Goal 1: final lecture on KPZ and directed polymers at finite dimension. We will show that for  a "glass transition" takes place.
 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   we found we found and a glassy regime present at all temperatures. Moreover, the stationary solution tell us that and a glassy regime present at all temperatures. Moreover, the stationary solution tell us that![{\displaystyle E_{\min }[x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c066b9d27e99b30b243587508ec15d396314ca63) is a Brownian motion in is a Brownian motion in . However this solution does not identify the actual distribution of . However this solution does not identify the actual distribution of for a given for a given . In particular we have no idea from where Tracy Widom comes from. . In particular we have no idea from where Tracy Widom comes from.
- In  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 ( 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 ( ). ).
Let's do replica!
To make progress in disordered systems we have to go through the moments of the  partition function. We recall that 
 is a Gaussian field with is a Gaussian field with
 
- From the Wick theorem, for a generic Gaussian  field we have field we have
![{\displaystyle {\overline {\exp(W)}}=\exp \left[{\overline {W}}+{\frac {1}{2}}({\overline {W^{2}}}-{\overline {W}}^{2})\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec766c1903396019121638fcaf787e48d74575af) 
The first moment of the partition function is
 ![{\displaystyle {\overline {Z_{t}[x_{1}]}}=\int {\cal {D}}x_{1}\exp \left[-{\frac {1}{T}}\int _{0}^{t}d\tau {\frac {1}{2}}(\partial _{\tau }x)^{2}\right]{\overline {\exp \left[-{\frac {1}{T}}\int d\tau V(x,\tau )\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f38626a78375187450126c52323b82bb0beaff0) 
Note that the term  has a short distance divergence due to the delta-function.  Hence we can write:
 has a short distance divergence due to the delta-function.  Hence we can write:
 ![{\displaystyle {\overline {Z_{t}[x_{1}]}}={\frac {1}{\sqrt {2\pi tT}}}\exp \left[-{\frac {x^{2}}{2tT}}\right]\exp \left[{\frac {Dt\delta (0)}{2T^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/867223da90de7c3a74d84de3248ded7b71e1b738) 
Exercise L4-A: the second moment
 ![{\displaystyle {\overline {Z[x_{t},t]^{2}}}=\exp \left[{\frac {Dt\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 {1}{T^{2}}}\delta [x_{1}(\tau )-x_{2}(\tau )]\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58f98fe9f51685b709903052ff9c35fba87aae01) 
Now you can change coordinate  and get:
 and get:
 Failed to parse (syntax error): {\displaystyle  \overline{Z[x_t,t]^2} = (\overline{Z[x_t,t]})^2 \right]\int_{u(0)=0}^{u(t)=0} {\cal D} u  \exp\left[-  \int_0^t d \tau  \frac{1}{4T}(\partial_\tau u)^2+ \frac{1}{T^2} \delta[x_1(u(\tau)]\right] }
Part 2: Structural glasses