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Fokker-Planck

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@ -1,8 +1,13 @@
#+TITLE: Quantum Monte Carlo #+TITLE: Quantum Monte Carlo
#+AUTHOR: Anthony Scemama, Claudia Filippi #+AUTHOR: Anthony Scemama, Claudia Filippi
#+SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup # SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup
# SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-bigblow.setup
#+STARTUP: latexpreview #+STARTUP: latexpreview
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#+HTML_HEAD: <link rel="alternate stylesheet" title="Zenburn" href="https://orgmode.org/worg/style/worg-zenburn.css" type="text/css" />
#+HTML_HEAD: <link rel="alternate stylesheet" title="Classic" href="https://orgmode.org/worg/style/worg-classic.css" type="text/css" />
* Introduction * Introduction
@ -14,10 +19,17 @@
computes a statistical estimate of the expectation value of the energy computes a statistical estimate of the expectation value of the energy
associated with a given wave function. associated with a given wave function.
Finally, we introduce the diffusion Monte Carlo (DMC) method which Finally, we introduce the diffusion Monte Carlo (DMC) method which
gives the exact energy of the H$_2$ molecule. gives the exact energy of the $H_2$ molecule.
Code examples will be given in Python and Fortran. Whatever language Code examples will be given in Python and Fortran. Whatever language
can be chosen. can be chosen.
We consider the stationary solution of the Schrödinger equation, so
the wave functions considered here are real: for an $N$ electron
system where the electrons move in the 3-dimensional space,
$\Psi : \mathbb{R}^{3N} \rightarrow \mathbb{R}$. In addition, $\Psi$
is defined everywhere, continuous and infinitely differentiable.
** Python ** Python
** Fortran ** Fortran
@ -45,7 +57,7 @@
$$ $$
when $a=1$, by checking that $\hat{H}\Psi(\mathbf{r}) = E\Psi(\mathbf{r})$ for 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 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})},
@ -54,6 +66,30 @@
is constant. is constant.
The probabilistic /expected value/ of an arbitrary function $f(x)$
with respect to a probability density function $p(x)$ is given by
$$ \langle f \rangle_p = \int_{-\infty}^\infty p(x)\, f(x)\,dx $$.
Recall that a probability density function $p(x)$ is non-negative
and integrates to one:
$$ \int_{-\infty}^\infty p(x)\,dx = 1 $$.
The electronic energy of a system is the expectation value of the
local energy $E(\mathbf{r})$ with respect to the $3N$-dimensional
electron density given by the square of the wave function:
\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}} \\
& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
= \langle E_L \rangle_{\Psi^2}
\end{eqnarray}
** Local energy ** Local energy
:PROPERTIES: :PROPERTIES:
:header-args:python: :tangle hydrogen.py :header-args:python: :tangle hydrogen.py
@ -63,9 +99,9 @@
The function accepts a 3-dimensional vector =r= as input arguments The function accepts a 3-dimensional vector =r= as input arguments
and returns the potential. and returns the potential.
$\mathbf{r}=\sqrt{x^2 + y^2 + z^2})$, so $\mathbf{r}=\sqrt{x^2 + y^2 + z^2}$, so
$$ $$
V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$ V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2}}
$$ $$
#+BEGIN_SRC python :results none #+BEGIN_SRC python :results none
@ -175,7 +211,7 @@ double precision function e_loc(a,r)
end function e_loc end function e_loc
#+END_SRC #+END_SRC
** Plot the local energy along the x axis ** Plot of the local energy along the $x$ axis
:PROPERTIES: :PROPERTIES:
:header-args:python: :tangle plot_hydrogen.py :header-args:python: :tangle plot_hydrogen.py
:header-args:f90: :tangle plot_hydrogen.f90 :header-args:f90: :tangle plot_hydrogen.f90
@ -184,7 +220,7 @@ end function e_loc
For multiple values of $a$ (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the For multiple values of $a$ (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the
local energy along the $x$ axis. local energy along the $x$ axis.
