35

XIV. CORRELATIONS AND SUSCEPTIBILITY

A. Correlations - saddle point approximation: the easy way out

Nopw it is time to face the menace - path integrals. We were tithering on the top of this ravine for long enough. We need to take the plunge. The meaning of the path integral is very simple - tofind the partition function of a system locallyfluctuating, we need to consider the all the possible patterns offlucutations, and the only way to do this is through a path integral. But is the path integral so bad? Well, if you believe that anyfluctuation is still quite costly, then the path integral, like the integral over the exponent of a quadratic function, is pretty much determined by where the argument of the exponent is a minimum, plus quadraticfluctuations. These belief is translated into equations via the saddle point approximation. Wefind the minimum of the free energy functional, and then expand upto second order in the exponent. 1 1 F [m(x),h] = ddx γ( m (x))2 +a(T T )m (x)2 + um (x)4 hm + ( m (x))2 ∂2F (219) L ∇ min − c min 2 min − min 2 − min � � � I write here deliberately the vague partial symbol, in fact, this should be the functional derivative. Does this look familiar? This should bring you back to the good old classical mechanics days. Where the Lagrangian is the only thing you wanted to know, and the Euler Lagrange equations were the only guidance necessary. Non of this non-commuting business, or partition function annoyance. Well, these times are back for a short time! In order tofind this, we do a variation:

m(x) =m min +δm (220) Upon assignment wefind: u F [m(x),h] = ddx γ( (m +δm)) 2 +a(T T )(m +δm) 2 + (m +δm) 4 h(m +δm) (221) L ∇ min − c min 2 min − min � � � Let’s expand this to second order:

d 2 2 d x γ (mmin) + δm 2γ m+γ( δm) → ∇ 2 ∇ · ∇ ∇ 2 +a (mmin) +δm 2am+aδm � � ∇ u 4 ·3 2 2 (222) + 2 mmin +2umminδm +3umminδm hm hδm] − min − Collecting terms in power ofδm, thefirst term is justF L[mmin]. The linear terms are, for now, much more important. They give us thefirst order variation. We say: δF [m(x)] F (1) = ddxδm(x) L = ddxδm(x) 2γ 2m+2am+2um 3 h (223) · δm(x) − ∇ − � � � � where we used integration by part to perform:

ddx δm m= ddx (δm m) δm 2m = ddxδm 2m + boundary terms. (224) ∇ ·∇ ∇· ·∇ − ∇ − ∇ � � � � � The thing in the brackets above is the functional derivative of the Landau free energy. The functional derivative is almost like a regular derivative, only it requires a bit of better normalization. For instance, if you want to know how muchm(x 1) influences the inntegral you would write: ∂F[m(x)] δF [m(x)] =d dx L (225) ∂m(x1) δm(x1) So that the functional derivatives also removes the integral sign: δ 1 ∂ = (226) δm(x) ddx ∂m(x) Equivalently, you could define the functional derivative such that: δ m(x�) =δ(x x �) (227) δm(x) − 36

this could be called the 3δ equation, I’m guessing. Equating the functional derivative to zero gives the Euler-Lagrange equations for the meanfieldm. Just like when functions have an extremum, derivative vanish, when functionals have an extremum, functional derivatives vanish. Thus, the Euler Lagrange equation is:

2γ 2m(x) 2am(x) + 2um(x) 3 =h(x) (228) ∇ − The second order term gives us the strength offluctuations, and it has the form:

1 δ2F [m(x)] F (2) = ddx ddx δm(x )δm(x ) L (229) 1 2 2 1 2 · δm(x )δm(x ) � � 1 2 with the second derivative, as expected:

δ2 = 2γ 2 + 2a + 12m2 (230) δm(x1)δm(x2) − ∇ Let’s leave this aside for now, and just deal with the minimization condition. First step is to ignore the laplacian term:

2am(x) + 2um(x)3 =h(x) (231) − This just gives us the self consistent codition form with which we started.

