On the critical end point in a two-flavor linear sigma model coupled to quarks

Abstract

We use the linear sigma model coupled to quarks to explore the location of the phase transition lines in the QCD phase diagram from the point of view of chiral symmetry restoration at high temperature and baryon chemical potential. We compute analytically the effective potential in the high- and low-temperature approximations up to sixth order, including the contribution of the ring diagrams to account for the plasma screening properties. We determine the model parameters, namely, the couplings and mass-parameter, from conditions valid at the first order phase transition at vanishing temperature and, using the Hagedorn limiting temperature concept applied to finite baryon density, for a critical baryochemical potential of order of the nucleon mass. We show that when using the set of parameters thus determined, the second order phase transition line (our proxy for the crossover transition) that starts at finite temperature and zero baryon chemical potential converges to the line of first order phase transitions that starts at zero temperature and finite baryon chemical potential to determine the critical end point to lie in the region $5.02<\mu _B^{\tiny {\text{ CEP }}}/T_c<5.18$, $0.14<T^{\tiny {\text{ CEP }}}/T_c<0.23$, where $T_c$ is the critical transition temperature at zero baryon chemical potential.

Appendices

Appendix A: Low temperature approximation

Here, we show the explicit computation of the effective potential at low temperature (Eq. (6)). We call $V_{\text {LT}}^{\text {b}}$ the boson piece and $V_{\text {LT}}^{\text {f}}$ the fermion piece. First we compute the boson piece. Let us start from the one-loop contribution

$$\begin{aligned} V_{\text {}}^{\text {b}}=T\sum _n\int \frac{d^3k}{(2\pi )^3} \ln D(\omega _n,\mathbf {k})^{1/2}, \end{aligned}$$

(A.1)

where

$$\begin{aligned} D(\omega _n,\mathbf {k})=\frac{1}{\omega _n^2+k^2+m_b^2}, \end{aligned}$$

(A.2)

is the Matsubara boson propagator with $m_b^2$ being the square of the boson mass and $\omega _n=2n\pi T$ the Matsubara frequencies for boson fields. Calculating the sums over Matsubara frequencies, we obtain

$$\begin{aligned} V_{\text {}}^{\text {b}}= & {} \frac{1}{2\pi ^2}\int _0^\infty dkk^2\left( \frac{\sqrt{k^2+m_b^2}}{2} \right. \nonumber \\&\left. + T \ln (1-e^{-\sqrt{k^2+m_b^2}/T})\right) \nonumber \\\equiv & {} V_{\text {}_{\text {vac}}}^{\text {b}}+ V_{\text {}_{\text {matt}}}^{\text {b}}, \end{aligned}$$

(A.3)

where we have separated the vacuum (vac) and matter (matt) contribution. First we calculate the vacuum contribution,

$$\begin{aligned} V_{\text {}_{\text {vac}}}^{\text {b}}=\frac{1}{2}\int \frac{d^3k}{(2 \pi )^3} \sqrt{k^2+m_b^2}, \end{aligned}$$

(A.4)

In order to carry out the calculation, we employ dimensional regularization. Using the well known expression

$$\begin{aligned} \int \frac{d^Dk}{(2\pi )^D} \frac{1}{(k^2-m_b^2)^n}=\text {i}(-1)^n\frac{(m^2)^{2-\epsilon -n}}{(4\pi )^{2-\epsilon }}\frac{\varGamma (n-2+\epsilon )}{\varGamma (n)}, \end{aligned}$$

(A.5)

with $D=d-2\epsilon$, this contribution can be written as

$$\begin{aligned} V_{\text {}_{\text {vac}}}^{\text {b}}=\frac{{\tilde{\mu }}^{3-d}}{2}\int \frac{d^dk}{(2\pi )^d} \sqrt{k^2+m_b^2}. \end{aligned}$$

(A.6)

In Eq. (A.6), we have explicitly $d=3$ and $n=-1/2$. Hence we have

$$\begin{aligned} V_{\text {}_{\text {vac}}}^{\text {b}}=-\frac{m_b^4}{32\pi ^2}\varGamma (\epsilon -2)\Big ( \frac{4\pi {\tilde{\mu }}^2}{m_b^2} \Big )^\epsilon . \end{aligned}$$

