LHPC : DW on Staggered

LHPC, Negele et al

m=350 MeV

Lattice Result for GMO

Electromagnetism

j ¹em =

23u°¹ u ¡

13d°¹ d ¡

13s°¹ s

= q Q °¹ q

Q =

0

@+2

30 0

0 ¡ 13

00 0 ¡ 1

3

1

A

Octet of SU(3)Singlet plus Triplet of SU(2)

Q = spurion field Q ! VQVy

Magnetic Moments in SU(3): Coleman-Glashow Relations

L = ¹ F Tr£

B¾¹ º F ¹ º [Q;B]¤

+ ¹ D Tr£

B¾¹ º F ¹ º fQ;Bg¤

Limit of exact SU(3) symmetry…. mu=md=ms

¹ N = ¹ F + 13

¹ D

¹ p ; ¹ n ; ¹ ¤ ; ¹ § + ; ¹ § ¡ ; ¹ ¥0 ; ¹ ¥ ¡ ; ¹ § ¤

Magnetic Moments : Coleman-Glashow Relations

Works as well as can be expected for SU(3) symmetry

¹ § + = ¹ p : 2:42§ 0:05 NM = 2:7928 NM

2¹ ¤ = ¹ n : ¡ 1:226§ 0:008 NM = ¡ 1:9130 NM

¹ ¥0 = ¹ n : ¡ 1:25§ 0:01 NM = ¡ 1:9130 NM

¹ § ¡ + ¹ n = ¡ ¹ p : ¡ 3:07§ 0:03 NM = ¡ 2:7928 NM

¹ ¥ ¡ = ¹ § ¡ : ¡ 0:6507§ 0:0025 NM = ¡ 1:16§ 0:03 NM

2¹ ¤ § 0 =p

3¹ n : 3:22§ 0:16 NM = 3:31 NM

L =e

4MN¹ i B¾¹ º F ¹ º B

Homework 2 : Explore the validity of the Coleman-Glashow

relations between the magnetic moments of the baryon octet.

Find analogous relations between the baryon decuplet, and find relations between the EM transition rates between the decuplet and octet assuming M1 transition.

Matrix Elements in Nucleon (1)

CONSTRAINTS

Similarly for the neutron

@¹ j ¹ = 0 ! F (p)3 (q2) = 0

F (p)1 (0) = +1

F (p)2 (0) = · p = ( 2:79 ¡ 1 ) NM

GE = F1 ¡ jQ2 j4M 2

NF2 GM = F1 + F2

hpjqQ°¹ qjpi = Up

·F (p)

1 °¹ + F (p)2 i¾¹ º qº

2MN+ F (p)

3 q¹

¸Up

F (p)i ´ F (p)

i (q2)

Proton : Q4 G(p)M / p

Perturbative QCD :

G(p)M / Q4

Perdisat et al

Dipole Form Factors for Nucleon !!GDipole(Q2) = 1

(1+Q2=0:71)2 Perdisat et al

Recent Comprehensive Lattice Study : S. Boinepalli et al., hep-lat/0604022

Clover on Quenched Not QCD (unfortunately), likely close to

nature from all previous experiences. Clover gives (a2) errors in the quarks Good step toward fully-dynamical

Disconnected diagrams evaluated phenomenologically…..computationally expensive

Lattice Contractions

Baryon Charge Radii

hr2E i = ¡ 6 d

dQ2 GE (Q2)¯¯Q2=0

Zanotti et al

Larger m¼ the smaller lattice can be !!

Baryon Magnetic Moments

Zanotti et al

Baryon Magnetic Radii

Zanotti et al

Alexandru et al ,hep-lat/0611008

Domain-Wall on Staggered Wilson on Quenched

Isovector-Vector Form Factors : Lattice

¹ GE =GM

GM (0)

Isovector-Vector Form Factor( George Fleming, LHPC )

Domain-Wall on Staggered

Just how Strange is

the Proton ?

