18 0 50 100 150 0 20 40 60 80 j c n=1 n=2 d = 1 c → → 0.1 1 10 100 1000 10000 1.1 1.2 1.3 1.5 n=1 n=2 d=3 FIG. 31: Free energy J Nk$_{B}$ as a function of the inverse tem- perature η for d = 1. We have taken µ = 1 < µ c . FIG. 30: Energy Λ as a function of α for the first two branches in d = 3. For α → $^{+}$∞, the energy of the first branch un- dergoes damped oscillations around the value Λ s ≃ 1 . 13. We have taken µ = 1. 80 100 120 140 160 180 200 -2.3 -2.2 -2.1 -2 j c → → n=1 n=2 d=1 c t Introducing the dimensionless variables defined previ- ously, recalling that r = ξR/α and µ = k$_{0}$R , and in- troducing the normalized energy (57), we obtain Using the expressions of κ and λ following Eq. (68), we find that so that, finally, This equation gives the relation between the energy Λ and α for the n -th branch. In Figs. 26, 27, 29 and 30, we plot the energy Λ as a function of α for the first two branches n = 1 and n = 2 in different dimensions of space d = 1 , 2 , 3. The discussion is similar to the one given in Sec. III D. We have also represented the branch corresponding to the homogeneous solution. Us- ing Eq. (87) and η = α $^{2}$/d , its equation is given by Λ = − d $^{2}$/ (2 α $^{2}$) + d/ (2 µ $^{2}$). Λ = − d 2 η + 1 2 η 2 1 α d − 2 ∫ α 0 ( dψ dξ ) 2 ξ d − $^{1}$dξ + µ 2 2 η 2 1 α d ∫ α 0 ( ψ + βm Φ$_{0}$) $^{2}$ξ d − $^{1}$dξ. (78) βm Φ$_{0}$ = − λ κ $_{2}$, (79) Λ = − d 2 η + 1 2 η 2 1 α d − 2 ∫ α 0 ( dψ dξ ) 2 ξ d − $^{1}$dξ + µ 2 2 η 2 1 α d ∫ α 0 ( ψ − λα 2 µ 2 ) 2 ξ d − $^{1}$dξ. (80) S = d 2 N k$_{B}$ ln T − k$_{B}$ ∫ ρ m ln ρ m d r . (81) FIG. 32: Free energy J Nk$_{B}$ as a function of the inverse tem- perature η for d = 1. We have taken µ = 10 > µ c . The free energies of the homogeneous phase and inhomogeneous phase become equal at η = η t ( µ ). This corresponds to a first order phase transition in the canonical ensemble marked by the discontinuity of the slope J $^{′}$( β ) = − E . FIG. 33: Free energy J Nk$_{B}$ as a function of the inverse tem- perature η for d = 2. We have taken µ = 1. E. The entropy and the free energy Finally, we relate α to the entropy S and to the free energy F . The entropy is given by 5 10 15 20 25 0 5 10 15 20 j $_{c}$=4 c  →  → d=2 n=1