(A.7)

Taking the limit $\epsilon \rightarrow 0$, we finally obtain

$$\begin{aligned} V_{\text {}_{\text {vac}}}^{\text {b}}=-\frac{m_b^4}{64\pi ^2}\Big [ \ln \Big ( \frac{4\pi {\tilde{\mu }}^2}{m_b^2}\Big ) -\gamma _E+\frac{3}{2}+\frac{1}{\epsilon }\Big ]. \end{aligned}$$

(A.8)

We use the Minimal Subtraction scheme (MS). After fixing the renormalization scale to ${\tilde{\mu }}=a e^{-1/2}$, the final expression for the vacuum contribution is given by

$$\begin{aligned} V_{\text {}_{\text {vac}}}^{\text {b}}=-\frac{m_b^4}{64\pi ^2}\Big [ \ln \Big ( \frac{4\pi a^2}{m_b^2}\Big )-\gamma _E+\frac{1}{2}\Big ]. \end{aligned}$$

(A.9)

Now, we calculate the matter contribution

$$\begin{aligned} V_{\text {}_{\text {matt}}}^{\text {b}}=\frac{T}{2\pi ^2}\int _0^\infty dk \ln \left( 1-e^{-\sqrt{k^2+m_b^2}/T}\right) , \end{aligned}$$

(A.10)

when the temperature is low (LT) we can expand the logarithm in a Taylor series of the form

$$\begin{aligned} \ln \left( 1-e^{-\sqrt{k^2+m_b^2}/T}\right) =-\sum _{n=1}^{\infty }\frac{e^{-n\sqrt{k^2+m_b^2}/T}}{n} , \end{aligned}$$

(A.11)

using Eq. (A.11) into Eq. (A.10), we obtain

$$\begin{aligned} V_{\text {LT}_{\text {matt}}}^{\text {b}}=-\frac{1}{2\pi ^2}\int _0^\infty dk \sum _{n=1}^{\infty }\frac{T e^{-n\sqrt{k^2+m_b^2}/T}}{n}. \end{aligned}$$

(A.12)

We can make the change of variable $\omega =\sqrt{k^2+m_b^2}$ to get

$$\begin{aligned} V_{\text {LT}_{\text {matt}}}^{\text {b}}= & {} -\frac{1}{2\pi ^2}\int _{m_b}^\infty d\omega \sum _{n=1}^{\infty }\frac{T e^{-n \omega /T}}{n}\nonumber \\= & {} -\frac{1}{2\pi ^2} \sum _{n=1}^{\infty } \left( \frac{T m_b}{n} \right) ^2 \left( \frac{T}{m_b n}\right) ^{1/2} K_2 \left( \frac{m_b n }{T}\right) ,\nonumber \\ \end{aligned}$$

(A.13)

where $K_2$ is a Modified Bessel function of the second kind. Now, using the asymptotic expansion

$$\begin{aligned} \lim _{{z\rightarrow \infty }}K_{\alpha }(z)=\sqrt{\frac{\pi }{2 z}} e^{-z}, \end{aligned}$$

(A.14)

we obtain

$$\begin{aligned} V_{\text {LT}_{\text {matt}}}^{\text {b}}=-\frac{\sqrt{\pi }}{2\pi ^2\sqrt{2}}\sum _{n=1}^{\infty } \left( \frac{T m_b}{n} \right) ^2 \left( \frac{T}{m_b n}\right) ^{1/2} e^{-m_b n/T}. \end{aligned}$$

(A.15)

Using that

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{1}{n^{5/2}}e^{-m_b/T}=\text {Li}_{5/2}(e^{-m_b/T}), \end{aligned}$$

(A.16)

where $\text {Li}$ is the polylogarithm function, we get

$$\begin{aligned} V_{\text {LT}_{\text {matt}}}^{\text {b}}=- T \left( \frac{m_b T}{2\pi } \right) ^{3/2} \text {Li}_{5/2}(e^{-m_b/T}). \end{aligned}$$

(A.17)