Flavor Structure of the Nucleon : Tree-Level

°;Z0

q

eV¹ and A¹

Relevant parts of the Standard Model

tanµw = g1g2

D¹ = @¹ + ieQA¹ + i esw cw

¡T3 ¡ Qs2

w

¢Z¹

D¹ = @¹ + ig2Wa¹ Ta + ig1

12Y B¹

YÁ = +1 hÁi =

µ0

v=p

2

¶

B¹ = 1pg2

1 +g22

(g1Z¹ + g2A¹ )

W3¹ = 1p

g21 +g2

2

(g1A¹ ¡ g2Z¹ )

Z0-couplings

Z0

u u

L int: = ¡ e4cw sw

u£¡

1¡ 83s2

w

¢°¹ ¡ °¹ °5

¤u Z¹

Flavor Structure of the Nucleon : EM

I = 0

I = 1

j ¹em =

23u°¹ u ¡

13d°¹ d ¡

13s°¹ s

= q Q °¹ q

=12

q

0

@1 0 00 ¡ 1 00 0 0

1

A °¹ q

+16

q

0

@1 0 00 1 00 0 ¡ 2

1

A °¹ q

Transforms as an octet under SU(3)

Flavor Structure of the Nucleon : Z0

I = 0I = 1

Vector Current

Axial-Vector Current

j ¹Z0 =

12

µ1¡

83s2

w

¶u°¹ u ¡

12

µ1¡

43

s2w

¶d°¹ d ¡

12

µ1¡

43s2

w

¶s°¹ s

+12u°¹ °5u ¡

12d°¹ °5d ¡

12s°¹ °5s

=12

¡1¡ 2s2

w

¢q

0

@1 0 00 ¡ 1 00 0 0

1

A °¹ q ¡13

s2wq

0

@1 0 00 1 00 0 ¡ 2

1

A °¹ q

¡12s°¹ s

+12q

0

@1 0 00 ¡ 1 00 0 0

1

A °¹ °5q ¡ s°¹ °5s

Matrix Elements in Nucleon (2)

CONSTRAINTS

hpjq

0

@1 0 00 1 00 0 ¡ 2

1

A °¹ qjpi = 3 Up

· ³F (p)

1 + F (n)1

´°¹ +

³F (p)

2 + F (n)2

´i¾¹ º qº

2MN

¸Up

hpjs°¹ sjpi = Up

·F (s)

1 (q2)°¹ + F (s)2 (q2)i¾¹ º qº

2MN

¸Up

hpjq

0

@1 0 00 ¡ 1 00 0 0

1

A °¹ qjpi = Up

· ³F (p)

1 ¡ F (n)1

´°¹ +

³F (p)

2 ¡ F (n)2

´i¾¹ º qº

2MN

¸Up

Isovector

Isoscalar

strange

F (s)1 (0) = 0

Tree-Level

°;Z0

q

eg(e)V » 1¡ 4s2

w , g(e)A » 1

Radiative Corrections

°;Z0

q

ee.g.

Hadronic Corrections

°;Z0

qq

e

Z0

Parity-violating vertex

Liu, McKeown and Ramsey-Musolf, arXiv:0706.0226v2

hN js°¹ sjN i is small !!

Q2 = 0.1 GeV2

G(s)E = F (s)

1 ¡ jQ2 j4M 2

NF (s)

2

G(s)M = F (s)

1 + F (s)2

Jlab and Bates

Strange Vector Form Factors

Axial-Current Matrix Elements in Nucleon (1)

CONSTRAINTS

hpju°¹ °5djni = Up

·g1(q2)°¹ °5 + g2(q2)i¾¹ º °5

qº

2MN+ g3(q2)°5q¹

¸Un

T and I ! g2(q2) = 0

g1(0) = gA ¯ ¡ decay

Neutron -decayn p

W-

e

g3 comes with a factor of me

gA = 1:26O » u°¹ (1¡ °5)d

¹ ¡ +p! n +º¹ sensitve to g3

PCAC

Aa¹ (x) = q°¹ °5Taq(x)

h0jAa¹ (x)j¼b(q)i = ¡ i f ¼ q¹ e¡ iq:x ±ab

h0j@¹ Aa¹ (x)j¼b(q)i = ¡ f¼ m2

¼ e¡ iq:x ±ab

@¹ Aa¹ (x) = ¡ f¼ m2

¼ ¼a(x)

A

Therefore @¹ A¹ is a good interpolating ¯eld for the pion.

