μ - Indico

Tılsımlı Baryonların
Elektromanyetik Özellikleri
1
Utku Can1,2
Orta Doğu Teknik Üniversitesi, Ankara
2 Özyeğin Üniversitesi, İstanbul
http://arxiv.org/abs/1306.0731 K. U. Can, G. Erkol, B. Isildak, M. Oka, T. T. Takahashi
Tuesday, July 23, 13
nt-like particle but in fact is composed of quarks
Motivasyon
Hadronların
iç yapısını anlamak
stributed •inside
the nucleon?
+
harge, but
does
it
have
a
+/core?
• Hapsoluş dinamiğini incelemek
- • Elektrik yükü dagılımı nasıl?
• Ağır kuarkların etkisi?
its experimentally observed properties
30% of the proton’s spin
tum?
Tuesday, July 23, 13
Kuantum Renk Dinamiği (KRD)
• Non-Abelian SU(3)c Ayar Teorisi
• Renk “yüklü” kuark, anti-quark ve
gluonların kuvvetli etkileşimi
Nf
SQCD [ , ¯, A] =
d x ¯(f ) (x) [/ + igA
/(x) + m(f ) ]
4
c
f =1
Fµ =
Tuesday, July 23, 13
µA
(x)
(f )
1
(x) +
c
2
Aµ (x) + ig[Aµ , A ]
d4 x T r[Fµ F µ ]
Kuantum Renk Dinamiği (KRD)
Fermion Eylemi
Çeşni sayısı
µ
µ
µ
Aµ
Çeşni indisi
N
LF [ , ¯, A] =
¯(f ) (x)[/ + igA
/(x) + m(f ) ]
,c
f =1
Uzay-zaman pozisyonu
(x)
c=
1
0 ,
0
u(p) c
0
0
1 , 0
0
1
8
a
Aµ Ta
Aµ (x) =
a=1
Ti : 3x3 Gell-Mann matrixleri
Tuesday, July 23, 13
Etkileşim sabiti
4
(f )
,c (x)
Fermiyon alanı (kuark)
Dirac spinör indisi
α = 1,2,3,4
Gauge alanı (gluon)
s
Ağaç düzeyi Etkileşimler
Renk indisi
c = 1,2,3
Kuantum Renk Dinamiği (KRD)
Ayar Eylemi
1
LG [Aµ (x)] = Tr[Fµ F µ ]
2
Fµ =
µA
Aµ + ig[Aµ , A ]
8
Aaµ Ta
Aµ (x) =
a=1
T a , T b = 2if abc T c
8
Fµ (x) =
a=1
{
a
µ A (x)
1
LG [Aµ (x)] =
4
Tuesday, July 23, 13
Aaµ (x)
Fµa (x)
8
Fµa Faµ
i=1
gf bca [Abµ , Ac ]}Ta
value emerges from the compilation of current determinations of αs :
αs (MZ2 ) = 0.1184 ± 0.0007 .
Metodlar
The results also provide a clear signature and proof of the energy dep
full agreement with the QCD prediction of Asymptotic Freedom. This i
Fig. 9.4, where results of αs (Q2 ) obtained at discrete energy scales Q, n
those based just on NLO QCD, are summarized and plotted.
• Kuark Modeli (KRD Öncesi)
• #QCD = 200 MeV, tedirgeme
kuramı tutarsız
• Tedirgeme dışı metodlar
• KRD Toplam Kuralları
• Kiral Tedirgeme Kuramı
• Örgü KRD
Tuesday, July 23, 13
Figure 9.4: Summary of measurements of αs as a function of the re
scale Q. The respective degree of QCD perturbation theory used in
of αs is indicated in brackets (NLO: next-to-leading order; NNLO: n
Örgü KRD
Nf
SQCD [ , ¯, A] =
d x ¯(f ) (x) [/ + igA
/(x) + m(f ) ]
4
c
f =1
Fµ =
µA
(x)
(f )
1
(x) +
c
2
d4 x T r[Fµ F µ ]
Aµ (x) + ig[Aµ , A ]
•
Ab initio metod
•
Teorideki sonsuzluklardan kurtulmak için iyi bir yöntem
•
Teoriyi sayısal olarak çözmeye olanak tanır
•
KRD aksiyonunu kesikli uzayda çözümlemek için değiştirmek gerek
Anahtar denklemler:
lim
T
Ô2 (t)Ô1 (0)
T
=
h
Ô2 (t)Ô1 (0) =
Tuesday, July 23, 13
D[ ]e
0| Ô2 | h h| Ô1 | 0 e
SE [ ]
Eh t
O2 [ (x, t)]O1 [ (x, 0)]
D[ ]e SE [ ]
Hadron özdurumları üzerinden toplam
(hadron kulesi)
iz integrali
Örgü KRD
$z integrali tanımlaması
Ô2 (t)Ô1 (0) =
D[ ]e
SE [ ]
O2 [ (x, t)]O1 [ (x, 0)]
D[ ]e SE [ ]
Importance Sampling
1
O = lim
N
N
• e
N
n=1
O[Un ]
Örgü, aslında KRD uzayının
anlık bir görünümü
(statistical ensemble)
Tuesday, July 23, 13
SE
, ağırlık fonksiyonu
• Minkowski
Öklid
• Wick dönüşümü, t
-i%
LATT
Lattice QCD = QCD in disc
Kesikli Uzay
QCD in discretized Euclidean sp
= {(n, n4 ) | n
(f )
(x)
(f )
c
3,
(an) ,
c
r
g
b
r
Uµrr
Uµrg
Uµrb
Uµ = g
Uµgr
Uµgg
Uµgb
b
Uµbr
Uµbg
Uµbb
µ
(n) =
(n + µ̂)
n4 = 0, 1, . . . , NT
¯(f ) (x)
¯(f ) (an)
c
Uµ (n) = exp (iaAµ (n))
(n
µ̂)
2a
(0, n2 , n3 , n4 ) = (N, n2 , n3 , n4 )
..
