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<h1 class="title">Quantum Monte Carlo</h1>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#org9c63dfa">1. Introduction</a></li>
<li><a href="#orgb3448ff">2. Numerical evaluation of the energy</a>
<ul>
<li><a href="#orgfd11035">2.1. Local energy</a>
<ul>
<li><a href="#org5b55d2e">2.1.1. Exercise 1</a></li>
<li><a href="#org267c2b4">2.1.2. Exercise 2</a></li>
<li><a href="#orgbb2e99c">2.1.3. Exercise 3</a></li>
<li><a href="#orgb93cbca">2.1.4. Exercise 4</a></li>
</ul>
</li>
<li><a href="#orge0c924a">2.2. Plot of the local energy along the \(x\) axis</a>
<ul>
<li><a href="#orgae1b757">2.2.1. Exercise</a></li>
</ul>
</li>
<li><a href="#org1d35c3d">2.3. Numerical estimation of the energy</a>
<ul>
<li><a href="#orgeee38dd">2.3.1. Exercise</a></li>
</ul>
</li>
<li><a href="#org0feb42b">2.4. Variance of the local energy</a>
<ul>
<li><a href="#orgb444cc3">2.4.1. Exercise (optional)</a></li>
<li><a href="#org11be305">2.4.2. Exercise</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org478ecf8">3. Variational Monte Carlo</a>
<ul>
<li><a href="#org279f738">3.1. Computation of the statistical error</a>
<ul>
<li><a href="#org9fc9374">3.1.1. Exercise</a></li>
</ul>
</li>
<li><a href="#org00499df">3.2. Uniform sampling in the box</a>
<ul>
<li><a href="#org0e626da">3.2.1. Exercise</a></li>
</ul>
</li>
<li><a href="#org3c75811">3.3. Metropolis sampling with \(\Psi^2\)</a>
<ul>
<li><a href="#org41c65bb">3.3.1. Exercise</a></li>
</ul>
</li>
<li><a href="#orgd11f044">3.4. Gaussian random number generator</a></li>
<li><a href="#orgd59955a">3.5. Generalized Metropolis algorithm</a>
<ul>
<li><a href="#orgae8d2ad">3.5.1. Exercise 1</a></li>
<li><a href="#orgbb7142f">3.5.2. Exercise 2</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org2042bc2">4. <span class="todo TODO">TODO</span> Diffusion Monte Carlo</a>
<ul>
<li><a href="#org964c160">4.1. Hydrogen atom</a></li>
<li><a href="#orgf75f8aa">4.2. Dihydrogen</a></li>
</ul>
</li>
</ul>
</div>
</div>
<div id="outline-container-org9c63dfa" class="outline-2">
<h2 id="org9c63dfa"><span class="section-number-2">1</span> Introduction</h2>
<div class="outline-text-2" id="text-1">
<p>
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 <i>local energy</i>.
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.
</p>
<p>
Code examples will be given in Python and Fortran. Whatever language
can be chosen.
</p>
<p>
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.
</p>
<p>
<b>Note</b>
</p>
<div class="important">
<p>
In Fortran, when you use a double precision constant, don't forget
to put d0 as a suffix (for example 2.0d0), or it will be
interpreted as a single precision value
</p>
</div>
</div>
</div>
<div id="outline-container-orgb3448ff" class="outline-2">
<h2 id="orgb3448ff"><span class="section-number-2">2</span> Numerical evaluation of the energy</h2>
<div class="outline-text-2" id="text-2">
<p>
In this section we consider the Hydrogen atom with the following
wave function:
</p>
<p>
\[
\Psi(\mathbf{r}) = \exp(-a |\mathbf{r}|)
\]
</p>
<p>
We will first verify that \(\Psi\) is an eigenfunction of the Hamiltonian
</p>
<p>
\[
\hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|}
\]
</p>
<p>
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
</p>
<p>
\[
E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
\]
</p>
<p>
is constant. We will also see that when \(a \ne 1\) the local energy
is not constant, so \(\hat{H} \Psi \ne E \Psi\).
</p>
<p>
The probabilistic <i>expected value</i> of an arbitrary function \(f(x)\)
with respect to a probability density function \(p(x)\) is given by
</p>
<p>
\[ \langle f \rangle_p = \int_{-\infty}^\infty p(x)\, f(x)\,dx \].
</p>
<p>
Recall that a probability density function \(p(x)\) is non-negative
and integrates to one:
</p>
<p>
\[ \int_{-\infty}^\infty p(x)\,dx = 1 \].
