Symmetry, Sliding Windows and Transfer Matrices.

Transcription

1 Symmetry, Sliding Windows and Transfer Matrices Alexander Shpunt Department of Physics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA (Dated: May 16, 2008) In this paper we study 1D k-neighbor Ising model Variational approach using modified nearestneighbor interaction strength is developed, but the optimization of the coupling constant appears at least as hard as the exact solution in the general case For the exact solution two formulations of transfer matrix are studied: the block-spin approach yielding matrix T and the sliding window approach yielding matrix T s Equivalence between the two is established with Ts k = T holding univesally Matrix T s is sparse and possesses apparent symmetries, giving hope for analytical computation of eigenvalues Special cases are worked out explicitly Finally, in the appendix we compute the exact partition function for the 2D Ising model on an anisotropic triangular lattice, using graphical techniques as shown in [1] PACS numbers: I ITRODUCTIO Despite its seeming simplicity, Ising models hold many theoretical challenges, especially when it comes to exact analytical calculations The significance of the obtaining such results lies in their Universality (cf eg [1]), meaning that the insights obtained from Ising model will be applicable to a variety of real physical systems, too complicated to study in their full generality The 1D nearest-neighbor Ising model has been solved exactly with arbitrary magnetic field Following the work of Onsager ([2]), and consequent simplifications of the treatment, the exact solution for 2D lattices has been achieved, with zero magnetic field In this paper, we investigate analytic approaches to k-neigbor Ising model (kim) In Section II A we develop a variational formulation for kim Following the apparent difficulty of analytical optimization of the parameters of the variational anzats, we investigate the connections between two alternative formulations of transfer matrices In Section II B we show that the two approaches are indeed equivalent and explore some of the consequences of the new formulation In Appendix we discuss the general case and conclude with the exact partition function for the 2D anisotropic triangular lattice II K-EIGHBOR 1D ISIG MODEL The Hamiltonian, with indices taken modulo is: βh = K j S i S i+j = K j S i S i+j A Variational Estimates Here we give an estimate of the free energy of the radius-k model through the variational principle, with the exactly solvable single parameter nearest-neighbor model as the trial distribution The well-known variational principle states βf β F min J {βf 0 + βh βh 0 0 }, (1) where 0 refer to averages taken with respect to the trial distribution Our trial distribution is the nearest neighbor Ising model with the Hamiltonian: βh 0 = J S i S i+1,

2 2 where again all the indices are understood modulo For such a model, denoting t = tanh J Moreover, Using (2), βf 0 = ln Z 0 = ( ln [2 cosh J] + ln [ 1 + t ]) = ln ln(1 t2 ) ln [ 1 + t ] βh 0 0 = J S i S i+j 0 = tj + t j 1 + t (2) S i S i+1 0 = J t + t t, Consequently, β F = ln 2 + min t We need to minimize βh 0 = K j S i S i+j 0 = 2 ln(1 t2 ) ln [ 1 + t ] t Ψ = 2 ln(1 t2 ) ln [ 1 + t ] t K j t j + t j 1 + t K j (t j + t j ) J(t + t 1 ) K j (t j + t j ) J(t + t 1 ) dψ dt = t 1 t 2 t t 2 t 1 (1 + t ) t K j (t j + t j ) J(t + t 1 ) K j (jt j 1 + ( j)t j 1 ) J(1 + ( 1)t 2 ) t + t 2 1 t 2 Unfortunately, it is impossible to solve for dψ dt = 0 analytically, since it involves solving for general roots of polynomial of degree One conceivable approximation would be small-coupling, large (such that one could neglect terms of the order of t etc), but this is not much simpler than solving the exact model B Exact Partition Function for 1D Ising Model Here we compute the exact partition function for the ext-earest-eighbor Cyclic 1D Ising model We do it in two different ways, both by the method of transfer matrices, differing in the way the transfer matrix is defined The second way is somewhat less orthodox and the purpose of the exercise in this Section is to convince ourselves that these approaches are indeed equivalent The benefit of the less orthodox approach is that the transfer matrix is sparse and has symmetries that could potentially enable exact computations for arbitrary k The Hamiltonian for the case (again all indices are modulo ) is βh = The topology of the model is shown in Figure 1 S i (K 1 S i+1 + K 2 S i+2 )

3 3 FIG 1: 1D Ising model 1 The Conventional Transfer Matrix Let us employ the conventional transfer matrix approach, having 1 or e αk 1+βk 2 as its entries Z = {S i} P k e Si KjSi+j = tr[ S 1 S 2 e K1(S1S2+S2S3)+K2(S1S3+S2S4) S 3 S 4 S 3 S 4 e K1(S3S4+S4S5)+K2(S3S5+S4S6) S 5 S 6 S 1 S e K1(S 1S +S S 1)+K 2(S 1 S 1+S S 2) S 1 S 2 ] = tr [ T /2 ], where T is a 4x4 transfer matrix with elements S 1 S 2 T S 3 S 4 = e K1(S1S2+S2S3)+K2(S1S3+S2S4), explicitly e 2(k 1+k 2 ) e 2k 1 1 e 2k 2 T = e 2k1 e 2k1+2k2 e 2k2 1 1 e 2k2 e 2k1+2k2 e 2k1 e 2k2 1 e 2k1 e 2(k1+k2) 2 The Sliding Window Transfer Matrix Alternatively, Z = {S i } P k e Si KjSi+j = tr[ S 1 S 2 e S 1(K 1 S 2 +K 2 S 3 ) S 2 S 3 S 2 S 3 e S2(K1S3+K2S4) S 3 S 4 S i S i+1 e Si(K1Si+1+K2Si+2) S i+1 S i+2 S S 1 e S (K 1S 1+K 2S 2) S 1 S 2 ] = tr [ T s ],

