L-4: Difference between revisions

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* Step 1:
* Step 1:
<center> <math>
<center> <math>
\overline{Z[x_t,t]^2 } = \exp\left[ \frac{D  t \delta(0)}{T^2}  \right]\int {\cal D} x_1\int  {\cal D} x_2 \exp\left[-  \int_0^t d \tau  \frac{1}{2T}[(\partial_\tau x_1)^2+ (\partial_\tau x_2)^2 ]+ \frac{D}{T^2} \delta[x_1(\tau)-x_2(\tau)]\right]
\overline{Z[x_t,t]^2 } = \exp\left[ \frac{D  t \delta_0}{T^2}  \right]\int {\cal D} x_1\int  {\cal D} x_2 \exp\left[-  \int_0^t d \tau  \frac{1}{2T}[(\partial_\tau x_1)^2+ (\partial_\tau x_2)^2 ]+ \frac{D}{T^2} \delta[x_1(\tau)-x_2(\tau)]\right]
</math></center>
</math></center>
Now you can change coordinate <math>X=(x_1+x_2)/2; \; u=x_1-x_2</math> and get:
Now you can change coordinate <math>X=(x_1+x_2)/2; \; u=x_1-x_2</math> and get:

Revision as of 17:14, 5 January 2024

Goal 1: final lecture on KPZ and directed polymers at finite dimension. We will show that for d>2 a "glass transition" takes place.

Goal 2: We will mention some ideas related to glass transition in true glasses.


Part 1: KPZ in finite dimension

  • In d=1 we found θ=1/3 and a glassy regime present at all temperatures. Moreover, the stationary solution tell us that Emin[x] is a Brownian motion in x. However this solution does not identify the actual distribution of Emin for a given x. In particular we have no idea from where Tracy Widom comes from.
  • In d>1 the exponents are not known. There is an exact solution for the Caley tree (infinite dimension) that predicts a freezing transition to an 1RSB phase (θ=0).

Let's do replica!

To make progress in disordered systems we have to go through the moments of the partition function. For simplicity we consider polymers starting in x0=0 and ending in xt. We recall that

  • V(x,τ) is a Gaussian field with
V(x,τ)=0,V(x,τ)V(x,τ)=Dδd(xx)δ(ττ)
  • From the Wick theorem, for a generic Gaussian W field we have
exp(W)=exp[W+12(W2W2)]

The first moment of the partition function is

Zt[xt,t]=x(0)=0x(t)=xt𝒟x1exp[1T0tdτ12(τx)2]exp[1TdτV(x,τ)]

Note that the term T2W2=dτ1dτ2V(x,τ1)V(x,τ2)=Dtδ0 has a short distance divergence due to the delta-function. Hence we can write:

Zt[x1]=1(2πtT)d/2exp[d2xt2tT]exp[Dtδ02T2]

Exercise L4-A: the second moment

  • Step 1:
Z[xt,t]2=exp[Dtδ0T2]𝒟x1𝒟x2exp[0tdτ12T[(τx1)2+(τx2)2]+DT2δ[x1(τ)x2(τ)]]

Now you can change coordinate X=(x1+x2)/2;u=x1x2 and get:

Z[xt,t]2=(Z[xt,t])2u(0)=0u(t)=0𝒟uexp[0tdτ14T(τu)2+DT2δ[u(τ)]]u(0)=0u(t)=0𝒟uexp[0tdτ14T(τu)2]

Part 2: Structural glasses