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Uniform sampling

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Anthony Scemama 2021-01-07 10:01:55 +01:00
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@ -1,9 +1,7 @@
#+TITLE: Quantum Monte Carlo
#+AUTHOR: Anthony Scemama, Claudia Filippi
#+SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup
#+STARTUP: latexpreview
#+STARTUP: indent
* Introduction
@ -33,162 +31,160 @@ can be chosen.
* Numerical evaluation of the energy
In this section we consider the Hydrogen atom with the following
wave function:
In this section we consider the Hydrogen atom with the following
wave function:
$$
\Psi(\mathbf{r}) = \exp(-a |\mathbf{r}|)
$$
$$
\Psi(\mathbf{r}) = \exp(-a |\mathbf{r}|)
$$
We will first verify that $\Psi$ is an eigenfunction of the Hamiltonian
We will first verify that $\Psi$ is an eigenfunction of the Hamiltonian
$$
\hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|}
$$
$$
\hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|}
$$
when $a=1$, by checking that $\hat{H}\Psi(\mathbf{r}) = E\Psi(\mathbf{r})$ for
all $\mathbf{r}$: we will check that the local energy, defined as
when $a=1$, by checking that $\hat{H}\Psi(\mathbf{r}) = E\Psi(\mathbf{r})$ for
all $\mathbf{r}$: we will check that the local energy, defined as
$$
E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
$$
$$
E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
$$
is constant.
is constant.
** Local energy
:PROPERTIES:
:header-args:python: :tangle hydrogen.py
:header-args:f90: :tangle hydrogen.f90
:header-args:f90: :tangle hydrogen.f90
:END:
*** Write a function which computes the potential at $\mathbf{r}$
The function accepts q 3-dimensional vector =r= as input arguments
and returns the potential.
The function accepts a 3-dimensional vector =r= as input arguments
and returns the potential.
$\mathbf{r}=\sqrt{x^2 + y^2 + z^2})$, so
$$
V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$
$$
$\mathbf{r}=\sqrt{x^2 + y^2 + z^2})$, so
$$
V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$
$$
#+BEGIN_SRC python
#+BEGIN_SRC python :results none
import numpy as np
def potential(r):
return -1. / np.sqrt(np.dot(r,r))
#+END_SRC
#+END_SRC
#+BEGIN_SRC f90
#+BEGIN_SRC f90
double precision function potential(r)
implicit none
double precision, intent(in) :: r(3)
potential = -1.d0 / dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )
end function potential
#+END_SRC
#+END_SRC
*** Write a function which computes the wave function at $\mathbf{r}$
The function accepts a scalar =a= and a 3-dimensional vector =r= as
input arguments, and returns a scalar.
The function accepts a scalar =a= and a 3-dimensional vector =r= as
input arguments, and returns a scalar.
#+BEGIN_SRC python
#+BEGIN_SRC python :results none
def psi(a, r):
return np.exp(-a*np.sqrt(np.dot(r,r)))
#+END_SRC
#+END_SRC
#+BEGIN_SRC f90
#+BEGIN_SRC f90
double precision function psi(a, r)
implicit none
double precision, intent(in) :: a, r(3)
psi = dexp(-a * dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
end function psi
#+END_SRC
#+END_SRC
*** Write a function which computes the local kinetic energy at $\mathbf{r}$
The function accepts =a= and =r= as input arguments and returns the
local kinetic energy.
The function accepts =a= and =r= as input arguments and returns the
local kinetic energy.
The local kinetic energy is defined as $$-\frac{1}{2}\frac{\Delta \Psi}{\Psi}$$.
The local kinetic energy is defined as $$-\frac{1}{2}\frac{\Delta \Psi}{\Psi}$$.
$$
\Psi(x,y,z) = \exp(-a\,\sqrt{x^2 + y^2 + z^2}).
$$
$$
\Psi(x,y,z) = \exp(-a\,\sqrt{x^2 + y^2 + z^2}).
$$
We differentiate $\Psi$ with respect to $x$:
We differentiate $\Psi$ with respect to $x$:
$$
\frac{\partial \Psi}{\partial x}
= \frac{\partial \Psi}{\partial r} \frac{\partial r}{\partial x}
= - \frac{a\,x}{|\mathbf{r}|} \Psi(x,y,z)
$$
$$
\frac{\partial \Psi}{\partial x}
= \frac{\partial \Psi}{\partial r} \frac{\partial r}{\partial x}
= - \frac{a\,x}{|\mathbf{r}|} \Psi(x,y,z)
$$
