Advanced Statistical Mechanics

Advanced Statistical Mechanics Victor Gurarie Fall 0 Week 3: D and D transverse field Ising models, exact solution

D Ising Model: Peierls Argument D Ising model is defined on a D lattice. Each spin of the D Ising model lives on a -dimensional space, thus it has two coordinates. Let s call them horizontal and vertical coordinates, and let s denote them x and τ. So each spin can be written as σ τ x. Suppose the Ising model has length N h in the horizontal direction and N v in the vertical direction, so all together it has N h N v spins. The energy of the Ising model can be written as E = and the partition function is x=,...,n h ; τ=,...,n v Z = σ i =± [ ] Jh σxσ τ x+ τ + J v σxσ τ x τ+. e E T.. For generality we introduce two distinct couplings, J h and J v in the horizontal and vertical directions, although in the classic Ising model J h = J v = J. We also need to specify what happens at the boundary. Let s take the periodic boundary conditions. To reduce clutter, let us adopt this common notations: σ Nv+ x σ x, σ τ N h + σ τ..3 In this way, the periodic boundary conditions are automatically encoded in.. Peierls came up with a hand-waving argument which supports that at temperature below some critical temperature T c, the Ising spins order. Let us go over this argument quickly in case where J h = J v = J, and J/T = K. A droplet of negative spins in the sea of positive spins has energy E 0 + JL,.4 where E 0 is the energy of all spins pointing up, or E 0 = JN v N h N v N h is the total number of bonds on the square lattice of our size. Here L is the length of the boundary between a droplet of spins pointing down, and the rest of the spins pointing up each bond where spin at one end points up and a spin at the other point points down has energy +J, which differs from its energy when both spins point up which is J by J. So J times the number of bonds with opposite spins at ends is the energy of the droplet, relative to E 0. There are L such bonds, so the total is E 0 + JL. Its contribution to Z is then e E 0 T KL. On the other hand, there are c L ways to position a droplet. This produces a contribution to Z of the order of E E0/T L ln c KL e

which, at large K small temperature, as K = J/T, will always be small. In other words, the free energy of the droplet is F = E T S = JL LT ln c E 0.5 and its minimum at T < J/ ln c is achieved at zero L. This means at T < J/ ln c, there is going to be no droplets around and the spins will all be positive. At T > J/ ln c the minimum of free energy is at very large L, in other words, these droplets will be around, they will be large, and eventually they will blur the difference between spin-ups and spin-downs, so the spins will not be ordered. The transition happens at critical temperature T c which this argument says is J/ ln c. This is just a rough estimate, in practice this temperature is proportional to J with some numerical coefficient in front. If J h J v, there is still a transition temperature T c which can be expressed in terms of these. This expression is known exactly, but its form is not illuminating and so won t be given here. Transfer Matrix and the Hamiltonian We rewrite the partition function as a transfer matrix. This is a little more involved than in D. In one dimension, the transfer matrix has two indices, σ and σ. In D, we designate all spins along one row of spins as one index of the transfer matrix, and all spins along an adjacent row as the other index. Thus a transfer matrix is a matrix of the size N h by N h. The transfer matrix takes the form T σ,σ,...,σ N h σ,σ,...,σ N h Nh = e x=k vσx σ x+ +Kvσ x σ x.. Here we introduced the natural notations K h = J h T, K v = J v T.. Again, it has two indices, labeling its rows and columns. Each index goes from to N h. However, we denote the index not by one integer going between these two values, but rather by a string of integers σ, σ, σ 3,..., σ N lower index and σ, σ, σ 3,..., σ N upper index, each σ taking values ±, all for Nv distinct strings of integers.

