Etmc2013 Talk

Baryon spectrum from Nf =2+1+1

Christos Kallidonis 4 April 2013

ETM Collaboration Meeting Nicosia, 3-6 April 2013

CaSToRC

OUTLINE o Simulation details o Overview of calculations o Interpolating elds o Results

o Chiral extrapolations xing the lattice spacing o Lattice artifacts Isospin symmetry breaking o Comparison between mixed action and unitary setup o Comparison with other collaborations

o Conclusions Summary

Mass spectrum from Nf = 2+1+1 4 April 2013

Mass spectrum from Nf = 2+1+1 4 April 2013 3

= 1.90, a = 0.0929(14) fm r0/a = 5.231(38)

203 48, L = 1.9 fma 0.0040No. of Confs 308m (GeV) 0.3166mL 2.99

243 48, L = 2.2 fma 0.0040 0.0060 0.0080 0.0100No. of Confs 523 479 449 501m (GeV) 0.3074 0.3664 0.4216 0.4729mL 3.48 4.15 4.77 5.35

323 64, L = 3.0 fma 0.0030 0.0040 0.0050No. of Confs 185 389 193m (GeV) 0.2629 0.3000 0.3351mL 3.97 4.53 5.05

= 1.95, a = 0.0815(11) fm, r0/a = 5.710(41)

243 48, L = 2.1 fma 0.0085No. of Confs 0.4689m (GeV) 0.4428mL 4.66

323 64, L = 2.6 fma 0.0025 0.0035 0.0055 0.0075No. of Confs 723 546 4644 521m (GeV) 0.2582 0.3047 0.3752 0.4357mL 3.42 4.03 4.97 5.77

= 2.10, a = 0.0641(8) fm r0/a = 7.538(58)

323 64, L = 2.1 fma 0.0045No. of Confs 952m (GeV) 0.3712mL 3.87

483 96, L = 3.1 fma 0.0015 0.002 0.003No. of Confs 606 186 113m (GeV) 0.2142 0.2471 0.3004mL 3.35 3.86 4.69

TABLE I. Input parameters (, L, ) of our lattice simulations and corresponding lattice spacing (a) and pion mass (m).Row Stat. refers to the number of gauge configurations used in the calculations. In row Abbr. we use an abbreviation for theensembles of the form Xl.L with X referring to the value used, in order to quickly refer to the ensembles later on. Thelattice spacing was determined using the nucleon mass in this work. NEEDS CORRECTION!!!

For consistency and cross check of the tuned masses, we have also performed two linear fits in constrained ranges ofah, one in the sector of the kaon mass and one in the sector of the D-meson mass, according to the expressions

a2M2PS(ah) = c1 + c2ah for the strange quark,

aMPS(ah) = d1 + d2ah for the charm quark. (11)

The linear fits are shown on the right panel of Fig. 1. We list the resulting polynomial fit parameters ai, the linearfit parameters ci and di as well as the tuned values of the strange and charm quark masses in Table II. As one cansee, the tuned values of the strange and charm quark masses are compatible among the polynomial and linear fits andthey are also in agreement with other similar calculations CITATION. Such agreements are satisfactory and showthat both fit procedures lead to the same tuned values within one standard deviation.

D. Interpolating fields

In order to extract the masses of the baryons in lattice QCD, we evaluate two-point functions. The baryon states arecreated from the vacuum with the use of interpolating fields which are constructed such that they have the quantumnumbers of the baryon in interest and reduce to the quark model wave functions in the non-relativistic limit. Thelow-lying baryons belonging to the octet and decuplet representation of SU(3) are given in Figs. 2 and 3 respectively.They are classified through the isospin, I, the third component of the isospin, Iz, the strangeness (s) and the electricalcharge (q). The corresponding interpolating fields for these baryons are collected in Tables III [15, 16] and IV [15, 17].

Overview of Calculations

8

As local interpolating fields are not optimal for suppressing excited state contributions, we apply Gaussian smearingto each quark field q(x, t) [19, 20]. The smeared quark field is given by qsmear(x, t) =

Py F (x,y;U(t))q(y, t), where

we have used the gauge invariant smearing function

F (x,y;U(t)) = (1 + H)n (x,y;U(t)), (12)

constructed from the hopping matrix understood as a matrix in coordinate, color and spin space,

H(x,y;U(t)) =3X

i=1

Ui(x, t)x,yai + U

i (x ai, t)x,y+ai

. (13)

In addition, we apply APE smearing to the spatial links that enter the hopping matrix. The parameters and nof the Gaussian and APE smearing at each value of are collected in Table IX.We need to explain why use these parameters?

= 1.95 = 2.1

APEn 20 50

0.5 0.5

Gaussiann 50 110

4 4

TABLE IX. Smearing parameters for the ensembles at = 1.95 and = 2.1.

E. Two-point correlators

The interpolating fields for the spin-3/2 baryons as defined in Tables IV,VI and VIII have overlap with their heavierspin-1/2 excitations as well. These overlaps can be removed with the incorporation of a spin-3/2 projector in thedefinitions of the interpolating fields

JX3/2 = P3/2JX . (14)

For non-zero momentum, P3/2 is defined by

P3/2 = 1

3 1

3p2( 6 pp + p 6 p) . (15)

In correspondance, the spin-1/2 interpolating field JX1/2 that has overlap only with the excited spin-1/2 state can

be obtained by acting with the spin-1/2 projector P1/2 = g P3/2 on JX . In this work we study the mass spectrum

of the baryons in the rest frame (~p = ~0) and therefore the last term of Eq. (15) vanishes. In order to extract masseswe consider two-point correlation functions defined by

CX(t, ~p = ~0) =1

4Tr (1 0)

Xxsink

hJX (xsink, tsink) JX (xsource, tsource)i, t = tsink tsource (16)