4 4 where T s is a 4x4 transfer matrix with elements S 1 S 2 T S 2 S 3 = e S1(K1S2+K2S3), naively written as e k1+k2 e k1 k2 T s = e k1+k2 e (k1+k2) e (k1+k2) e k1+k2 e k1 k2 e k1+k2 It is really tempting to rearrange (relabeling) the rows of T s, toobtain a block-diagonal matrix e k 1 k 2 T s (b) = e (k 1+k 2 ) e k 1+k 2 e k 1+k 2 e (k 1+k 2 ) e k 1 k 2 However, unfortunately, in the general case, T s and T s (b) connected by similarity transformation will not have the same eigenvalues, since they are not 3 Comparison of the Two Approaches Examining the two formulations, it becomes obvious that they are completely equivalent provided T = Ts 2 and it indeed holds: holds, T 2 s = e k 1 k 2 e k 1+k 2 e (k 1+k 2 ) e (k 1+k 2 ) e k 1+k 2 e k 1 k 2 2 = e 2(k 1+k 2 ) e 2k 1 1 e 2k 2 e 2k 1 e 2k 1+2k 2 e 2k e 2k2 e 2k1+2k2 e 2k1 e 2k 2 1 e 2k 1 e 2(k 1+k 2 ) = T Since we have shown equivalence at the level of transfer matrices, it is cleat that all derived quantities will agree too Moreover, the derivations are by no means unique to the model and hold for a general k-neighbors also We thus have established the equivalence of the two approaches for transfer matrix computations To compute the exact partition function, lets use T s The eigenvalues of the matrix T s are: ) λ 1± = e (cosh K2 K 1 ± (sinh K 1 ) 2 + e 4K 2, Consequently, ) λ 2± = e (sinh K2 K 1 ± (cosh K 1 ) 2 e 4K 2 [ ( ( Z = e K 2 cosh K 1 + (sinh K 1 ) 2 + e 2) 4K + cosh K 1 (sinh K 1 ) 2 + e 2) 4K + + ( ( ) ] sinh K 1 + (cosh K 1 ) 2 e 2) 4K + sinh K 1 (cosh K 1 ) 2 e 4K 2 III DISCUSSIO AD FUTURE DIRECTIOS In this paper we considered analytical approaches to Ising models with long-range interactions The symmetry structure of the newly developed sliding window transfer matrix is promising and merits further research It is shown for instance, that partition function is computable exactly using these symmetries It would be interesting to continue the investigation of the symmetry properties and ways to use them to obtain non-trivial exact results

5 5 IV APPEDIX A Why the Exact Partition Function for the 1D Ising Model With K neighbors is Still ot Computed ow that we have convinced ourselves through simple example that the two alternative transfer matrix formulations are indeed equivalent, the treatment for the k-neighbor Ising model follows as a simple generalization of the k = 2 case Z = {S i} P k P k e Si KjSi+j = tr[ S 1 S 2 S k e S1 KjS1+j S 2 S 3 S k+1 S 2 S 3 S k+1 e S 2 P k K js 2+j S 3 S 4 S k+2 S i S i+1 S k+i 1 e S i P k K js i+j S i+1 S i+2 S k+i S S 1 S k 1 e S P k K js +j S 1 S 2 S k ] = tr [ T s(k) ], where T s(k) is a 2 k x2 k sliding window transfer matrix with entries { P S 1 S 2 S k T s(k) S 1S 2 S k = e S k 1 KjS j If (S 2, S 3,, S k ) = (S 1, S 2,, S k 1 ), 0 Otherwise (since mutually inconsistent) It is important to note that each row and column of T s(k) has exactly 2 entries S 1 S 2 S k T s(k) S 2 S 3 S k+1 = e S P k 1 K js 1+j, and the rest are zeros The matrix T s(k) is sparse, with the number of non-zero entries linear in matrix size For instance, for model, we have the following matrix (denote κ i = e K i ): κ 1 κ 2 κ 3 κ 1 κ 2 /κ κ 1 κ 3 /κ 2 κ 1 /(κ 2 κ 3 ) + + κ 2 κ 3 /κ 1 κ 2 /(κ 1 κ 3 ) + κ 3 /(κ 1 κ 2 ) 1/(κ 1 κ 2 κ 3 ) + + 1/(κ 1 κ 2 κ 3 ) κ 3 /(κ 1 κ 2 ) + κ 2 /κ 1 κ 3 (κ 2 κ 3 )/κ 1 + κ 1 /(κ 2 κ 3 ) κ 1 κ 3 /κ 2 κ 1 κ 2 /κ 3 κ 1 κ 2 κ 3 The matrix has a pronounced symmetry, which gives hope for an analytical solution for the eigenvalues Actually, the solution for is possible, since it involves quartic polynomials Due to a severe lack of time we have not been able to conclude our calculations for Unfortunately, tempting as it is, in spite of the fact that T s(k) can be cast into a block-diagonal term by the cyclic shift of one of it s coordinates, this will change the eigenvalues κ 1 κ 2 κ 3 κ 1 κ 2 /κ /(κ 1 κ 2 κ 3 ) κ 3 /(κ 1 κ 2 ) + + κ 1 κ 3 /κ 2 κ 1 /(κ 2 κ 3 ) + κ 2 /κ 1 κ 3 (κ 2 κ 3 )/κ κ 2 κ 3 /κ 1 κ 2 /(κ 1 κ 3 ) + κ 1 /(κ 2 κ 3 ) κ 1 κ 3 /κ 2 + κ 3 /(κ 1 κ 2 ) 1/(κ 1 κ 2 κ 3 ) κ 1 κ 2 /κ 3 κ 1 κ 2 κ 3 The determinant, i λ i is correct and can be computed easily Since we were not sure of the utility of the calculation, we did not pursue this explicitly