.
(n1 , n2 , n3 , 0) = (n1 , n2 , n3 , NT )
Tuesday, July 23, 13
c
1}
d4 x
a4
Uµ (N, n2 , n3 , n4 ) = Uµ (0, n2 , n3 , n4 )
..
B. Musch (TUM)
.
lattice spacing:
Uµ (n1 , n2 , n3 , NT ) = Uµ (n1 , n2 , n3 , 0)
a→0
limits / extr
Extended Gauge
lattic
L
Örgü KRD
Naif kesikleme
Ayar Eylemi
1
SG [Aµ (x)] =
2
Fµ = µ A
4
d xT r[Fµ F
µ
]
Aµ + ig[Aµ , A ]
Wilson Ayar Eylemi
34
2 QCD on the lattice
n − µ̂
n
U− µ (n) ≡ Uµ (n −
n +µ̂
n
µ̂)†
Uµ(n)
2 QCD on the lattice
Fig. 2.2. The link variables Uµ (n) and U−µ (n)
points from n to n − µ̂ and is related to the positively oriented link variable
Uµ (n − µ̂) via the definition
U−µ (n) ≡ Uµ (n − µ̂) .
†
(2.34)
In Fig. 2.2 we illustrate the geometrical setting of the link variables on the
lattice. From the definitions (2.34) and (2.33) we obtain the transformation
properties of the link in negative direction
"
U−µ (n) → U−µ
(n) = Ω(n) U−µ (n) Ω(n − µ̂)† .
SG [U ] =
3
n
& = 2 Nc/g2
(2.35)
Note that we have introduced the gluon fields Uµ (n) as elements of the gauge
group
SU(3),
as elementswhich
of the Lie
algebra
were used in the
3. The
four
linknotvariables
build
upwhich
the plaquette
Uµνcon(n).
the gauge transformations (2.33) and (2.35) also the
s thetinuum.
order According
that thetolinks
are run through in SU(3).
the plaquette
transformed link variables µ
are elements of the group
µ
Having introduced the link variables and their properties under gauge
transformations, we can2now
generalize
!
! the free fermion action (2.29) to the
so-called naive
for fermions
in an
SG [Ufermion
] = action
Re tr
[1external
− Uµνgauge
(n)]field
. U:
2
g"
Tuesday, July 23, 13
#
n∈Λ µ<ν
The circle
†
†
U
(n)
=
U
(n)U
(n
+
µ̂)U
(n
+
ˆ
)U
(n)
Plaket:
µ
(2.49)
µ<
Re {T r[1
Uµ (n)]}
Örgü KRD
Naif kesikleme
Fermiyon Eylemi
SF [ , ¯, A] =
d4 x ¯(x)[/ + igA
/(x) + m] (x)
4
¯(n)
SF [ , ¯] = a4
µ
Uµ (n)
4
¯(n)
SF [ , ¯, U ] = a
n
Tuesday, July 23, 13
Uµ (n) =
Uµ (n) (n + µ̂)
µ
µ=1
µ̂)
+ m (n)
Ayar dönüşümleri altında değişken
Bağlantı değişkenlerini kullan!