Space-time reflection symmetries of the action and the anti-periodic boundary conditions in the temporal directionfor the quark fields imply, for zero three-momentum correlators, that C+X(t) = CX(T t). Therefore in order todecrease errors we average correlators in the forward and backward direction and define

CX(t) = C+X(t) CX(T t) . (17)


Calculation of two-point correlation functions

Anti-periodic boundary conditions in t-direction imply

Average to gain statistics

Nucleon case average proton and neutron correlation functions

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

m = m0 4cm2

g216f2

m3

m = m0 4cm2

25

27

g216f2

m3

m = m0 4cm2

10

9

g216f2

m3

J++ = abcuTa ub

uc

J 0 = abcsTaC5ub

sc

J 0 = abcsTa ub

sc

J 0 =1p3abc

2sTaCub

sc +

sTaCsb

uc

C+X(t) = CX(T t)

Overview of Calculations Mass spectrum from Nf = 2+1+1 4 April 2013

Eective mass

9

In addition, the source location is chosen randomly on the whole lattice for each configuration, in order to decreasecorrelation among measurements. Then, masses can be extracted from eective mass calculations. The eective massis defined by

mXe(t) = log

CX(t)

CX(t+ 1)

= mX + log

1 +

P1i=1 cie

it

1 +P1

i=1 ciei(t+1)

!t!1mX (18)

where i = mi mX is the mass dierence of the excited state i with respect to the ground mass mX .All results in this work have been extracted from correlators where Gaussian smearing is applied both at the source

and sink. In general, eective masses from local correlators are expected to have the same value in the large timelimit, but smearing suppresses excited state contaminations, therefore yielding a plateau region at earlier source-sinktime separations and better accuracy in the extraction of the mass. Our fitting procedure to extract mX is as follows:The sum over excited states in the eective mass given in Eq. (18) is truncated, keeping only the first excited state.The upper time slice boundary is kept fixed, and, allowing a separation of a couple of time slices the eective massis fitted to the form given in Eq. (19). This yields an estimation for the parameters c1 and 1 = m1 mX . Thena constant fit is carried out, increasing the lower time slice boundary (hence decreasing the fitting range) until thecontribution to mX due to the first excited state is less than 50% of the statistical error of mX from the constant fit.This criterion is in most cases in agreement with 2/d.o.f. < 1. In the cases in which this criterion is not satisfied acareful examination of the eective mass is made to ensure that the fit range is in the plateau region.

mXe(t) mX + log

1 + c1e1t

1 + c1e1(t+1)

!t!1mX (19)

We show representative results of the eective masses of baryons considered in this work in Fig. 4. The error bandson the constant fits are obtained using jackknife analysis.

FIG. 4. Representative eective mass plots for = 1.95,al = 0.0055 where both the constant and the exponential fits aredisplayed.

9


mXe(t) = log

CX(t)

CX(t+ 1)

= mX + log

1 +

P1i=1 cie

it

1 +P1

i=1 ciei(t+1)

!t!1mX (18)



mXe(t) mX + log

1 + c1e1t

1 + c1e1(t+1)

!t!1mX (19)



9


mXe(t) = log

CX(t)

CX(t+ 1)

= mX + log

1 +

P1i=1 cie

it

1 +P1

i=1 ciei(t+1)

!t!1mX (18)



mXe(t) mX + log

1 + c1e1t

1 + c1e1(t+1)

!t!1mX (19)



9


mXe(t) = log

CX(t)

CX(t+ 1)

= mX + log

1 +

P1i=1 cie

it

1 +P1

i=1 ciei(t+1)

!t!1mX (18)



mXe(t) mX + log

1 + c1e1t

1 + c1e1(t+1)

!t!1mX (19)



6

Charm Strange BaryonQuark

Interpolating field I Izcontent

c = 0 s = 2?0 uss 1p

3abc

2sTaCub

sc +

sTaCsb

uc

1/2 +1/2

? dss 1p3abc

2sTaCdb

sc +

sTaCsb

dc

1/2 -1/2

c = 1 s = 1?+c usc

q23 abc

uTaCsb

cc +

sTaCcb

uc +

cTaCub

sc

1/2 +1/2

?0c dscq

23 abc

dTaCsb

cc +

sTaCcb

dc +

cTaCdb

sc

1/2 -1/2

s = 2 ?0c ssc1p3abc

2sTaCcb

sc +

sTaCsb

cc

0 0

c = 2 s = 0?++cc ucc

1p3abc

2cTaCub

cc +

cTaCcb

uc

1/2 +1/2

?+cc dcc1p3abc

2cTaCdb

cc +

cTaCcb

dc

1/2 -1/2

s = 1 ?+cc scc1p3abc

2cTaCsb

cc +

cTaCcb

sc

0 0

TABLE VI. Alternative interpolating fields for the spin-3/2 baryons.

= 1.95 = 2.1

APEn 20 50

0.5 0.5

Gaussiann 50 110

4 4

TABLE VII. Smearing parameters for the ensembles at = 1.95 and = 2.1.

mXe(t) = log

CX(t)

CX(t+ 1)

= mX + log

1 +

P1i=1 cie

it

1 +P1

i=1 ciei(t+1)

!t!1mX (17)



mXe(t) mX + log

1 + c1e1t

1 + c1e1(t+1)

!t!1mX (18)


mcX mX12 (m

cX +mX)

12mcXCriterion:

Interpolating elds Mass spectrum from Nf = 2+1+1 4 April 2013

Interpolating elds are constructed such that they have the quantum numbers of the baryon in interest

4 quark avours Baryons (qqq)

Outline Lattice QCD Charmed mesons Charmed baryons Summary

SU(4) representations+c

+c0c

udc

uds

usc

dsc

Flavour symmetry is not respected.

Simplest way to see which baryons should exist.

SU(4): 4 4 4 = 20 20 20 4!!! = !!! !!

!!!!

!!!

Gunnar Bali (Uni Regensburg) Charmed hadrons 22 / 30

Outline Lattice QCD Charmed mesons Charmed baryons Summary

SU(4) representations+c

+c0c

udc

uds

usc

dsc

Flavour symmetry is not respected.

