common tangent construction, we plot in Fig. 8 the free- energy landscape as a function of amplitudes A and B . The free-energy landscape exhibits two minima that cor- respond to the stable liquid and solid phases. The above amplitude equation calculation shows that the two-mode PFC model describes well solid-liquid coexistence with a well-defined free-energy barrier between solid and liquid. We have only treated here the case where the ampli- tudes are spatially uniform to characterize the bulk free- energy landscape. A more general free-energy functional that includes gradient terms would be necessary to treat the case where the amplitudes are spatially varying. Such a functional could then be used to compute the excess free-energy of the solid-liquid interface and its anisotropy, as done previously for bcc [14]. Those computations will be presented elsewhere. In general, the above free-energy density is a multi- variate function of the fourteen amplitudes A 0 i and B 0 i of different density waves, and cannot be represented graph- ically in a simple way. However, the free-energy barrier between solid and liquid can be made explicit by assum- ing that all the 〈 111 〉 and all the 〈 200 〉 density waves have the same amplitude (i.e., A 0 i = A and B 0 i = B , respec- tively). In this isotropic approximation (see [14] for the bcc analog), the free-energy density becomes This expression can also be obtained by evaluating di- rectly the difference between the solid and liquid free- energy densities, Δ f AE = ϵ − $^{2}$( f$_{s}$ − f$_{l}$ ), with f$_{l}$ and f$_{s}$ given by Eqs. (26) and (28), respectively, and the substi- tutions ¯ ψ$_{s}$ = ¯ ψ$_{l}$ = ψ$_{c}$ϵ 1 / $^{2}$, A$_{s}$ = Aϵ 1 / 2 and B$_{s}$ = Bϵ 1 / $^{2}$. For the parameters R = 0 and ψ$_{c}$ calculated from the So far, we have only examined the possibility of fcc- liquid coexistence. However, the phase-diagram of Fig. 1 shows that bcc can have a lower free-energy than fcc for small enough ϵ if R$_{1}$ is finite. We now use the amplitude equations to understand the relative stability of fcc and bcc. As a first step, it is useful to re-examine the scaling of the mean density that is controlled by the parameter ψ$_{c}$ . We computed previously the equilibrium solid and liquid densities using the common tangent construction, from which we obtained the scalings ( ¯ ψ$_{s}$ + ¯ ψ$_{l}$ ) / 2 = ψ$_{c}$ϵ 1 / $^{2}$, which defines ψ$_{c}$ , and ψ$_{s}$ − ¯ ψ$_{l}$ ∼ ϵ 3 / $^{2}$, which shows that the size of the density difference between solid and liq- C. Relative stability of fcc and bcc Δ f AE = 4( − 1 + 3 ψ$_{c}$ $^{2}$) A 2 +3( − 1 + 3 ψ$_{c}$ 2 + R ( − Q 2 1 + 1) $^{2}$) B 2 +54 A 4 + 45 2 B 4 + 144 A $^{2}$B2 + 72 ψ$_{c}$A $^{2}$B. (45) A 0 0.05 0.1 0.15 0.2 0.05 0.1 0.15 0.2 Student Version of MATLAB FIG. 8: (Color online) Free-energy landscape defined by Eq. (45) as a function of the amplitudes A and B of 〈 111 〉 and 〈 200 〉 density waves, respectively, for R = 0 and the corre- sponding coexistence value ψ$_{c}$ = − 0 . 