Then we can write the partition function in the following way σ=± e x=,...,nh ; τ=,...,nv [K hσxσ τ x+ τ +Kvστ xσx τ+ ] = T σ...σ N h T σ3...σ3 Nh σ σ=±...σ N σ h...σ N h This in turn can be interpreted as... T σ...σ Nh σ Nv...σ Nv N h.3 Z = Tr T Nv = n λ Nv n λ Nv 0,.4 where λ n are the eigenvalues of T there are going to be N h of those because this matrix is N h by N h and λ0 is the largest of these eigenvalues. So the task is to diagonalize the matrix T, written explicitly in.. This is a complicated matrix and it s not easy to diagonalize. However, there is a particular limit in which the matrix considerably simplifies. This is the limit K v, K h. Let us derive the transfer matrix in this limit. To do that, we first of all split the transfer matrix into a product of two matrices, the one responsible for the horizontal part of the transfer matrix, and the one responsible for the vertical part, in the following way T σ,σ,...,σ N h σ,σ,...,σ N h = σ=± [T h ] σ, σ,..., σ N h σ,σ,...,σ N h [T v ] σ,σ,...,σ Nh σ, σ,..., σ N h..5 Here and [T h ] σ, σ,..., σ N h σ,σ,...,σ N h = e K Nh h [T v ] σ,σ,...,σ N h σ, σ,..., σ N h x= σ xσx+ δ σ δ σ σ σ... δ σ Nv σ Nv.6 = e Kv Nh x= σ xσ x..7 It should be clear that multiplying T h by T v, as is done in.5, gives T, as in.. Next we employ the following strategy. We will show that, under conditions K v and K h, each of these two matrices can be written as T h = e H h, T v = e Hv.8 where H h and H v are very small matrices. Now in general for two matrices A and B, e A e B e A+B. But for two very small matrices e H h e Hv H h H v H h H v e H h H v..9 So we will be able to derive T in this way. 3

Let s follow this route. Take T h first. Suppose K v. Then we can Taylor-expand the exponential: [T ] σ, σ,..., σ N h h = δ σ σ,σ,...,σ N σ h δ σ... δ σ N h σ σ N h +K h x= [ δ σ δ σ σ σ... δ σ x σ x σ x δ σ x σ x ] σ x+δ σ x+ δ σ x+... δ σ N h σx+ σx+ σn h.0 We rewrite this with the help of the Pauli matrices [ τ 3 ] σ x 0 x = = σ xδ σ x 0 σ.. x This gives σ x [T ] σ, σ,..., σ N h h = δ σ σ,σ,...,σ N σ h δ σ... δ σ N h σ σn h + K h x= [ δ σ δ σ σ σ ]... δ σ x τ σ xτ 3 x+δ σ 3 x+... δ σ N h x σx+ σn h. Now the first term in this sum is nothing but an identity matrix a very large, N h N h identity matrix. The second term represents a sum of terms, each is a product of many little identities acting on all spins except σ x and σ x+, which are acted upon by Pauli matrices instead. A shorthand notation is often introduced, where all these little Kronecker deltas are omitted just as multiplication by is usually omitted, which leads to the following simplified notation yet completely equivalent. [T h ] σ, σ,..., σ N h σ,σ,...,σ N h Here I stands for this very large identity matrix. The final step is to recall that K h, so we can write N = I + K h τxτ 3 x+. 3.3 x= T h = e K h N x= τ 3 xτ 3 x+..4 So we accomplished the task of writing T h in the form e H h, where H h = K h N x= by τ 3 xτ 3 x+..5 Now on to T v. T v will be transformed in the following way. We first write [T v ] σ,σ,...,σ N h σ, σ,..., σ N h = Let us study each matrix in this product. N h x= e Kv σ x σ x..6 e Kv σ xσ x.7 4