</p>
<p>
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:
</p>
\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*}
</div>
<div id="outline-container-orgfd11035" class="outline-3">
<h3 id="orgfd11035"><span class="section-number-3">2.1</span> Local energy</h3>
<div class="outline-text-3" id="text-2-1">
</div>
<div id="outline-container-org5b55d2e" class="outline-4">
<h4 id="org5b55d2e"><span class="section-number-4">2.1.1</span> Exercise 1</h4>
<div class="outline-text-4" id="text-2-1-1">
<div class="exercise">
<p>
Write a function which computes the potential at \(\mathbf{r}\).
The function accepts a 3-dimensional vector <code>r</code> as input arguments
and returns the potential.
</p>
</div>
<p>
\(\mathbf{r}=\left( \begin{array}{c} x \\ y\\ z\end{array} \right)\), so
\[
V(\mathbf{r}) = -\frac{1}{\sqrt{x^2 + y^2 + z^2}}
\]
</p>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">import</span> numpy <span style="color: #a020f0;">as</span> np
<span style="color: #a020f0;">def</span> <span style="color: #0000ff;">potential</span>(r):
<span style="color: #a020f0;">return</span> -1. / np.sqrt(np.dot(r,r))
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #228b22;">double precision </span><span style="color: #a020f0;">function</span><span style="color: #a0522d;"> </span><span style="color: #0000ff;">potential</span><span style="color: #000000; background-color: #ffffff;">(r)</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> r(3)</span>
potential = -1.d0 / dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">potential</span>
</pre>
</div>
</div>
</div>
<div id="outline-container-org267c2b4" class="outline-4">
<h4 id="org267c2b4"><span class="section-number-4">2.1.2</span> Exercise 2</h4>
<div class="outline-text-4" id="text-2-1-2">
<div class="exercise">
<p>
Write a function which computes the wave function at \(\mathbf{r}\).
The function accepts a scalar <code>a</code> and a 3-dimensional vector <code>r</code> as
input arguments, and returns a scalar.
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">def</span> <span style="color: #0000ff;">psi</span>(a, r):
<span style="color: #a020f0;">return</span> np.exp(-a*np.sqrt(np.dot(r,r)))
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #228b22;">double precision </span><span style="color: #a020f0;">function</span><span style="color: #a0522d;"> </span><span style="color: #0000ff;">psi</span><span style="color: #000000; background-color: #ffffff;">(a, r)</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a, r(3)</span>
psi = dexp(-a * dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">psi</span>
</pre>
</div>
</div>
</div>
<div id="outline-container-orgbb2e99c" class="outline-4">
<h4 id="orgbb2e99c"><span class="section-number-4">2.1.3</span> Exercise 3</h4>
<div class="outline-text-4" id="text-2-1-3">
<div class="exercise">
<p>
Write a function which computes the local kinetic energy at \(\mathbf{r}\).
The function accepts <code>a</code> and <code>r</code> as input arguments and returns the
local kinetic energy.
</p>
</div>
<p>
The local kinetic energy is defined as \[-\frac{1}{2}\frac{\Delta \Psi}{\Psi}.\]
</p>
<p>
We differentiate \(\Psi\) with respect to \(x\):
</p>
<p>
\[\Psi(\mathbf{r}) = \exp(-a\,|\mathbf{r}|) \]
\[\frac{\partial \Psi}{\partial x}
= \frac{\partial \Psi}{\partial |\mathbf{r}|} \frac{\partial |\mathbf{r}|}{\partial x}
= - \frac{a\,x}{|\mathbf{r}|} \Psi(\mathbf{r}) \]
</p>
<p>
and we differentiate a second time:
</p>
<p>
\[
\frac{\partial^2 \Psi}{\partial x^2} =
\left( \frac{a^2\,x^2}{|\mathbf{r}|^2} -
\frac{a(y^2+z^2)}{|\mathbf{r}|^{3}} \right) \Psi(\mathbf{r}).