6901 where the solid and liquid minima have the same height. For the case where the amplitudes are spatially uniform, we obtain the free-energy density Δ F $^{AE}$/V ≡ Δ f AE (44) = ( − 1 + 3 ψ$_{c}$ $^{2}$)( | A 0 111 | 2 + | A 0 11 ¯ 1 | 2 + | A 0 1 ¯ 11 | 2 + | A 0 1 ¯ ¯ 1 | $^{2}$) + ( − 1 + 3 ψ$_{c}$ 2 + R ( − Q 2 1 + 1) $^{2}$) ( | B 0 200 | 2 + | B 0 020 | 2 + | B 0 002 | $^{2}$) + 3 2 ( | A 0 111 | 4 + | A 0 11 ¯ 1 | 4 + | A 0 1 ¯ 11 | 4 + | A 0 1 ¯ 1 | 4 + | B 0 200 | 4 + | B 0 020 | 4 + | B 0 002 | $^{4}$) +6( | A 0 111 | $^{2}$| A 0 1 ¯ 11 | 2 + | A 0 111 | $^{2}$| A 0 11 ¯ 1 | 2 + | A 0 111 | $^{2}$| A 0 1 ¯ ¯ 1 | 2 + | A 0 1 ¯ 11 | $^{2}$| A 0 11 ¯ 1 | 2 + A 0 1 ¯ 11 A 0 1 ¯ ¯ 1 2 + A 0 11 ¯ 1 A 0 1 ¯ 1 $^{2}$) +6( | B 0 200 | $^{2}$| B 0 020 | 2 + | B 0 200 | $^{2}$| B 0 002 | 2 + | B 0 020 | $^{2}$| B 0 002 | $^{2}$) + 6( | A 0 111 | 2 + | A 0 11 ¯ 1 | 2 + | A 0 1 ¯ 11 | 2 + A 0 1 ¯ ¯ 1 $^{2}$)( | B 0 200 | 2 + | B 0 020 | 2 + | B 0 002 | $^{2}$) +6 ψ$_{c}$ ( A 0 $_{111}$A 0 1 ¯ 1 $_{1}$B 0 ¯ 200 + A 0 $_{111}$A 0 ¯ 11 ¯ $_{1}$B 0 0 ¯ 20 + A 0 $_{111}$A 0 ¯ 1 $_{11}$B 0 00 ¯ 2 + A 0 11 ¯ $_{1}$A 0 1 ¯ $_{11}$B 0 ¯ 200 + A 0 11 ¯ $_{1}$A 0 ¯ $_{111}$B 0 0 ¯ 20 + A 0 11 ¯ $_{1}$A 0 ¯ 1 ¯ 1 $_{1}$B 0 002 + A 0 1 ¯ $_{11}$A 0 ¯ $_{111}$B 0 00 ¯ 2 + A 0 1 ¯ $_{11}$A 0 ¯ 1 ¯ 1 $_{1}$B 0 020 + A 0 1 ¯ $_{1}$A 0 ¯ 11 ¯ $_{1}$B 0 002 + A 0 1 ¯ $_{1}$A 0 ¯ 1 ¯ $_{11}$B 0 020 + A 0 ¯ $_{111}$A 0 ¯ 1 ¯ $_{1}$B 0 200 + A 0 ¯ 11 ¯ $_{1}$A 0 ¯ 1 ¯ $_{11}$B 0 $_{200}$) +6 A 0 11 ¯ $_{1}$A 0 1 ¯ $_{11}$A 0 ¯ $_{111}$A 0 ¯ 1 ¯ 1 + 6 A 0 ¯ 1 $_{11}$A 0 ¯ 11 ¯ $_{1}$A 0 1 ¯ $_{1}$A 0 111 +6 A 0 ¯ 1 $_{11}$A 0 ¯ 1 ¯ $_{1}$B 0 $_{200}$B 0 020 + 6 A 0 ¯ 11 ¯ $_{1}$A 0 ¯ 1 ¯ $_{1}$B 0 $_{200}$B 0 002 +6 A 0 1 ¯ ¯ $_{1}$A 0 ¯ 1 ¯ $_{1}$B 0 $_{002}$B 0 020 + 6 A 0 ¯ $_{111}$A 0 $_{111}$B 0 00 ¯ $_{2}$B 0 0 ¯ 20 +6 A 0 11 ¯ $_{1}$A 0 $_{111}$B 0 ¯ $_{200}$B 0 0 ¯ 20 + 6 A 0 1 ¯ $_{11}$A 0 $_{111}$B 0 ¯ $_{200}$B 0 00 ¯ 2 +6 A 0 ¯ $_{111}$A 0 ¯ 11 ¯ $_{1}$B 0 $_{200}$B 0 0 ¯ 20 + 6 A 0 1 ¯ $_{1}$A 0 1 ¯ $_{11}$B 0 ¯ $_{200}$B 0 020 +6 A 0 ¯ $_{111}$A 0 ¯ 1 $_{11}$B 0 $_{200}$B 0 00 ¯ 2 + 6 A 0 11 ¯ $_{1}$A 0 1 ¯ $_{1}$B 0 ¯ $_{200}$B 0 002 +6 A 0 1 ¯ $_{11}$A 0 ¯ 1 $_{11}$B 0 00 ¯ $_{2}$B 0 020 + 6 A 0 11 ¯ $_{1}$A 0 ¯ 11 ¯ $_{1}$B 0 $_{002}$B 0 0 ¯ $_{20}$. B 〈 9