Finally, we obtain for the boson case

$$\begin{aligned} V_{\text {LT}}^{\text {b}}= & {} -\frac{m_b^4}{64\pi ^2}\Big [ \ln \Big ( \frac{4\pi a^2}{m_b^2}\Big )-\gamma _E+\frac{1}{2}\Big ]. \nonumber \\- & {} T \left( \frac{m_b T}{2\pi } \right) ^{3/2} \text {Li}_{5/2}(e^{-m_b/T}). \end{aligned}$$

(A.18)

Now, we proceed to calculate the fermion case. The general expression for the one-loop correction at finite temperature for a fermion with mass $m_f$ is given by

$$\begin{aligned} V^{\text {f}}=-T\sum _n\int \frac{d^3k}{(2\pi )^3} \text {Tr}[\ln S({\tilde{\omega }}_n-i\mu _q,\mathbf {k})^{-1}], \end{aligned}$$

(A.19)

where

$$\begin{aligned} S({\tilde{\omega }}_n,\mathbf {k})=\frac{1}{\gamma _0 {\tilde{\omega }}_n+k\!\!/+m_f}, \end{aligned}$$

(A.20)

is the Matsubara fermion propagator, ${\tilde{\omega }}_n=(2n+1)\pi T$ are the Matsubara frequencies for fermion fields and $\mu _q$ is quark chemical potential. We calculate the sum over the Matsubara frequencies to obtain

$$\begin{aligned} V_{\text {}}^{\text {f}}= & {} -2 \int \frac{d^3k}{(2 \pi ^3)}\Bigg (\sqrt{k^2+m_f^2}\nonumber \\&+ T \ln \left( 1+e^{-(\sqrt{k^2+m_f^2}-\mu _q)/T}\right) \nonumber \\&+ T \ln \left( 1+e^{-(\sqrt{k^2+m_b^2}+\mu _q)/T}\right) \Bigg ) \nonumber \\\equiv & {} V_{\text {}_{\text {vac}}}^{\text {f}}+ V_{\text {}_{\text {mattI}}}^{\text {f}}+ V_{\text {}_{\text {mattII}}}^{\text {f}}, \end{aligned}$$

(A.21)

where we have separated the vacuum (vac) and the matter (mattI and mattII) parts. First we calculate the vacuum part. We notice that the fermion case differs from the boson case by an overall factor $-4$. Therefore if we multiply Eq. (A.9) by $-4$, we get the vacuum one-loop contribution from one fermion field

$$\begin{aligned} V^{\text {f}}_{\text {}_{\text {vac}}}=\frac{m_f^4}{16\pi ^2}\left[ \ln \left( \frac{4\pi a^2}{m_f^2}\right) -\gamma _E+\frac{1}{2}\right] . \end{aligned}$$

(A.22)

Now we calculate the matter contribution, which we separate into two parts

$$\begin{aligned} V_{\text {}_{\text {mattI}}}^{\text {f}} =-2 \int \frac{d^3k}{(2 \pi ^3)}T \ln \left( 1+e^{-(\sqrt{k^2+m_f^2}-\mu _q)/T}\right) , \end{aligned}$$

(A.23)

and

$$\begin{aligned} V_{\text {}_{\text {mattII}}}^{\text {f}} =-2 \int \frac{d^3k}{(2 \pi ^3)}T \ln \left( 1+e^{-(\sqrt{k^2+m_f^2}+\mu _q)/T}\right) . \end{aligned}$$

(A.24)

We have to analyse two cases. For the case at low temperature (LT), the first case is when $\sqrt{k^2+m_f^2}>\mu _q$. In this case $V_{\text {}_{\text {mattI}}}^{\text {f}}\rightarrow 0$ and $V_{\text {}_{\text {mattII}}}^{\text {f}}\rightarrow 0$. The second case is when $\sqrt{k^2+m_f^2}<\mu _q$. In this case $V_{\text {}_{\text {mattII}}}^{\text {f}}\rightarrow 0$, and

$$\begin{aligned} V_{\text {LT}_{\text {mattI}}}^{\text {f}} =2 \int \frac{d^3k}{(2 \pi ^3)} \left( \sqrt{k^2+m_f^2}-\mu _q\right) \theta \left( \mu _q-\sqrt{k^2+m_f^2}\right) , \end{aligned}$$

