1. Scaling. First carry out a scale transformation, where we zoom out such that all distances appear smaller by a factor b > 1: Δx Δx = Δx/b (307)

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1 45 XVIII. RENORMALIZATION GROUP TECHNIQUE The alternative technique to man field has to be significantly different, and rely on a new idea. This idea is the concept of scaling. Ben Widom, Leo Kadanoff, Michael Fisher, and particularly Ken Wilson, figured out that the fact that there is a diverging length scale at a critical point is crcial. It means, they argued, that the critical point is invariant under rescaling, but that as youconsider an off-critical system, and you zoom out, the system would appear less and less critical. Each RG transformation has two steps: 1. Scaling. First carry out a scale transformation, where we zoom out such that all distances appear smaller by a factor b > 1: Δx Δx = Δx/b (307) 2. Decimation. When we do this, though, the lattice we are considering becomes much more dense. To deal with that, each in their own way, said: how about we eliminate, or decimate, or truncate, or do something violent to the system that will result in the elimination of b d 1 sites, such that the new system looks the same as the old one. If the scaling and decimation together bring us back the sysem we started with, but with differnet paramters, then we can identify a critical point with an unstable fixed point of the mapping. Why? Away from criticality we have: ξ = ξ/b (308) which becomes less critical the more we renormalize. A stable fixed point, on the other hand, is identidified with stabel phases. It is where reduced correlations are at a minimum, and the sysmtem is either symmetry broken, with a single domain and no fluctuations ( T = 0 ), or a completely disordered fixed point ( T = ). The more complex the RG you can imagine that other monsters of fixed points could occur. A. 1d Ising Model The harmonic oscillator of statitical mechanics is teh 1d Ising model. H = J i S i S i+1 (309) What we are going to do is rescale by b = 2 - and then eliminate every other site. This will result in the same model exactly, but with a transformed J/T, we ll find. The rescaling part doesn t really do anything except for relabeling lengths: ξ = ξ/2 (310) clearly. We zoom out, and the lengths change units such that what used to be Δx = 1m is now Δx = 0.5m. The crucual step is the elimination step. This is the tricky part. There is no specific way to do it. To some degree it is an art. The 1d Ising model provides a good illustration. We start by writing the partition function as: Z = exp[βj(s 2n S 2n+1 + S 2n+1 S 2n+2 )] (311) {S 2n=±1} {S 2n+1=±1} n Now we carry out the sum over the odd spins. This can be done exaclty for each term in the product: exp[βj(s 2n S 2n+1 + S 2n+1 S 2n+2 )] = 2 cosh[βj(s 2n + S 2n+2 )] (312) S 2n=±1 The two steps of this RG transformation are described in fig. 5.