and we differentiate a second time:
and we differentiate a second time:
$$
\frac{\partial^2 \Psi}{\partial x^2} =
\left( \frac{a^2\,x^2}{|\mathbf{r}|^2} - \frac{a(y^2+z^2)}{|\mathbf{r}|^{3}} \right) \Psi(x,y,z).
$$
$$
\frac{\partial^2 \Psi}{\partial x^2} =
\left( \frac{a^2\,x^2}{|\mathbf{r}|^2} - \frac{a(y^2+z^2)}{|\mathbf{r}|^{3}} \right) \Psi(x,y,z).
$$
The Laplacian operator $\Delta = \frac{\partial^2}{\partial x^2} +
\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$
applied to the wave function gives:
The Laplacian operator $\Delta = \frac{\partial^2}{\partial x^2} +
\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$
applied to the wave function gives:
$$
\Delta \Psi (x,y,z) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(x,y,z)
$$
$$
\Delta \Psi (x,y,z) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(x,y,z)
$$
So the local kinetic energy is
$$
-\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right)
$$
So the local kinetic energy is
$$
-\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right)
$$
#+BEGIN_SRC python
#+BEGIN_SRC python :results none
def kinetic(a,r):
return -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))
#+END_SRC
#+END_SRC
#+BEGIN_SRC f90
#+BEGIN_SRC f90
double precision function kinetic(a,r)
implicit none
double precision, intent(in) :: a, r(3)
kinetic = -0.5d0 * (a*a - (2.d0*a) / &
dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ) )
end function kinetic
#+END_SRC
#+END_SRC
*** Write a function which computes the local energy at $\mathbf{r}$
The function accepts =x,y,z= as input arguments and returns the
local energy.
The function accepts =x,y,z= as input arguments and returns the
local energy.
$$
E_L(x,y,z) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) + V(x,y,z)
$$
$$
E_L(x,y,z) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) + V(x,y,z)
$$
#+BEGIN_SRC python
#+BEGIN_SRC python :results none
def e_loc(a,r):
return kinetic(a,r) + potential(r)
#+END_SRC
#+END_SRC
#+BEGIN_SRC f90
#+BEGIN_SRC f90
double precision function e_loc(a,r)
implicit none
double precision, intent(in) :: a, r(3)
double precision, external :: kinetic, potential
e_loc = kinetic(a,r) + potential(r)
end function e_loc
#+END_SRC
#+END_SRC
** Plot the local energy along the x axis
:PROPERTIES:
:header-args:python: :tangle plot_hydrogen.py
:header-args:f90: :tangle plot_hydrogen.f90
:END:
:LOGBOOK:
CLOCK: [2021-01-03 Sun 17:48]
:header-args:f90: :tangle plot_hydrogen.f90
:END:
For multiple values of $a$ (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the
local energy along the $x$ axis.
#+begin_src python
#+begin_src python :results output
import numpy as np
import matplotlib.pyplot as plt
@ -212,6 +208,8 @@ plt.legend()
plt.savefig("plot_py.png")
#+end_src
#+RESULTS:
[[./plot_py.png]]
@ -275,36 +273,35 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
** Compute numerically the average energy
:PROPERTIES:
:header-args:python: :tangle energy_hydrogen.py
:header-args:f90: :tangle energy_hydrogen.f90
:header-args:f90: :tangle energy_hydrogen.f90
:END:
We want to compute
We want to compute
\begin{eqnarray}
E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} \\
& = & \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\
& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
\end{eqnarray}
\begin{eqnarray}
E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} \\
& = & \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\
& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
\end{eqnarray}
If the space is discretized in small volume elements
$\delta x\, \delta y\, \delta z$, this last equation corresponds
to a weighted average of the local energy, where the weights are
the values of the square of the wave function at $(x,y,z)$
multiplied by the volume element:
If the space is discretized in small volume elements $\delta
\mathbf{r}$, this last equation corresponds to a weighted average of
the local energy, where the weights are the values of the square of