Simplest way to see which baryons should exist.

SU(4): 4 4 4 = 20 20 20 4!!! = !!! !!

!!!!

!!!

Gunnar Bali (Uni Regensburg) Charmed hadrons 22 / 30

20 = 8 6 3 320 =10 6 31

SU(3) sub-groups of SU(4)

20plet of spin-1/2 baryons 20plet of spin-3/2 baryons

Interpolating elds Mass spectrum from Nf = 2+1+1 4 April 2013

Interpolating elds of spin-3/2 baryons, depending on structure, can have overlap with heavier spin-1/2 excited states spin-1/2 ground states

Overlaps are removed with the incorporation of projectors

Projector to 3/2 Projector to 1/2 8

JX1/2 = P1/2JX .

P1/2 = g P3/2

II. LATTICE RESULTS

A. Isospin breaking

The twisted mass action breaks isospin explicitly to O(a2) and the size of O(a2) terms determines how large thisbreaking would be. It is expected that the splitting is zero in the continuum limit. In general, isospin symmetrybreaking manifests itself as a mass splitting between baryons belonging to the same multiplets. Therefore, it would beinteresting to examine at first the degree of isospin splitting between baryons belonging to the same isospin multipletsdue to lattice artifacts. We begin this analysis by presenting the eective masses of the light baryons in the decupletrepresentation of SU(3), namely the ++, , + and 0 shown in Fig. 4.

FIG. 4. Eective mass of ++, (left) and +, 0 (right) for = 1.95,al = 0.0055 (circles) and = 2.1,al = 0.0015(squares).

As shown in Fig. 4, the masses of the baryons show consistency within errors, as one would expect, since thelight quarks u and d are in fact degenerate. Therefore, no isospin splitting eects are observed for the light baryons.However, the strange baryons of the octet representation of SU(3) is a case where some degree of isospin splitting

is clearly visible for = 1.95, whereas for = 2.1 the splitting is decreased as expected. To demonstrate this, weshow results on the masses for the strange particles and in Fig. 5. A more quantitative determination of thedegree of isospin breaking will follow.It is of interest to notice that the same behaviour is not observed for the corresponding spin-3/2 strange baryon

states and . Here, the splitting eect is not apparent within errors, neither for = 1.95, nor for = 2.1. Thisis shown in Fig. 6, where we plot the eective masses of the strange spin-3/2 baryons of the decuplet. NOTE: Ineed some help for commenting on thisWe continue our analysis by studying the isospin breaking on the charm baryons. To this end we concentrate on

the results obtained for the charm c baryons, containing light quarks and a charm quark as well as the charm cbaryons, which contain a light, a strange and a charm quark. As before, we plot results for = 1.95 and = 2.1 tocompare between dierent lattice spacings.As the left plot of Fig. 7 suggests, no isospin splitting among the three states of the charm c baryons is observed,

both for = 1.95 and = 2.1, which is in contrast with the corresponding strange states as we have discussedabove. On the other hand, this eect, although in a smaller extent, is still present on the c baryons for = 1.95 ascan be seen from the right plot of Fig. 7. As expected, one can see from Fig. 8 that neither of the three spin-3/2 cand the two c states show this eect for both = 1.95 and = 2.1. All these results could have led to a generalconclusion that the isospin splitting is a common cut-o eect between the strange spin-1/2 baryons belonging to thesame multiplets. What comes in contrast to this is the behaviour of the charm 0c resonance states. Despite havingthe same quark content as the c, Fig. 9 clearly shows that no splitting can be observed for these states, neither for = 1.95, nor for = 2.1.

3

FIG. 1. The low-lying baryons of the octet representationlabeled by the value of I3 and hyper-charge

FIG. 2. The low-lying baryons of the decuplet representationlabeled by the value of I3 and hyper-charge


H(x,y;U(t)) =3X

i=1

Ui(x, t)x,yai + U

i (x ai, t)x,y+ai

. (12)

In addition, we apply APE smearing to the spatial links that enter the hopping matrix. The parameters and nof the Gaussian and APE smearing at each value of are collected in Table VII.We need to explain why use these parameters?


The interpolating fields for the spin-3/2 baryons as defined in Tables II,IV and VI have overlap with their heavierspin-1/2 excitations as well. These overlaps can be removed with the incorporation of a spin-3/2 projector in thedefinitions of the interpolating fields

JX3/2 = P3/2JX . (13)


P3/2 = 1

3 1

3p2( 6 pp + p 6 p) . (14)




CX(t, ~p = ~0) =1

4Tr (1 0)

Xxsink



CX(t) = C+X(t) CX(T t) . (16)


3

FIG. 1. The low-lying baryons of the octet representationlabeled by the value of I3 and hyper-charge

FIG. 2. The low-lying baryons of the decuplet representationlabeled by the value of I3 and hyper-charge


H(x,y;U(t)) =3X

i=1

Ui(x, t)x,yai + U

i (x ai, t)x,y+ai

. (12)

In addition, we apply APE smearing to the spatial links that enter the hopping matrix. The parameters and nof the Gaussian and APE smearing at each value of are collected in Table VII.We need to explain why use these parameters?


