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Added drifted diffusion

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Anthony Scemama 2021-01-20 19:12:05 +01:00
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@ -988,111 +988,126 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
: E = -0.49478505004797046 +/- 2.0493795299184956E-004
: A = 0.51737800000000000 +/- 4.1827406733181444E-004
** Gaussian random number generator
** Sampling with $\Psi^2$
We will now use the square of the wave function to make the sampling:
\[
P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
\]
The expression for the energy will be simplified to the average of
the local energies, each with a weight of 1.
$$
E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)
$$
*** Importance sampling
:PROPERTIES:
:header-args:python: :tangle vmc.py
:header-args:f90: :tangle vmc.f90
:END:
To generate the probability density $\Psi^2$, we consider a
diffusion process characterized by a time-dependent density
$[\Psi(\mathbf{r},t)]^2$, which obeys the Fokker-Planck equation:
\[
\frac{\partial \Psi^2}{\partial t} = \sum_i D
\frac{\partial}{\partial \mathbf{r}_i} \left(
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
[\Psi(\mathbf{r},t)]^2.
\]
$D$ is the diffusion constant and $F_i$ is the i-th component of a
drift velocity caused by an external potential. For a stationary
density, \( \frac{\partial \Psi^2}{\partial t} = 0 \), so
To obtain Gaussian-distributed random numbers, you can apply the
[[https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform][Box Muller transform]] to uniform random numbers:
\begin{eqnarray*}
0 & = & \sum_i D
\frac{\partial}{\partial \mathbf{r}_i} \left(
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
[\Psi(\mathbf{r})]^2 \\
0 & = & \sum_i D
\frac{\partial}{\partial \mathbf{r}_i} \left(
\frac{\partial [\Psi(\mathbf{r})]^2}{\partial \mathbf{r}_i} -
F_i(\mathbf{r})\,[\Psi(\mathbf{r})]^2 \right) \\
0 & = &
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} -
\frac{\partial F_i }{\partial \mathbf{r}_i}[\Psi(\mathbf{r})]^2 -
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
z_1 &=& \sqrt{-2 \ln u_1} \cos(2 \pi u_2) \\
z_2 &=& \sqrt{-2 \ln u_1} \sin(2 \pi u_2)
\end{eqnarray*}
we search for a drift function which satisfies
Below is a Fortran implementation returning a Gaussian-distributed
n-dimensional vector $\mathbf{z}$. This will be useful for the
following sections.
*Fortran*
#+BEGIN_SRC f90 :tangle qmc_stats.f90
subroutine random_gauss(z,n)
implicit none
integer, intent(in) :: n
double precision, intent(out) :: z(n)
double precision :: u(n+1)
double precision, parameter :: two_pi = 2.d0*dacos(-1.d0)
integer :: i
call random_number(u)
if (iand(n,1) == 0) then
! n is even
do i=1,n,2
z(i) = dsqrt(-2.d0*dlog(u(i)))
z(i+1) = z(i) * dsin( two_pi*u(i+1) )
z(i) = z(i) * dcos( two_pi*u(i+1) )
end do
else
! n is odd
do i=1,n-1,2
z(i) = dsqrt(-2.d0*dlog(u(i)))
z(i+1) = z(i) * dsin( two_pi*u(i+1) )
z(i) = z(i) * dcos( two_pi*u(i+1) )
end do
z(n) = dsqrt(-2.d0*dlog(u(n)))
z(n) = z(n) * dcos( two_pi*u(n+1) )
end if
end subroutine random_gauss
#+END_SRC
** Generalized Metropolis algorithm
:PROPERTIES:
:header-args:python: :tangle vmc_metropolis.py
:header-args:f90: :tangle vmc_metropolis.f90
:END:
One can use more efficient numerical schemes to move the electrons.
But in that case, the Metropolis accepation step has to be adapted
accordingly: the acceptance
probability $A$ is chosen so that it is consistent with the
probability of leaving $\mathbf{r}_n$ and the probability of
entering $\mathbf{r}_{n+1}$:
\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1,
\frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}
\right)
\]
where $T(\mathbf{r}_n \rightarrow \mathbf{r}_{n+1})$ is the
probability of transition from $\mathbf{r}_n$ to
$\mathbf{r}_{n+1}$.