B. Correlations and Susceptibility from the saddle point

Let’s assume that we are abve the critical point. In this case we can neglect the cubic term, and then we have:

2γ 2m(x) 2a(T T )m(x)=h(x) (232) ∇ − − c This equation allows us so much more than what we could achieve before. For instance, before, we couldfind the susceptibility for a uniformfield. Now we canfind the susceptibility to a magneticfield wihich is any function of space. Since the equation is linear we can start with just a delta function. Let’s define the susceptibility function:

δm(x) χ(x, x�) = (233) δh(x�) This may still look stragne to those of you who are not so familiar with functional derivaties, but note that this is just the differential form of:

d m(x) = d x�χ(x, x�)h(x�) (234) � In particular, assume:

h(x) =hδ(x x �) (235) − and we obtain:

m(x) =χ(x, x �)h. (236)

This simple relation allows us to almost immediatelyfind the susceptibility. By exploiting the assumed translational invariance of the system, we note that:

χ(x, x�) =χ(x x �) (237) − and thus we have: a(T T ) h 2m(x) − c m(x) = δ(x) (238) ∇ − γ − 2γ 37

To warm up, let’s solve this in 1d.In most of space, there is no magneticfield, and therefore we need to solve the homogeneous equation: a 2m(x) m(x) = 0 (239) ∇ − γ which implies:

x/ξ m(x) = Ae± (240)

with γ ξ= (241) a(T T ) � − C Now, because we assume that the magnetization decays rather than explode at , we can restrict ourselves to: ±∞ x /ξ m(x) = Ae−| | (242)

Where we also use symmetry about thex = 0 point. TofindA we need to carefully integrate over the singularity: dm dm h = 2A/ξ= (243) dx − dx − − 2γ �x=� �x= � � � − � � which yields: � � hξ A= (244) 4γ Thus we obtain:

ξ r /ξ χ(r)= e−| | (245) 4γ So indeedξ is the corrletion length, whcih obeys: 1 ξ . (246) ∼ √t To obtain the uniform susceptibility, all we need is to integrate over space:

∞ 2 d ξ χh= d x�χ(x, x�)h(x�) = dxχ(x)h= (247) 2γ � � −∞ which yields:

γc = 2ν. (248)

C. Connection with correlations

How does the above result for the susceptibility exactly relates to correlation functions. As I will now show you, they are nearly exactly the same: 1 1 ξ(x, x�) = G(x, x�) = ( m m m m ) (249) T T � i j� − � i�� j� In order to show this rather intuitive result, we’ll make use of the path integral - consider this as good practice for the future. In the path integral formulation, again neglecting the fourth order term: 1 1 m(x) = D[m]m(x) exp ddx γ( m)2 +a(T T )m2 h(x)m(x) (250) � � Z −T ∇ − c − � � � � � � 38

andZ is the partition function, which is just the integral without the m(x) in front. Now, what is the functional derivative of this thing with respect toh(x �)? So it is just like the partial derivative with respect toh(x �) - this will d take down from the exponent everything that multipliesh(x �), including thed x. But the functional derivative bit, just cancels the differential. What is the stuff multiplyingh(x �)? it ism(x �)/T . We get: δ 1 m(x ) 1 m(x) = D[m]m(x) � exp ddx γ( m)2 +a(T T )m2 h(x)m(x) (251) δh(x )� � Z T −T ∇ − c − � � � � � � � But, we are acutally not done. If we have a non-zero average magnetization, as we do below the transition temperature, then there is yet another term:

1 δ 1 m(x�) 2 m(x) Z= 2 m(x) � � (252) −Z � �δh(x�) − Z � � T and everything together is: δ 1 χ(x, x�) = m(x) = ( mimj mi mj ) (253) δh(x�)� � T � � − � �� � Now, when we calculated the susceptibility above, we didn’t use stationary pahse, or the Euler Lagrange equations. Instead, we used the path integral directly. How do they relate to eachother? I’ll leave it as an exercise for you to show that when thefield theory that we are using is quadratic, than the stationary phase solution of thefield theory is exact. This stops being tru if the fourth order term is important.