The interpolating fields for the spin-3/2 baryons as defined in Tables II,IV and VI have overlap with their heavierspin-1/2 excitations as well. These overlaps can be removed with the incorporation of a spin-3/2 projector in thedefinitions of the interpolating fields

JX3/2 = P3/2JX . (13)


P3/2 = 1

3 1

3p2( 6 pp + p 6 p) . (14)




CX(t, ~p = ~0) =1

4Tr (1 0)

Xxsink



CX(t) = C+X(t) CX(T t) . (16)


8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

II. LATTICE RESULTS

A. Isospin breaking







both for = 1.95 and = 2.1, which is in contrast with the corresponding strange states as we have discussedabove. On the other hand, this eect, although in a smaller extent, is still present on the c baryons for = 1.95 ascan be seen from the right plot of Fig. 7. As expected, one can see from Fig. 8 that neither of the three spin-3/2 cand the two c states show this eect for both = 1.95 and = 2.1. All these results could have led to a generalconclusion that the isospin splitting is a common cut-o eect between the strange spin-1/2 baryons belonging to thesame multiplets. What comes in contrast to this is the behaviour of the charm 0c resonance states. Despite having

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

II. LATTICE RESULTS

A. Isospin breaking







both for = 1.95 and = 2.1, which is in contrast with the corresponding strange states as we have discussedabove. On the other hand, this eect, although in a smaller extent, is still present on the c baryons for = 1.95 ascan be seen from the right plot of Fig. 7. As expected, one can see from Fig. 8 that neither of the three spin-3/2 c

2-point function for spin-3/2 baryons

Projecting to 3/2 and 1/2 for the eective mass of ++


Interpolating elds

1/2 projection gives a higher mass as expected

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

m = m0 4cm2

g216f2

m3

m = m0 4cm2

25

27

g216f2

m3

m = m0 4cm2

10

9

g216f2

m3

J++ = abcuTa ub

uc

II. LATTICE RESULTS

A. Isospin breaking

The twisted mass action breaks isospin explicitly to O(a2) and the size of O(a2) terms determines how large thisbreaking would be. It is expected that the splitting is zero in the continuum limit. In general, isospin symmetrybreaking manifests itself as a mass splitting between baryons belonging to the same multiplets. Therefore, it would beinteresting to examine at first the degree of isospin splitting between baryons belonging to the same isospin multipletsdue to lattice artifacts. We begin this analysis by presenting the eective masses of the light baryons in the decupletrepresentation of SU(3), namely the ++, , + and 0 shown in Fig. 4.As shown in Fig. 4, the masses of the baryons show consistency within errors, as one would expect, since the

light quarks u and d are in fact degenerate. Therefore, no isospin splitting eects are observed for the light baryons.However, the strange baryons of the octet representation of SU(3) is a case where some degree of isospin splitting


states and . Here, the splitting eect is not apparent within errors, neither for = 1.95, nor for = 2.1. Thisis shown in Fig. 6, where we plot the eective masses of the strange spin-3/2 baryons of the decuplet. NOTE: Ineed some help for commenting on this

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

C 12(t) =

1

3Tr[C] 1

3

3Xi 6=j

ijCij

II. LATTICE RESULTS

A. Isospin breaking





states and . Here, the splitting eect is not apparent within errors, neither for = 1.95, nor for = 2.1. This

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

C 12(t) =

1

3Tr[C] 1

3

3Xi 6=j

ijCij

II. LATTICE RESULTS

A. Isospin breaking






D15.48


Interpolating elds

eective mass of ++ is hardly aected by including the 3/2-projector

the overlap of this interpolating eld with the excited spin-1/2 state is small

Projecting to 3/2 and 1/2 for the eective mass of ++

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

C 12(t) =

1

3Tr[C] 1

3

3Xi 6=j

ijCij

II. LATTICE RESULTS

A. Isospin breaking






8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

II. LATTICE RESULTS

A. Isospin breaking






55.32, 200 statistics

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

m = m0 4cm2

g216f2

m3

m = m0 4cm2

25

27

g216f2

m3

m = m0 4cm2

10

9

g216f2

m3

J++ = abcuTa ub

uc

II. LATTICE RESULTS

A. Isospin breaking


light quarks u and d are in fact degenerate. Therefore, no isospin splitting eects are observed for the light baryons.However, the strange baryons of the octet representation of SU(3) is a case where some degree of isospin splitting




Interpolating elds Projecting to 3/2 and 1/2 for the eective mass of *0

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

m = m0 4cm2

g216f2

m3

m = m0 4cm2

25

27

g216f2

m3

m = m0 4cm2

10

9

g216f2

m3

J++ = abcuTa ub

uc

J 0 = abcsTa ub

sc

J 0 =1p3abc

2sTaCub

sc +

sTaCsb

uc

II. LATTICE RESULTS

A. Isospin breaking


light quarks u and d are in fact degenerate. Therefore, no isospin splitting eects are observed for the light baryons.

55.32, 200 statistics

eective mass of *0 is systematically higher when using the 3/2-projector for interpolating eld

this interpolating eld has overlap with spin-1/2 state

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

m = m0 4cm2

g216f2

m3

m = m0 4cm2

25

27

g216f2

m3

m = m0 4cm2

10

9

g216f2

m3

J++ = abcuTa ub

uc

J 0 = abcsTa ub

sc

J 0 =1p3abc

2sTaCub

sc +

sTaCsb

uc

II. LATTICE RESULTS

A. Isospin breaking



8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

C 12(t) =

1

3Tr[C] 1

3

3Xi 6=j

ijCij

II. LATTICE RESULTS

A. Isospin breaking






8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

II. LATTICE RESULTS

A. Isospin breaking






8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

C 12(t) =

1

3Tr[C] 1

3

3Xi 6=j

ijCij

II. LATTICE RESULTS

A. Isospin breaking







Interpolating elds Projecting to 3/2 and 1/2 for the eective mass of *0

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

m = m0 4cm2

g216f2

m3

m = m0 4cm2

25

27

g216f2

m3

m = m0 4cm2

10

9

g216f2

m3

J++ = abcuTa ub

uc

J 0 = abcsTa ub

sc

J 0 =1p3abc

2sTaCub

sc +

sTaCsb

uc

II. LATTICE RESULTS

A. Isospin breaking



Introduce

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

C 12(t) =

1

3Tr[C] 1

3

3Xi 6=j

ijCij

II. LATTICE RESULTS

A. Isospin breaking






8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ijC 3

2(t) +

1

3ijC 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

C 12(t) =

1

3Tr[C] 1

3

3Xi 6=j

ijCij

II. LATTICE RESULTS

A. Isospin breaking






consistent results on the mass of *0 (projection to 3/2) in analogy with the case, the 1/2 projection of this eld gives an

excited spin-1/2 state

8

JX1/2 = P1/2JX .

