This is a unitary setup κsea=κvalence
P5P5 = 0.2358(36) χ2/dof= 883.81
V0P5 = 0.01329(25) χ2/dof= 8479.2
A0P5 = -0.00475(12) χ2/dof= 10.826
V0P5_insP = -0.313(22) χ2/dof= 5.3516
MPS= 167.3(2.1) MeV
M_{PS} = 0.07698(95) χ2/dof= 0.43318
Here we compute the variation of MPS respect to the bare mass m
DeltaM = -2.09(14) χ2/dof= 0.37252
DeltaMt = 18(41) χ2/dof= 0.33376
mpcac = -0.001079(78) χ2/dof= 0.79115
deltam = 0.0120(14) χ2/dof= 2.1906
deltam_sub = 0.00713(29) χ2/dof= 0.38428
Expanding in taylor series in m the correlator and observing that ∂m⟨O⟩=⟨OP5⟩ (there is a factor two in the correlator that in the code is called ⟨V0P5P5⟩ since insertion of P5 is done in only one quark line)
⟨V0P5⟩(mcr)=⟨V0P5⟩(m)+iδm⟨V0P5P5⟩(m) since ⟨V0P5⟩(mcr)=0 then
δm=⟨V0P5⟩i⟨V0P5P5⟩ using the fact that the numerator is purely immaginary and the denominator real we have δm=Im(⟨V0P5⟩)Re(⟨V0P5P5⟩)
form the relation κ=1/(2m+8) we have
κcr=κ(1+2δmκ). Since the correlator ⟨V0P5P5⟩ is not flat in time but linear when we are far from maximal twist, we can subtract the linear part ⟨V0P5⟩=B(e−Mt−e−M(T−t))∂m⟨V0P5⟩=i⟨V0P5P5⟩=(∂mB)(e−Mt−e−M(T−t))−(∂mM)B(te−Mt−(T−t)e−M(T−t)) with ∂mM computed in the section before
## Current $kappa=0.140064$
## $kappa_{cr}$= 0.139595(54)
## $kappa_{cr,sub}$= 0.139785(11)
This is not a unitary setup κsea=0.140064
MPS= 143.3(1.6) MeV
M_{PS} = 0.06596(72) χ2/dof= 0.25425
DeltaM = 1.07(13) χ2/dof= 0.23577
DeltaMt = 39(20) χ2/dof= 0.031808
mpcac = 0.000431(44) χ2/dof= 0.90198
deltam = -0.00225(18) χ2/dof= 0.55268
deltam_sub = -0.00212(15) χ2/dof= 0.47876
## Current $kappa=0.139779$
## $kappa_{cr}$= 0.1398671(71)
## $kappa_{cr,sub}$= 0.1398617(58)
This is not a unitary setup κsea=0.140064
M_{PS} = 0.06428(68) χ2/dof= 0.41313
MPS= 139.7(1.5) MeV
DeltaM = 0.05(11) χ2/dof= 0.25218
DeltaMt = 1.4(4.9) χ2/dof= 2.2876
mpcac = 2.7(53.4)e-6 χ2/dof= 0.74192
deltam = -0.00024(17) χ2/dof= 0.36804
deltam_sub = -0.00024(17) χ2/dof= 0.36914
## Current $kappa=0.1398563$
## $kappa_{cr}$= 0.1398656(65)
## $kappa_{cr,sub}$= 0.1398656(65)