D. Higher dimensions

I trust that you all can follow the procedure for solving the differential equation that yields the magnetizaiton. But I don’t trust that you will actually go home and do it on your own... So I’ll do it for you. The result is very important, becvause it’ll tell us something very fundamental about thefield m scaling dimension itself. In d dimensions, the mean-field differential equation that defines the susceptibility is the following: ∂2m(r) d 1 ∂m(r) 1 1 + − + (angle derivatives)m(r) m(x) = δ d(�x) (254) ∂r2 r ∂r r2 − ξ2 − The d-dimensional delta is this sharp function that is only nonzero at zero, and if we integrate over a small sphere in d-dimensions around zero we get 1. Now, we assume that our system is still quite simple. We leave complicated geometries and pathological setups for homework sets. So we still have rotational symmetry, at least at large length scales. Therefore, we expect that the angle-dependent part of the laplaccian vanishes. Therefore we get: ∂2m(r) d 1 ∂m(r) 1 + − m(x) = δ d(�x) (255) ∂r2 r ∂r − ξ2 − One can solve this equations exactly, by exactly I meant that the solution will be given by a rather unscrutible expression whicch will invariably have the word Bessel in it, and with rather high likelyhood, also the word associated. A much simple approach would be to just guess what the asymptotic form is going to be at the limit of interest, and just solve for that. The limit of interest is the far limit -r ξ. What do you think this function might be? well, it has to� have the exponential decay in it. Otherwise there wouldn’t be afinite correlation length. Apart from that, we are open to the idea of a power law in front of the exponent:

1 r/ξ m(r) e− (256) ∼ rn If we put this into the equation we get:

1 r/ξ 1 1 n(n + 1) d 1 1 n 1 e− + 2 + − + = 0 (257) rn ξ2 rξ r2 − r ξ r − ξ2 � � � � and since we are far away from the origin the delta function is zero. Also, since we are far away from the origin, we can forget about higher powers of 1/r. They will give us subleading corrections to the assymptotics. The leading power of 1/r is : 1 2n d 1 − = 0 (258) rn+1 ξ − ξ � � 39

Which yields: d 1 n= − (259) 2 Which is true as long asξ= . Now, what is the scaling� ∞ dimention ofm? By this we mean what is the dependence ofm on the length scale? In other words, we are doing dimensional analysis! Looking back at the differential equation we see that the RHS has d dimensions of length. The terms that multiplym on the other hand are divided by length squared. So m should have− dimensions of: 1 m d 2 (260) ∼ [L] − But the r dependence ofm just gives us (d 1)/2 what do we have at our disposal to make up for this deficiency? justξ. So here is thefinal answer: −

1 r/ξ m(r) d 1 d 3 e− (261) ∼ r −2 ξ −2 Now this has the right units! Here is the question I’ve been waiting for: What happens whenξ ? Looking back at our above results, all the leading 1/rn+1 terms are now afflicted with thtis diverging denominator→∞ - and thus vanish. But the next order term survives. It reads: 1 (n(n + 1) (d 1)n) = 0 (262) rn+2 − − and this has to vanish now. This gives: n=d 2 (263) − but we could have guessed that! Whenξ diverges, we nno longer have any length scale in the meanfield problem except the radius, and the microscopic cutoff. But the cutoff does not appear in this differential equation! And therefore me must have by the dimensional analisys above that atT C : 1 m(r) d 2 (264) ∼ r − And thus the correlations at the critical point also fall off as: 1 G(r) d 2 . (265) ∼ r − This is acutally quite fundamental: this exponent depends only on the dimension - d - and therefore is universal. The uiversal scaling exponent that joins the pantheon ofα,β,γ,δ,ν is nowη, which is defined as: 1 G(r) d 2+η . (266) ∼ r − And in the mean-field approximation, or saddle-point approximation that we are making, it is just: η=0 (267) In some senseη is the exponent that measures how much we screw up by doing saddle point. If it is actually non-zero in reality, there are severe deviations from meanfield! Then we say that thefield - in this casem - has obtained an anomalous scaling dimension.

XV. COMMENT ON LANDAU MEAN FIELD THEORY

In order to address the unwieldy path-integral, we used the saddle point approximation. This approximation is exact when thefield-theory is quadratic. The Isingfield theory is NOT quadratic, and that’s why in the equation we got: γ 2m a(T T )m bm 3 = 2h(x) (268) ∇ − − c − − 40

we have anm 3 term. No problem - we pretend that our theory is quadratic by perturbing around the minimum ofm:

m=m 0 +Δm (269) with

2 m0 = at/b (270) which solves the uniform part of the saddle point equation. Plugging this in, we get:

γ 2Δm a(T T )m bm3 atΔm 3bm 2Δm . . .= 2h(x) (271) ∇ − − c 0 − 0 − − 0 − them 0 only terms cancel, and we neglect higher order terms inΔm. This gives two equations: t >0γ 2Δm atΔm ∇ − = 2h(x) (272) t <0γ 2Δm 4atΔm − ∇ − � Mean-field was used to reduce the order of the Hamiltonian to quadratic only. This is good, as long asfluctuations are small. But sometimes - they are just not!