P1/2 = P3/2 =

1

3

Cij(t) =

ij 1

3ij

C 3

2(t) +

1

3ij

C 1

2(t)

C 32(t) =

1

3Tr[C] +

1

6

3Xi 6=j

ijCij

Cw/p(t) =1

3Tr[C]

m = m0 4cm2

g216f2

m3

m = m0 4cm2

25

27

g216f2

m3

m = m0 4cm2

10

9

g216f2

m3

J++ = abcuTa ub

uc

J 0 = abcsTa ub

sc

J 0 =1p3abc

2sTaCub

sc +

sTaCsb

uc

II. LATTICE RESULTS

A. Isospin breaking



D15.48

Chiral extrapolations Nucleon

17

one-loop order of the pion mass dependence for the nucleon mass in HBPT is given by

mN = m0N 4c1m2

3g2A32f2

m3 (20)

where m0N is the nucleon mass at the chiral limit and together with c1 are treated as fit parameters. This lowest orderresult for the nucleon in HBPT was first derived in Ref. [22] and successfully fitted lattice data are also discussed inprevious studies [23, 24]. This result is the same in HBPT with dimensional and infra-red regularization as well aswhen the degree of freedom is explicitly included. It is also the same in manifestly Lorentz-invariant formulationwith infra-red regularization. Therefore we will use this result to fix the lattice spacing using the nucleon mass asinput. This can also provide a non-trivial check of our lattice formulation. The lattice spacings a=1.90, a=1.95 anda=2.10 are considered as additional independent fit parameters in a combined fit of our data at = 1.90, = 1.95and = 2.10. We constrain our fit so that the fitted curve passes through the physical point by fixing the value of c1,hence the lattice spacings are determined using the nucleon mass at the physical point as the only input. The physicalvalues of f and gA are used in the fits, namely f = 0.092419(7)(25) and gA = 1.2695(29), which is common practicein chiral fits to lattice data on the nucleon mass [2527]. The left plot of Fig. 18 shows the fit to the O(p3) result ofEq. (20) on the nucleon mass. The error band and the errors on the fit parameters are obtained from super-jackknifeanalysis. As can be seen, the O(p3) result provides a very good fit to our lattice data, which in fact confirms thatcut-o and finite volume eects are small for the -values used. The results on the fit parameter m0N and the latticespacings a=1.90, a=1.95 and a=2.10 are collected in Table XI. The physical spatial lengths of the lattices used inthis calculation are obtained from the resulting lattice spacings of the fit and their values are labeled in Fig. ??.We perform the same analysis with higher order chiral corrections to Eq. (20). These O(p4) corrections are known

within several expansion schemes. In the so called small scale expansion (SSE) scheme [27], the -degress of freedomare explicitly included in covariant baryon PT by introducing as an additional counting parameter the -nucleonmass splitting, m mN , taking O(/mN ) O(m/mN ). In SSE the nucleon mass is given by

mN = m0N 4c1m2

3g2A32f2

m3 4E1()m4 3g2A + 3c

2A

642f2m

0N

m4 3g2A + 10c

2A

322f2m

0N

m4 logm

c

2A

32f2

1 +

2m0N

4m2 +

3 3

2m2

log

m2

+2 m2

R (m)

(21)

where R (m) = pm2 2 cos1

m

for m > and R (m) =

p2 m2 log

m

+q2

m2 1

for m < .

We take the cut-o scale = 1 GeV, c1 = 1.127 [27] and treat the counter-term E1 as an additional fit parameter.As in the O(p3) case we use the physical values of gA and f. The corresponding plot is shown on the right panel ofFig. 18. The error band as well as the errors on the fit parameters are obtained using super-jackknife analysis. Onecan see that this formulation provides a good description of the lattice data as well and yields values of the latticespacings and m0N which are consistent with those obtained in O(p3) of HBPT. The resulting parameters of this fitare given in Table XI.

FIG. 18. The nucleon mass as a function of m2 for = 1.90, = 1.95 and = 2.10 in O(p3) HBPT (left) and in O(p4) ofSSE scheme (right). The physical nucleon mass is shown with the black asterisk.

17

one-loop order of the pion mass dependence for the nucleon mass in HBPT is given by

mN = m0N 4c1m2

3g2A32f2

m3 (20)

where m0N is the nucleon mass at the chiral limit and together with c1 are treated as fit parameters. This lowest orderresult for the nucleon in HBPT was first derived in Ref. [22] and successfully fitted lattice data are also discussed inprevious studies [23, 24]. This result is the same in HBPT with dimensional and infra-red regularization as well aswhen the degree of freedom is explicitly included. It is also the same in manifestly Lorentz-invariant formulationwith infra-red regularization. Therefore we will use this result to fix the lattice spacing using the nucleon mass asinput. This can also provide a non-trivial check of our lattice formulation. The lattice spacings a=1.90, a=1.95 anda=2.10 are considered as additional independent fit parameters in a combined fit of our data at = 1.90, = 1.95and = 2.10. We constrain our fit so that the fitted curve passes through the physical point by fixing the value of c1,hence the lattice spacings are determined using the nucleon mass at the physical point as the only input. The physicalvalues of f and gA are used in the fits, namely f = 0.092419(7)(25) and gA = 1.2695(29), which is common practicein chiral fits to lattice data on the nucleon mass [2527]. The left plot of Fig. 18 shows the fit to the O(p3) result ofEq. (20) on the nucleon mass. The error band and the errors on the fit parameters are obtained from super-jackknifeanalysis. As can be seen, the O(p3) result provides a very good fit to our lattice data, which in fact confirms thatcut-o and finite volume eects are small for the -values used. The results on the fit parameter m0N and the latticespacings a=1.90, a=1.95 and a=2.10 are collected in Table XI. The physical spatial lengths of the lattices used inthis calculation are obtained from the resulting lattice spacings of the fit and their values are labeled in Fig. ??.We perform the same analysis with higher order chiral corrections to Eq. (20). These O(p4) corrections are known

