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Anthony Scemama 2021-02-03 00:05:04 +01:00
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@ -2758,349 +2758,6 @@ gfortran hydrogen.f90 qmc_stats.f90 pdmc.f90 -o pdmc
: A = 0.98963533333333342 +/- 6.3052128284666221E-005 : A = 0.98963533333333342 +/- 6.3052128284666221E-005
** H_2
We will now consider the H_2 molecule in a minimal basis composed of the
$1s$ orbitals of the hydrogen atoms:
$$
\Psi(\mathbf{r}_1, \mathbf{r}_2) =
\exp(-(\mathbf{r}_1 - \mathbf{R}_A)) +
$$
where $\mathbf{r}_1$ and $\mathbf{r}_2$ denote the electron
coordinates and $\mathbf{R}_A$ and $\mathbf{R}_B$ the coordinates of
the nuclei.
* Old sections to be removed :noexport:
:PROPERTIES:
:header-args:python: :tangle none
:header-args:f90: :tangle none
:END:
** Gaussian sampling :noexport:
:PROPERTIES:
:header-args:python: :tangle qmc_gaussian.py
:header-args:f90: :tangle qmc_gaussian.f90
:END:
We will now improve the sampling and allow to sample in the whole
3D space, correcting the bias related to the sampling in the box.
Instead of drawing uniform random numbers, we will draw Gaussian
random numbers centered on 0 and with a variance of 1.
Now the sampling probability can be inserted into the equation of the energy:
\[
E = \frac{\int P(\mathbf{r})
\frac{\left|\Psi(\mathbf{r})\right|^2}{P(\mathbf{r})}\,
\frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left|\Psi(\mathbf{r}) \right|^2}{P(\mathbf{r})} d\mathbf{r}}
\]
with the Gaussian probability
\[
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right).
\]
As the coordinates are drawn with probability $P(\mathbf{r})$, the
average energy can be computed as
$$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \frac{\left|\Psi(\mathbf{r}_i)\right|^2}{P(\mathbf{r}_i)} \delta \mathbf{r}
$$
*** Exercise
#+begin_exercise
Modify the program of the previous section to sample with
Gaussian-distributed random numbers. Can you see an reduction in
the statistical error?
#+end_exercise
**** Python
#+BEGIN_SRC python :results output
#!/usr/bin/env python3
from hydrogen import *
from qmc_stats import *
norm_gauss = 1./(2.*np.pi)**(1.5)
def gaussian(r):
return norm_gauss * np.exp(-np.dot(r,r)*0.5)
def MonteCarlo(a,nmax):
E = 0.
N = 0.
for istep in range(nmax):
r = np.random.normal(loc=0., scale=1.0, size=(3))
w = psi(a,r)
w = w*w / gaussian(r)
N += w
E += w * e_loc(a,r)
return E/N
a = 1.2
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+RESULTS:
: E = -0.49511014287471955 +/- 0.00012402022172236656
**** Fortran
#+BEGIN_SRC f90
double precision function gaussian(r)
implicit none
double precision, intent(in) :: r(3)
double precision, parameter :: norm_gauss = 1.d0/(2.d0*dacos(-1.d0))**(1.5d0)
gaussian = norm_gauss * dexp( -0.5d0 * (r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
end function gaussian
subroutine gaussian_montecarlo(a,nmax,energy)
implicit none
double precision, intent(in) :: a
integer*8 , intent(in) :: nmax
double precision, intent(out) :: energy
integer*8 :: istep
double precision :: norm, r(3), w
double precision, external :: e_loc, psi, gaussian
energy = 0.d0
norm = 0.d0
do istep = 1,nmax
call random_gauss(r,3)
w = psi(a,r)
w = w*w / gaussian(r)
norm = norm + w
energy = energy + w * e_loc(a,r)
end do
energy = energy / norm
end subroutine gaussian_montecarlo
program qmc
implicit none
double precision, parameter :: a = 1.2d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
integer :: irun
double precision :: X(nruns)
double precision :: ave, err
do irun=1,nruns
call gaussian_montecarlo(a,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 qmc_gaussian.f90 -o qmc_gaussian
./qmc_gaussian
#+end_src
#+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:
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, 2D \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
\]
where $\chi$ is a Gaussian random variable with zero mean and
variance $\delta t\,2D$.
**** 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
#!/usr/bin/env python3
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,dt,nmax):
sq_dt = np.sqrt(dt)
# 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 + dt * d_old + chi*sq_dt
N += 1.
E += e_loc(a,r_new)
r_old = r_new
return E/N
a = 1.2
nmax = 100000
dt = 0.2
X = [MonteCarlo(a,dt,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,dt,nmax,energy)
implicit none
double precision, intent(in) :: a, dt
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_dt, chi(3)
double precision, external :: e_loc
sq_dt = dsqrt(dt)
! 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(:) + dt * d_old(:) + chi(:)*sq_dt
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 = 1.2d0
double precision, parameter :: dt = 0.2d0
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,dt,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
* Project * Project