(A.25)

where $\theta (x)$ is the Heaviside step function. We can make the change of variable $\omega =\sqrt{k^2+m_b^2}$ to obtain

$$\begin{aligned} V_{\text {LT}_{\text {mattI}}}^{\text {f}}= & {} \frac{1}{\pi ^2} \int _{m_f}^{\mu _q} d\omega \omega \sqrt{\omega ^2-m_f^2}\nonumber \\= & {} \frac{1}{24\pi ^2} \biggl [ \mu _q (5 m_f^2-2\mu _q^2) \sqrt{\mu _q^2-m_f^2} \nonumber \\&+ 3 m_f^4 \ln \left( \frac{m_f}{\mu _c+\sqrt{\mu _c^2-m_f^2}}\right) \biggr ]. \end{aligned}$$

(A.26)

Finally, we obtain for the fermion case

$$\begin{aligned} V_{\text {LT}}^{\text {f}}= & {} \frac{m_f^4}{16\pi ^2}\left[ \ln \left( \frac{4\pi a^2}{\left( \mu _q+\sqrt{\mu _q^2-m_f^2}\right) ^2} \right) \right. \nonumber \\&-\left. \gamma _E+\frac{1}{2}\right] -\frac{\mu _f\sqrt{\mu _q^2-m_f^2}}{24\pi ^2}(2\mu _q^2-5m_q^2)\nonumber \\\equiv & {} V_0^{\text {f}}. \end{aligned}$$

(A.27)

To calculate the next order at low temperature, we make a Taylor expansion around $T=0$ [73]

$$\begin{aligned} V_{\text {LT}}^{\text {f}}&=V_0^{\text {f}}(\mu _q+xT)\Big |_{T=0}\nonumber \\&\quad +\frac{\pi ^2 T^2}{6}\frac{\partial ^2}{\partial (xT)^2}V_0^{\text {f}}(\mu _q+xT)\Big |_{T=0}\nonumber \\&\quad +\frac{\pi ^4 T^4}{360}\frac{\partial ^4}{\partial (xT)^4}V_0^{\text {f}}(\mu _q+xT)\Big |_{T=0} \nonumber \\&\quad +\frac{31\pi ^6 T^6}{15120}\frac{\partial ^4}{\partial (xT)^4}V_0^{\text {f}}(\mu _q+xT)\Big |_{T=0}+\cdots , \end{aligned}$$

(A.28)

Thus, we get

$$\begin{aligned} V_{\text {LT}}^{\text {f}}= & {} \frac{m_q^4}{16\pi ^2}\Biggl [\ln \left( \frac{4\pi a^2}{\left( \mu _q+\sqrt{\mu _q^2-m_q^2}\right) ^2}\,\, \right) \nonumber \\&-\left. \gamma _E+\frac{1}{2}\right] -\frac{\mu _q\sqrt{\mu _q^2-m_q^2}}{24\pi ^2}(2\mu _q^2-5m_q^2)\nonumber \\&-\frac{T^2\mu _q}{6}\sqrt{\mu _q^2-m_q^2}-\frac{7\pi ^2T^4\mu _q}{360}\frac{(2\mu _q^2-3m_q^2)}{(\mu _q^2-m_q^2)^{3/2}}\nonumber \\&+\frac{31\pi ^4 \mu _q m_q^4 T^6}{1008(\mu _q^2-M_q^2)^{7/2}}. \end{aligned}$$

(A.29)

Appendix B: High temperature approximation

Here, we show the explicit computation for the effective potential at high temperature, Eq. (7). It has two pieces, bosonic and fermionic contributions. The effective potential is computed beyond the mean field approximation, up to order ring diagrams, within the imaginary time formalism. The computation is performed in the same fashion as “Appendix A”. We begin with the boson piece, the one loop contribution is

$$\begin{aligned} V_{\text {}}^{\text {b}}=T\sum _n\int \frac{d^3k}{(2\pi )^3} \ln D(\omega _n,\mathbf {k})^{1/2}, \end{aligned}$$

