3  B48

M_{PS}_5 = 0.05703(14) $\chi^2/dof=$ 0.13582

M_{PS}_4 = 0.09716(11) $\chi^2/dof=$ 0.093013

M_{PS}_3 = 0.12436(10) $\chi^2/dof=$ 0.12299

M_{PS}_2 = 0.14639(10) $\chi^2/dof=$ 0.15091

M_{PS}_1 = 0.16990(11) $\chi^2/dof=$ 0.45464

M_{K}_5 = 0.19982(13) $\chi^2/dof=$ 0.27299

M_{K}_4 = 0.20704(11) $\chi^2/dof=$ 0.17173

M_{K}_3 = 0.21407(11) $\chi^2/dof=$ 0.20041

M_{K}_2 = 0.22092(11) $\chi^2/dof=$ 0.20844

M_{K}_1 = 0.22923(11) $\chi^2/dof=$ 0.20696

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The correction to the pion correlator is $$C(t)= C_0(t) + e^2\delta_e C(t)$$

We define the discrete second derivative $\nabla_e^2f(e)=(f(e)-2f(0)+f(-e))/e^2$. due to the different quark charges we need to compute the derivative respect the two quark lines separately $$\delta_e C = \frac{1}{2}[e_u^2 \nabla^2_{e1} C + e_d^2 \nabla^2_{e2} C + e_u e_d(\nabla^2_{e} C - \nabla^2_{e1} C -\nabla^2_{e2} C) ]$$ with $\nabla_{e1}$ is the derivative respect only one leg. To compute the correction respect to the effective mass $M(e)=M_0+\delta M$ we use the formula $$\delta M(t)=\left( \frac{\delta C(t+1)}{C_0(t)} -\frac{\delta C(t)}{C_0(t) } \right)\frac{1}{[(T/2-t-1) \tanh(M_0(T/2-t-1))-(T/2-t) \tanh(M_0(T/2-t)) ]} \tag{3.1}$$

Delta_e_M_{PS}_5 = 0.010(11) $\chi^2/dof=$ 0.073818

Delta_e_M_{PS}_1 = -0.0328(33) $\chi^2/dof=$ 0.069971

Delta_e_M_{Kp}_5 = -0.0684(95) $\chi^2/dof=$ 0.13614

Delta_e_M_{Kp}_1 = -0.0397(26) $\chi^2/dof=$ 0.086328

Delta_e_M_{K0}_5 = -0.0246(24) $\chi^2/dof=$ 0.12349

Delta_e_M_{K0}_1 = -0.01733(80) $\chi^2/dof=$ 0.096857

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3.1 exchange only

Below we compute only $$\delta_{exc} C\frac{1}{2}e_u e_d(\nabla^2_{e} C - \nabla^2_{e1} C -\nabla^2_{e2} C)$$ and we plot $\delta_{exc} C/C$

Delta_e_exc_fit_M_{PS}_5 = 0.010207(48) -0.7670(12) $\chi^2/dof=$ 0.00068649

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using formula (Equation 3.1) using $M_0(t)$ and $M_0$ from the plateau fit we get

Delta_e_exc_M_{PS}_5 = 0.010207(47) $\chi^2/dof=$ 0.20755

Delta_e_exc_mefft_M_{PS}_5 = 0.010216(45) $\chi^2/dof=$ 0.29331

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3.2 mass

The correction to the pion correlator due to a change of mass $\mu_u= \mu_{iso}+\delta \mu_u$ and $\mu_d=\mu_{iso}+\delta \mu_d$ $$C(t)= C_0(t) + \delta\mu_u\delta_{\mu_u} C(t) +\delta\mu_d\delta_{\mu_d} C(t)$$ since $\delta_{\mu_u} C = \delta_{\mu_d} C \equiv\delta_{\mu} C$ we car rewrite $$C(t)= C_0(t) + 2(\mu_{ud} -\mu_{iso} ) \delta_{\mu} C(t)$$ where $\delta_\mu C$ is computed with the insertions