XVI. FAILURE OF LANDAU MEAN-FIELD THEORY

I keep talking about this looming doom - the failure of Landau meanfield theory... scary! When does it actually fail? How do wefind out? Here is what Ginzburg reaoned when faced with this problem. Meanfield fails iffluctuations - according to itself - overcome the average magnetization it predicts. So that if the system undergoes a phase transition, meanfield is actually at a loss as to where it happened... Butfluctuations on what scale? On the macroscopic scale there is a lot offluctuations - spin can just be up or down... Of course on the scale of the whole system - or in teh thermodynamic limit there are nofluctuations around the average. The only scale that makes sense is the correlation length. The medium of spins in a region the size of the correlation length is mostly uniform - that unit of the system mayfluctuate, and it is itsfluctuations that we would like to understand. We need to calculate: 2 1 d d d xm(x) m (273) � ξ d −� � � � �ξ � and compare it to:

m 2 (274) � � The latter is really easy: a (T T ) m 2 = | − C | t 2β (275) � � 2b ∼| | what about the awful average? It is also not that tough:

2 1 d 1 2 d d xm(x) = 2d dx dx� m(x)m(x�) m (276) � ξ d � ξ � � − � � � �ξ � � � � � but this is just an integral over the susceptibility:

1 d νd νd 1 νd γ d xG(x, x�) t χ= t = t − (277) → ξd ∼| | | | t γ | | � | | Now, we need to compare the two, which gives:

νd γ fluctuations t − νd γ 2β d/2 2 | | t − − t − 1 (278) magnetization squared ∼ t 2β ∼| | ∼| | � | | 41

Sincet does all but vanishes, in order for this gruesome combination to vanish near the transition - verifying our saddle point approach - we need:

d/2 2>0 (279) − and:

d >4 (280)

This is the Ginzburg criterion, it getsfilfilled above the upper critical dimension. So the theory as we know it so far is sure to fail when the dimensionality is smaller than 4. For instance - 3. How does it fail?η= 0, the heat capacity has a divergence - so thatα= 0. Does this so called Ginzburg criterion tell us something about� what to do? No. � The only thing that really could happen, is thatfluctuations necessitate dealing with the fullm 4 theory, not sweeping things under the carpet anymore. How can we deal with this? The answer is RG. In the next class Waheb will show you this new method of dealing with stat. mech. models. and after that, will come back and review the landau theory ith new eyes.

XVII. DOMAIN WALLS AND THE LOWER CRITICAL DIMENSION

In class, we discussed domain-wall excitations as the biggest threat to order and symmetry breaking. We said that if the free energy of a domain wall is negative, i.e., introducing a domain wall reduces the free energy of an ordered system by:

ΔF =U TS , (281) DW DW − DW then domain walls proliferate, and order is destroyed. In these notes we derive the estimates discussed in class for the entoropy, we define the lower critical dimension (LCD) from the scaling properties ofΔF DW , and discuss how to obtain the LCD from the Landau free energy approach.

A. Entropy (over)estimate of a domain wall ind dimensions

In a d-dimensional system with discrete symmetries, such as the Ising model, order might be hampered by domain walls. These are best described by a transition from an ordered state in one direction at one side of the system, x1 = L/2:m= σ < 0, to a state ordered in a different direction,m= σ > 0, in the other edge,x 1 =L/2 (assume− all sides� of� the system are of lengthL, and the system is a d-dimensional� � ’square’ lattice, with lattice constant set toa = 1). A domain wall must separate between these two ordered states. Since the symmetry of the system is discrete, a domain wall must be abrupt. Therefore in a two-dimensional system we assumed it is a line that meanders from the top of the system to the bottom, and estimated the entropy as:

eSDW (z 1) L (282) ∼ − This is an overestimate, but if we use it to show that domain walls can not proliferate above the lower critical dimensiond LCD, then this can suffice.

