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.