2 46 FIG. 5: The two steps of the RG transformation (not in order). Scaling down by a factor b = 2, and decimation of the odd spins. The resulting chain looks the same as the original chain but with coupling J. At this point we need to pause. We wanted something that looks like what we had before. This looks nothing like that. What happened? Well, it is not that differnet. cosh(βj(s 2n + S 2n+2 )) has only two possible results: { 2 cosh(2βj) S2n = S cosh(βj(s 2n + S 2n+2 )) = 2n+2 (313) 2 S 2n = S 2n+2 But whetehr the spins are the same or not is also simply reflected by the product S 2n S 2n+2. Hence we could try to find a function of the product (the domain wall variables... ) that is the same as the cosh. This could have the form: Ae βj S 2nS 2n+2 = 2 cosh[βj(s 2n + S 2n+2 )] (314) The main thing here is the renormalization rule for (βj). A here is not important. But A does play a role: it is the contribution to the partition function from the fluctuation of the spins eliminated. We would care for it if we were interested in the full free energy of the system. But at this point, we are not. So we simply divide the above equation for the tow cases: And we obtain the RG rule: Ae βj (1) Ae = 2 cosh(2βj) βj ( 1) 2 (315) e 2βJ = cosh(2βj) (316) Note that there is no reason to seperate J and β = 1/T. Thy always appear together. You can think of the RG rule as renormalizing J or T, or simply their product. The RG rule Eq. (316) has two limiting behaviors: (βj) (betaj) 2 βj 1 e 2(βJ) = 1 2 e2βj βj 1. (317) The RG rule gives rise to a flow diagram shown in Fig. 6. The flow is always towards βj = 0. There are two fixed points: βj = 0 - stable fixed point, which is also the disordered infinite-t point, and βj =, which is the FM fixed point. Each stable fixed point corresponds to a stable phase. FM and paramagentic are fair examples. Each unstable fixed point is a critical point. You flow away from it as you zoom out, as we know that the correlation length shrinks as we carry out the RG. Why is the FM fixed point here unstable? Because symmetry breaking is forbidden at T > 0 in 1d. Mermin Wagner Thm. Funny enough, we can already see from the RG what the correlation length is. Near the unstable fixed point, we see that the RG transformation of βj in the small T limit coincides with the RG for ξ if: ξ = e 2βJ (318) you can see that this indeed fits with our other results for the 1d Ising model. So what did we obtain? The two fixed points of the model, and the correlation length. This constitute a solution of the model as well, since we know when we have a broken symmtry, and what is the scaling behavior around it.

3 47 FIG. 6: The RG flow of βj for the 1d Ising model. An unstable fixed point exists at the βj = side. The arrows indicate the flow direction upon scale transformation. This is the zero-t FM fixed point. It is also unstable - this is a manifestation of the 1d nature - there should be no broken symmtery, so the FM fixed point is unstable. This is atypical. The βj = 0 point is also a fixed point, which corresponds to the T = disordered fixed point. FIG. 7: The 2d triangular lattice of spins could be RG d by creating block spins in this particular fashion. Note that the block spins, marked with tilde d letters (ĩ, j...) have an internal interaction marked by Ĥĩ and containing the red bonds, and a nearest neighbor interaction denoted by Vĩ j - maked by the dashed blue lines. B. RG of the triangular-lattice 2d Ising model Quite a bit more exciting is the 2d Ising model. It is a great demonstration of the more general structure of the real-space RG method. In 2d, the elimination trick doesn t work easily, since any spin eliminated produces coupling between all of its neighbors. So we need to do somehting else. What we do is termed coarse-graining. We replace a bunch of spins with a single block-spin which has the same character of the original spins. Could be ±1. Let s look more into the specifics. In the triangular lattice there are z = 6 nearest neighbor to each spin. So our expectation is that T c = 6J or β c J = 1 6 from mean field. We aim to do better. For that we need RG. We need two steps - rescaling and RG. For both, we need to decide how we will carry out the decimation. It seems so natural to group spins in 3 s, and replace each triangle with a block spin. But we can t just group them in any groups of three. The lattice emerging for the block spins must be the same as the original lattice. By playing around a bit, you can see that the system described in Fig. 7. Each triangle in the figure, should be described by a single spin. How would we do that? Majority rule. If the three spins in a block are S 1 + S 2 + S 3 = 1, 3 we would have M = 1 and S 1 + S 2 + S 3 = 1, 3 will correspond to 1, see Fig. 8. Let us denote the blocks with tilde d letters, and the three spins within a triangle ĩ as Sĩn with n = 1, 2, 3. Thus we have: Mĩ = sign(sĩ1 + Sĩ2 + Sĩ3) (319) Rescaling. Now that we know that we are going to make 3 spins into 1, we can figure out the scale b. We are