(n) = (n) (n)
¯ (n) = ¯(n) † (n)
4
(n
2a
µ=1
n
(n)
¯(n)
(n + µ̂)
(n)Uµ (n)
Uµ† (n
2a
†
(n + µ̂)
µ̂) (n
µ̂)
+ m (n)
Lattice QCD
Geliştirilmiş Wilson Fermiyon Eylemi
•Naif fermiyonların kopyaları var (fiziksel değiller)
• p-uzayındaki Dirac operatörüne ∝ cos terimi ekleyerek
kopyalardan kurtulmak mümkün
• Ters Fourier Dönüşümü ile x-uzayı Dirac operatörü elde edilir
W
SF [
4
¯
, , U] = a
¯(n)D(n|m) (m)
n,m
D(n|m) =
Tuesday, July 23, 13
4
m+
a
4
n,m
1
(1 +
2a µ=1
µ ) Uµ (n)
n+µ̂,m + (1
†
)
U
µ
µ (n)
n µ̂,m
Örgü KRD
Geliştirilmiş Eylemler
Iwasaki Ayar Eylemi
SG =
6
Wµ1
c0
1
PhysRevD.79.034503
Wµ1
(x) + c1
x,µ<
2
(x)
x,µ<
c0 = 1 - 8c1 = 3.648
c1 = -0.331
1×1
Wµν
(x)
Clover Fermion Eylemi
SFC
=
SFW
+
q
cSW a
Commun.Math.Phys, 97, pp.59,1985
5
n
csw = 1.715
µ<
1
¯(n)
2
PhysRevD.79.034503
1
Fµ (n) =
Qµ (n)
8i
Tuesday, July 23, 13
1×2
Wµν
(x)
µ
†
Qµ
µ
1
= [
2
(n)
Fµ (n) (n)
9.1 The Symanzik improvem
µ,
]
n
ν
µ
Fig. 9.1. Graphical representation of the sum Q
(n) of plaque
İş akışı
9 / 32
(n)¯ (m) =
Z=
1
Z
D[U ]e
D[U ]e
SGLattice
[U ]
QCD and meson-baryon1 interactions 1
det[D
Du (n|m) Dd (m|n)
u ] det[Dd ] T r
SIMULATION PARAMETERS
SG
[U ] lattice with two flavors of dynamical quarks
• 163x32
det[Du ] det[Dd ]
(generated by CP-PACS)
•
• The renormalization-group improved gauge action
Farklı örgüler üret (Importance
sampling)
at !=1.95
•
quark action (unquenched)
• Wilson clover
det[D] terimleri deniz-kuarklarını
modelliyor
•
•
Lattice size = (2.5 fm)3x(5.0 fm)
Valans kuark propagatorlerini
hesapla
Hopping parameter κ
sea,
•
•
İdeali, Her noktadan her
• Up- and down-quark masses 150, 100, 90, 60, 35 MeVm0 ,
her noktaya. Hesap süresi fazla•
•
κval=0.1375, 0.1390, 0.1393, 0.1400, 0.1410
•
or Smeared source S
Kolaylık: Tek noktadan her
(x2,t2)
F (U ) =
Hadron özelliklerini hesapla
F (U ), N
e
0
aa0
# of iter.
(x
2 1,t1)
, gaussian(0,0)
smearing
0, wall smearing
26
STRANGE HADRON SYSTEMS AND QUANTUM CHROMODYNAMICS
3
2
Uj (n, nt ) (n + ĵ, m) + Uj† (n
=
j=1
Tuesday, July 23, 13
=
i
her noktaya
•
0 ,a0
(m) = (m m0 )
Smeared source and sink operators separated by 8 lattice units in theatemporal direction.
N
Statistical errors with jackknife analysis
n0 , 0 ,a0
P(x1)
Choose a Dirac Delta (Point) S0
ĵ, nt ) (n
ĵ, nt )
İş akışı
• Yeterli istatistiği topla (Importance sampling)
• Hadron özelliğine özgü analizi yap. (istatistiksel hatalar
1/ N )
• Farklı kuark kütleleri için işlemleri tekrarla
• Fiziksel kuark kütlesindeki değeri hesapla (chiral limit)
• Farklı ögrü aralıkları ve boyutları için işlemleri tekrarla
• Sistematik hataları öngör
Tuesday, July 23, 13
Simülasyon Detayları
•
323 x 64, β = 1.9, 2+1-çeşnili örgüler
•
Gauge action: Iwasaki, Fermion action: Clover
•
a = 0.0907(13)fm, a-1 = 2.176(31)GeV
•
45, 50, 70, 90 istatisik
•
mπ
∼
700, 570, 410, 300 MeV
•
Smeared operatorler
•
Yaratma-yoketme aralığı, t=12a
Tuesday, July 23, 13
Elektromanyetik Yapı Faktörleri
momentum insertions: (|px |, |py |, |pz |) = (0, 0, 0), (1, 0, 0), (2, 0, 0), (3, 0, 0). W
momentum in other directions and using the isotropy of space we average ov
j
k
= ijk
[cT i (x)C
momenta for both D and D (x)
in order
to increase
the statistics.