within several expansion schemes. In the so called small scale expansion (SSE) scheme [27], the -degress of freedomare explicitly included in covariant baryon PT by introducing as an additional counting parameter the -nucleonmass splitting, m mN , taking O(/mN ) O(m/mN ). In SSE the nucleon mass is given by

mN = m0N 4c1m2

3g2A32f2

m3 4E1()m4 3g2A + 3c

2A

642f2m

0N

m4 3g2A + 10c

2A

322f2m

0N

m4 logm

c

2A

32f2

1 +

2m0N

4m2 +

3 3

2m2

log

m2

+2 m2

R (m)

(21)

where R (m) = pm2 2 cos1

m

for m > and R (m) =

p2 m2 log

m

+q2

m2 1

for m < .

We take the cut-o scale = 1 GeV, c1 = 1.127 [27] and treat the counter-term E1 as an additional fit parameter.As in the O(p3) case we use the physical values of gA and f. The corresponding plot is shown on the right panel ofFig. 18. The error band as well as the errors on the fit parameters are obtained using super-jackknife analysis. Onecan see that this formulation provides a good description of the lattice data as well and yields values of the latticespacings and m0N which are consistent with those obtained in O(p3) of HBPT. The resulting parameters of this fitare given in Table XI.

FIG. 18. The nucleon mass as a function of m2 for = 1.90, = 1.95 and = 2.10 in O(p3) HBPT (left) and in O(p4) ofSSE scheme (right). The physical nucleon mass is shown with the black asterisk.

O(p3)

O(p4 )

HBPT (leading order)

SSE Scheme


Omega O(p2 ) HBPT (leading order)

18

a=1.90 a=1.95 a=2.10 m0N E1()

2/d.o.f

O(p3) HBPT 0.0934(14) 0.0819(11) 0.0645(8) 0.8681(15) 1.4314O(p4) SSE 0.0972(19) 0.0858(17) 0.0672(12) 0.8826(48) -2.5849(2530) 0.9303

TABLE XI. Fit parameters a=1.90, a=1.95, a=2.10 in fm, m0N in GeV and E1() in GeV

3 from O(p3) HBPT and O(p4)SSE using the nucleon mass.

B. mass

Having determined that the tuning of the bare strange quark mass in the mixed action approach reproduces themass of the in the unitary setup, we perform a similar analysis as for the nucleon mass to fix the lattice spacingusing the mass as input. This could provide a cross-check of the results when using the nucleon mass and alsoconfirm that the tuning of the strange quark mass is optimal. In the case of the , the leading one-loop order HBPTresult reduces to the simple form

m = m0 4c1m2 (22)

where m0 is the mass at the chiral limit and as in the nucleon mass case, the value of c1 is kept fixed such as thefitted curve passes through the physical point. In Fig. 19 we show the fit on the result of Eq. (22) as a function of m2.In addition, we provide the resulting parameters of this fit in Table XII. The physical spatial lengths of the latticesused are obtained from the resulting lattice spacings of the fit and their values are shown in Fig. 19. As one can seewhen comparing the results from the nucleon and the , the lattice spacings a=1.95 and a=2.10 show consistency,while for = 1.90 the value is larger in the case and its outside error bars, when compared to the value extractedfrom the nucleon. To this end, we conclude that fixing the lattice spacing using the mass is not as reliable as usingthe nucleon, since the case encompasses assumptions which in principle may not be accurate. Such assumptionsinclude deviations of the physical kaon mass from the simulated kaon mass, which is used as input to perform thestrange quark tuning.

FIG. 19. The mass as a function of m2 for = 1.90, = 1.95 and = 2.10 in SU(2) HBPT. The physical mass is shownwith the black asterisk.

4 April 2013 Mass spectrum from Nf = 2+1+1 4 April 2013

Chiral extrapolations

Chiral extrapolations - Results Nucleon results

analysis for Lm > 3.5 to check nite volume eects

Omega results


16

a=1.90 a=1.95 a=2.10 m0N E1()

2/d.o.f

O(p3) HBPT 0.0929(14) 0.0815(11) 0.0641(8) 0.8681(15) 1.4314O(p4) SSE 0.0967(19) 0.0854(17) 0.0668(12) 0.8826(48) -2.5849(2530) 0.9303

TABLE VIII. Fit parameters a=1.90, a=1.95, a=2.10 in fm, m0N in GeV and E1() in GeV


a=1.90 a=1.95 a=2.10 m0N E1()

2/d.o.f

O(p3) HBPT 0.0916(17) 0.0805(14) 0.0631(11) 0.8667(15) 1.8290O(p4) SSE 0.0978(27) 0.0866(26) 0.0678(19) 0.8859(72) -2.7476(3732) 1.2580

TABLE IX. Fit parameters a=1.90, a=1.95, a=2.10 in fm, m0N in GeV and E1() in GeV


when comparing the results from the nucleon and the , the lattice spacings a=1.95 and a=2.10 show consistency,while for = 1.90 the value is larger in the case and its outside error bars, when compared to the value extractedfrom the nucleon. To this end, we conclude that fixing the lattice spacing using the mass is not as reliable as usingthe nucleon, since the case encompasses assumptions which in principle may not be accurate. Such assumptionsinclude mixed action approach and unitary setup deviations, as well as deviations of the physical kaon mass from thesimulated kaon mass, which is used to perform the strange quark tuning.