4 48 FIG. 8: A block spin pointing up could be the result of four different configuratoins. These are given here along with their boltzmann factor, e β Ĥ. going to reduce the density of spins by a factor ρ /ρ = 1/3. But ρ 1 a and therefore should scale like b 2. So shriking 2 things, or zooming out, by a factor b = 3 will bring us back to the original lattice. Decimation. This is the moment of truth. How do we decimate? We make extensive use of the Kronecker-δ: ( ) δ Mĩ sign(sĩ1 + Sĩ2 + Sĩ3) This function will allow us to project the spin configurations in the original lattice onto the new lattice. Explicitly, we write the partition function in terms of two sums - one over the S s and the other over the M s: Z = {Mĩ=±1} {SĨn =±1} ĩ [ ( )] βvĩ j e βĥĩ ĩ j δ Mĩ sign(sĩ1 + Sĩ2 + Sĩ3) e. (320) We did some funny things with this partition function. To read it, let s go over notation. We define: Ĥĩ = J(Sĩ1Sĩ2 + Sĩ2Sĩ3 + Sĩ3Sĩ1) as the hamiltonian with triangle ĩ. The we define Vĩ j as the interaction between two nearest neighbor triangles. Take a look at Fig. 7. We see that generically there are two bonds connecting two nn triangles. So we can write: O {Mĩ} = Vĩ j = JSĩ1(S j 2 + S j 3 ). The rest of the partition function becomes more obvious now. For each triangle we also need to impose the constraint. No problem. We stick it into the product. Finally: the RG consists of carrying out the S n sum. Block averaging. The partition function, Eq. (320), looks like basically averaging the interaction term over the hamiltonian of the individual triangles. So it is going to be necessary to necessary to know how to carry out an averaging for a given configuration of the block spins. Formally, consider an operator O which si a function of some spins {S n }. We define an average in a given configuration {Mĩ} as: [ ( )] e βĥĩ δ Mĩ sign(sĩ1 + Sĩ2 + Sĩ3) O {SĨn =±1} {SĨn =±1} ĩ [ ĩ ( ) ] (321) e βĥĩ δ Mĩ sign(sĩ1 + Sĩ2 + Sĩ3 ) Notice that there is an O in the numerator. A good example, which is going to be super useful, is a single spin. Consider Sĩ1. Clearly it is only affected by Mĩ. So all the terms that pertain to other triangles, other blocks, are going to cancel in the formula for the average, and the only thing remaining will be: S 1 M = S n=±1 n=1,2,3e β Ĥδ [M sign(s 1 + S 2 + S 3 )] S 1 Z M (322) Where we drop the ĩ indices since we are just thinking of a single triangle, and ZM = n=1,2,3e β Ĥδ [M sign(s 1 + S 2 + S 3 )] = e 3βJ + 3e βj is the partition of a triangle given that M is its S n=±1

5 block spin. Let s think that M = 1. Then on the top, there are three possibilities for S 1 = 1 - all S n = 1 - Boltzmann weight of e 3βJ since all bonds are satisfied, and two possibilities of S 2 = S 3, with weight e βj. There is one possibility for S 1 = 1, and for that S 2 = S 3 = 1 so that M = 1. So: S 1 M=1 = e3βj + 2e βj e βj e 3βJ + 3e βj It is easy to see that for M = 1, we ll just get the opposite answer. Therefore we have: Indeed, S 1 averages to be in the M direction, but a bit smaller than = e3βj + e βj e 3βJ + 3e βj (323) S 1 M=1 = e3βj + e βj e 3βJ M. (324) + 3e βj

6 50 C. Comment on RG transformation linearization near a fixed point - useful for the problem set If we have an RG transformation, with rescaling b: J = f(j) (325) which has an unstable fixed point: J = f(j ) (326) Then we can assume that: ξ 1 J J ν (327) Near the fixed point, though, we can ask, how does the distance δj = J J scales: δj = f(j) J f(j) δj (328) Where we used the assumption that f(j) is differentiable, and it always is. This amounts to a linear expansion of the transformation for the detuning from the fixed point. Here is the cool part: C δj ν = ξ = C bξ = b δj ν = b taking the right and left of this long relation we get: and f(j) ν = J=J f(j) C (329) δj ν J=J ν = b (330) J=J ln b ln f(j) J=J. (331)