¯ ψ s − ¯ ψ l ϵ 10 − 2 10 − 1 − 4 10 − 3 10 − 2 R=0 R=3 R=5 8 Since ( ∇ 2 + 1) $^{2}$( ∇ 2 + Q 2 $_{1}$)$^{2}$ψ$_{2}$ gives a vanishing contri- bution for all density waves associated with sets { K$_{i}$ } and { K ′ j } , all remaining terms ∼ e i ⃗ K$_{i}$ · ⃗ r and e i ⃗ K$_{j}$ $^{′}$· ⃗ r must balance each other in order for a solution of Eq. (39) to exist. F or example, the condition that the coefficients of e i ⃗ K$_{111}$ · ⃗ r balance each other yields and requiring that the coefficients of e i ⃗ K$_{200}$ · ⃗ r balance each other yields in turn and Next, to carry out the amplitude expansion, we start from the equilibrium equation δF/δψ = µ$_{E}$ , where µ$_{E}$ is the equilibrium value of the chemical potential. With F defined by Eqs. (19) and (20), we obtain The above solvability condition must be satisfied inde- pendently for each reciprocal lattice vector. This yields a set of fourteen coupled amplitude equations that are straightforward to obtain. From those amplitude equa- tions, it is useful to express the free-energy of the system measured from its constant value in the liquid, defined as the difference Δ F $^{AE}$, as a functional of the density wave amplitudes A 0 i and B 0 $_{i}$. This quantity can be ex- pressed solely in terms of the amplitudes of density waves owing to the property that the size of the coexistence region ( ∼ ϵ 3 / $^{2}$) is much smaller than the mean density ( ∼ ϵ 1 / $^{2}$) in the small ϵ limit. Since the amplitudes are not conserved order parameters, the equilibrium state simply corresponds to a minimum of this free-energy. Hence the amplitude equations must be equivalent to We substitute the small ϵ expansion of the density field (29) into Eq. (34) and collect terms with the same power of ϵ . We find at the order ϵ 1 / 2 which has the solution where the summations are over 〈 111 〉 and 〈 200 〉 RLVs, respectively, and | ⃗ K$_{i}$ | = 1 , | ⃗ $^{$_{K$_{j}$}$′}$| = $^{√}$4 / 3, in our scaled units. At order ϵ , we obtain which has the solution and collecting the terms at order ϵ 3 / 2 yields Saturday, December 19, 2009 FIG. 7: (Color Online) Plots of ¯ ψ$_{s}$ − ¯ ψ$_{l}$ versus ϵ for the same values of R as in Fig. 6, with ¯ ψ$_{s}$ and ¯ ψ$_{l}$ calculated from the common tangent construction. The solid black line has a slope of exactly 3 / 2 on this log-log plot showing that the density difference between solid and liquid ∼ ϵ 3 / $^{2}$. ( − 1 + 3 ψ$_{c}$ $^{2}$) A 0 111 + (3 | A 0 111 | 2 + 6 | A 0 1 ¯ 11 | 2 + 6 | A 0 11 ¯ 1 | 2 +6 | A 0 1 ¯ 1 | 2 + 6 | B 0 200 | 2 + 6 | B 0 020 | 2 + 6 | B 0 002 | $^{2}$) A 0 111 +6 ψ$_{c}$ ( A 0 ¯ $_{111}$B 0 200 + A 0 1 ¯ $_{11}$B 0 020 + A 0 11 ¯ $_{1}$B 0 $_{002}$) +6 A 0 11 ¯ $_{1}$A 0 1 ¯ $_{11}$A 0 ¯ 111 + 6 A 0 ¯ 1 ¯ $_{11}$B 0 $_{200}$B 0 020 +6 A 0 ¯ 11 ¯ $_{1}$B 0 $_{200}$B 0 002 + 6 A 0 1 ¯ $_{1}$B 0 $_{002}$B 0 020 = 0 , (40) ( − 1 + 3 ψ$_{c}$ 2 + R ( − Q 2 1 + 1) $^{2}$) B 0 200 + (6 | A 0 111 | 2 + 6 | A 0 1 ¯ 11 | 2 +6 | A 0 11 ¯ 1 | 2 + 6 | A 0 1 ¯ ¯ 1 | 2 + 3 | B 0 200 | 2 + 6 | B 0 020 | 2 +6 B 0 002 $^{2}$) B 0 200 + 6 ψ$_{c}$ ( A 0 11 ¯ $_{1}$A 0 1 ¯ 11 + A 0 $_{111}$A 0 1 ¯ $_{1}$) +6( B 0 $_{020}$A 0 1 ¯ $_{11}$A 0 1 ¯ 1 + B 0 $_{002}$A 0 1 ¯ $_{1}$A 0 11 ¯ 1 + B 0 0 ¯ $_{20}$A 0 11 ¯ $_{1}$A 0 111 + B 0 00 ¯ $_{2}$A 0 1 ¯ $_{11}$A 0 $_{111}$) = 0 . (41) ψ$_{l}$$_{1}$ = ψ$_{s}$$_{1}$ = 0 . (33) µ$_{E}$ = − ϵψ + ( ∇ 2 + 1) $^{[}$( ∇ 2 + Q 2 $_{1}$) 2 + ϵR $^{]}$ψ + ψ $^{3}$. (34) ( ∇ 2 + 1) $^{2}$( ∇ 2 + Q 2 $_{1}$)$^{2}$ψ$_{0}$ = Q 4 $_{1}$ψ$_{c}$ , (35) ψ$_{0}$ = ψ$_{c}$ + ∑ i A 0 $_{i}$e i ⃗ K$_{i}$ · ⃗ r + ∑ j B 0 $_{j}$e i ⃗ K$_{j}$ $^{′}$· ⃗ $^{r}$, (36) ( ∇ 2 + 1) $^{2}$( ∇ 2 + Q 2 $_{1}$)$^{2}$ψ$_{1}$ = 0 , (37) ψ$_{1}$ = ∑ i A 1 $_{i}$e i ⃗ K$_{i}$ · ⃗ r + ∑ j B 1 $_{j}$e i ⃗ K$_{j}$ $^{′}$· ⃗ $^{r}$, (38) − ψ$_{0}$ + ( ∇ 2 + 1) $^{2}$( ∇ 2 + Q 2 $_{1}$)$^{2}$ψ$_{2}$ + R ( ∇ 2 + 1) $^{2}$ψ$_{0}$ + ( ψ$_{0}$ ) 3 = − ψ$_{c}$ + Q 4 $_{1}$ψ$_{l}$$_{2}$ + Rψ$_{c}$ + ( ψ$_{c}$ ) $^{3}$. (39) δ Δ F AE δA 0 ∗ i = 0 , (42) δ Δ F AE δB 0 ∗ i = 0 . (43) and