PCAC : Goldberger-Treiman (1) (1958)

In chiral limit,

hN jAa¹ (x)jN i = U [ g1°¹ °5 + g3 q¹ °5 ] Ta U e¡ iq:x

hN j@¹ Aa¹ (x)jN i = ¡ iU

£g1q¹ °¹ °5 + g3 q2°5

¤Ta U e¡ iq:x

@¹ Aa¹ = 0 hence 2MN g1(q2) + q2g3(q2) = 0

PCAC : Goldberger-Treiman (2)

L = i g¼N N N°5TbN¼b

hN jAa¹ (x)jN i = ¡ U

·g¼N N f¼

q¹

q2°5

¸Ta U e¡ iq:x

In chiral limit,

g3(q2) = ¡g¼N N f ¼

q2 + non-pole

PCAC : Goldberger-Treiman (3)

In chiral limit,

g3(q2) = ¡g¼N N f ¼

q2

2MN g1(q2) + q2g3(q2) = 0

g1(q2) =g¼N N f ¼

2MN

Away from the chiral limit, g3(q2) = ¡

g¼N N f¼

q2 ¡ m2¼

g1(q2) =g¼N N f¼

2MN+ O(

m2¼

¤2Â

)

PCAC : Goldberger-Treiman (4)

g3(q2) = ¡g¼N N f¼

q2 ¡ m2¼

g1(q2) =g¼N N f¼

2MN+ O(

m2¼

¤2Â

)

At the physical point,

gA = 1:2654§ 0:0042

g¼N N = 13:12 ; 13:02

1¡2MN gA

f¼g¼N N= 0:023 ; 0:015

Sid Coon,Nucl-th/9906011

gA from Lattice QCD

Axial Charges : D.I.S.(Deep Inelastic Scattering)

2

~

Large Q2

N

N NOperator-Product Expansion

Axial Charges (2)

Related by Isospin to gA Related by SU(3) to Hyperon Decays

Measure 12

³1 ¡ ®s (Q2)

¼

´hpjq Q2 °¹ °5qjpi in DIS

R10

dx g1(x;Q2)

Q2 =16

0

@1 0 00 ¡ 1 00 0 0

1

A +118

0

@1 0 00 1 00 0 ¡ 2

1

A +29

0

@1 0 00 1 00 0 1

1

A

Axial Charges (3)Including SU(3)-breaking

¡ 0:35< ¢ s < 0

¡ 0:1< ¢ u+¢ d+¢ s < +0:3

2 ¢ q Up s¹ Up = hpjq°¹ °5qjpi

Nucleon -Term (1)H (mq) jN (mq)i = E(mq)jN (mq)i

hN(mq)jH (mq) jN (mq)i = E(mq)

mq@

@mqE (mq) = hN(mq)jmq

@@mq

H (mq) jN (mq)i

L QCD (mq) = L QCD (0) ¡X

i

mi qi qi

mi@

@miH (mq) = mi qi qi

H (mq) = H (0) +X

i

mi@

@miH (mq)

Feynman-Hellman Thm

Nucleon -Term (2) : SU(2)

Note that : hN (mq)jN (mq)i = 1 : conventional to use = 2MN

See later

¾N » 45 MeV from scattering

¾N =X

i

mi@MN

@mi= m2

¼@MN

@m2¼

+ ::

= hN (mq)j muuu + mddd jN (mq)i

= m hN (mq)j uu + dd jN (mq)i

Nucleon -Term (3) : SU(3)

SU(3) Singlet

¾N = hN(mq)j muuu + mddd + msss jN (mq)i

=13

(2m+ms) hN (mq)j uu + dd + ss jN (mq)i

+13

(m¡ ms) hN (mq)j uu + dd ¡ 2 ss jN (mq)i

SU(3) Octet

Nucleon -Term (4) : strangeness

Using¾N » 45MeV gives

m hN(mq)juu+dd¡ 2ssjN (mq)i » 35§ 5 MeV

=34

m2¼

m2K ¡ m2

¼

·(M¥ ¡ MN ) ¡

12

(M § ¡ M¤ )¸

2hN (mq)jssjN (mq)ihN (mq)juu+ddjN (mq)i

» 0:2! 0:4

Nucleon -Term (5)

Strange quarks (non-valence) play a nontrivial role on the structure of the Nucleon

hN (mq)j H (0)jN (mq)i » 764 MeV

hN(mq)j ms ss jN (mq)i » 130 MeV

-Term from the Lattice

Two methods used presently :1. Compute MN and take numerical derivatives … poor

precision…many configs (QCD)

2. Compute 3-pt function

Lattice Results: MN vs m

StaggeredClover

Physical value used in fit not included in fit

MN (GeV)

M2 GeV2

m ~ 235 MeV

Physical point

Galletly et al, hep-lat/0607024

Lattice Results: MN vs m

Overlap fermions

Lattice Results (2): MN vs m

N is derivative of curve--- Much larger uncertainties

Dilatations

Energy-Momentum Tensor

T¹ º (y) = 2p¡ g

±±g¹ º (y)

Rd4x

p¡ g L

Improved Energy-Momentum Tensor and Scale (Dilatation) Current

O¹ º = T ¹ º + surface terms

@¹ O¹ º = 0

Scale-CurrentS¹ = O¹ º xº

@¹ S¹ = O®®

Scale Transformation

x ! x0 = e® x

S =

Zd4x j@¹ Á(x)j2 !