#+begin_src python :results output #+BEGIN_SRC python :results none
import numpy as np import numpy as np
import matplotlib.pyplot as plt import matplotlib.pyplot as plt
@ -270,72 +306,66 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
#+RESULTS: #+RESULTS:
[[file:plot.png]] [[file:plot.png]]
** Compute numerically the average energy ** Compute numerically the expectation value of the energy
:PROPERTIES: :PROPERTIES:
:header-args:python: :tangle energy_hydrogen.py :header-args:python: :tangle energy_hydrogen.py
:header-args:f90: :tangle energy_hydrogen.f90 :header-args:f90: :tangle energy_hydrogen.f90
:END: :END:
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}
If the space is discretized in small volume elements $\delta If the space is discretized in small volume elements $\delta
\mathbf{r}$, this last equation corresponds to a weighted average of \mathbf{r}$, the expression of \langle E_L \rangle_{\Psi^2}$ becomes
the local energy, where the weights are the values of the square of a weighted average of the local energy, where the weights are the
the wave function at $\mathbf{r}$ multiplied by the volume element: values of the probability density at $\mathbf{r}$ multiplied
by the volume element:
$$ $$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\; \langle E \rangle_{\Psi^2} \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} 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 In this section, we will compute a numerical estimate of the
$50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$. 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: Note: the energy is biased because:
- The energy is evaluated only inside the box - The volume elements are not infinitely small (discretization error)
- The volume elements are not infinitely small - The energy is evaluated only inside the box (incompleteness of the space)
#+BEGIN_SRC python :results output :exports both #+BEGIN_SRC python :results none
import numpy as np import numpy as np
from hydrogen import e_loc, psi from hydrogen import e_loc, psi
interval = np.linspace(-5,5,num=50) interval = np.linspace(-5,5,num=50)
delta = (interval[1]-interval[0])**3 delta = (interval[1]-interval[0])**3
r = np.array([0.,0.,0.]) r = np.array([0.,0.,0.])
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]: for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
E = 0. E = 0.
norm = 0. norm = 0.
for x in interval: for x in interval:
r[0] = x r[0] = x
for y in interval: for y in interval:
r[1] = y r[1] = y
for z in interval: for z in interval:
r[2] = z r[2] = z
w = psi(a,r) w = psi(a,r)
w = w * w * delta w = w * w * delta
E += w * e_loc(a,r) E += w * e_loc(a,r)
norm += w norm += w
E = E / norm E = E / norm
print(f"a = {a} \t E = {E}") print(f"a = {a} \t E = {E}")
#+end_src #+end_src
#+RESULTS: #+RESULTS:
: a = 0.1 E = -0.24518438948809218 : a = 0.1 E = -0.24518438948809218
: a = 0.2 E = -0.26966057967803525 : a = 0.2 E = -0.26966057967803525
: a = 0.5 E = -0.3856357612517407 : a = 0.5 E = -0.3856357612517407
: a = 0.9 E = -0.49435709786716214 : a = 0.9 E = -0.49435709786716214
: a = 1.0 E = -0.5 : a = 1.0 E = -0.5
: a = 1.5 E = -0.39242967082602226 : a = 1.5 E = -0.39242967082602226
: a = 2.0 E = -0.08086980667844901 : a = 2.0 E = -0.08086980667844901
#+begin_src f90 #+begin_src f90
@ -367,7 +397,6 @@ program energy_hydrogen
r(3) = x(l) r(3) = x(l)
w = psi(a(j),r) w = psi(a(j),r)
w = w * w * delta w = w * w * delta
energy = energy + w * e_loc(a(j), r) energy = energy + w * e_loc(a(j), r)
norm = norm + w norm = norm + w
end do end do
@ -380,7 +409,7 @@ program energy_hydrogen
end program energy_hydrogen end program energy_hydrogen
#+end_src #+end_src
To compile and run: To compile the Fortran and run it:
#+begin_src sh :results output :exports both #+begin_src sh :results output :exports both
gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
@ -401,20 +430,23 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
:header-args:f90: :tangle variance_hydrogen.f90 :header-args:f90: :tangle variance_hydrogen.f90
:END: :END:
The variance of the local energy measures the magnitude of the The variance of the local energy is a functional of $\Psi$
fluctuations of the local energy around the average. If the local which measures the magnitude of the fluctuations of the local
energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$) energy associated with $\Psi$ around the average:
the variance is zero.
$$ $$
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[ \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}} E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
$$ $$
If the local energy is constant (i.e. $\Psi$ is an eigenfunction of
$\hat{H}$) the variance is zero, so the variance of the local
energy can be used as a measure of the quality of a wave function.
Compute a numerical estimate of the variance of the local energy Compute a numerical estimate of the variance of the local energy
in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$. 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 none
import numpy as np import numpy as np
from hydrogen import e_loc, psi from hydrogen import e_loc, psi
@ -451,7 +483,6 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
s2 += w * (El - E)**2 s2 += w * (El - E)**2
s2 = s2 / norm s2 = s2 / norm
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}") print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
#+end_src #+end_src
#+RESULTS: #+RESULTS:
@ -541,6 +572,8 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
* Variational Monte Carlo * Variational Monte Carlo
Numerical integration with deterministic methods is very efficient
in low dimensions. When the number of dimensions becomes larger than
Instead of computing the average energy as a numerical integration Instead of computing the average energy as a numerical integration
on a grid, we will do a Monte Carlo sampling, which is an extremely 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 efficient method to compute integrals when the number of dimensions is
@ -582,7 +615,7 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
Write a function returning the average and statistical error of an Write a function returning the average and statistical error of an
input array. input array.
