μ - Indico
Transkript
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