16

a=1.90 a=1.95 a=2.10 m0N E1()

2/d.o.f

O(p3) HBPT 0.0929(14) 0.0815(11) 0.0641(8) 0.8681(15) 1.4314O(p4) SSE 0.0967(19) 0.0854(17) 0.0668(12) 0.8826(48) -2.5849(2530) 0.9303





a=1.90 a=1.95 a=2.10 m0

2/d.o.f

SU(2) HBPT 0.1018(5) 0.0857(4) 0.0660(4) 1.6673(16) 2.0925

TABLE IX. Fit parameters a=1.90, a=1.95, a=2.10 in fm and m0 in GeV from SU(2) HBPT using the mass.

16

a=1.90 a=1.95 a=2.10 m0N E1()

2/d.o.f

O(p3) HBPT 0.0929(14) 0.0815(11) 0.0641(8) 0.8681(15) 1.4314O(p4) SSE 0.0967(19) 0.0854(17) 0.0668(12) 0.8826(48) -2.5849(2530) 0.9303





a=1.90 a=1.95 a=2.10 m0

2/d.o.f

SU(2) HBPT 0.1018(5) 0.0857(4) 0.0660(4) 1.6673(16) 2.0925

TABLE IX. Fit parameters a=1.90, a=1.95, a=2.10 in fm and m0 in GeV from SU(2) HBPT using the mass.

Isospin symmetry breaking Mass spectrum from Nf = 2+1+1 4 April 2013

Twisted mass action breaks isospin symmetry explicitly to Size of terms determine how large the breaking will be It is expected to be zero in the continuum limit Manifests itself as mass splitting between baryons belonging to the

same isospin multiplets due to lattice artifacts

Masses of light baryons calculated on the lattice are not expected to exert any degree of isospin symmetry breaking

O(a2 )O(a2 )

No isospin splitting for light baryons (u and d quarks are degenerate)

D15.48


baryons

B55.32

Strange baryons

10

II. LATTICE RESULTS

A. Isospin breaking




is clearly visible for = 1.95, whereas for = 2.1 the splitting is decreased as expected. To demonstrate this, weshow results on the masses for the strange particles and in Fig. 6. A more quantitative determination of thedegree of isospin breaking will follow.

FIG. 6. Left: Eective mass of +, 0 and for = 1.95,al = 0.0055 (circles) and = 2.1,al = 0.0015 (squares). Right:Eective mass of 0 and for = 1.95,al = 0.0055 (circles), = 1.95,al = 0.0035 (triangles) and = 2.1,al = 0.0015(squares).

It is of interest to notice that the same behaviour is not observed for the corresponding spin-3/2 strange baryonstates and . Here, the splitting eect is not apparent within errors, neither for = 1.95, nor for = 2.1. Thisis shown in Fig. 7, where we plot the eective masses of the strange spin-3/2 baryons of the decuplet. NOTE: Ineed some help for commenting on thisWe continue our analysis by studying the isospin breaking on the charm baryons. To this end we concentrate on

the results obtained for the charm c baryons, containing light quarks and a charm quark as well as the charm c

11

FIG. 7. Left: Eective mass of +, 0 and for = 1.95,al = 0.0055 (circles) and = 2.1,al = 0.0015 (squares). Right:Eective mass of 0 and for = 1.95,al = 0.0055 (circles), = 1.95,al = 0.0035 (triangles) and = 2.1,al = 0.0015(squares).

baryons, which contain a light, a strange and a charm quark. As before, we plot results for = 1.95 and = 2.1 tocompare between dierent lattice spacings.

FIG. 8. Left: Eective mass of ++c , +c and

0c for = 1.95,al = 0.0055 (circles) and = 2.1,al = 0.0015 (squares). Right:

Eective mass of +c and 0c for = 1.95,al = 0.0055 (circles), = 1.95,al = 0.0035 (triangles) and = 2.1,al = 0.0015

(squares).

As the left plot of Fig. 8 suggests, no isospin splitting among the three states of the charm c baryons is observed,both for = 1.95 and = 2.1, which is in contrast with the corresponding strange states as we have discussedabove. On the other hand, this eect, although in a smaller extent, is still present on the c baryons for = 1.95 ascan be seen from the right plot of Fig. 8. As expected, one can see from Fig. 9 that neither of the three spin-3/2 cand the two c states show this eect for both = 1.95 and = 2.1. All these results could have led to a generalconclusion that the isospin splitting is a common cut-o eect between the strange spin-1/2 baryons belonging to thesame multiplets. What comes in contrast to this is the behaviour of the charm 0c resonance states. Despite havingthe same quark content as the c, Fig. 10 clearly shows that no splitting can be observed for these states, neither for = 1.95, nor for = 2.1.NOTE: I need some help for the explanation of this.We further expand this analysis and focus on the double charm baryons ++cc and

+cc. Our results on the eective

masses of these baryons presented in Fig. 11 indicate that they dont show any degree of isospin splitting, as expected,since they dont contain a strange quark.A more quantitative method to determine the degree of isospin breaking where it is present, is to plot the mass

dierence of baryons belonging to the same multiplet as a function of a2 for all pion masses. Such plots are presentedin Fig. 12 for +, and 0, . For completeness and comparison purposes, we also plot for the spin-3/2 +, and 0, . As can be seen, the mass dierence is consistent with zero for the spin-3/2 and for alllattice spacings, while for and there are small non-zero values and the mass dierence is smaller for the smallerlattice spacing as expected. Therefore, the general conclusion is that isospin breaking is indeed small at the values oflattice spacing considered in this work and it vanishes at the continuum limit.At this stage, dependence of the degree of the isospin breaking on the pion mass can also be examined for the

Strange spin-1/2 states

splitting between baryons belonging

to same isospin multiplets

Circles: B55.32 Squares: D15.48

Circles: B55.32 Squares: D15.48

Strange spin-3/2 states

No isospin splitting


12


0c for = 1.95,al = 0.0055 (circles) and = 2.1,al = 0.0015 (squares).