d How do we generalize this estimate to higher dimensions, with the coordinates marked as x i i=1? A domain wall becomes a (d 1) dimensional object, which we can paramterize as a function of the coordinates{ }x withi=2...d, − i where the function gives the locationx 1 of the domain wall (see Fig. 4):

x1 =x 1(x2, x3, . . . , xd) (283)

The domain wall may cross the line (x2, x3, . . . , xd) more than once (i.e., have folds in it), but let us neglect this possibility. The location of the domain wall at two adjacent points,x 1(x2, x3, . . . , xd) and, for instance,x 1(x2 + 1, x3, . . . , xd) should not be too different. Let us assume that it is bounded by a numberc of order 1:

x (x , x , . . . , x ) x (x + 1, x , . . . , x ) c. (284) | 1 2 3 d − 1 2 3 d |≤ 42

L m m x 1 (x 2) x 2

x 1

L

FIG. 4: A domain wall can be described as a curvex 1(x2, . . . , xd). To estimate the number of configurations of a domain wall, SDW e , we note that the at each point on the domain wall, e.g.,x 1(x2), the differencex 1(x2 + 1) x 1(x2)

The domain wall is by and large characterized by a single point, and the set of differencesx (x , x , . . . , x ) 1 2 3 d − x1(x2� , x3� , . . . , xd� ), where the two sets of coordinates, primed and unprimed, are nearest neighbors. Rougly speaking, there are 2c + 1 possibilities for each difference,x 1(x2, x3, . . . , xd) x 1(x2� , x3� , . . . , xd� ), and we need to specify all − d 1 possible steps between nearest neighbors on the domain wall manifold - 2(d 1) L − . Thus we can estimate the number of possiblities for constructing a domain wall by: − ·

d 1 S 2(d 1)L − e DW L (2c + 1) − (285) ∼ ·

where thefirst factor ofL is the reference point,x 1(0,0,0,..., 0). The entropy estimate we obtain is thus:

d 1 S lnL + 2(d 1)L − ln(2c + 1). (286) DW ∼ − In this estimate we neglected the constraints that require the sum of differences along a closed loop to vanish, and therefore this is a gross overestimate. In addition, we did not discriminate against domain walls that are highly costly energetically, which again means that we are over counting.

B. Comparison of entropy to energy

The energy of a domain wall is clearly proportional to its area, (in higher dimensions than 3, the volume of the hypersurface the domain wall describes). Therefore:

d 1 U > J L − (287) DW · whereJ is approximately the magnetic coupling between nearest neighbors. In fact, Eq. (287) uses the energy of a planar domain wall as a lower bound, and neglects the possibility of a domain wall meandering and folding. Because we now have an overestimate for the entropy, and an underestimate of the energy, we can write:

d 1 d 1 ΔF > J L − T lnL + 2(d 1)L − ln(2c + 1) (288) DW · − − � � 43

We see that for any dimensiond> 1 the lnL in the entropy can be neglected, and for: J T < (289) 2(d 1) ln(2c + 1) − we have:

ΔFDW >0, (290)

and because the free energy of domain walls is positive, they do not disorder the symmetry-broken state. d 1 From above we see that for any dimensiond> 1, the energy of a domain wall diverges asL − . In fact, as soon as a domain wall becomes an extended object, both its entropy and energy are extensive quantities, that depend on its d 1 (hypersurface) volume:S ,U L − . And sinceU > 0, there is always someT > 0, such that forT

UDW > T SDW . (291)

And asL ,ΔF =U TS + . →∞ DW DW − DW → ∞

C. Lower critical dimension

The lower critical dimension,d LCD, roughly speaking, is the dimension above which the free energy of a domain fall diverges, and therefore, at sufficiently low temperature (see Eq. 291), it isΔF DW = + . Furthermore, quite generally: ∞

d d ΔF L − LCD . (292) DW ∼

As you can see from the arguments above, for any system with discrete symmetries, we would haved LCD = 1.