5 (x)]c (x)
(Baryonlar)
⇤
cc
2
2
1
We consider point-split lattice vector
current
Vµ = c µ c + u µ u
d µd
3
3
3
Vµ = 1/2[q(x + µ)Uµ† (1 + µ )q(x) q(x)Uµ (1
• Spin - 1/2 baryon
B(p)|Vµ |B(p ) = ū(p)
µ )q(x
+ µ)],
which is conserved by Wilson fermions. Therefore it does not require any renor
q
2
F
(q
)
+
i
F
(q
) u(p)
the
lattice,
where
the
renormalization
factors
are
computed
in a perturbative
µ 1,B
2,B
2mB
the lattice. The local axial-vector
current, on the other hand, needs to be ren
µ
2
III.
• Sachs Yapı Faktörleri
RESULTS AND DISCUSSION
We begin our discussion with the D⇤ D⇡ coupling constant. We find that
contribution to 2
the coupling constant comes from the first term in Eq. (8).
q
F
(q
) and minor contributions to the coup
GE,B (q ) = F1,B (qlisted) +
2,B
2
in Table
I.
We
give
the
dominant
4mB
2
2
2 including the ratio G2/G1 contributes
term
to the coupling around 10%. O
individually.
GM,B (q 2 ) = F1,B (q 2 ) + F2,B (q 2 ) 
ud
G1 (q 2 = 0) G2 /G1
gD⇤ D⇡
0.13700 14.15(1.58) 0.09(2) 15.45(1.78)
0.13727 13.63(1.57) 0.12(4) 15.24(1.81)
0.13754 12.76(1.43) 0.15(7) 15.54(2.08)
0.13770 15.46(2.17) 0.07(6) 16.44(2.41)
Lin. Fit
16.23(1.71)
Quad. Fit
17.09(3.23)
TABLE I: Dominant and minor contributions to the coupling gD⇤ D⇡ at each sea-q
Tuesday, July 23, 13
Elektromanyetik Yapı Faktörleri
(Baryonlar)
• Spin - 1/2
C B (t; p;
cc
e
ip·x
vac|T [
4
x
C BVµ B (t2 , t1 ; p , p; ) =
i
B (x)¯B
ip·x2 iq·x1
e
e
[cT i (x)C
5
j
(x)]ck (x)
C BVµ B (t2 , t1 ; p , p; )
R(t2 , t1 ; p , p; ; µ) =
C B (t2 ; p ; 4 )
R(t2 , t1 ; p , p; ; µ)
t1
C B (t2
C B (t2
a
t2 t1
a
t1 ; p;
t1 ; p ;
4)
4)
B (x2 )Vµ (x1 )¯B
C B (t1 ; p ;
C B (t1 ; p;
1/2
1
2EB (EB + mB )
(0)]|vac
B
4 ) C (t2 ; p ; 4 )
B
4 ) C (t2 ; p; 4 )
(p , p; ; µ)
(EB + mB )
(0, q; 4 ; µ = 4) =
2EB
j ; µ = i) =
(0)]|vac
vac|T [
x2 ,x1
Tuesday, July 23, 13
ijk
cc
4) =
(0, q;
(x) =
GE,B (q 2 )
1/2
2
q
G
(q
)
ijk k M,B
1/2
Elektrik & Manyetik yük yarıçapları
Manyetik Moment
•
2
hrE,M
i
=
rms Yük Yarıçapı
6
d
2
G
(Q
)
E,M
2
GE,M (0) dQ
, ansatz
Q2 =0
GE,M (0)
GE,M (Q ) =
(1 + Q2 / 2E,M )2
2
böylece yük yarıçapı:
2
rE,M
•
=
2
E,M
Manyetik Moment
Doğal birimler
µ⌅cc = GM (0)
Tuesday, July 23, 13
12
✓
e
2m⌅cc
Nuclear magneton
◆
= GM (0)
✓
mN
m⌅cc
◆
µN
III.
Sonuçlar
RESULTS AND DISCUSSION
R(t1,t2;0,p;𝛤4 ; μ = 4)
(plato bölgeleri, elektrik)
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
𝜅ud = 0.13700
R(t1,t2;0,p;𝛤4 ; μ = 4)
0.4
0
𝜅c = 0.1224
2
4
6
8
10
0
2
4
6
8
10
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
𝜅ud = 0.13754
0.4
2
0
𝜅c = 0.1224
2
4
6
8
10
0
2
4
6
8
10
6
8
10
𝜅ud = 0.13770
0.4
𝜅c = 0.1224
0
𝜅ud = 0.13727
0.4
4
t1
6
8
10
𝜅c = 0.1224
0
2
4
t1
FIG. 1: The ratio in Eq. (4) as function of the current insertion time, t1 , for all the quark masses
we consider and first nine four-momentum insertions. The horizontal lines denote the plateau
regions as determined by using a p-value criterion (see text).
