5  0.6  0.8  1  1.2  1.4  1.6  1.8  0.2  0.4  0.6  0.8  1  1.2  1.4 C/N T/D (II) -2.2 -2 -1.8 -1.6 -1.4 -1.2 E/N (I) case (a) case (b) case (c) FIG. 1: (Color online) Temperature dependence of (I) inter- nal energy E/N and (II) heat capacity C/N in the purely dipolar system. Simulated annealing method is used for the measurement. The size N is 10 $^{3}$. Measurements are done in the three cases (see text). where the first terms and the second terms in the right hand sides denote contributions from long-range inter- actions and those from short-range interactions, respec- tively. As we have already mentioned, the first terms are given by eqs. (31), (32), (33), and (42), respectively. On the other hand, by substituting ˜ V$_{kl}$ = V (S) kl and Δ V ∗ kl = 0 into eqs. (27), (28), (29), and (40), we readily obtain MC simulations of a three dimensional magnetic dipolar system on a L 3 simple cubic lattice. The boundary condi- tion is open. The Hamiltonian of the system is described as where S$_{i}$ is a classical Heisenberg spin of | S$_{i}$ | = 1, r$_{ij}$ is the vector spanned from a site i to j in the unit of the lattice constant a , r$_{ij}$ = | r$_{ij}$ | , and D = ( gµ$_{B}$S ) /a $^{3}$. We hereafter call the system purely dipolar system . In simulations with the SCO method, we regard each term in the right hand side of eq. (45) as V$_{ij}$ . Since ˜ V$_{ij}$ = 0 in the SCO method, Δ V$_{ij}$ defined by eq. (3) is equal to V$_{ij}$ . Interaction V$_{ij}$ has the maximum value 2 D/r 3 ij when S$_{i}$ and S$_{j}$ are anti-parallel along r$_{ij}$ . We therefore set Δ V ∗ $_{ij}$, which should be equal to or greater than Δ V$_{ij}$ over all S$_{i}$ and S$_{j}$ , to 2 D/r 3 $_{ij}$. By substituting these into eq. (2), we obtain H = H dip = D ∑ i<j [ S$_{i}$ · S$_{j}$ − 3( S$_{i}$ · r$_{ij}$ )( S$_{j}$ · r$_{ij}$ ) r 3 ij ] , (45) P$_{ij}$ = exp [ βD { S$_{i}$ · S$_{j}$ − 3( S$_{i}$ · r$_{ij}$ )( S$_{j}$ · r$_{ij}$ ) − 2 r 3 ij }] , (46) VI. NUMERICAL TESTS A. Internal energy and heat capacity measurements In order to check the validity of the formulae for inter- nal energy and heat capacity measurements, we perform where we have used the fact that all the potentials are switched to ˜ V$_{kl}$ . Note that P$_{kl}$ = 1 when ˜ V$_{kl}$ = V$_{kl}$ and Δ V ∗ ij = 0. These results are quite natural because they coincide with the results in the usual MC procedure. E (S) const = 0 , (44a) ˜ E (S) = − ∑ 〈 kl $_{〉}$V (S) kl , (44b) ∂ ˜ E (S) ∂β = 0 , (44c) X (S) SPS = ∏ 〈 kl $_{〉}$exp[( β$_{m}$ − β$_{n}$ )( V$_{kl}$ ( S$_{k}$ , S$_{l}$ ) − V$_{kl}$ ( S ′ $_{k}$, S ′ $_{l}$))] = X (S) B , (44d) When long-range interactions { V (L) ij } and short-range interactions { V (S) ij } coexist, we do not need to use the SCO method for short-range interactions because it does not reduce the computational time significantly. In other words, we can switch V (L) ij to 0 or ¯ V (L) ij with V (S) ij un- changed. This can be realized by setting ˜ V (L) ij = 0 and ˜ V (S) ij = V (S) ij . In this case, we can decompose E$_{const}$ , ˜ E , ∂ ˜ E $_{∂β}$, and X$_{SPS}$ in eqs. (27), (28), (29), and (40) as V. THE CASE WHEN LONG-RANGE INTERACTIONS AND SHORT-RANGE INTERACTIONS COEXIST Since X$_{SPS}$ in this formula is calculated only from the pairs whose graph variables are one, it can be calculated with very short computational time as ˜ E ′ and ∂ ˜ E ′ ∂β in eqs. (32) and (33) are. E$_{const}$ = E (L) const + E (S) $_{const}$, (43a) ˜ E = ˜ E (L) + ˜ E $^{(S)}$, (43b) ∂ ˜ E ∂β = ∂ ˜ E (L) ∂β + ∂ ˜ E (S) ∂β , (43c) X$_{SPS}$ = X (L) SPS × X (S) $_{SPS}$, (43d)