is a two by two matrix, which explicitly takes the familiar from D Ising model form [ ] e K v σ xσ x σ x = σ x e K v e Kv e Kv This matrix is close to identity if K v. In this case we can write it as e Kv.8 e K v e Kv e K v = e Kv = e Kv δ + e K v τ e Kv e e Kv τ..9 e Kv e Kv e Kv Here the natural notations are Therefore, we can write τ = 0, δ = 0 0..0 0 [ ] e K v σ x σ σ x x = e Kv e e Kv τx.. σ x Here the index x on τ x serves as a reminder that τ 3 is applied to the x-th spin. Now we can write [T v ] σ,σ,...,σ N h σ, σ,..., σ N h = N h x= Finally, we recall that K v, so e Kv into a big sum. So we succeeded in writing [T v ] σ,σ,...,σ N h σ, σ,..., σ N h e Kv σ x σ x = e N h K v N h x= e e Kv τ x.., so all these exponentials can be combined = e N hk v e e Lv N h x= τ x..3 T v = e KvN h e Hv, H v = e Lv x= τ x..4 This completes the derivation of T v. Finally, we put everything together. We introduce convenient notations γ = e Kv, β = K h..5 This gives T = e KvN h e H,.6 where H = γ x= τ x β 5 x= τ 3 xτ 3 x+..7

This Hamiltonian is called the D transverse field Ising model. Indeed, this is a quantum mechanical Hamiltonian, which acts on quantum spins aligned in a D chain. τ x are spin operators Pauli matrices acting on x-th specific spin. We have a term τxτ 3 x+ 3 which describes two spins trying to align in the same direction zdirection and a term γτx which corresponds to a magnetic field acting in the x direction. So it s a D quantum spin chain with a magnetic field in the transverse direction. Our goal is now to study this H and find its ground and excited states. This will solve the D Ising model, at least in the regime where K v, K h. 3 Ordered and Disordered Regimes 3. Ordered regime If γ β, then the Hamiltonian is approximately neglect the γ term Then it has two ground states, 0 = N h x= H = β 0 x x=, or 0 = τ 3 xτ 3 x+. 3. N h x= 0 i. 3. The notations here correspond to a wave function a spinor for spins on sites x. These are spins all pointing up or down. In the spirit of spontaneous symmetry breaking the spins choose one of these two states and ignore the other. suppose the ground state corresponds to all spins up. The energy is E 0 = βn h. 3.3 An excited states would correspond to taking one spin and making it point down. This would correspond to the energy E = βn h + β. 3.4 Of course, there are many such states, since any spin out of N h spins, can be made to point down. But this does not matter. We see that E E 0 = β. Now since T = e H, this leads to the eigenvalues of the transfer matrix λ 0 = e E 0, λ = e E, 3.5 6

and the correlation length is It is finite. ξ = ln λ 0 λ = = E E 0 β. 3.6 3. Disordered regime Suppose now γ β. Neglect β to obtain The ground state takes the form H = γ 0 = N h x= x= τ x. 3.7 x. 3.8 This is because, is the eigenvector for τ with the eigenvalue. This corresponds to the energy E = γn h. 3.9 An excited state would correspond to taking one spin and replacing its spinor by, again, there are many such excited states. eigenvalue. So this gives This is an eigenvector of τ but with an E = γn h + γ. 3.0 So we see that ξ = = E E 0 γ. 3. In the next section, we are going to solve our problem exactly. We will see that the exact expression is ξ = γ β. 3. This matches 3.6 and 3. in the limits γ β and β γ respectively. 4 Jordan-Wigner transformation Now we will solve the Hamiltonian.7 exactly. First of all we do a unitary transformation so that H = γ x= τ 3 x β 7 x= τ xτ x+. 4.

Corresponding transformation is given by U = + iτ, and Then, it is convenient to introduce U τ 3 U = τ, U τ U = τ 3. 4. b x = τ x + iτx, b x = τ x iτx 4.3 They satisfy b x = b x = 0, b xb x + b x b x = 4.4 In this way they look like creation and annihilation operators. The transformations expressing τ in terms of b read τx = b x + b x, τx = ib x b x, τx 3 = i [ ] τ x, τx = bx b x b xb x. 4.5 Now we introduce the Jordan-Wigner string a x = e iπ y<x b yb y b x, a x = e iπ y<x b yb y b x. 4.6 Then not only a i obey creation and annihilation relations on a given site, but also a xa y + a y a x = 0, x y, 4.7 and so do a x, a y and a x, a y. Therefore, these are now fermions, called Jordan-Wigner fermions. Therefore, the Hamiltonian becomes taking into account the Jordan-Wigner string which leads to the minus sign in the term a x a x H = γ x= a x a x a x a x N h β x= a x a x a x+ + a x+ 4.8 This is the so-called Bogoliubov Hamiltonian, studied in superconductivity. This Hamiltonian can now be diagonalized using standard methods. 5 Diagonalizing the Hamiltonian - the straightforward method To diagonalize this Hamiltonian we go to the plane wave basis a x = Nh k e ikx a k, a x = Nh 8 k e ikx a k 5.