In the previous example, we were using uniform random
numbers. Hence, the transition probability was
\[
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
[\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
\text{constant}
\]
to obtain a second derivative on the left, we need the drift to be
of the form
So the expression of $A$ was simplified to the ratios of the squared
wave functions.
Now, if instead of drawing uniform random numbers
choose to draw Gaussian random numbers with mean 0 and variance
$\tau$, the transition probability becomes:
\[
F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\tau} \right]
\]
\[
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
[\Psi(\mathbf{r})]^2 \frac{\partial
g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} +
[\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2
\Psi^2}{\partial \mathbf{r}_i^2} +
\frac{\partial \Psi^2}{\partial \mathbf{r}_i}
g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
\]
$g = 1 / \Psi^2$ satisfies this equation, so
To sample even better the density, we can "push" the electrons
into in the regions of high probability, and "pull" them away from
the low-probability regions. This will mechanically increase the
acceptance ratios and improve the sampling.
To do this, we can add the drift vector
\[
F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
\]
\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}
\].
In statistical mechanics, Fokker-Planck trajectories are generated
by a Langevin equation:
The numerical scheme becomes a drifted diffusion:
\[
\frac{\partial \mathbf{r}(t)}{\partial t} = 2D \frac{\nabla
\Psi(\mathbf{r}(t))}{\Psi} + \eta
\]
where $\eta$ is a normally-distributed fluctuating random force.
Discretizing this differential equation gives the following drifted
diffusion scheme:
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau\, 2D \frac{\nabla
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
\]
where $\chi$ is a Gaussian random variable with zero mean and
variance $\tau\,2D$.
**** Exercise 1
where $\chi$ is a Gaussian random variable with zero mean and
variance $\tau$.
The transition probability becomes:
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\tau} \right]
\]
*** Exercise 1
#+begin_exercise
Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
@ -1117,161 +1132,11 @@ subroutine drift(a,r,b)
end subroutine drift
#+END_SRC
**** TODO Exercise 2
*** Exercise 2
#+begin_exercise
Sample $\Psi^2$ approximately using the drifted diffusion scheme,
with a diffusion constant $D=1/2$. You can use a time step of
0.001 a.u.
#+end_exercise
*Python*
#+BEGIN_SRC python :results output
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,tau,nmax):
sq_tau = np.sqrt(tau)
# Initialization
E = 0.
N = 0.
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
for istep in range(nmax):
d_old = drift(a,r_old)
chi = np.random.normal(loc=0., scale=1.0, size=(3))
r_new = r_old + tau * d_old + chi*sq_tau
N += 1.
E += e_loc(a,r_new)
r_old = r_new
return E/N
a = 0.9
nmax = 100000
tau = 0.2
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+RESULTS:
: E = -0.4858534479298907 +/- 0.00010203236131158794
*Fortran*
#+BEGIN_SRC f90
subroutine variational_montecarlo(a,tau,nmax,energy)
implicit none
double precision, intent(in) :: a, tau
integer*8 , intent(in) :: nmax
double precision, intent(out) :: energy
integer*8 :: istep
double precision :: norm, r_old(3), r_new(3), d_old(3), sq_tau, chi(3)
double precision, external :: e_loc
sq_tau = dsqrt(tau)
! Initialization
energy = 0.d0
norm = 0.d0
call random_gauss(r_old,3)
do istep = 1,nmax
call drift(a,r_old,d_old)
call random_gauss(chi,3)
r_new(:) = r_old(:) + tau * d_old(:) + chi(:)*sq_tau
norm = norm + 1.d0
energy = energy + e_loc(a,r_new)
r_old(:) = r_new(:)
end do
energy = energy / norm
end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: tau = 0.2
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
integer :: irun
double precision :: X(nruns)
double precision :: ave, err
do irun=1,nruns
call variational_montecarlo(a,tau,nmax,X(irun))
enddo
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
end program qmc
#+END_SRC
#+begin_src sh :results output :exports both
gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
./vmc
#+end_src
#+RESULTS:
: E = -0.48584030499187431 +/- 1.0411743995438257E-004
*** Generalized Metropolis algorithm
:PROPERTIES:
:header-args:python: :tangle vmc_metropolis.py
:header-args:f90: :tangle vmc_metropolis.f90
:END:
Discretizing the differential equation to generate the desired
probability density will suffer from a discretization error
leading to biases in the averages. The [[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][Metropolis-Hastings
sampling algorithm]] removes exactly the discretization errors, so
large time steps can be employed.