\]
</p>
<p>
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:
</p>
<p>
\[
\Delta \Psi (\mathbf{r}) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(\mathbf{r})
\]
</p>
<p>
So the local kinetic energy is
\[
-\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right)
\]
</p>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">def</span> <span style="color: #0000ff;">kinetic</span>(a,r):
<span style="color: #a020f0;">return</span> -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #228b22;">double precision </span><span style="color: #a020f0;">function</span><span style="color: #a0522d;"> </span><span style="color: #0000ff;">kinetic</span><span style="color: #000000; background-color: #ffffff;">(a,r)</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a, r(3)</span>
kinetic = -0.5d0 * (a*a - (2.d0*a) / <span style="color: #a020f0;">&amp;</span>
dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ) )
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">kinetic</span>
</pre>
</div>
</div>
</div>
<div id="outline-container-orgb93cbca" class="outline-4">
<h4 id="orgb93cbca"><span class="section-number-4">2.1.4</span> Exercise 4</h4>
<div class="outline-text-4" id="text-2-1-4">
<div class="exercise">
<p>
Write a function which computes the local energy at \(\mathbf{r}\).
The function accepts <code>x,y,z</code> as input arguments and returns the
local energy.
</p>
</div>
<p>
\[
E_L(\mathbf{r}) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) + V(\mathbf{r})
\]
</p>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">def</span> <span style="color: #0000ff;">e_loc</span>(a,r):
<span style="color: #a020f0;">return</span> kinetic(a,r) + potential(r)
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #228b22;">double precision </span><span style="color: #a020f0;">function</span><span style="color: #a0522d;"> </span><span style="color: #0000ff;">e_loc</span><span style="color: #000000; background-color: #ffffff;">(a,r)</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a, r(3)</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> kinetic, potential</span>
e_loc = kinetic(a,r) + potential(r)
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">e_loc</span>
</pre>
</div>
</div>
</div>
</div>
<div id="outline-container-orge0c924a" class="outline-3">
<h3 id="orge0c924a"><span class="section-number-3">2.2</span> Plot of the local energy along the \(x\) axis</h3>
<div class="outline-text-3" id="text-2-2">
</div>
<div id="outline-container-orgae1b757" class="outline-4">
<h4 id="orgae1b757"><span class="section-number-4">2.2.1</span> Exercise</h4>
<div class="outline-text-4" id="text-2-2-1">
<div class="exercise">
<p>
For multiple values of \(a\) (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the
local energy along the \(x\) axis.
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">import</span> numpy <span style="color: #a020f0;">as</span> np
<span style="color: #a020f0;">import</span> matplotlib.pyplot <span style="color: #a020f0;">as</span> plt
<span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> e_loc
<span style="color: #a0522d;">x</span>=np.linspace(-5,5)
<span style="color: #a020f0;">def</span> <span style="color: #0000ff;">make_array</span>(a):
<span style="color: #a0522d;">y</span>=np.array([ e_loc(a, np.array([t,0.,0.]) ) <span style="color: #a020f0;">for</span> t <span style="color: #a020f0;">in</span> x])
<span style="color: #a020f0;">return</span> y
plt.figure(figsize=(10,5))
<span style="color: #a020f0;">for</span> a <span style="color: #a020f0;">in</span> [0.1, 0.2, 0.5, 1., 1.5, 2.]:
<span style="color: #a0522d;">y</span> = make_array(a)
plt.plot(x,y,label=f<span style="color: #8b2252;">"a={a}"</span>)
plt.tight_layout()
plt.legend()
plt.savefig(<span style="color: #8b2252;">"plot_py.png"</span>)
</pre>
</div>
<div class="figure">
<p><img src="./plot_py.png" alt="plot_py.png" />
</p>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">program</span> <span style="color: #0000ff;">plot</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> x(50), energy, dx, r(3), a(6)</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> i, j</span>
a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
dx = 10.d0/(<span style="color: #a020f0;">size</span>(x)-1)
<span style="color: #a020f0;">do</span> i=1,<span style="color: #a020f0;">size</span>(x)
x(i) = -5.d0 + (i-1)*dx
<span style="color: #a020f0;">end do</span>
r(:) = 0.d0
<span style="color: #a020f0;">do</span> j=1,<span style="color: #a020f0;">size</span>(a)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'# a='</span>, a(j)
<span style="color: #a020f0;">do</span> i=1,<span style="color: #a020f0;">size</span>(x)
r(1) = x(i)
energy = e_loc( a(j), r )
<span style="color: #a020f0;">print</span> *, x(i), energy
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">''</span>
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">''</span>