Right: Eective mass of +c and 0c for = 1.95,al = 0.0055 (circles), = 1.95,al = 0.0035 (triangles) and = 2.1,al =

0.0015 (squares).

FIG. 10. Left: Eective mass of 0+c and 00c for = 2.1,al = 0.0015. Right: Eective mass of

0+c and

00c for = 1.95,al =

0.0055 (circles), = 1.95,al = 0.0035 (triangles) and = 1.95,al = 0.0025 (squares).

= 1.95 simulations, where we have three pion masses. As Fig. 12 suggests, the results for dierent pion masses atthe same lattice spacings are consistent. Therefore, we conclude that the isospin breaking does not depend on thepion mass within our statistical accuracy.

12


0c for = 1.95,al = 0.0055 (circles) and = 2.1,al = 0.0015 (squares).

Right: Eective mass of +c and 0c for = 1.95,al = 0.0055 (circles), = 1.95,al = 0.0035 (triangles) and = 2.1,al =

0.0015 (squares).

FIG. 10. Left: Eective mass of 0+c and 00c for = 2.1,al = 0.0015. Right: Eective mass of

0+c and

00c for = 1.95,al =

0.0055 (circles), = 1.95,al = 0.0035 (triangles) and = 1.95,al = 0.0025 (squares).

= 1.95 simulations, where we have three pion masses. As Fig. 12 suggests, the results for dierent pion masses atthe same lattice spacings are consistent. Therefore, we conclude that the isospin breaking does not depend on thepion mass within our statistical accuracy.

Mass spectrum from Lattice QCD 4 April 2013

Charm c resonance states

No isospin symmetry breaking for spin-1/2 c resonance states

D15.48

Circles: B55.32 Triangles: B35.32 Squares: B25.32

13

FIG. 11. Eective mass of ++cc , +cc (left) and

++cc ,

+cc (right) for = 1.95,al = 0.0055 (circles), = 1.95,al = 0.0035

(triangles) and = 2.1,al = 0.0015 (squares).

FIG. 12. Mass dierence of baryons belonging to the same multiplets versus a2 for m = 209 GeV (red), m = 243 GeV(green), m = 288 GeV (magenta) and m = 354 GeV (blue).

B. Eective masses from various interpolating fields

The second part of this work is devoted to check whether alternative interpolating fields for some of the strangeand charm baryons also used in other recent works, e.g. [18], lead to consistent results regarding the eective massesof these baryons. These interpolating fields are tabulated in Tables VII and VIII. We also check to see whether thetwo-point correlators and therefore the eective masses of the spin-3/2 baryon states, when projected to their spin-1/2component, are consistent with their corresponding spin-1/2 states. These additional contractions were performedfor the simulations = 1.95,al = 0.0035 and = 2.1,al = 0.0015. We begin by displaying results on the charm 0cbaryon, shown in Fig. 13. The abbreviation alt. is used on the plot labels to distinguish between the interpolatingfields from one another.

FIG. 13. Eective mass of 0c , 0c for = 1.95,al = 0.0035 (left) and = 2.1,al = 0.0015 (right).

13

FIG. 11. Eective mass of ++cc , +cc (left) and

++cc ,

+cc (right) for = 1.95,al = 0.0055 (circles), = 1.95,al = 0.0035

(triangles) and = 2.1,al = 0.0015 (squares).

FIG. 12. Mass dierence of baryons belonging to the same multiplets versus a2 for m = 209 GeV (red), m = 243 GeV(green), m = 288 GeV (magenta) and m = 354 GeV (blue).

B. Eective masses from various interpolating fields

The second part of this work is devoted to check whether alternative interpolating fields for some of the strangeand charm baryons also used in other recent works, e.g. [18], lead to consistent results regarding the eective massesof these baryons. These interpolating fields are tabulated in Tables VII and VIII. We also check to see whether thetwo-point correlators and therefore the eective masses of the spin-3/2 baryon states, when projected to their spin-1/2component, are consistent with their corresponding spin-1/2 states. These additional contractions were performedfor the simulations = 1.95,al = 0.0035 and = 2.1,al = 0.0015. We begin by displaying results on the charm 0cbaryon, shown in Fig. 13. The abbreviation alt. is used on the plot labels to distinguish between the interpolatingfields from one another.

FIG. 13. Eective mass of 0c , 0c for = 1.95,al = 0.0035 (left) and = 2.1,al = 0.0015 (right).

Charm cc baryons Mass spectrum from Nf = 2+1+1 4 April 2013

No isospin symmetry breaking for charm spin-1/2 AND spin-3/2 baryons

Circles: B55.32 Triangles: B35.32 Squares: D15.48

Circles: B55.32 Triangles: B35.32 Squares: D15.48

Mass spectrum from Lattice QCD 4 April 2013

Isospin splitting as a function of lattice spacing

Isospin splitting eects are small and reduce for smaller values of the lattice spacing for spin-1/2 baryons

consistent with zero for spin-3/2 baryons (as for the baryons)

15

compared to those of their corresponding spin-1/2 states dier either just in the structure (C instead of C5)or in the structure and their formation as well. The general conclusion to which all the above results lead is thatwhenever the interpolating fields of the spin-1/2 and the spin-3/2 baryons dier in the formation, the results on thespin-1/2 component of the spin-3/2 field are more noisy and give a higher value of the eective mass, when comparedto those of the spin-1/2 baryons field.

C. mass in unitary and mixed action setup

Consistency between the mass extracted from unitary setup and mixed action setup calculations has been exam-ined. For this purpose, in addition to the mixed action setup, we performed calculations for simulations with latticesize 32364 at = 1.90, a = 0.004 and = 1.95 at a = 0.0035, a = 0.0055 and a = 0.0075 in the unitary setup,to extract the eective mass for this approach. The rest of the details of the simulations are provided in Table I,while the numbers of gauge configurations for each ensemble used in the unitary setup are collected in Table X.

= 1.90 = 1.95

a 0.004 0.0035 0.0055 0.0075

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