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#+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
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
:PROPERTIES:
:header-args:python: :tangle hydrogen.py
: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.
$\mathbf{r}=\sqrt{x^2 + y^2 + z^2})$, so
$$
V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$
$$
#+BEGIN_SRC python
import numpy as np
def potential(r):
return -1. / np.sqrt(np.dot(r,r))
#+END_SRC
#+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
*** 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.
#+BEGIN_SRC python
def psi(a, r):
return np.exp(-a*np.sqrt(np.dot(r,r)))
#+END_SRC
#+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
*** 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 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
$$
-\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right)
$$
#+BEGIN_SRC python
def kinetic(a,r):
return -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))
#+END_SRC
#+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
*** 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.
$$
E_L(x,y,z) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) + V(x,y,z)
$$
#+BEGIN_SRC python
def e_loc(a,r):
return kinetic(a,r) + potential(r)
#+END_SRC
#+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
** 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]
: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
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")
#+end_src
[[./plot_py.png]]
#+begin_src f90
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 plot
#+end_src
To compile and run:
#+begin_src sh :exports both
gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
./plot_hydrogen > data
#+end_src
#+RESULTS:
To plot the data using gnuplot"
#+begin_src gnuplot :file plot.png :exports both
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'
#+end_src
#+RESULTS:
[[file:plot.png]]
** Compute numerically the average energy
:PROPERTIES:
:header-args:python: :tangle energy_hydrogen.py
:header-args:f90: :tangle energy_hydrogen.f90
: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
$\delta x\, \delta y\, \delta z$, 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:
$$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta x\, \delta y\, \delta z
$$
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
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}")
#+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
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_hydrogen
#+end_src
To compile and run:
#+begin_src sh :results output :exports both
gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
./energy_hydrogen
#+end_src
#+RESULTS:
: a = 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
:PROPERTIES:
:header-args:python: :tangle variance_hydrogen.py
: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.
$$
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
$$
Compute an numerical estimate of the variance of the local energy
in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
#+BEGIN_SRC python :results output :exports both
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}")
#+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
#+begin_src f90
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_hydrogen
#+end_src
To compile and run:
#+begin_src sh :results output :exports both
gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
./variance_hydrogen
#+end_src
#+RESULTS:
: a = 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 in large dimensions.
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.
#+BEGIN_SRC python :results output
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))
#+END_SRC
#+RESULTS:
** Uniform sampling in the box
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.
Then, 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 $a=0.9$.
#+BEGIN_SRC python
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
#+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
#+RESULTS:
: E = -0.4952626284319677 +/- 0.0006877988969872546
** 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
$$
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r})} \delta x\, \delta y\, \delta z
$$
#+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
#+RESULTS:
#+BEGIN_SRC python
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
#+END_SRC
#+RESULTS:
#+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
#+RESULTS:
: 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 $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
#+BEGIN_SRC python
def drift(a,r):
ar_inv = -a/np.sqrt(np.dot(r,r))
return r * ar_inv
#+END_SRC
#+RESULTS:
#+BEGIN_SRC python
def 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/N
#+END_SRC
#+RESULTS:
#+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
#+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:
$$
\Psi(\mathbf{r}_1, \mathbf{r}_2) =
\exp(-(\mathbf{r}_1 - \mathbf{R}_A)) +
$$
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.

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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_hydrogen

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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}")

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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
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
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
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

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import numpy as np
def potential(r):
return -1. / np.sqrt(np.dot(r,r))
def psi(a, r):
return np.exp(-a*np.sqrt(np.dot(r,r)))
def kinetic(a,r):
return -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))
def e_loc(a,r):
return kinetic(a,r) + potential(r)

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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 plot

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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")

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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_hydrogen

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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}")