Tuesday, July 23, 13
We give the ratio in Eq. (4) for the electric (magnetic) form factors as functions of the
Sonuçlar
elektrik yapi faktörü
𝜅ud =0.13700
𝜅ud =0.13727
1
𝜅ud =0.13754
𝜅ud =0.13770
GE(Q2)/GE(0)
0.9
0.8
0.7
0.6
𝜅ud
𝜒 2/d.o.f
p
0.5
0.13700
0.13727
0.13754
0.13770
0.15
0.73
0.85
0.45
.99
.67
.56
.89
0.4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Q2 [GeV2]
2
FIG. 3: The electric form factor GE (Q2 ) of ⌅++
cc as a function of Q and as normalized with its
electric charge, for all the quark masses we consider. The dots mark the lattice data and the curves
show the best fit to the dipole form in Eq. (13).
To extract the coupling constant to the ⇢-meson we use the VMD approach [27, 34]:
Tuesday, July 23, 13
2
Sonuçlar
R(t1,t2;0,p;𝛤j ; μ = i)
(plato bölgeleri, manyetik)
1.6
1.6
1.4
1.4
1.2
1.2
1
1
0.8
0.8
𝜅ud = 0.13700
0.6
R(t1,t2;0,p;𝛤j ; μ = i)
1.8
0
0
𝜅c = 0.1224
𝜅ud = 0.13727
2
4
6
8
10
2
4
6
8
10
0.6
1.8
1.6
1.6
1.4
1.4
1.2
1.2
1
1
0.8
0.8
𝜅ud = 0.13754
0.6
0
0
2
4
6
8
10
2
4
6
8
10
6
8
10
𝜅c = 0.1224
0.4
0
2
𝜅ud = 0.13770
0.6
𝜅c = 0.1224
0.4
𝜅c = 0.1224
4
t1
6
8
10
0
2
4
t1
FIG. 2: Same as Fig. 1 but for the magnetic form factors.
We use a dipole form to describe the data at finite momentum transfers and to extrapolate:
Tuesday, July 23, 13
Sonuçlar
(manyetik yapı faktörü)
𝜅ud = 0.13700
𝜅ud = 0.13727
1.8
𝜅ud = 0.13754
𝜅ud = 0.13770
1.6
GM(Q2)
1.4
1.2
1
𝜅ud
0.13700
0.13727
0.13754
0.13700
0.8
𝜒 2/d.o.f
0.21
0.26
0.11
0.55
p
.96
.93
.99
.74
0.6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Q2 [GeV2]
FIG. 4:
2
The magnetic form factor GM (Q2 ) of ⌅+
cc as a function of Q for all the quark masses
we consider. The dots mark the lattice data and the curves show the best fit to the dipole form in
Eq. (13).
Tuesday, July 23, 13
+
ud
stat.
0.2
u,d
val
lin. quad.
0.18
𝜒 2/ d.o.f.
1.37
0.32
p
.25
.57
0.16
cc
<r 2 E, ++ > [fm2 ]
𝛯
Sonuçlar
13770|13754|13727|13700
170 150
100 100
0.14
0.12
0.1
0
0.02
0.04
cc
<r 2 E, + > [fm2 ]
𝛯
0.1
0.08
𝜒 2/ d.o.f.
2.07
0.17
p
.13
.68
cc
<r 2 M, + > [fm2 ]
𝛯
0.1
0.12
0.14
0.02
0.04
0.06
0.08
0.1
0.12
0.14
lin. quad.
0.22
𝜒 2/ d.o.f.