Zd4x j@¹

¡e®dÁ Á(e®x)

¢j2

= e2®(dÁ ¡ 1)Z

d4x0 j@0¹ Á(x0)j2

S0 = e2®(dÁ ¡ 1) S

dÁ = 1 ; dà = 32

Require scale-invariant when massless

Masses Break Scale-Invariance

¡Z

d4xm2¼jÁ(x)j2 ! ¡ e®(2dÁ ¡ 4)

Zd4x0m2

¼jÁ(x0)j2

@¹ S¹ = 2m2¼ jÁj2

¡Z

d4x mN N N ! ¡ e®(2dÁ ¡ 4)

Zd4x0 mN N N

@¹ S¹ = mN N N

Gauge FieldsRenormalization Scale ¹ related to coordinates via ¹ » 1=x

x ! x0 = e® x

QCD -function

L = ¡1

4g2Tr

£G2

¤

±L±® =

¯2g3

Tr£G2

¤= @¹ S¹

g = g(Q2=¹ 2) ! g(e¡ 2®Q2=¹ 2))

g(¹ ) ! g(e®¹ )

Nucleon Mass

Anomalous dimension = quantum corrections

hN jO®® jN i = MN

= hNj¯

2g3Tr

£G2

¤jN i +

hN j(1¡ °u)muuu + (1¡ °d)mddd + (1¡ °s)msssjN i

Ademollo-Gatto Theorem (1964) Corrections to the matrix elements of a

charge operator between states in the same irreducible representation first occur as the square of the symmetry breaking parameter True if matrix element is analytic function of

breaking parameter NOT valid for vector current matrix elements in light

hadrons due to IR behavior of QCD True for heavy quark symmetry..Luke’s Theorem

Vector Current Matrix elements between members of SU(3), SU(2) irreps are protected from symmetry breaking effects,

since they are the charge operators

Ademollo-Gatto Theorem (1964)

Qab =

Zd3x qa(x)°0qb(x) =

Zd3x qay°0qb

hQus ; Qsu

i= Quu ¡ Qss

hK 0jh

Qus ; QsuijK 0i = hK 0jQuu ¡ Qss jK 0i

Pn

³hK 0jQusjnihnjQsu jK 0i ¡ hK 0jQsujnihnjQus jK 0i

´= 0 ¡ (¡ 1)

Pn

³jhnjQsu jK 0i j2 ¡ jhnjQus jK 0i j2

´= 1

1¡ h¼¡ jQsujK 0i = O(¸2) = SU(3) breaking parameter

= 0AND transitions outsideoctet are O(¸)

Baryon Resonance Spectrum So far just discussed extracting the ground states

from lattice calculations. What about excitations

If stable, the correlation function has simple exponential form If unstable, volume dependence required…

(see later)

Flavour, Orbital and RadialStructure

• States classified according to SU(2) Flavor• Spatial and radial structure explored using displaced-source (sink) quark propagators

Classified wrt transformation under hypercubic group … the symmetry group of the lattice

Methodology: Luscher-Wolff

min ( En – Ei)

• Eigenvalues ! Energies = masses of stable particles, (or energy of scattering state for unstable particles)

• Eigenvectors ! “wave functions”

Compute correlation matrix from the r sources and r sinks

C® (t;t0) = h0jO®(t) O¯ (t0)j0i

The eigenvalues of

are

A =1

pC(t0)

C(t)1

pC(t0)

¸ i ! e¡ E i (t¡ t0)³1 + e¡ ¢ E (t¡ t0)

´

Glimpsing nucleon spectrum

Adam Lichtl, PhD 2006

Spectroscopy Group ... JLab

Summary Huge amount of phenomenology … traditionally the

domain of nuclear physics (t > ~ 1970…QCD) Flavor structure Interactions Excitations

Far fewer lattice calculations than for mesons Correlator falls much faster Signal degrades exponentially faster Requires significantly more effort … people-power and

computers Relatively straighforward procedure to follow

Go forth and compute the properties of the building blocks of nuclei from QCD !