22.1 Gaussian Smearing

Following (Montvay and Urbach 2012) we can define the smeared field $$S(t,x)=\sum_{y_i:|x_i,y_i|<R} \phi(t,y) e^{-\rho \sum_i |x_i,y_i|^2 }$$

with $|x,y|=\min\{ |x-y|, |x-y+L|, |x-y-L|\}$. The projected smeared field to zero momentum $$\tilde S(t,P=0)=\sum_{x_i}\sum_{y_i:|x_i,y_i|<R} \phi(t,y) e^{-\rho \sum_i |x_i,y_i| }=\sum_{\vec x}\sum_{|\vec\mu|<R} \phi(t,\vec x+\vec\mu) e^{-\rho | \vec \mu| }\\ = \sum_{|\vec\mu|<R}\sum_{\vec x} \phi(t, \vec x+\vec\mu) e^{-\rho | \vec \mu| } = \sum_{|\vec\mu|<R} \tilde\phi(t, p=0) e^{-\rho | \vec \mu| }\\ =\tilde\phi(t, p=0) \sum_{|\vec\mu|<R} e^{-\rho | \vec \mu| }$$ where we introduce the vector $\vec\mu=| x , y |$ and we use the fact that $\sum_{\vec x} \phi(t,\vec x +\vec \mu)=\sum_{\vec x} \phi(t,\vec x )$.

Thus we have that correlation function of the smeared field at zero momentum are proportional to the correlation function of the non-smeared field

$$\langle \tilde S(t,P=0)\tilde S(0,P=0)\rangle =N \langle \tilde \phi(t,P=0)\tilde \phi(0,P=0)\rangle$$ Below we compare the effective mass of the correlators

• $\langle \tilde \phi(t,P=0)\tilde \phi(0,P=0)\rangle$ with label E1_0
• $\langle \tilde S(t,P=0)\tilde S(0,P=0)\rangle$ with label sE1_0

in a L4T16 lattice

22.1.0.1 ../../g0.025/out/G2t_T64_L28_msq0-4.900000_msq1-4.650000_l02.500000_l12.500000_mu5.000000_g0.025000_rep0_output

E1_0(L28T64) = 0.127807(50) $\chi^2/dof=$ 0.81852

sE1_0(L28T64) = 0.127807(50) $\chi^2/dof=$ 0.81852

E1_0_p1(L28T64) = 0.257235(17) $\chi^2/dof=$ 7.2542

sE1_0_p1(L28T64) = 0.257235(17) $\chi^2/dof=$ 7.2542

22.1.0.2 ../../momentum/out/G2t_T128_L24_msq0-4.900000_msq1-4.650000_l02.500000_l12.500000_mu5.000000_g0.000000_rep0_output

GEVP_phi0_phi03_phi1_p1_meffl0(L24T128) = 0.290220(31) $\chi^2/dof=$ 1.6069

GEVP_phi0_phi03_phi1_p1_meffl1(L24T128) = 0.515128(60) $\chi^2/dof=$ 0.61451

GEVP_phi0_phi03_phi1_p1_meffl2(L24T128) = 0.56689(19) $\chi^2/dof=$ 1.1024

E1_0_p1(L24T128) = 0.290237(13) $\chi^2/dof=$ 4.9672

E1_1_px(L24T128) = 0.515041(44) $\chi^2/dof=$ 0.8219

sE1_0_p1(L24T128) = 0.290237(13) $\chi^2/dof=$ 4.9672

22.1.0.3 ../../momentum/out/G2t_T128_L24_msq0-4.900000_msq1-4.650000_l02.500000_l12.500000_mu5.000000_g0.000000_rep0_output

E2_0(L24T128) = 0.264920(85) $\chi^2/dof=$ 4.6562

sE2_0(L24T128) = 0.264911(85) $\chi^2/dof=$ 3.7711

E2_0_p0(L24T128) = 1.3060(19) $\chi^2/dof=$ 1572.8

References

Montvay, I., and C. Urbach. 2012. “Exploratory Investigation of Nucleon-Nucleon Interactions Using Euclidean Monte Carlo Simulations.” The European Physical Journal A 48 (3). https://doi.org/10.1140/epja/i2012-12038-1.