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Hadrons and Nuclei : Single Hadrons
Lattice Summer SchoolLattice Summer School
Martin Savage
Summer 2007
University of Washington
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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¡
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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
+
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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
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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
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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
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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 )
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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+ ::
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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®
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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
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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
+
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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
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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.
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, 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´
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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¤
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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
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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
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Latest Lattice results : LHPC : DW on Staggered
LHPC, Negele et al
m=350 MeV
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Lattice Result for GMO
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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
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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 § ¤
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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
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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.
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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)
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Proton : Q4 G(p)M / p
Perturbative QCD :
G(p)M / Q4
Perdisat et al
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Dipole Form Factors for Nucleon !!GDipole(Q2) = 1
(1+Q2=0:71)2 Perdisat et al
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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
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Lattice Contractions
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Baryon Charge Radii
hr2E i = ¡ 6 d
dQ2 GE (Q2)¯¯Q2=0
Zanotti et al
Larger m¼ the smaller lattice can be !!
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Baryon Magnetic Moments
Zanotti et al
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Baryon Magnetic Radii
Zanotti et al
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Alexandru et al ,hep-lat/0611008
Domain-Wall on Staggered Wilson on Quenched
Isovector-Vector Form Factors : Lattice
¹ GE =GM
GM (0)
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Isovector-Vector Form Factor( George Fleming, LHPC )
Domain-Wall on Staggered
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Just how Strange is
the Proton ?
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Flavor Structure of the Nucleon : Tree-Level
°;Z0
q
eV¹ and A¹
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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 )
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Z0-couplings
Z0
u u
L int: = ¡ e4cw sw
u£¡
1¡ 83s2
w
¢°¹ ¡ °¹ °5
¤u Z¹
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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)
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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
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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
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Tree-Level
°;Z0
q
eg(e)V » 1¡ 4s2
w , g(e)A » 1
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Radiative Corrections
°;Z0
q
ee.g.
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Hadronic Corrections
°;Z0
e
Z0
Parity-violating vertex
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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
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Strange Vector Form Factors
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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
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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
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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.
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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
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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
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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
)
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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
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gA from Lattice QCD
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Axial Charges : D.I.S.(Deep Inelastic Scattering)
2
~
Large Q2
N
N NOperator-Product Expansion
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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
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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
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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
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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
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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
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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
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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
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-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
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Lattice Results: MN vs m
StaggeredClover
Physical value used in fit not included in fit
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MN (GeV)
M2 GeV2
m ~ 235 MeV
Physical point
Galletly et al, hep-lat/0607024
Lattice Results: MN vs m
Overlap fermions
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Lattice Results (2): MN vs m
N is derivative of curve--- Much larger uncertainties
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Dilatations
Energy-Momentum Tensor
T¹ º (y) = 2p¡ g
±±g¹ º (y)
Rd4x
p¡ g L
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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®®
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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
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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
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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 )
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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
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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
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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(¸)
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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)
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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
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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)
´
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Glimpsing nucleon spectrum
Adam Lichtl, PhD 2006
Spectroscopy Group ... JLab
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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 !