After the new position $\mathbf{r}_{n+1}$ has been computed, an
additional accept/reject step is performed. The acceptance
probability $A$ is chosen so that it is consistent with the
probability of leaving $\mathbf{r}_n$ and the probability of
entering $\mathbf{r}_{n+1}$:
\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1,
\frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}
\right)
\]
where $T(\mathbf{r}_n \rightarrow \mathbf{r}_{n+1})$ is the
probability of transition from $\mathbf{r}_n$ to $\mathbf{r}_{n+1})$.
In our Hydrogen atom example, $P$ is $\Psi^2$ and $g$ is a
solution of the discretized Fokker-Planck equation:
\begin{eqnarray*}
P(r_{n}) &=& \Psi^2(\mathbf{r}_n) \\
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
\frac{1}{(4\pi\,D\,\tau)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} - 2D \frac{\nabla
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{4D\,\tau} \right]
\end{eqnarray*}
The accept/reject step is the following:
- Compute $A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$.
- Draw a uniform random number $u$
- if $u \le A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, accept
the move
- if $u>A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, reject
the move: set $\mathbf{r}_{n+1} = \mathbf{r}_{n}$, but *don't remove the sample from the average!*
The /acceptance rate/ is the ratio of the number of accepted step
over the total number of steps. The time step should be adapted so
that the acceptance rate is around 0.5 for a good efficiency of
the simulation.
**** Exercise
#+begin_exercise
Modify the previous program to introduce the accept/reject step.
You should recover the unbiased result.
Adjust the time-step so that the acceptance rate is 0.5.
Modify the previous program to introduce the drifted diffusion scheme.
(This is a necessary step for the next section).
#+end_exercise
*Python*
@ -1407,7 +1272,6 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
: E = -0.49499990423528023 +/- 1.5958250761863871E-004
: A = 0.78861366666666655 +/- 3.5096729498002445E-004
* TODO Diffusion Monte Carlo
:PROPERTIES:
:header-args:python: :tangle dmc.py
@ -1577,11 +1441,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
the nuclei.
* Appendix
* Appendix :noexport:
** Gaussian sampling :noexport:
:PROPERTIES:
@ -1595,48 +1455,6 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
Instead of drawing uniform random numbers, we will draw Gaussian
random numbers centered on 0 and with a variance of 1.
To obtain Gaussian-distributed random numbers, you can apply the
[[https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform][Box Muller transform]] to uniform random numbers:
\begin{eqnarray*}
z_1 &=& \sqrt{-2 \ln u_1} \cos(2 \pi u_2) \\
z_2 &=& \sqrt{-2 \ln u_1} \sin(2 \pi u_2)
\end{eqnarray*}
Here is a Fortran implementation returning a Gaussian-distributed
n-dimensional vector $\mathbf{z}$;
*Fortran*
#+BEGIN_SRC f90 :tangle qmc_stats.f90
subroutine random_gauss(z,n)
implicit none
integer, intent(in) :: n
double precision, intent(out) :: z(n)
double precision :: u(n+1)
double precision, parameter :: two_pi = 2.d0*dacos(-1.d0)
integer :: i
call random_number(u)
if (iand(n,1) == 0) then
! n is even
do i=1,n,2
z(i) = dsqrt(-2.d0*dlog(u(i)))
z(i+1) = z(i) * dsin( two_pi*u(i+1) )
z(i) = z(i) * dcos( two_pi*u(i+1) )
end do
else
! n is odd
do i=1,n-1,2
z(i) = dsqrt(-2.d0*dlog(u(i)))
z(i+1) = z(i) * dsin( two_pi*u(i+1) )
z(i) = z(i) * dcos( two_pi*u(i+1) )
end do
z(n) = dsqrt(-2.d0*dlog(u(n)))
z(n) = z(n) * dcos( two_pi*u(n+1) )
end if
end subroutine random_gauss
#+END_SRC
Now the sampling probability can be inserted into the equation of the energy:
\[
@ -1755,3 +1573,192 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
#+RESULTS:
: E = -0.49517104619091717 +/- 1.0685523607878961E-004
** Improved sampling with $\Psi^2$ :noexport:
*** Importance sampling
:PROPERTIES:
:header-args:python: :tangle vmc.py
:header-args:f90: :tangle vmc.f90
:END:
To generate the probability density $\Psi^2$, we consider a
diffusion process characterized by a time-dependent density
$[\Psi(\mathbf{r},t)]^2$, which obeys the Fokker-Planck equation:
\[
\frac{\partial \Psi^2}{\partial t} = \sum_i D
\frac{\partial}{\partial \mathbf{r}_i} \left(
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
[\Psi(\mathbf{r},t)]^2.