Substituting this into 4.8 gives H = γ k a k a k a k a k β k a k a k a k + a k e ik. 5. Here the sum over k goes from π to π, as usual this is called the Brillouin zone, in steps of π/n h. This Hamiltonian clearly splits into terms coupling the modes k and k for each positive k. Therefore, we can write it as H = H k, 5.3 k>0 where H k = γ β cos k a k a k a k a k H k can be written in a convenient matrix form iβ sin k a k a k + iβ sin k a ka k. 5.4 H k = a γ β cos k iβ sin k ak k a k iβ sin k γ + β cos k a. 5.5 k This matrix has the Bogoliubov-de-Gennes form from the theory of superconductivity. It is a little inconvenient that it has imaginary entries, but these imaginary entries can be completely eliminated if one does a transformation a k = iã k, a k = iã k. 5.6 The new variable ã k are as good a fermionic variable as the old one, and the Hamiltonian becomes H k = a γ β cos k β sin k ak k ã k β sin k γ + β cos k ã. 5.7 k Now the Hamiltonian can be diagonalized by a Bogoliubov transformation. This transformation reads a k = cosφ k c k + sinφ k c k, ã k = sinφ k c k + cosφ k c k. 5.8 Here k > 0. It is clear from this that φ k = φ k. Importantly, these transformations preserve the fermionic anticommutation relations. For example, { ak, a } { k = cosφk c k + sinφ k c k, sinφ k c k + cosφ k c } k = cos φ k + sin φ k =, 5.9 9

{a k, ã k } = { cosφ k c k + sinφ k c k, cosφ k c k sinφ k c k, } = cos φ k sin φ k +cos φ k sin φ k = 0, 5.0 and so on to compute the left hand side, we assume that c k satisfy the fermionic anticommutation relations, in other words, if c k are fermions, then so are a k. This Bogoliubov transformation is nothing but the orthogonal rotation of the matrix, in other words, ak cos φk sin φ ã = k ck k sin φ k cos φ k c k The Hamiltonian, when expressed in terms of c k, becomes H k = c γ β cos k β sin k k c k U ck U β sin k γ + β cos k c, U = k. 5. cos φk sin φ k. sin φ k cos φ k 5. We can now choose the orthogonal matrix U in such a way that the Hamiltonian becomes diagonal. Its eigenvalues are h k and h k where h k = γ β cos k + β sin k = γ + β γβ cos k = γ β + 4γβ sin k. This brings the Hamiltonian to the form Finally, this means H = k>0 5.3 H k = c hk 0 ck k c k 0 h k c. 5.4 k γ β + 4γβ sin k c k c k c k c k. 5.5 This is a very interesting Hamiltonian. In its ground state all fermionic states are empty. The ground state energy is then E 0 = k>0 γ β + 4γβ sin k = N h π π dk π γ β + 4γβ sin k. 5.6 To create an excitation, we populate one fermionic state, either with positive momentum q or with negative momentum q. That increases the energy by E q E 0 = γ β + 4γβ sin q. 5.7 The minimum energy increase occurs if we populate the state with q = 0. This gives E q=0 E 0 = γ β, 5.8 0