Delta_mu_u_M_{PS}_5 = -19.337(43) $\chi^2/dof=$ 0.40255

Delta_mu_u_M_{PS}_1 = -6.0712(28) $\chi^2/dof=$ 0.25454

Delta_mu_u_M_{Kp}_5 = -5.073(48) $\chi^2/dof=$ 0.64198

Delta_mu_u_M_{Kp}_1 = -4.5459(39) $\chi^2/dof=$ 0.19412

Delta_mu_u_M_{K0}_5 = -5.073(48) $\chi^2/dof=$ 0.64198

Delta_mu_u_M_{K0}_1 = -4.5459(39) $\chi^2/dof=$ 0.19412

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3.3 critical mass

The correction to the pion correlator due to a change of critical mass $m_0$ $$C(t)= C_0(t) + \delta m_0^u\delta_{m_0} C(t)+\delta m_0^d\delta_{m_0} C(t)$$ where $\delta_{m_0} C$ is computed with the insertions

Delta_m0_u_M_{PS}_5 = -0.059(91) $\chi^2/dof=$ 0.065068

Delta_m0_u_M_{PS}_1 = -0.329(25) $\chi^2/dof=$ 0.048132

Delta_m0_u_M_{Kp}_5 = -0.601(91) $\chi^2/dof=$ 0.14434

Delta_m0_u_M_{Kp}_1 = -0.392(23) $\chi^2/dof=$ 0.040707

Delta_m0_u_M_{K0}_5 = -0.601(91) $\chi^2/dof=$ 0.14434

Delta_m0_u_M_{K0}_1 = -0.392(23) $\chi^2/dof=$ 0.040707

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3.4 critical mass determination

We consider the correlator VP:

$$C^{VP}=\langle \, \left( \bar u_+ \gamma_0 d_- \right) (x) \, \left( \bar d_- \gamma_5 u_+ \right) (0) \, \rangle$$ expanding in the counterterms, after noticing that it does not depend on $\mu$, we get $$C^{VP} =C^{VP}_0 + e^2 \delta_e C^{VP}+2(m_0^{ud}-m_0^{iso})\delta_{m0}C^{VP}\,.$$ From the requirement $C^{VP}=0$ we compute $$m_0^{ud}-m_0^{iso} = - e^2 \frac{\delta_e C^{VP}}{2\delta_{m0}C^{VP}}$$

Delta_e_VP_5 = 0.05997(32) $\chi^2/dof=$ -nan

Delta_e_VP_1 = 0.026551(37) $\chi^2/dof=$ inf

Delta_m0u_VP_5 = -0.4323(23) $\chi^2/dof=$ inf

Delta_m0u_VP_1 = -0.19595(26) $\chi^2/dof=$ inf

dm0_cr_nabla_5 = 0.0064760(21) $\chi^2/dof=$ 0.9496

dm0_cr_nabla_1 = 0.0064672(25) $\chi^2/dof=$ 0.85433

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3.5 Approximated System

Here we want to solve the system taking the lattice spacing from ISOQCD: $$\begin{pmatrix} \delta_{\mu_u}M_{\pi^+} & \delta_{\mu_d}M_{\pi^+} &0\\ \delta_{\mu_u}M_{K^+} & 0 &\delta_{\mu_s}M_{K^+} \\ 0 & \delta_{\mu_d}M_{K^0} &\delta_{\mu_s}M_{K^0} \\ \end{pmatrix} \begin{pmatrix} \delta{\mu_u}\\ \delta{\mu_d}\\ \delta{\mu_s} \end{pmatrix} = \begin{pmatrix} M_{\pi^+}^{exp}-M_{\pi^+}^{iso}-e^2\delta_e M_{\pi^+}\\ M_{K^+}^{exp}-M_{K}^{iso}-e^2\delta_e M_{K^+}\\ M_{K^0}^{exp}-M_{K}^{iso}-e^2\delta_e M_{K^0}\\ \end{pmatrix} - \begin{pmatrix} \delta_{m_u}M_{\pi^+} & \delta_{m_d}M_{\pi^+} &0\\ \delta_{m_u}M_{K^+} & 0 &\delta_{m_s}M_{K^+} \\ 0 & \delta_{m_d}M_{K^0} &\delta_{m_s}M_{K^0} \\ \end{pmatrix} \begin{pmatrix} \delta{m_u}\\ \delta{m_d}\\ \delta{m_s} \end{pmatrix}$$

3.6 Fit from large $M_\pi$

$\chi^2/dof=$ 37.3647

P	value
P[0]	-0.000196(31)
P[1]	-0.38162(46)

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