\]
$D$ is the diffusion constant and $F_i$ is the i-th component of a
drift velocity caused by an external potential. For a stationary
density, \( \frac{\partial \Psi^2}{\partial t} = 0 \), so
\begin{eqnarray*}
0 & = & \sum_i D
\frac{\partial}{\partial \mathbf{r}_i} \left(
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
[\Psi(\mathbf{r})]^2 \\
0 & = & \sum_i D
\frac{\partial}{\partial \mathbf{r}_i} \left(
\frac{\partial [\Psi(\mathbf{r})]^2}{\partial \mathbf{r}_i} -
F_i(\mathbf{r})\,[\Psi(\mathbf{r})]^2 \right) \\
0 & = &
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} -
\frac{\partial F_i }{\partial \mathbf{r}_i}[\Psi(\mathbf{r})]^2 -
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
\end{eqnarray*}
we search for a drift function which satisfies
\[
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
[\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
\]
to obtain a second derivative on the left, we need the drift to be
of the form
\[
F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
\]
\[
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
[\Psi(\mathbf{r})]^2 \frac{\partial
g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} +
[\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2
\Psi^2}{\partial \mathbf{r}_i^2} +
\frac{\partial \Psi^2}{\partial \mathbf{r}_i}
g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
\]
$g = 1 / \Psi^2$ satisfies this equation, so
\[
F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
\]
In statistical mechanics, Fokker-Planck trajectories are generated
by a Langevin equation:
\[
\frac{\partial \mathbf{r}(t)}{\partial t} = 2D \frac{\nabla
\Psi(\mathbf{r}(t))}{\Psi} + \eta
\]
where $\eta$ is a normally-distributed fluctuating random force.
Discretizing this differential equation gives the following drifted
diffusion scheme:
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau\, 2D \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
\]
where $\chi$ is a Gaussian random variable with zero mean and
variance $\tau\,2D$.
**** Exercise 2
#+begin_exercise
Sample $\Psi^2$ approximately using the drifted diffusion scheme,
with a diffusion constant $D=1/2$. You can use a time step of
0.001 a.u.
#+end_exercise
*Python*
#+BEGIN_SRC python :results output
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,tau,nmax):
sq_tau = np.sqrt(tau)
# Initialization
E = 0.
N = 0.
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
for istep in range(nmax):
d_old = drift(a,r_old)
chi = np.random.normal(loc=0., scale=1.0, size=(3))
r_new = r_old + tau * d_old + chi*sq_tau
N += 1.
E += e_loc(a,r_new)
r_old = r_new
return E/N
a = 0.9
nmax = 100000
tau = 0.2
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+RESULTS:
: E = -0.4858534479298907 +/- 0.00010203236131158794
*Fortran*
#+BEGIN_SRC f90
subroutine variational_montecarlo(a,tau,nmax,energy)
implicit none
double precision, intent(in) :: a, tau
integer*8 , intent(in) :: nmax
double precision, intent(out) :: energy
integer*8 :: istep
double precision :: norm, r_old(3), r_new(3), d_old(3), sq_tau, chi(3)
double precision, external :: e_loc
sq_tau = dsqrt(tau)
! Initialization
energy = 0.d0
norm = 0.d0
call random_gauss(r_old,3)
do istep = 1,nmax
call drift(a,r_old,d_old)
call random_gauss(chi,3)
r_new(:) = r_old(:) + tau * d_old(:) + chi(:)*sq_tau
norm = norm + 1.d0
energy = energy + e_loc(a,r_new)
r_old(:) = r_new(:)
end do
energy = energy / norm
end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: tau = 0.2
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
integer :: irun
double precision :: X(nruns)
double precision :: ave, err
do irun=1,nruns
call variational_montecarlo(a,tau,nmax,X(irun))
enddo
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
end program qmc
#+END_SRC
#+begin_src sh :results output :exports both
gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
./vmc
#+end_src
#+RESULTS:
: E = -0.48584030499187431 +/- 1.0411743995438257E-004