6 6 FIG 2: Direction labels for 2D triangular lattice B Exact Partition Function for the Ising Model on 2D Triangular Lattice Here we develop the exact partition function for the Ising model on 2D triangular lattice, using graphic techniques It is easy to see that the same approach that worked for the square lattice works also here Denoting t i = tanh(k i ), and repeating the steps in [1], we get ln Z 3 = ln(2 cosh(k i )) 1 3 tr ln(1 T (q)) = ln(2 cosh(k i )) q d 2 q (2π) 2 ln det(1 T (q)), where in the last equality we used the fact that tr ln(1 T (q)) = ln det(1 T (q)) The directions for building the T matrix are shown in Figure 2 Using these, we have the following directions map Consequently, t 1 e iqx t 1 e i(qx+π/3) t 1 e i(qx+2π/3) 0 t 1 e i(qx 2π/3) t 1 e i(qx π/3) t 2 e i(qx+qy π/3) t 2 e i(qx+qy) t 2 e i(qx+qy+π/3) t 2 e i(qx+qy+2π/3) 0 t 2 e i(qx+qy 2π/3) T (q) = t 3 e i(qy 2π/3) t 3 e i(qy π/3) t 3 e iqy t 3 e i(qy+π/3) t 3 e i(qy+2π/3) 0 0 t 1 e i(qx+2π/3) t 1 e i(qx+π/3) t 1 e iqx t 1 e i(qx π/3) t 1 e i(qx 2π/3) t 2 e i(qx+qy 2π/3) 0 t 2 e i(qx+qy+2π/3) t 2 e i(qx+qy+π/3) t 2 e i(qx+qy) t 2 e i(qx+qy π/3) t 3 e i(qy π/3) t 3 e i(qy 2π/3) 0 t 3 e i(qy+2π/3) t 3 e i(qy+π/3) t 3 e iqy Denoting X e iq x, Y e iq y, Ω e iπ/3, t 1 X 1 t 1 X 1 Ω 1 t 1 X 1 Ω 2 0 t 1 X 1 Ω 2 t 1 X 1 Ω t 2 (XY ) 1 Ω t 2 (XY ) 1 t 2 (XY ) 1 Ω 1 t 2 (XY ) 1 Ω 2 0 t 2 (XY ) 1 Ω 2 T (q) = t 3 Y 1 Ω 2 t 3 Y 1 Ω t 3 Y 1 t 3 Y 1 Ω 1 t 3 Y 1 Ω t 1 XΩ 2 t 1 XΩ t 1 X t 1 XΩ 1 t 1 XΩ 2 t 2 XY Ω 2 0 t 2 XY Ω 2 t 2 XY Ω t 2 XY t 2 XY Ω 1 t 3 Y Ω 1 t 3 Y Ω 2 0 t 3 Y Ω 2 t 3 Y Ω t 3 Y Using Mathematica, we arrive at the following expression 3 det(1 T (q)) = cosh(2k i) 3 sinh(2k i) sinh(2k 1 ) cos(q x ) sinh(2k 2 ) cos(q y ) sinh(2k 3 ) cos(q x + q y ) cosh 2 (K 1 ) cosh 2 (K 2 ) cosh 2 (K 3 )

7 Summarizing, ln Z 2 = 2 2π 0 dq 2 xdq 2 y (2π) 2 [ 3 ln cosh(2k i ) ] 3 sinh(2k i ) sinh(2k 1 ) cos(q x ) sinh(2k 2 ) cos(q y ) sinh(2k 3 ) cos(q x + q y ) 7 [1] M Kardar, Statistical Physics of Fields, Cambridge University Press, 2007 [2] L Onsager, Phys Rev 65, 117 (1944)