and leads to the previously advertised result for the correlation length. ξ = E q=0 E 0 = γ β. 5.9 The point where γ = β is the transition point. At this point the correlation length is infinity. Interestingly, at small q the single fermion excitation energy has a relativistic form E q E 0 /4 = γ β + γβq. 5.0 Here γ β plays the role of a relativistic mass. That s why the critical point γ = β is often called massless, while the phases with gaps in the spectrum are often called massive. Finally, we observe that this Hamiltonian 5.5 represents the Majorana fermions, not the usual Dirac fermions. That s because it represents the positive momentum particles whose energy increases with k, so they move to the right. But also it has negative momentum particles whose energy increases as k becomes more negative, so they move to the left. Real-world Dirac fermions would ve been moving either left and/or right at all k, not just at positive k. These kinds of special fermions which have their momentum restricted to either positive values c k moving to the right or negative values c k moving to the left are called Majorana fermions. 6 Diagonalizing the Hamiltonian - alternative scheme Here is another way to diagonalize the Hamiltonian, faster but involving a more convoluted mathematical construction. To diagonalize it, we employ the Bogoliubov transformations, c a A B a c = U a = a, c c = a a U. 6. B A the anticommutation relations. This is shorthand notations, in components this looks like this: U ij is a matrix, and N h [ ] c x = Axy a y + B xy a N h [ ] y, c x = B xy a y + A xya y. 6. y= Let us prove that U must be a unitary matrix. This is a consequence of the anticommutation relations. Let us introduce ψ j = y= a a. 6.3

This is again a shorthand notation. Here j goes from to N h. If j N h, then ψ j = a j. If N h < j N h, then ψ j = a j N h. We also introduce φ j = c c. 6.4 Now it should be clear that ψ jψ m + ψ m ψ j = δ ij. 6.5 Indeed, if m = j N h, this amounts to anticommutations between a m and a j. If m > N h, n > N h, this is again an anticommutator between a m N h, a n Nh. If m > N h and j < N h, this is an anticommutator between a m N h and a j, which is zero. Finally, if m N h, j > N h then this is an anticommutator between a m and a j Nh, which is also zero. In the same way, φ jφ m + φ m φ j = δ mj. 6.6 Now let us use that φ j = m U jm ψ m, φ j = m ψ mu mj. 6.7 Substitute these into 6.6, we find [ ψ n U nju ml ψ l + U ml ψ l ψ nu ] nj = δmj. 6.8 nl Now we note that ψ nψ l + ψ l ψ n = δ nl. 6.9 This gives This means UU = I, or U is a unitary matrix. Let us write the Hamiltonian 4.8 in the following way H = a a U mn U nj = δ mj. 6.0 n γ + β β β Γ Γ γ β a a = ψ Hψ. 6. Here H is a N h by N h matrix. We can now employ the Bogoliubov transformation, to write ψ = U φ. 6. Then H = φ UHU φ. 6.3

We choose U so that UHU is diagonal. Then in terms of c the Hamiltonian becomes H = H = k h k [ c k c k c k c k], 6.4 Here h k are eigenvalues of H. So the program is then to find the eigenvalues of H. We write γ + β H = where and Γ are matrices such that and Σ 3, Σ are Pauli matrices. Σ 3 + i β ΓΣ, 6.5 xy = δ x,y+ + δ x,y, Γ xy = δ x,y δ x,y+, 6.6 To find h k, we will use a series of unitary transformations. We first rotate Σ 3 into Σ by employing U = + iσ and H = U HU. We find H = γ + β Σ + i β ΓΣ 0 γ + β + Γ = γ + β. 6.7 Γ 0 The eigenvalue equations become These can be solved in terms of plane waves, hs x = γt x + βt x+, 6.8 ht x = γs x + βs x. 6.9 s x = s k e ikx, t x = t k e ikx. 6.0 Here x varies from to N h. k must vary from 0 to π because x is integer if k > π, then e ixk = e ixk π, so k larger than π is equivalent to k π. We now find This is equivalent to finding the eigenvalues of k = 0, π N h, 4π N h,..., π π N h. 6. hs k = γ + βe ik t k, ht k = γs k + βe ik s k. 6. 0 γ + βe ik γ + βe ik 0 3. 6.3

