Contrastive Divergence by Accelerated Langevin Dynamics . . . - PowerPoint PPT Presentation

. Contrastive Divergence by Accelerated Langevin Dynamics . . . Masayuki Ohzeki Kyoto University 2015/08/11 Japan-France Joint Seminar “New Frontiers in Non-equilibrium Physics of Glassy Materials” This work is in collaboration with M. Yasuda (Yamagata Univ.) and A. Ichiki (Nagoya Univ.) . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 1 / 20

. . . Accelerated Langevin dynamics 1 Formulation Example: double-valley potential Example: XY model . . . Boltzmann Machine Learning 2 Basic Contrastive divergence Preliminary result . . . Summary 3 . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 2 / 20

What is the accelerated stochastic dynamics? . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 3 / 20

. Ordinary Langevin dynamics . . The over-damped N -dimensional Langevin dynamics is given by √ d x = − ∂E ( x ) + 2 Td W , ∂ x where T is the temperature and W is the Wiener process. . . . . Equilibrium distribution . . The equilibrium state is P eq ( x ) = 1 ( − E ( x ) ) Z exp . T . . . Why do you use this dynamics? Investigation of the probability distribution in the dynamics Simulation of the natural stochastic dynamics . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 4 / 20

. Ordinary Langevin dynamics . . The over-damped N -dimensional Langevin dynamics is given by √ d x = − ∂E ( x ) + 2 Td W , ∂ x where T is the temperature and W is the Wiener process. . . . . Equilibrium distribution . . The equilibrium state is P eq ( x ) = 1 ( − E ( x ) ) Z exp . T . . . Why do you use this dynamics? Investigation of the probability distribution in the dynamics Simulation of the natural stochastic dynamics . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 4 / 20

. Ordinary Langevin dynamics . . The over-damped N -dimensional Langevin dynamics is given by √ d x = − ∂E ( x ) + 2 Td W , ∂ x where T is the temperature and W is the Wiener process. . . . . Equilibrium distribution . . The equilibrium state is P eq ( x ) = 1 ( − E ( x ) ) Z exp . T . . . Why do you use this dynamics? Investigation of the probability distribution in the dynamics Simulation of the natural stochastic dynamics . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 4 / 20

. Ordinary Langevin dynamics . . The over-damped N -dimensional Langevin dynamics is given by √ d x = − ∂E ( x ) + 2 Td W , ∂ x where T is the temperature and W is the Wiener process. . . . . Equilibrium distribution . . The equilibrium state is P eq ( x ) = 1 ( − E ( x ) ) Z exp . T . . . Why do you use this dynamics? Investigation of the probability distribution in the dynamics Simulation of the natural stochastic dynamics . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 4 / 20

In order to evaluate the distribution quickly, we do not necessarily use the natural force . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 5 / 20

. Accelerated Langevin dynamics . . Let us find the accelerated Langevin dynamics with the simple form of √ d x = − ∂E ( x ) + F ( x ) + 2 Td W , ∂ x where T is the temperature and d W is the Wiener process. . . . . Condition . . The steady state has the Gibbs-Boltzmann distribution P ss ( x ) = 1 ( − E ( x ) ) Z exp T . . . What force can hold the same steady state? . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 6 / 20

. Accelerated Langevin dynamics . . Let us find the accelerated Langevin dynamics with the simple form of √ d x = − ∂E ( x ) + F ( x ) + 2 Td W , ∂ x where T is the temperature and d W is the Wiener process. . . . . Condition . . The steady state has the Gibbs-Boltzmann distribution P ss ( x ) = 1 ( − E ( x ) ) Z exp T . . . What force can hold the same steady state? . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 6 / 20

. Accelerated Langevin dynamics . . Let us find the accelerated Langevin dynamics with the simple form of √ d x = − ∂E ( x ) + F ( x ) + 2 Td W , ∂ x where T is the temperature and d W is the Wiener process. . . . . Condition . . The steady state has the Gibbs-Boltzmann distribution P ss ( x ) = 1 ( − E ( x ) ) Z exp T . . . What force can hold the same steady state? . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 6 / 20

. Nontrivial force [M.Ohzeki and A. Ichiki (2015)] . . Find solution of the Fokker-Planck equation ( ) ∂P t ( x ) = − ∂ − ∂E ( x ) + F ( x ) − T ∂ P t ( x ) ∂t ∂ x ∂ x ∂ x The condition is reduced to 0 = − ∂ ∂ x ( F ( x ) P ss ( x )) . . . Equilibrium force F ( x ) = 0 Exponential force F ( x ) ∝ γ exp ( E ( x ) /T ) Rotational force ([ ∂E ( x ) ) ] [ ∂E ( x ) ] [ F ( x )] P ( i ) = γ − ∂ x ∂ x P ( i − 1) P ( i +1) where P ( i ) is the permutation of indices. . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 7 / 20