#+BEGIN_SRC python #+BEGIN_SRC python :results none
from math import sqrt from math import sqrt
def ave_error(arr): def ave_error(arr):
M = len(arr) M = len(arr)
@ -638,23 +671,23 @@ end subroutine ave_error
from hydrogen import * from hydrogen import *
from qmc_stats import * from qmc_stats import *
def MonteCarlo(a, nmax): def MonteCarlo(a, nmax):
E = 0. E = 0.
N = 0. N = 0.
for istep in range(nmax): for istep in range(nmax):
r = np.random.uniform(-5., 5., (3)) r = np.random.uniform(-5., 5., (3))
w = psi(a,r) w = psi(a,r)
w = w*w w = w*w
N += w N += w
E += w * e_loc(a,r) E += w * e_loc(a,r)
return E/N return E/N
a = 0.9 a = 0.9
nmax = 100000 nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)] X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X) E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}") print(f"E = {E} +/- {deltaE}")
#+END_SRC #+END_SRC
#+RESULTS: #+RESULTS:
: E = -0.4956255109300764 +/- 0.0007082875482711226 : E = -0.4956255109300764 +/- 0.0007082875482711226
@ -868,7 +901,12 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
#+RESULTS: #+RESULTS:
: E = -0.49606057056767766 +/- 1.3918807547836872E-004 : E = -0.49606057056767766 +/- 1.3918807547836872E-004
** Sampling with $\Psi^2$ ** Sampling with $\Psi^2$
:PROPERTIES:
:header-args:python: :tangle vmc.py
:header-args:f90: :tangle vmc.f90
:END:
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:
@ -876,19 +914,79 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
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 The expression for the energy will be simplified to the average of
average of the local energies, each with a weight of 1. 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 To generate the probability density $\Psi^2$, we consider a
diffusion scheme: 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
\[
F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
\]
following drifted diffusion scheme:
\[ \[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
\Psi(r)}{\Psi(r)} + \eta \sqrt{\tau} \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \eta \sqrt{\tau}
\] \]
where $\eta$ is a normally-distributed Gaussian random number. where $\eta$ is a normally-distributed Gaussian random number.
@ -902,7 +1000,66 @@ def drift(a,r):
return r * ar_inv return r * ar_inv
#+END_SRC #+END_SRC
#+RESULTS: #+BEGIN_SRC f90
subroutine drift(a,r,b)
implicit none
double precision, intent(in) :: a, r(3)
double precision, intent(out) :: b(3)
double precision :: ar_inv
ar_inv = -a / dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3))
b(:) = r(:) * ar_inv
end subroutine drift
#+END_SRC
Now we can write the Monte Carlo sampling
#+BEGIN_SRC f90
subroutine variational_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 variational_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 vmc.f90 -o vmc
./vmc
#+end_src
#+BEGIN_SRC python #+BEGIN_SRC python
def MonteCarlo(a,tau,nmax): def MonteCarlo(a,tau,nmax):
@ -932,11 +1089,8 @@ def MonteCarlo(a,tau,nmax):
N += 1. N += 1.
E += e_loc(a,r_old) E += e_loc(a,r_old)
return E/N return E/N
#+END_SRC
#+RESULTS:
#+BEGIN_SRC python :results output
nmax = 100000 nmax = 100000
tau = 0.1 tau = 0.1
X = [MonteCarlo(a,tau,nmax) for i in range(30)] X = [MonteCarlo(a,tau,nmax) for i in range(30)]
@ -950,15 +1104,15 @@ print(f"E = {E} +/- {deltaE}")
* Diffusion Monte Carlo * Diffusion Monte Carlo
We will now consider the H_2 molecule in a minimal basis composed of the We will now consider the H_2 molecule in a minimal basis composed of the
$1s$ orbitals of the hydrogen atoms: $1s$ orbitals of the hydrogen atoms:
$$ $$
\Psi(\mathbf{r}_1, \mathbf{r}_2) = \Psi(\mathbf{r}_1, \mathbf{r}_2) =
\exp(-(\mathbf{r}_1 - \mathbf{R}_A)) + \exp(-(\mathbf{r}_1 - \mathbf{R}_A)) +
$$ $$
where $\mathbf{r}_1$ and $\mathbf{r}_2$ denote the electron where $\mathbf{r}_1$ and $\mathbf{r}_2$ denote the electron
coordinates and \mathbf{R}_A$ and $\mathbf{R}_B$ the coordinates of coordinates and $\mathbf{R}_A$ and $\mathbf{R}_B$ the coordinates of
the nuclei. the nuclei.