1.79
0.90
p
.17
.34
0.19
0.16
0.13
0.1
0
0.02
0.04
0.51
0.06
0.08
0.1
0.12
𝜒 2/ d.o.f. 0.06
p
.94
0.05
.81
cc
μ𝛯+
0.36
0.06
0.08
(a m𝜋)2
Tuesday, July 23, 13
0.1
0.12
2
2
rE,
am
+
cc
cc
2
[fm ]
0.052(10)
1.779(6)
0.13727
2.033(71)
4.183(644)
0.113(8)
0.027(8)
1.748(6)
0.13754
1.970(75)
4.723(955)
0.120(9)
0.021(8)
1.726(7)
0.13770
Lin. Fit
1.801(81)
1.697(63)
3.555(500)
3.600(306)
0.144(13)
0.135(11)
0.037(10)
0.020(7)
1.706(8)
1.697(6)
Quad. Fit
1.476(103)
2.409(698)
0.164(18)
0.049(12)
1.696(12)
M,
+
cc
2
rM,
+
cc
0.13700
[GeV]
1.829(86)
[fm2 ]
0.139(13)
0.13727
1.936(88)
0.13754
GM,
+
cc
(0)
µ
g
+
cc
cc
3.693(13)
GeV
cc
1.573(65)
[µN ]
0.381(16)
6.555(428)
0.124(11)
1.600(41)
0.394(10)
5.651(211)
1.940(94)
0.124(12)
1.629(62)
0.407(15)
5.721(263)
0.13770
Lin. Fit
1.706(97)
1.788(67)
0.160(18)
0.135(12)
1.674(83)
1.671(61)
0.423(22)
0.424(15)
5.536(230)
5.700(245)
Quad. Fit
1.444(136)
0.173(25)
1.695(106)
0.430(27)
6.098(380)
+
cc
0.39
0.04
++
cc
[fm ]
0.136(12)
++
cc
0.42
0.02
2
rE,
[GeV]
2.988(280)
<r2>E, = 0.138 fm2
<r2>E, = 0.020 fm2
*<r2> = 0.770 fm2
E,p
0.45
0
+
cc
0.14
lin. quad.
0.48
E,
[GeV]
1.853(79)
0.04
0.02
++
cc
0.13700
u,d
val
0.06
0.25
μN ]
0.08
lin. quad.
0
[
0.06
lin. fit
quad. fit
E,
SELEX Collab.
3.518(9) GeV
0.14
<r2>M, = 0.173 fm2
*<r2>
2
M,p = 0.604 fm
# = 0.424 #N
*# = 2.793 #
p
N
PACS-CS a m! cc = 1.656(10)
our value deviates by ~ 2%
Clover action has O(a mq) discretization errors
mass comparison is a way to estimate sys. errors
+
cc
+
cc
* PDG değerleri
Sonuçlar
ud
stat.
13770|13754|13727|13700
170 150
100 100
QCDSF 2 flavor Clover fermions, Phys.Rev. D84, 074507 (2011)
The value
Lagrangia
30], F =
constants
lin. quad.
𝜒 2/ d.o.f. 1.37
0.18
p
.25
0.32
.57
0.16
cc
<r 2 E, ++ > [fm2 ]
𝛯
0.2
Hall et.al, arXiv:1305.3984 [hep-lat]
0.14
0.12
0.1
0
0.02
cc
<r 2 E, + > [fm2 ]
𝛯
0.1
0.04
0.08
p
.13
cc
<r 2 M, + > [fm2 ]
𝛯
0.1
0.12
0.14
.68
0.06
0.04
0.02
0.02
0.04
0.25
0.06
0.08
0.1
0.12
0.14
lin. quad.
0.22
𝜒 2/ d.o.f.
1.79
0.90
p
.17
.34
FIG. 1: (color online). Lattice QCD results for !r2 "E from QCDSF [10],
0.19
0.16
0.13
0.1
0
m𝜋20.51
~
μN ]
0.08
lin. quad.
𝜒 2/ d.o.f. 2.07 0.17
0
0.02
0.04
0.06
0.08
0.1
0.09 0.169
0.325 0.49
lin. quad.
𝜒 2/ d.o.f. 0.06
0.48
p
.94
0.12
0.14
GeV2
0.05
.81
0.45
μ𝛯+
cc
[
0.06
lin. fit
quad. fit
0.42
0.39
0.36
0
Tuesday, July 23, 13
0.02
0.04
0.06
0.08
(a m )2
0.1
0.12
0.14
In FRR
teristic m
the loop i
they satis
calculatio
tice simul
In this inv
following
using the Ansatz from Eq. (31), and the experimental value as marked [22,
Nucleon:
sadece
kuarklardan
23]. The lattice
results hafif
satisfy: L
> 1.5 fm and mπ Loluşuyor
> 3. A naı̈ve linear
trend-line is also included, which does not reach the experimental value. The
: Kuark kütlesine göre davranış Nucleon’a göre
ccphysical point is shown with a vertical dotted line.
•
farklı
•
While co
• Fiziksel bir olayın işareti olabilir mi?
remains e
2
2
c - cmomentum
sisteminitransfer
bozuyor
• Ağırlaşan
where Q is hafif
definedquark
as positive
Q =olabilir
in modell
2
!