the wave function at $\mathbf{r}$ multiplied by the volume element:
$$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta x\, \delta y\, \delta z
$$
$$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta \mathbf{r}
$$
We now compute an numerical estimate of the energy in a grid of
$50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
We now compute an numerical estimate of the energy in a grid of
$50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
Note: the energy is biased because:
- The energy is evaluated only inside the box
- The volume elements are not infinitely small
#+BEGIN_SRC python :results output :exports both
Note: the energy is biased because:
- The energy is evaluated only inside the box
- The volume elements are not infinitely small
#+BEGIN_SRC python :results output :exports both
import numpy as np
from hydrogen import e_loc, psi
@ -314,31 +311,32 @@ delta = (interval[1]-interval[0])**3
r = np.array([0.,0.,0.])
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
E = 0.
norm = 0.
for x in interval:
r[0] = x
for y in interval:
r[1] = y
for z in interval:
r[2] = z
w = psi(a,r)
w = w * w * delta
E += w * e_loc(a,r)
norm += w
E = E / norm
print(f"a = {a} \t E = {E}")
#+end_src
#+RESULTS:
: a = 0.1 E = -0.24518438948809218
: a = 0.2 E = -0.26966057967803525
: a = 0.5 E = -0.3856357612517407
: a = 0.9 E = -0.49435709786716214
: a = 1.0 E = -0.5
: a = 1.5 E = -0.39242967082602226
: a = 2.0 E = -0.08086980667844901
E = 0.
norm = 0.
for x in interval:
r[0] = x
for y in interval:
r[1] = y
for z in interval:
r[2] = z
w = psi(a,r)
w = w * w * delta
E += w * e_loc(a,r)
norm += w
E = E / norm
print(f"a = {a} \t E = {E}")
#+end_src
#+RESULTS:
: a = 0.1 E = -0.24518438948809218
: a = 0.2 E = -0.26966057967803525
: a = 0.5 E = -0.3856357612517407
: a = 0.9 E = -0.49435709786716214
: a = 1.0 E = -0.5
: a = 1.5 E = -0.39242967082602226
: a = 2.0 E = -0.08086980667844901
#+begin_src f90
program energy_hydrogen
@ -400,23 +398,23 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
** Compute the variance of the local energy
:PROPERTIES:
:header-args:python: :tangle variance_hydrogen.py
:header-args:f90: :tangle variance_hydrogen.f90
:header-args:f90: :tangle variance_hydrogen.f90
:END:
The variance of the local energy measures the intensity of the
fluctuations of the local energy around the average. If the local
energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$)
the variance is zero.
The variance of the local energy measures the magnitude of the
fluctuations of the local energy around the average. If the local
energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$)
the variance is zero.
$$
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
$$
$$
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
$$
Compute an numerical estimate of the variance of the local energy
in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
Compute a numerical estimate of the variance of the local energy
in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
#+BEGIN_SRC python :results output :exports both
#+BEGIN_SRC python :results output :exports both
import numpy as np
from hydrogen import e_loc, psi
@ -450,20 +448,20 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
w = psi(a, r)
w = w * w * delta
El = e_loc(a, r)
s2 += w * (El - E)**2
s2 += w * (El - E)**2
s2 = s2 / norm
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
#+end_src
#+end_src
#+RESULTS:
: a = 0.1 E = -0.24518439 \sigma^2 = 0.02696522
: a = 0.2 E = -0.26966058 \sigma^2 = 0.03719707
: a = 0.5 E = -0.38563576 \sigma^2 = 0.05318597
: a = 0.9 E = -0.49435710 \sigma^2 = 0.00577812
: a = 1.0 E = -0.50000000 \sigma^2 = 0.00000000
: a = 1.5 E = -0.39242967 \sigma^2 = 0.31449671
: a = 2.0 E = -0.08086981 \sigma^2 = 1.80688143
#+RESULTS:
: a = 0.1 E = -0.24518439 \sigma^2 = 0.02696522
: a = 0.2 E = -0.26966058 \sigma^2 = 0.03719707
: a = 0.5 E = -0.38563576 \sigma^2 = 0.05318597
: a = 0.9 E = -0.49435710 \sigma^2 = 0.00577812
: a = 1.0 E = -0.50000000 \sigma^2 = 0.00000000