The eigenvalues are then h = γ + βe ik γ + βe ik = γ + β + γβ cosk = γ β + γβ cosk + 6.4 k hk = ± γ β + 4γβ cos. 6.5 It is convenient to introduce q = π k. As k varies from 0 to π, q varies from π to π. Then hq ± Thus the Hamiltonian becomes H = q γ β + 4γβ sin q γ β + 4γβ sin q Now we can construct the ground and excited states. fermions, then. 6.6 c qc q c q c q. 6.7 If we fill all the states with c qc q c q c q =. 6.8 This gives the minimal possible energy, that is the ground state. The ground state energy is thus this q, E 0 = q γ β + 4γβ sin q = N h π π dq q γ β π + 4γβ sin. 6.9 To create an excitation, we take a fermion with momentum q and remove it. Then for This increases energy by E q E 0 = c qc q c q c q =. 6.30 γ β + 4γβ sin q. 6.3 This gives excitation energy. It s also possible to remove more than one fermion, to get even higher energy. Interesting to note that at small q it has a relativistic form, This is the spectrum of particles of mass m = γ β. h q = γ β + γβq. 6.3 4

7 Critical exponent The lowest excited state is at q = 0, giving E q=0 E 0 = γ β. 7. This is the gap in the spectrum. The correlation length is the inverse gap and is given by ξ = γ β. 7. It diverges at γ = β. It is customary, if the correlation length is divergent at some critical T c, to denote the power of its divergence as ν: ξ We see that for the D Ising model ν =. Indeed, recall that β = K h = J h T, γ = e Kv T T c ν. 7.3 = e Jv T. 7.4 So β γ = J h T Jv e T. 7.5 The critical temperature T c is when J h e Jv Tc = 0. 7.6 T c This is when the gap vanishes and the correlation length is infinity. If T is close to T c, then with some slope α. So β γ J h T c + J v e Jv Tc T T c = αt T c 7.7 T c ξ ν = is one of Onsager s famous exact results. T T c. 7.8 Another result for the free energy easily follows as well. The free energy is given by F = T log Z. 7.9 Since Z is proportional to exp N v E 0 where E 0 is the ground state energy and N v is the length of the Ising model in the time direction, that gives F = T N v E 0 7.0 5

In other words, the ground state energy in quantum mechanics is the same as the free energy in statistical mechanics. In this problem π dq q E 0 N h γ β π π + 4γβ sin 7. So π dq q F = T N v N h γ β π π + 4γβ sin. 7. Notice that F is proportional to N v N h, which is the total number of sites in our D Ising model, which is like a volume of the system. So F is extensive proportional to volume, as we should have expected. The integral in the expression for F is a known integral, whose value is related to elliptic functions. We are not interested in its exact value. Rather we are only interested in the specific heat c = T F dt 7.3 and only at T close to T c. To find it we need to differentiate F with respect to T, remembering that γ and β are both temperature dependent. We recall that at T close to T c, β γ αt T c. So we can write π dq q F = T N v N h α π π T T c + 4γT βt sin. 7.4 Differentiating with respect to T once gives F π T = T N dq α T T c + vn γβ T sin q h π π α T T c + 4γT βt sin q π dq q N v N h α π π T T c + 4γT βt sin. 7.5 Computing the second derivative gives F = T N vn π h dq α T T c + γβ T sin q T π π α T T c + 4γT βt sin 3 q π dq α + γβ sin q T T N v N h π π α T T c + 4γT βt sin q N v N h π π dq π α T T c + T γβ sin q α T T c + 4γT βt sin q 6. 7.6

As T T c, the first and the third term remain finite. The second term can be estimated to behave as log T T c. So we find c log. 7.7 T T c This is another of Onsager s famous results. A standard notation in statistical mechanics is From here it follows that α = 0. c T T c α 7.8 7