22 KiB
Quantum Monte Carlo
- Introduction
- Numerical evaluation of the energy
- Variational Monte Carlo
- Diffusion Monte Carlo
Introduction
We propose different exercises to understand quantum Monte Carlo (QMC) methods. In the first section, we propose to compute the energy of a hydrogen atom using numerical integration. The goal of this section is to introduce the local energy. Then we introduce the variational Monte Carlo (VMC) method which computes a statistical estimate of the expectation value of the energy associated with a given wave function. Finally, we introduce the diffusion Monte Carlo (DMC) method which gives the exact energy of the H$_2$ molecule.
Code examples will be given in Python and Fortran. Whatever language can be chosen.
Python
Fortran
- 1.d0
- external
- r(:) = 0.d0
- a = (/ 0.1, 0.2 /)
- size(x)
Numerical evaluation of the energy
In this section we consider the Hydrogen atom with the following wave function:
$$ \Psi(\mathbf{r}) = \exp(-a |\mathbf{r}|) $$
We will first verify that $\Psi$ is an eigenfunction of the Hamiltonian
$$ \hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|} $$
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
$$ E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}, $$
is constant.
Local energy
Write a function which computes the potential at $\mathbf{r}$
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 $$ V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$ $$
import numpy as np
def potential(r):
return -1. / np.sqrt(np.dot(r,r))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 potentialWrite 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.
def psi(a, r):
return np.exp(-a*np.sqrt(np.dot(r,r)))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 psiWrite 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 local kinetic energy is defined as $$-\frac{1}{2}\frac{\Delta \Psi}{\Psi}$$.
$$ \Psi(x,y,z) = \exp(-a\,\sqrt{x^2 + y^2 + z^2}). $$
We differentiate $\Psi$ with respect to $x$:
$$ \frac{\partial \Psi}{\partial x} = \frac{\partial \Psi}{\partial r} \frac{\partial r}{\partial x} = - \frac{a\,x}{|\mathbf{r}|} \Psi(x,y,z) $$
and we differentiate a second time:
$$ \frac{\partial^2 \Psi}{\partial x^2} = \left( \frac{a^2\,x^2}{|\mathbf{r}|^2} - \frac{a(y^2+z^2)}{|\mathbf{r}|^{3}} \right) \Psi(x,y,z). $$
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:
$$ \Delta \Psi (x,y,z) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(x,y,z) $$
So the local kinetic energy is $$ -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) $$
def kinetic(a,r):
return -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))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 kineticWrite a function which computes the local energy at $\mathbf{r}$
The function accepts x,y,z as input arguments and returns the
local energy.
$$ E_L(x,y,z) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) + V(x,y,z) $$
def e_loc(a,r):
return kinetic(a,r) + potential(r)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_locPlot the local energy along the x axis
For multiple values of $a$ (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the local energy along the $x$ axis.
import numpy as np
import matplotlib.pyplot as plt
from hydrogen import e_loc
x=np.linspace(-5,5)
def make_array(a):
y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
return y
plt.figure(figsize=(10,5))
for a in [0.1, 0.2, 0.5, 1., 1.5, 2.]:
y = make_array(a)
plt.plot(x,y,label=f"a={a}")
plt.tight_layout()
plt.legend()
plt.savefig("plot_py.png")
program plot
implicit none
double precision, external :: e_loc
double precision :: x(50), energy, dx, r(3), a(6)
integer :: i, j
a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do
r(:) = 0.d0
do j=1,size(a)
print *, '# a=', a(j)
do i=1,size(x)
r(1) = x(i)
energy = e_loc( a(j), r )
print *, x(i), energy
end do
print *, ''
print *, ''
end do
end program plotTo compile and run:
gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
./plot_hydrogen > dataTo plot the data using gnuplot"
set grid
set xrange [-5:5]
set yrange [-2:1]
plot './data' index 0 using 1:2 with lines title 'a=0.1', \
'./data' index 1 using 1:2 with lines title 'a=0.2', \
'./data' index 2 using 1:2 with lines title 'a=0.5', \
'./data' index 3 using 1:2 with lines title 'a=1.0', \
'./data' index 4 using 1:2 with lines title 'a=1.5', \
'./data' index 5 using 1:2 with lines title 'a=2.0'
Compute numerically the average energy
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 $\delta \mathbf{r}$, 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:
$$ E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\; w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta \mathbf{r} $$
We now compute an numerical estimate of the energy in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
Note: the energy is biased because:
- The energy is evaluated only inside the box
-
The volume elements are not infinitely small
import numpy as np from hydrogen import e_loc, psi interval = np.linspace(-5,5,num=50) 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}")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
program energy_hydrogen
implicit none
double precision, external :: e_loc, psi
double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm
integer :: i, k, l, j
a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do
delta = dx**3
r(:) = 0.d0
do j=1,size(a)
energy = 0.d0
norm = 0.d0
do i=1,size(x)
r(1) = x(i)
do k=1,size(x)
r(2) = x(k)
do l=1,size(x)
r(3) = x(l)
w = psi(a(j),r)
w = w * w * delta
energy = energy + w * e_loc(a(j), r)
norm = norm + w
end do
end do
end do
energy = energy / norm
print *, 'a = ', a(j), ' E = ', energy
end do
end program energy_hydrogenTo compile and run:
gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
./energy_hydrogena = 0.10000000000000001 E = -0.24518438948809140 a = 0.20000000000000001 E = -0.26966057967803236 a = 0.50000000000000000 E = -0.38563576125173815 a = 1.0000000000000000 E = -0.50000000000000000 a = 1.5000000000000000 E = -0.39242967082602065 a = 2.0000000000000000 E = -8.0869806678448772E-002
Compute the variance of the local energy
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.