: a = 1.5 E = -0.39242967 \sigma^2 = 0.31449671
: a = 2.0 E = -0.08086981 \sigma^2 = 1.80688143
#+begin_src f90
program variance_hydrogen
@ -521,7 +519,7 @@ program variance_hydrogen
s2 = s2 / norm
print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
end do
end program variance_hydrogen
#+end_src
@ -543,15 +541,21 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
* Variational Monte Carlo
Instead of computing the average energy as a numerical integration
on a grid, we will do a Monte Carlo sampling, which is an extremely
efficient method to compute integrals in large dimensions.
Instead of computing the average energy as a numerical integration
on a grid, we will do a Monte Carlo sampling, which is an extremely
efficient method to compute integrals when the number of dimensions is
large.
Moreover, a Monte Carlo sampling will alow us to remove the bias due
to the discretization of space, and compute a statistical confidence
interval.
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
:header-args:f90: :tangle qmc_stats.f90
:END:
To compute the statistical error, you need to perform $M$
independent Monte Carlo calculations. You will obtain $M$ different
@ -579,7 +583,8 @@ $$
Write a function returning the average and statistical error of an
input array.
#+BEGIN_SRC python :results output
#+BEGIN_SRC python
from math import sqrt
def ave_error(arr):
M = len(arr)
assert (M>1)
@ -588,28 +593,53 @@ def ave_error(arr):
return (average, sqrt(variance/M))
#+END_SRC
#+RESULTS:
#+BEGIN_SRC f90
subroutine ave_error(x,n,ave,err)
implicit none
integer, intent(in) :: n
double precision, intent(in) :: x(n)
double precision, intent(out) :: ave, err
double precision :: variance
if (n == 1) then
ave = x(1)
err = 0.d0
else
ave = sum(x(:)) / dble(n)
variance = sum( (x(:) - ave)**2 ) / dble(n-1)
err = dsqrt(variance/dble(n))
endif
end subroutine ave_error
#+END_SRC
** Uniform sampling in the box
:PROPERTIES:
:header-args:python: :tangle qmc_uniform.py
:header-args:f90: :tangle qmc_uniform.f90
:END:
Write a function to perform a Monte Carlo calculation of the
average energy. At every Monte Carlo step,
In this section we write a function to perform a Monte Carlo
calculation of the average energy.
At every Monte Carlo step:
- Draw 3 uniform random numbers in the interval $(-5,-5,-5) \le
(x,y,z) \le (5,5,5)$
- Compute $\Psi^2 \times E_L$ at this point and accumulate the
result in E
- Compute $\Psi^2$ at this point and accumulate the result in N
- Draw 3 uniform random numbers in the interval $(-5,-5,-5) \le
(x,y,z) \le (5,5,5)$
- Compute $\Psi^2 \times E_L$ at this point and accumulate the
result in E
- Compute $\Psi^2$ at this point and accumulate the result in N
Once all the steps have been computed, return the average energy
computed on the Monte Carlo calculation.
Once all the steps have been computed, return the average energy
computed on the Monte Carlo calculation.
Then, write a loop to perform 30 Monte Carlo runs, and compute the
average energy and the associated statistical error.
In the main program, write a loop to perform 30 Monte Carlo runs,
and compute the average energy and the associated statistical error.
Compute the energy of the wave function with $a=0.9$.
Compute the energy of the wave function with $a=0.9$.
#+BEGIN_SRC python
#+BEGIN_SRC python :results output
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a, nmax):
E = 0.
N = 0.
@ -620,53 +650,103 @@ def MonteCarlo(a, nmax):
N += w
E += w * e_loc(a,r)
return E/N
#+END_SRC
#+BEGIN_SRC python
a = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+END_SRC
#+RESULTS:
: E = -0.4952626284319677 +/- 0.0006877988969872546
#+RESULTS:
: E = -0.4956255109300764 +/- 0.0007082875482711226
#+BEGIN_SRC f90
subroutine uniform_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
energy = 0.d0
norm = 0.d0
do istep = 1,nmax
call random_number(r)
r(:) = -5.d0 + 10.d0*r(:)
w = psi(a,r)
w = w*w
norm = norm + w
energy = energy + w * e_loc(a,r)
end do
energy = energy / norm