. Nontrivial force [M.Ohzeki and A. Ichiki (2015)] . . Find solution of the Fokker-Planck equation ( ) ∂P t ( x ) = − ∂ − ∂E ( x ) + F ( x ) − T ∂ P t ( x ) ∂t ∂ x ∂ x ∂ x The condition is reduced to 0 = − ∂ ∂ x ( F ( x ) P ss ( x )) . . . Equilibrium force F ( x ) = 0 Exponential force F ( x ) ∝ γ exp ( E ( x ) /T ) Rotational force ([ ∂E ( x ) ) ] [ ∂E ( x ) ] [ F ( x )] P ( i ) = γ − ∂ x ∂ x P ( i − 1) P ( i +1) where P ( i ) is the permutation of indices. . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 7 / 20

. Nontrivial force [M.Ohzeki and A. Ichiki (2015)] . . Find solution of the Fokker-Planck equation ( ) ∂P t ( x ) = − ∂ − ∂E ( x ) + F ( x ) − T ∂ P t ( x ) ∂t ∂ x ∂ x ∂ x The condition is reduced to 0 = − ∂ ∂ x ( F ( x ) P ss ( x )) . . . Equilibrium force F ( x ) = 0 Exponential force F ( x ) ∝ γ exp ( E ( x ) /T ) Rotational force ([ ∂E ( x ) ) ] [ ∂E ( x ) ] [ F ( x )] P ( i ) = γ − ∂ x ∂ x P ( i − 1) P ( i +1) where P ( i ) is the permutation of indices. . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 7 / 20

. Nontrivial force in duplicate system [M.Ohzeki and A. Ichiki (2015)] . . Find solution of the Fokker-Planck equation for a duplicate system ( ) ∂P t ( x 1 , x 2 ) − ∂ − ∂E ( x 1 ) + F 1 ( x 1 , x 2 ) − T ∂ = P t ( x 1 , x 2 ) ∂t ∂ x 1 ∂ x 1 ∂ x 1 ( ) − ∂ − ∂E ( x 2 ) + F 2 ( x 1 , x 2 ) − T ∂ P t ( x 1 , x 2 ) ∂ x 2 ∂ x 2 ∂ x 2 The condition is reduced to 0 = − ∂ ∂ ( F 1 ( x 1 , x 2 ) P ss ( x 1 ) P ss ( x 2 )) − ( F 1 ( x 1 , x 2 ) P ss ( x 1 ) P ss ( x 2 )) ∂ x 1 ∂ x 2 . . . Nontrivial force in the duplicate system γ ∂E ( x 2 ) F 1 ( x 1 , x 2 ) = ∂ x 2 − γ ∂E ( x 1 ) F 2 ( x 1 , x 2 ) = . ∂ x 1 . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 8 / 20

. What does the nontrivial force yield? . . Violation of the detailed balance condition ( γ � = 0 ) Convergence to nonequilibrium steady state Faster convergence than equilibrium system in analytical way by matrix analysis [A. Ichiki and M. Ohzeki (2013)] . . . . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 9 / 20

. Example: double-valley potential [M. Ohzeki and A. Ichiki (2015)] . . We set N = 1000 particles in a double-valley potential E ( x ) = − 1 2 x 2 + 1 4 x 4 . . . 2 1.5 1 E(x) 0.5 x eq 0 x 0 −0.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 x . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 10 / 20

. Example: double-valley potential [M. Ohzeki and A. Ichiki (2015)] . . We set N = 1000 particles in a double-valley potential E ( x ) = − 1 2 x 2 + 1 4 x 4 . . . at t = 5 in T = 1 . γ = 0 (red) vs γ = 1 (blue and purple). 1.2 (x 0 ,x 0 ) 1 1 0.8 0.8 0.6 0.6 <x> <x 2 > 0.4 0.4 0.2 (x eq ,x eq ) 0.2 0 0 −0.2 −0.4 −0.2 0 0.5 1 0 1 2 3 4 5 <x 1 > time . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 10 / 20

. Example: double-valley potential [M. Ohzeki and A. Ichiki (2015)] . . We set N = 1000 particles in a double-valley potential E ( x ) = − 1 2 x 2 + 1 4 x 4 . . . at t = 5 in T = 1 . γ = 0 (red) vs γ = 2 (blue and purple). 1.2 (x 0 ,x 0 ) 1 1 0.8 0.8 0.6 0.6 <x> <x 2 > 0.4 0.4 0.2 (x eq ,x eq ) 0.2 0 0 −0.2 −0.4 −0.2 0 0.5 1 0 1 2 3 4 5 <x 1 > time . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 10 / 20

. Example: double-valley potential [M. Ohzeki and A. Ichiki (2015)] . . We set N = 1000 particles in a double-valley potential E ( x ) = − 1 2 x 2 + 1 4 x 4 . . . at t = 5 in T = 1 . γ = 0 (red) vs γ = 5 (blue and purple). 1.2 (x 0 ,x 0 ) 1 1 0.8 0.8 0.6 0.6 <x> <x 2 > 0.4 0.4 0.2 (x eq ,x eq ) 0.2 0 0 −0.2 −0.4 −0.2 0 0.5 1 0 1 2 3 4 5 <x 1 > time . . . . . . M. Ohzeki (KU) Japan-France Joint Seminar 2015/08/11 10 / 20