(B.30)

where $D(\omega _n,\mathbf {k})$ is the boson propagator and $\omega _n$ the boson Matsubara frequencies. After the sum over n is performed, we get

$$\begin{aligned} V_{\text {}}^{\text {b}}= & {} \frac{1}{2\pi ^2}\int _0^\infty dkk^2\left( \frac{\sqrt{k^2+m_b^2}}{2} \right. \nonumber \\&\left. + T \ln \left( 1-e^{-\sqrt{k^2+m_b^2}/T}\right) \right) \nonumber \\\equiv & {} V_{\text {}_{\text {vac}}}^{\text {b}}+ V_{\text {}_{\text {matt}}}^{\text {b}}, \end{aligned}$$

(B.31)

where the one-loop contribution is written as the sum of two terms, the vacuum and matter pieces. The vacuum piece was computed in “Appendix A” with the result

$$\begin{aligned} V_{\text {}_{\text {vac}}}^{\text {b}}=-\frac{m_b^4}{64\pi ^2}\left[ \ln \left( \frac{4\pi a^2}{m_b^2}\right) -\gamma _E+\frac{1}{2}\right] . \end{aligned}$$

(B.32)

The matter contribution is now computed in the high temperature approximation, which means that T is the hard scale. Also, we consider $m_b/T \ll 1$. Then the matter contribution is

$$\begin{aligned} V_{\text {}_{\text {matt}}}^{\text {b}}=\frac{T}{2\pi ^2}\int _0^\infty dk \ln \left( 1-e^{-\sqrt{k^2+m_b^2}/T}\right) , \end{aligned}$$

(B.33)

and after expand Eq. (B.33) in terms of powers of $m_b/T$, the first five terms of the series are given by

$$\begin{aligned} V_{\text {HT}_{\text {matt}}}^{\text {b}}&= -\frac{m_b^4}{64\pi ^2}\ln \Bigg ( \frac{m_b^2}{(4\pi T)^2}\Bigg )-\frac{\pi ^2 T^4}{90}\nonumber \\&\quad +\frac{m_b^2 T^2}{24}-\frac{m_b^3 T}{12\pi }-\frac{\zeta (3)m_b^6}{96\pi ^4T^2}. \end{aligned}$$

(B.34)

Therefore, the boson one-loop contribution becomes

$$\begin{aligned} V_{\text {HT}}^{\text {b}}&=-\frac{m_b^4}{64\pi ^2}\Bigg [ \ln \Big ({\frac{a^2}{4\pi T^2}\Big )}-\gamma _E+\frac{1}{2}\Bigg ] -\frac{\pi ^2 T^4}{90}\nonumber \\&\quad +\frac{m_b^2 T^2}{24}-\frac{m_b^3 T}{12\pi }-\frac{\zeta (3)m_b^6}{96\pi ^4T^2}. \end{aligned}$$

(B.35)

In order to go beyond the mean field approximation, we consider the plasma screening effects. These can be accounted for by means of the ring diagrams. Since we are working in the high temperature approximation, we notice that the lowest Matsubara frequency is the most dominant term. Therefore we do not need to compute the other modes and it becomes

$$\begin{aligned} V^{\text {Ring}}&=\frac{T}{2}\int \frac{d^3k}{(2\pi )^3}\ln \left( 1+\varPi _b^{HT}D(\mathbf {k})\right) \nonumber \\&=\frac{T}{4\pi ^2}\int dk \ k^2 \Big \{ \ln \left( k^2+m_b^2+\varPi _B^{HT}\right) \nonumber \\&\quad -\ln \left( k^2+m_b^2\right) \Big \}. \end{aligned}$$

(B.36)

From Eq. (B.36), we see that both integrands are almost the same except that one is modified by the self-energy and the other one is not. Thus, after integration, we obtain that the ring diagrams contribution is

$$\begin{aligned} V^{\text {Ring}}=\frac{T}{12\pi }\left( m_b^3-(m_b^2+\varPi _b^{HT})^{3/2}\right) . \end{aligned}$$

(B.37)