2
−q = −(p − et.al’ın
p) . Lattice
QCD resultsqQQ
are often
constructed
Analyses
çalışması
sistemlerinde
Q -Q
• Yamamoto
from an alternative representation, using the form factors F1
sible form
arası
gerilimin
hafif
kuarktan
ötürü
azaldığını
and F2 : the Dirac and Pauli form factors, respectively. The
Sachs formarxiv:0708.3610
factors are simply
linear combinations of F1 and
gösteriyor.
[hep-lat]
F2 ,
2
Sadete geldik
• Ağır kuarkın varlığı baryonu küçültüyor (EM bazında)
•
arXiv:hep-ph/1101.1983, isospin splitting hesaplarına dayalı öngörüyle uyumlu
• Hafif ve ağır baryonların hapsoluş dinamiğinde
farklılıklar var, incelemeye devam...
Tuesday, July 23, 13
Pek Yakında
• Elektromanyetik Yapı Faktörleri
• ⌃c (uuc, ddc), ⌦c (ssc)
• Tek tılsımlı quark içeren baryonların özellikleri
•
⌦cc (scc)
• Xi_cc ile karşılaştırınca hapsoluş dinamiği ile ilgili önemli
bilgiler elde edilebilir.
Tuesday, July 23, 13
TEŞEKKÜRLER!
Tuesday, July 23, 13
BACKUP SLIDES
Tuesday, July 23, 13
Lattice QCD
Fermion Doubling
Story:
•
Naive discretization ⟶ (2d - 1) unphysical fermions
•
Add extra term to discrete action
Let’s find the doublers! Compute fermion propagator,
SF [ , ¯, U ] = a4
¯(n)D(n|m) (m)
n,m
4
D(n|m) =
Uµ (n)
µ
µ=1
Uµ† (n)
n+µ̂,m
n µ̂,m
2a
+m
n,m
Consider trivial gauge field, Uμ=1 and massless fermions
Using,
n,m
=
Tuesday, July 23, 13
4
1
| |
e
n,m
iakµ (n m)
D̃(n|m) =
n,m
i
sin(kµ a)
a µ=1
µ
Lattice QCD
Fermion Doubling
4
4
D̃(n|m) =
n,m
i
sin(kµ a)
a µ=1
i
D(k) =
sin(kµ a)
a µ=1
µ
D̃(n|m)D(k)
1
=
µ
n,m
Propagator is the inverse of the Dirac term
(D̃(n|m))
D(k)
1
=
sin term in the denominator
leads to unphysical fermions
Tuesday, July 23, 13
ia
a
1
= (D(k))
1
2
µ
1
sin(kµ a)
µ
2 (k a)
sin
µ
µ
kµ = ( , 0, 0, 0), (0, , 0, 0), . . . , ( , , , )
a
a
a a a a
al choice is where
the basis
of the
eigenstates
H.(1.15)
Denoting
in the
second
step we of
used
and the
the eigenstates
normalization of the eigenwhere
in
the
second
step
we
used
(1.15)
and
the
normalization
of the eigenas |n!, the corresponding
eigenvalue
equation
reads
states, #n|n! = 1. The partition function ZT is simply a sum over the expostates, #n|n! = 1. The partition function ZT is simply a sum over the exponentials of all energy
eigenvalues En .
!
nentials of all energy
eigenvalues
H |n! = En |n! . En .
(1.15)
The Euclidean correlator #O2 (t) O1 (0)!T can be evaluated in a similar way.
The Euclidean correlator #O2 (t) O1 (0)!T can be evaluated in a similar way.
When E
computing
the traceand,
as before
and discrete
insertingspectrum,
the unit operator in the
nergy eigenvalues
are
real
numbers
assuming
n
When computing
the trace as !before and inserting the unit operator in the
form
(1.10)
to the right !of O2 we obtain
der the form
states
such
that
(1.10)
to the right of O
2 we obtain
"
1
! !
! !
−(T −t) H
−t H
"
1
E
≤
E
≤
E
≤
E
.
.
.
(1.16)
#O (t) O1 (0)!
! ! O2 |n!#n|e
! ! O1 |m!
1 T =2
3 #m|e
−(T −t) H
−t H
O
O1 |m!
#O2 (t)20O1 (0)!
=
#m|e
|n!#n|e
ZT m,n
2
T
ZT m,n
ndex n labels all the combinations of1quantum
numbers describing the
"
−(T −t) Em
−t En
"
!
!1 |m! . (1.18)
1
=
e
#m|
O
|n!
e
#n|
O
2
such that their energies are ordered
according
−(T −t)to
Em(1.16).
−t
E
n
!2 |n! e
!1 |m! . (1.18)
=
e
#m|O
#n|O
ZT m,n
ZT m,n
hen using the basis |n! for evaluating
the trace in (1.13) one finds
"
!