$$ \sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[ E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} $$
Compute a numerical estimate of the variance of the local energy in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
import numpy as np
from hydrogen import e_loc, psi
interval = np.linspace(-5,5,num=50)
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
El = e_loc(a, r)
E += w * El
norm += w
E = E / norm
s2 = 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
El = e_loc(a, r)
s2 += w * (El - E)**2
s2 = s2 / norm
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")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
program variance_hydrogen
implicit none
double precision, external :: e_loc, psi
double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm, s2
integer :: i, k, l, j
a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
dx = 10.d0/(size(x)-1)
do i=1,size(x)
x(i) = -5.d0 + (i-1)*dx
end do
delta = dx**3
r(:) = 0.d0
do j=1,size(a)
energy = 0.d0
norm = 0.d0
do i=1,size(x)
r(1) = x(i)
do k=1,size(x)
r(2) = x(k)
do l=1,size(x)
r(3) = x(l)
w = psi(a(j),r)
w = w * w * delta
energy = energy + w * e_loc(a(j), r)
norm = norm + w
end do
end do
end do
energy = energy / norm
s2 = 0.d0
norm = 0.d0
do i=1,size(x)
r(1) = x(i)
do k=1,size(x)
r(2) = x(k)
do l=1,size(x)
r(3) = x(l)
w = psi(a(j),r)
w = w * w * delta
s2 = s2 + w * ( e_loc(a(j), r) - energy )**2
norm = norm + w
end do
end do
end do
s2 = s2 / norm
print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
end do
end program variance_hydrogenTo compile and run:
gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
./variance_hydrogena = 0.10000000000000001 E = -0.24518438948809140 s2 = 2.6965218719733813E-002 a = 0.20000000000000001 E = -0.26966057967803236 s2 = 3.7197072370217653E-002 a = 0.50000000000000000 E = -0.38563576125173815 s2 = 5.3185967578488862E-002 a = 1.0000000000000000 E = -0.50000000000000000 s2 = 0.0000000000000000 a = 1.5000000000000000 E = -0.39242967082602065 s2 = 0.31449670909180444 a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814270851303
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 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.
Computation of the statistical error
To compute the statistical error, you need to perform $M$ independent Monte Carlo calculations. You will obtain $M$ different estimates of the energy, which are expected to have a Gaussian distribution by the central limit theorem.
The estimate of the energy is
$$ E = \frac{1}{M} \sum_{i=1}^M E_M $$
The variance of the average energies can be computed as
$$ \sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_M - E)^2 $$
And the confidence interval is given by
$$ E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}} $$
Write a function returning the average and statistical error of an input array.
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))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_errorUniform sampling in the box
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 Once all the steps have been computed, return the average energy computed on the Monte Carlo calculation. 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 $a=0.9$.
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}")E = -0.4956255109300764 +/- 0.0007082875482711226
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 qmcgfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
./qmc_uniformE = -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.
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 \[ P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right) \]
As the coordinates are drawn with probability $P(\mathbf{r})$, the average energy can be computed as
$$ E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\; w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r}_i)} \delta \mathbf{r} $$
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/Na = 0.9
nmax = 100000
X = [MonteCarlo(a,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")E = -0.4952488228427792 +/- 0.00011913174676540714
Sampling with $\Psi^2$
We will now use the square of the wave function to make the sampling:
\[ P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2 \]
Now, the expression for the energy will be simplified to the average of the local energies, each with a weight of 1.
$$ E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)} $$
To generate the probability density $\Psi^2$, we can use a drifted diffusion scheme:
\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla \Psi(r)}{\Psi(r)} + \eta \sqrt{\tau} \]
where $\eta$ is a normally-distributed Gaussian random number.
First, write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
def drift(a,r):
ar_inv = -a/np.sqrt(np.dot(r,r))
return r * ar_invdef MonteCarlo(a,tau,nmax):
E = 0.
N = 0.
sq_tau = sqrt(tau)
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
d_old = drift(a,r_old)
d2_old = np.dot(d_old,d_old)
psi_old = psi(a,r_old)
for istep in range(nmax):
eta = np.random.normal(loc=0., scale=1.0, size=(3))
r_new = r_old + tau * d_old + sq_tau * eta
d_new = drift(a,r_new)
d2_new = np.dot(d_new,d_new)
psi_new = psi(a,r_new)
# Metropolis
prod = np.dot((d_new + d_old), (r_new - r_old))
argexpo = 0.5 * (d2_new - d2_old)*tau + prod
q = psi_new / psi_old
q = np.exp(-argexpo) * q*q
if np.random.uniform() < q:
r_old = r_new
d_old = d_new
d2_old = d2_new
psi_old = psi_new
N += 1.
E += e_loc(a,r_old)
return E/Nnmax = 100000
tau = 0.1
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
E, deltaE = ave_error(X)
print(f"E = {E} +/- {deltaE}")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:
$$ \Psi(\mathbf{r}_1, \mathbf{r}_2) = \exp(-(\mathbf{r}_1 - \mathbf{R}_A)) + $$ 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.