On the other hand, in the same way that we compute the one-loop boson piece, we now calculate the fermion one. Then, we begin with

$$\begin{aligned} V^{\text {f}}=-T\sum _n\int \frac{d^3k}{(2\pi )^3} \text {Tr}\left[ \ln S({\tilde{\omega }}_n-i\mu _q,\mathbf {k})^{-1}\right] , \end{aligned}$$

(B.38)

where $S({\widetilde{\omega }}_n,\mathbf {k})$ is the fermion propagator, ${\widetilde{\omega }}_n$ the fermion Matsubara frequencies and $\mu _q$ the quark chemical potential. When we perform the sum over n, we get

$$\begin{aligned} V_{\text {}}^{\text {f}}= & {} -2 \int \frac{d^3k}{(2 \pi ^3)}\left( \sqrt{k^2+m_f^2}\right. \nonumber \\&\left. + T \ln \left( 1+e^{-\left( \sqrt{k^2+m_f^2}-\mu _q\right) /T}\right) \right. \nonumber \\&\left. + T \ln \left( 1+e^{-\left( \sqrt{k^2+m_b^2}+\mu _q\right) /T}\right) \right) \nonumber \\\equiv & {} V_{\text {}_{\text {vac}}}^{\text {f}}+ V_{\text {}_{\text {mattI}}}^{\text {f}}+ V_{\text {}_{\text {mattII}}}^{\text {f}}, \end{aligned}$$

(B.39)

In Eq. (B.39), we have one piece from the vacuum contribution, it was computed in “Appendix A” and the result is

$$\begin{aligned} V^{\text {f}}_{\text {}_{\text {vac}}}=\frac{m_f^4}{16\pi ^2}\left[ \ln \left( \frac{4\pi a^2}{m_f^2}\right) -\gamma _E+\frac{1}{2}\right] , \end{aligned}$$

(B.40)

and two matter pieces, one for the particle and other one for the antiparticle. In the high temperature approximation, where $m_f/T\ll 1$, we proceed in a fashion entirely analogous to the boson case and obtain

$$\begin{aligned} V_{\text {HT}_{\text {matt}}}^{\text {f}}&= \frac{1}{16\pi ^2}\Bigg \{m_f^4\Bigg [\ln \Big ( \frac{m_f^2}{(4\pi T)^2}\Big )\nonumber \\&\quad -\psi ^{(0)}\bigg (\frac{1}{2}+\frac{i\mu _q}{2\pi T}\bigg )-\psi ^{(0)}\bigg (\frac{1}{2}-\frac{i\mu _q}{2\pi T}\bigg )\Bigg ]\nonumber \\&\quad -8m_f^2T^2\big [\text {Li}_2(-e^{\mu _q/T})+\text {Li}_2(-e^{-\mu _q/T})\big ]\nonumber \\&\quad +32T^4\big [\text {Li}_4(-e^{\mu _q/T})+\text {Li}_4(-e^{-\mu _q/T})\big ]\nonumber \\&\quad +\frac{m_f^6}{6T^2}\left[ \psi ^{(2)}\bigg (\frac{3}{2}+\frac{\mathrm {i}\mu _q}{2\pi T}\bigg )+\psi ^{(2)}\bigg (\frac{3}{2}-\frac{\mathrm {i}\mu _q}{2\pi T}\bigg )\right] \Bigg \}. \end{aligned}$$

(B.41)

Therefore, the fermion one-loop contribution is

$$\begin{aligned} V_{\text {HT}}^{\text {f}}&= \frac{1}{16\pi ^2}\Bigg \{m_f^4\Bigg [\ln \Big ( \frac{a^2}{4\pi T^2}\Big )-\gamma _E+\frac{1}{2}\nonumber \\&\quad -\psi ^{(0)}\bigg (\frac{1}{2}+\frac{i\mu _q}{2\pi T}\bigg )-\psi ^{(0)}\bigg (\frac{1}{2}-\frac{i\mu _q}{2\pi T}\bigg )\Bigg ]\nonumber \\&\quad -8m_f^2T^2\big [\text {Li}_2(-e^{\mu _q/T})+\text {Li}_2(-e^{-\mu _q/T})\big ]\nonumber \\&\quad +32T^4\big [\text {Li}_4(-e^{\mu _q/T})+\text {Li}_4(-e^{-\mu _q/T})\big ]\nonumber \\&\quad +\frac{m_f^6}{6T^2}\left[ \psi ^{(2)}\bigg (\frac{3}{2}+\frac{\mathrm {i}\mu _q}{2\pi T}\bigg )+\psi ^{(2)}\bigg (\frac{3}{2}-\frac{\mathrm {i}\mu _q}{2\pi T}\bigg )\right] \Bigg \}. \end{aligned}$$

