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Gaussian sampling

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Anthony Scemama 2021-01-07 11:07:18 +01:00
parent 8594cdfa39
commit 45bdaf2d41
6 changed files with 516 additions and 392 deletions
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QMC.org
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@ -550,7 +550,6 @@ Moreover, a Monte Carlo sampling will alow us to remove the bias due
to the discretization of space, and compute a statistical confidence
interval.
** Computation of the statistical error
:PROPERTIES:
:header-args:python: :tangle qmc_stats.py
@ -635,7 +634,6 @@ At every Monte Carlo step:
Compute the energy of the wave function with $a=0.9$.
#+BEGIN_SRC python :results output
from hydrogen import *
from qmc_stats import *
@ -714,13 +712,57 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
: E = -0.49588321986667677 +/- 7.1758863546737969E-004
** Gaussian sampling
: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
equation for the energy is changed into
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*}
#+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 equation for the energy is changed into
\[
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}}
@ -738,15 +780,14 @@ 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}
$$
#+BEGIN_SRC python
#+BEGIN_SRC python :results output
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)
#+END_SRC
#+RESULTS:
#+BEGIN_SRC python
def MonteCarlo(a,nmax):
E = 0.
N = 0.
@ -757,11 +798,7 @@ def MonteCarlo(a,nmax):
N += w
E += w * e_loc(a,r)
return E/N
#+END_SRC
#+RESULTS:
#+BEGIN_SRC python :results output
a = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
@ -770,8 +807,67 @@ print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+RESULTS:
: E = -0.4952488228427792 +/- 0.00011913174676540714
: E = -0.49507506093129827 +/- 0.00014164037765553668
#+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 * dsqrt(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 , 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 = 0.9
integer , 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 = -0.49606057056767766 +/- 1.3918807547836872E-004
** Sampling with $\Psi^2$
We will now use the square of the wave function to make the sampling:

28
qmc_stats.f90
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@ -13,3 +13,31 @@ subroutine ave_error(x,n,ave,err)
err = dsqrt(variance/dble(n))
endif
end subroutine ave_error
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