end subroutine uniform_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 uniform_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_uniform.f90 -o qmc_uniform
./qmc_uniform
#+end_src
#+RESULTS:
: E = -0.49588321986667677 +/- 7.1758863546737969E-004
** Gaussian sampling
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.
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
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
\[
E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}}
\]
with
\[
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right)
\]
\[
E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}}
\]
with
\[
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right)
\]
As the coordinates are drawn with probability $P(\mathbf{r})$, the
average energy can be computed as
As the coordinates are drawn with probability $P(\mathbf{r})$, the
average energy can be computed as
$$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r})} \delta x\, \delta y\, \delta z
$$
$$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r}_i)} \delta \mathbf{r}
$$
#+BEGIN_SRC python
#+BEGIN_SRC python
norm_gauss = 1./(2.*np.pi)**(1.5)
def gaussian(r):
return norm_gauss * np.exp(-np.dot(r,r)*0.5)
#+END_SRC
#+END_SRC
#+RESULTS:
#+RESULTS:
#+BEGIN_SRC python
#+BEGIN_SRC python
def MonteCarlo(a,nmax):
E = 0.
N = 0.
@ -677,58 +757,58 @@ def MonteCarlo(a,nmax):
N += w
E += w * e_loc(a,r)
return E/N
#+END_SRC
#+END_SRC
#+RESULTS:
#+RESULTS:
#+BEGIN_SRC python :results output
#+BEGIN_SRC python :results output
a = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+END_SRC
#+RESULTS:
: E = -0.4952488228427792 +/- 0.00011913174676540714
#+RESULTS:
: E = -0.4952488228427792 +/- 0.00011913174676540714
** Sampling with $\Psi^2$
We will now use the square of the wave function to make the sampling:
We will now use the square of the wave function to make the sampling:
\[
P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
\]
\[
P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
\]
Now, the expression for the energy will be simplified to the
average of the local energies, each with a weight of 1.
Now, the expression for the energy will be simplified to the
average of the local energies, each with a weight of 1.
$$
E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)}
$$
$$
E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)}
$$
To generate the probability density $\Psi^2$, we can use a drifted
diffusion scheme:
To generate the probability density $\Psi^2$, we can use a drifted
diffusion scheme:
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
\Psi(r)}{\Psi(r)} + \eta \sqrt{\tau}
\]
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
\Psi(r)}{\Psi(r)} + \eta \sqrt{\tau}
\]
where $\eta$ is a normally-distributed Gaussian random number.
where $\eta$ is a normally-distributed Gaussian random number.
First, write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
First, write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
#+BEGIN_SRC python
#+BEGIN_SRC python
def drift(a,r):
ar_inv = -a/np.sqrt(np.dot(r,r))
return r * ar_inv
#+END_SRC
#+END_SRC
#+RESULTS:
#+RESULTS:
#+BEGIN_SRC python
#+BEGIN_SRC python
def MonteCarlo(a,tau,nmax):
E = 0.
N = 0.
@ -753,36 +833,36 @@ def MonteCarlo(a,tau,nmax):
d_old = d_new
d2_old = d2_new
psi_old = psi_new
N += 1.
E += e_loc(a,r_old)
N += 1.
E += e_loc(a,r_old)
return E/N
#+END_SRC
#+END_SRC
#+RESULTS:
#+RESULTS:
#+BEGIN_SRC python :results output
#+BEGIN_SRC python :results output
nmax = 100000
tau = 0.1
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")
#+END_SRC
#+END_SRC
#+RESULTS:
: E = -0.4951783346213532 +/- 0.00022067316984271938
#+RESULTS:
: E = -0.4951783346213532 +/- 0.00022067316984271938
* Diffusion Monte Carlo
We will now consider the H_2 molecule in a minimal basis composed of the
$1s$ orbitals of the hydrogen atoms:
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.
$$
\Psi(\mathbf{r}_1, \mathbf{r}_2) =
\exp(-(\mathbf{r}_1 - \mathbf{R}_A)) +
$$
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.