(B.42)

With the Eqs. (B.35), (B.37) and (B.42) at hand, we can write the effective potential up to the ring diagrams contribution in the high temperature approximation.

Appendix C: Fixing the set of parameters

Following the idea in Sect. 3, the set of equations, to determine the free parameters in the model, is

$$\begin{aligned}&V^{\text {eff}}\Big |_{v=0,T=0, \mu _q=\mu _B^c/3}=V^{\text {eff}}\Big |_{v=v_0',T=0, \mu _q=\mu _B^c/3} \nonumber \\&\frac{\partial }{\partial v}V(v, \lambda , g)\Big |_{v={\tilde{v}}_{0},T=0, \mu _q=\mu _B^c/3} =0 \nonumber \\&\frac{\partial }{\partial v}V(v, \lambda , g)\Big |_{v=v_0',T=0, \mu _q=\mu _B^c/3} =0 \nonumber \\&a=\sqrt{\frac{4}{3}\lambda (4 {\tilde{v}}^2_{0}-v_0'^2)}\nonumber \\ \end{aligned}$$

(C.43)

Note the last of Eq. (C.43) comes from the level rule, where $m_{\pi }|_{v={\tilde{v}}_0}=\frac{1}{2} m_{\pi }|_{v=v_0'}$. Also, we use $\lambda$ as an input. Hence, we have four equations for four unknowns (a, g, ${\tilde{v}}_{0}$, $v_0'$) from where we get the solution. It is remarkable that this solution exists only for a narrow window of $\lambda$ values.

Appendix D: Vacuum stability conditions

The vacuum stability conditions are introduced to ensure that $v_0$ and the sigma-mass maintain their tree level values, even after including the vacuum pieces stemming from the one-loop corrections. These conditions are

$$\begin{aligned} \frac{1}{2v}\frac{dV^{\text {vac}}}{dv} \Big |_{v=v_0}&=0,\nonumber \\ \frac{d^2V^{\text {vac}}}{dv^2} \Big |_{v=v_0}&=2a^2+3m_\pi ^2, \end{aligned}$$

(D.44)

where $V^{\text {vac}}$ is the one-loop vacuum piece of the effective potential. The solution for the counterterms $\delta a^2$ and $\delta \lambda$ is given by

$$\begin{aligned} \delta a^2= & {} \frac{m_\pi ^2}{2} -\frac{1}{16\pi ^2 \lambda }\left\{ \frac{}{} 3\lambda ^2(6 a^2+4 m_\pi ^2) -8 g^4(a^2+m_\pi ^2) \right. \nonumber \\&\quad +\left. 3 a^2 \lambda ^2 \Big [\ln \Big (\frac{m_\pi ^2}{a^2}\Big )+ \ln \Big (\frac{ 2a^2+3m_\pi ^2}{a^2}\Big )\Big ]\right\} , \end{aligned}$$

(D.45)

$$\begin{aligned} \delta \lambda= & {} \frac{\lambda }{2}\left( \frac{ m_\pi ^2}{a^2+ m_\pi ^2}\right) \nonumber \\&\quad -\frac{1}{16\pi ^2} \left\{ \frac{}{}-16 g^4+24\lambda ^2\right. -8 g^4 \ln \left( g^2 \frac{(a^2+m_\pi ^2)}{a^2\lambda }\right) \nonumber \\&\quad +\left. 3 \lambda ^2\left[ \ln \left( \frac{m_\pi ^2}{a^2}\right) +3 \ln \left( \frac{ 2a^2+3m_\pi ^2}{a^2}\right) \right] \right\} . \end{aligned}$$

(D.46)