The next "
step is to
the
expression
(1.17) for ZT into the last equation
−Tinsert
H
−T En
ZTstep
= is #n|e
,
(1.17)
The next
to insert |n!
the=expression
(1.17)
for ZT into
the last equation
−T Ee0
both in the numerator and in the denomiand to pull out a factor of e
and to pull outn a factor of e−T E0n both in the numerator and in the denominator. This leads to
nator. This leads to
in the second step we used (1.15)
of ∆E
the eigen# and the! normalization
−t
n −(T −t) ∆Em
!1 |m! e
#
#m|
O
|n!
#n|
O
eexpo2
−t
∆E
−(T
−t) ∆Em
m,n
n
!
!
is
simply
a
sum
over
the
, #n|n! = 1. The
partition
function
Z
O1 |m! e
e
#O2 (t) O1 (0)!T =m,n #m|TO2 |n! #n|
, (1.19)
−T
∆E
−T
∆E
1
2
, (1.19)
1 −T
+lattice
e∆E1
+ e∆E2
+ ...
ls of all energy
1 (0)!path
n.
T =Eintegral
6 #O2 (t)eigenvalues
1OThe
on
the
−T
1+e
+e
+ ...
e Euclidean correlator #O2 (t) O1 (0)!T can be evaluated in a similar way.
where we defined
where
wetrace
defined
computing
the
as before
and inserting
the
unit
operatori.e.,
in only
the the contributions
numerator
only
those terms
where
E
0E
survive,
mE=
∆E
=
−
.
(1.20)
n
n
0
|m>
=
0,
E
=
0
!
m
∆En = En − E0 .
(1.20)
obtain
1.10) to where
the right
2 weWe
|m!of=O|0!.
thus obtain
Thus, the Euclidean correlator #O1 (0) O2 (t)!T depends only on the energies
" correlator! #O1 (0) O!
Thus, the Euclidean
depends only on the energies
2 (t)!
1
!T vacuum.
−(T
−t)energy
H !
−t
H
−t Enthese energy
normalized
relative
to
the
E
of
the
It
is|0!
exactly
!
"
"
0
O
O
#O2 (t)normalized
O1 (0)!T = relative
|n!#n|e
|m!
lim#m|e
"O
(t)
O
(0)!
=
"0|
O
|n!"n|
O
e
. energy (1.21)
2
1
2 energy
1 E0T of the vacuum.
2 It is exactly
1
to
the
these
Z
Tm,n
→∞
differencesTthat
can be measured in annexperiment. For convenience from now
differences that can be measured in an experiment.
For convenience from now
to denote the energy differences relative to the vacuum, i.e., we
on we use1En"
−(T the
−t) Eenergy
−t En relative
differences
to.the
vacuum, i.e., we
on we use=En to denote
m
!
!1 |m!
e
#m|
O
|n!
e
#n|
O
(1.18)
2
The
which
we
have
derived
here
will
be
central
the interpreof
∆E
.
This
implies
that
the
energy
E
of
thefor
vacuum
|0! is
useexpression
En instead
n
0
Z
T of
∆En . This implies that the energy E0 of the vacuum |0! is
use E instead
Tuesday, July 23, 13 n
m,n
Ground State Dominance
Wall-Smearing Method
CD
Aµ D
(t2 , t1 ; p , p) =
i
e
ip·x2 iq·x1
e
Tr[
µ
Su (0, x1 )
µ 5
Su (x1 , x2 )
5
Sc (x2 , 0)]
x2 ,x1
While point-to-all propagators Su (0, x1 ) and Sc (x2 , 0) can be easily obtained,
the computation of all-to-all propagator Su (x1 , x2 ) is a formidable task. One
common method is to use a sequential source composed of Su (0, x1 ) and Sc (x2 , 0)
for the Dirac matrix and invert it in order to compute Su (x1 , x2 ). However, this
method requires to fix either the inserted current or the sink momentum.
An approach that does not require to fix any of the above is the wall-smearing
method, where a summation over the spatial sites at the sink time point, x2 , is
made before the inversion. This corresponds to having a wall source or sink:
D Aµ D
CSW
(t2 , t1 ; 0, p) =
eiq·x1 Tr[
i
µ
Su (0, x1 )
µ 5
Su (x1 , x2 )
5
Sc (x2 , 0)]
x2 ,x2 ,x1
where the propagator (instead of the hadron state) is projected on to definite
momentum (S and W are smearing labels for shell and wall ). The wall method
has the advantage that one can first compute the shell and wall propagators and
then contract these to obtain the three-point correlator, avoiding any sequential
inversions.
Tuesday, July 23, 13