28
energy_hydrogen.py
View File

@ -7,17 +7,17 @@ delta = (interval[1]-interval[0])**3
r = np.array([0.,0.,0.])
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
E = 0.
norm = 0.
for x in interval:
r[0] = x
for y in interval:
r[1] = y
for z in interval:
r[2] = z
w = psi(a,r)
w = w * w * delta
E += w * e_loc(a,r)
norm += w
E = E / norm
print(f"a = {a} \t E = {E}")
E = 0.
norm = 0.
for x in interval:
r[0] = x
for y in interval:
r[1] = y
for z in interval:
r[2] = z
w = psi(a,r)
w = w * w * delta
E += w * e_loc(a,r)
norm += w
E = E / norm
print(f"a = {a} \t E = {E}")

15
qmc_stats.f90 Normal file
View File

@ -0,0 +1,15 @@
subroutine ave_error(x,n,ave,err)
implicit none
integer, intent(in) :: n
double precision, intent(in) :: x(n)
double precision, intent(out) :: ave, err
double precision :: variance
if (n == 1) then
ave = x(1)
err = 0.d0
else
ave = sum(x(:)) / dble(n)
variance = sum( (x(:) - ave)**2 ) / dble(n-1)
err = dsqrt(variance/dble(n))
endif
end subroutine ave_error

7
qmc_stats.py Normal file
View File

@ -0,0 +1,7 @@
from math import sqrt
def ave_error(arr):
M = len(arr)
assert (M>1)
average = sum(arr)/M
variance = 1./(M-1) * sum( [ (x - average)**2 for x in arr ] )
return (average, sqrt(variance/M))

41
qmc_uniform.f90 Normal file
View File

@ -0,0 +1,41 @@
subroutine uniform_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
energy = 0.d0
norm = 0.d0
do istep = 1,nmax
call random_number(r)
r(:) = -5.d0 + 10.d0*r(:)
w = psi(a,r)
w = w*w
norm = norm + w
energy = energy + w * e_loc(a,r)
end do
energy = energy / norm
end subroutine uniform_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 uniform_montecarlo(a,nmax,X(irun))
enddo
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
end program qmc

19
qmc_uniform.py Normal file
View File

@ -0,0 +1,19 @@
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a, nmax):
E = 0.
N = 0.
for istep in range(nmax):
r = np.random.uniform(-5., 5., (3))
w = psi(a,r)
w = w*w
N += w
E += w * e_loc(a,r)
return E/N
a = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")

2
variance_hydrogen.f90
View File

@ -53,5 +53,5 @@ program variance_hydrogen
s2 = s2 / norm
print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
end do
end program variance_hydrogen

2
variance_hydrogen.py
View File

@ -31,6 +31,6 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
w = psi(a, r)
w = w * w * delta
El = e_loc(a, r)
s2 += w * (El - E)**2
s2 += w * (El - E)**2
s2 = s2 / norm
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")