D. Finding the critical dimension from the Landaufield theory

One way offinding the lower critical dimension is to ask: what is the free energy difference between a system which is periodic in thex coordinate [m(x = L/2) =m(x =L/2], and a system with antiperiodic boundary conditions 1 1 − 1 in thex 1 dimension [m(x1 = L/2) = m(x 1 =L/2]. In the latter case, we forced a domain wall to exist in the system. This free energy difference− can− be found from the saddle point equations of the Landau free energy. The Landau free energy functional for a second order Ising phase transition in the ordered phase is:

1 [m(x)] = ddx γ( m)2 a(T T ) m2 + um4 (293) FL ∇ −| − c | 2 � � � and we omit the symmetry breaking term. The saddle point equation is:

γ 2m a(T T ) m+ um 3 = 0. (294) − ∇ −| − c | Everywhere in this sectionx is a d-dimensional vector, andm=m(x) - the Landau order parameter. In the Ising case,m is real. To obtain the free energy, we’ll solve for the saddle point, and then substitute the functionm(x) we’ll find into the free energy functional, Eq. (293), and obtain the free energy itself (the minimum of the functional). In the case of periodic boundary conditions, the solution is:

m = m = a(T T ) /b. (295) P er ± 0 ± | − c | For the antiperiodic case, we can follow the prescription in PS� 2 forfinding the approximate profile of the magnetization with a domain wall: x m m tanh 1 (296) Anti ≈± 0 ξ

2 γ withξ = 2 a , the correlation length. In this case we set the domain wall at 0 - allowing the domain wall to appear elsewhere will| | increase its entropy only as lnL (as is the entropy of a domain wall in a 1-d chain), but we are looking d d for behavior that trumps the logarithm, withF L − LCD . DW ∼ 44

Now: ΔF =F F [m ] [m ] (297) DW Anti − P er ≈F L Anti −F L P er And this, after trivial manipulation, we can write as:

d = d x(γ (m Anti m P er) (m Anti +m P er) ∇ − ·∇ (298) 1 2 2 a(T T c) (mAnti �m P er)(mAnti +m P er) + u(mAnti m P er) (mAnti +m P er) −| − | − 2 − But whenx ξ, we have: � 1 � (m m )(m +m ) =m 2 m 2 0 Anti − P er Anti P er Anti − P er ≈ and therefore:

ξ FDW dx1 dx2 . . . dxd (γ (mAnti m P er) (m Anti +m P er) ≈ ξ ∇ − ·∇ − (299) a(T T ) (m � m� )(m +m ) + 1 u(m m )2(m +m )2 −| − c | Anti − P er Anti P er 2 Anti − P er Anti P er d 1 ξL − � ∼ The bottom contains the answer for the lower critical dimension:

disc dLCD = 1, (300) which applies to all systems with discrete symmetries described by a Landau theory.

E. Lower critical dimension of a system with a broken continuous symmetry

According to the Mermin-Wagner theorem, a system with continuous symmetry will have a higher LCD than a system with a discrete symmetry. Using the same definition ofd LCD as above, let us now demonstrate this. To be concrete, consider an X-Y model, which is described by a complex order parameter,ψ(x) =m x + imy. Consider, once more, a d-dimensional system. The Landau functional is: 1 [m(x)] = ddx γ ψ 2 a(T T ) ψ 2 + u ψ 4 . (301) FL |∇ | −| − c || | 2 | | � � � The saddle point equation is: γ 2ψ 2 a(T T ) ψ(ψ2 ψ 2) = 0. (302) − ∇ − | − c | 0 −| | 2 withψ 0 = b/ a , the magnitude of the uniform magnetization. As before, let us construct the periodic and anti-periodic solutions for| this| saddle point equation:

ψP er =ψ 0, (303) and:

i (x +L/2) ψ ψ e L 1 . (304) Anti ≈ 0 We assumed that atx 1 = L/2,ψ is positive and real in both cases. Note thatψ P er,ψ Anti both satisfy the grad-free part of the saddle point equations,− and therefore:

L/2 2 FDW dx2 . . . dxd dx1γ ψAnti ≈ L/2 |∇ | − � L/2 � (305) d 1 1 d 2 =L − γ L2 dx1 L − L/2 ∼ − � From here we see that the lower critical dimension for a system with a continuous symmetry is:

cont. dLCD = 2 (306)