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end program</span> <span style="color: #0000ff;">plot</span>
</pre>
</div>
<p>
To compile and run:
</p>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
./plot_hydrogen &gt; data
</pre>
</div>
<p>
To plot the data using gnuplot:
</p>
<div class="org-src-container">
<pre class="src src-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'
</pre>
</div>
<div class="figure">
<p><img src="plot.png" alt="plot.png" />
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-org1d35c3d" class="outline-3">
<h3 id="org1d35c3d"><span class="section-number-3">2.3</span> Numerical estimation of the energy</h3>
<div class="outline-text-3" id="text-2-3">
<p>
If the space is discretized in small volume elements \(\mathbf{r}_i\)
of size \(\delta \mathbf{r}\), the expression of \(\langle E_L \rangle_{\Psi^2}\)
becomes a weighted average of the local energy, where the weights
are the values of the probability density at \(\mathbf{r}_i\)
multiplied by the volume element:
</p>
<p>
\[
\langle E \rangle_{\Psi^2} \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta \mathbf{r}
\]
</p>
<div class="note">
<p>
The energy is biased because:
</p>
<ul class="org-ul">
<li>The volume elements are not infinitely small (discretization error)</li>
<li>The energy is evaluated only inside the box (incompleteness of the space)</li>
</ul>
</div>
</div>
<div id="outline-container-orgeee38dd" class="outline-4">
<h4 id="orgeee38dd"><span class="section-number-4">2.3.1</span> Exercise</h4>
<div class="outline-text-4" id="text-2-3-1">
<div class="exercise">
<p>
Compute a numerical estimate of the energy in a grid of
\(50\times50\times50\) points in the range \((-5,-5,-5) \le
\mathbf{r} \le (5,5,5)\).
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">import</span> numpy <span style="color: #a020f0;">as</span> np
<span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> e_loc, psi
<span style="color: #a0522d;">interval</span> = np.linspace(-5,5,num=50)
<span style="color: #a0522d;">delta</span> = (interval[1]-interval[0])**3
<span style="color: #a0522d;">r</span> = np.array([0.,0.,0.])
<span style="color: #a020f0;">for</span> a <span style="color: #a020f0;">in</span> [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
<span style="color: #a0522d;">E</span> = 0.
<span style="color: #a0522d;">norm</span> = 0.
<span style="color: #a020f0;">for</span> x <span style="color: #a020f0;">in</span> interval:
<span style="color: #a0522d;">r</span>[0] = x
<span style="color: #a020f0;">for</span> y <span style="color: #a020f0;">in</span> interval:
<span style="color: #a0522d;">r</span>[1] = y
<span style="color: #a020f0;">for</span> z <span style="color: #a020f0;">in</span> interval:
<span style="color: #a0522d;">r</span>[2] = z
<span style="color: #a0522d;">w</span> = psi(a,r)
<span style="color: #a0522d;">w</span> = w * w * delta
<span style="color: #a0522d;">E</span> += w * e_loc(a,r)
<span style="color: #a0522d;">norm</span> += w
<span style="color: #a0522d;">E</span> = E / norm
<span style="color: #a020f0;">print</span>(f<span style="color: #8b2252;">"a = {a} \t E = {E}"</span>)
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">program</span> <span style="color: #0000ff;">energy_hydrogen</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> x(50), w, delta, energy, dx, r(3), a(6), norm</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> i, k, l, j</span>
a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
dx = 10.d0/(<span style="color: #a020f0;">size</span>(x)-1)
<span style="color: #a020f0;">do</span> i=1,<span style="color: #a020f0;">size</span>(x)
x(i) = -5.d0 + (i-1)*dx
<span style="color: #a020f0;">end do</span>
delta = dx**3
r(:) = 0.d0
<span style="color: #a020f0;">do</span> j=1,<span style="color: #a020f0;">size</span>(a)
energy = 0.d0
norm = 0.d0
<span style="color: #a020f0;">do</span> i=1,<span style="color: #a020f0;">size</span>(x)
r(1) = x(i)
<span style="color: #a020f0;">do</span> k=1,<span style="color: #a020f0;">size</span>(x)
r(2) = x(k)
<span style="color: #a020f0;">do</span> l=1,<span style="color: #a020f0;">size</span>(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
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end do</span>
energy = energy / norm
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'a = '</span>, a(j), <span style="color: #8b2252;">' E = '</span>, energy
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end program</span> <span style="color: #0000ff;">energy_hydrogen</span>
</pre>
</div>
<p>
To compile the Fortran and run it:
</p>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
./energy_hydrogen
</pre>
</div>
<pre class="example">
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
</pre>
</div>
</div>
</div>
<div id="outline-container-org0feb42b" class="outline-3">
<h3 id="org0feb42b"><span class="section-number-3">2.4</span> Variance of the local energy</h3>
<div class="outline-text-3" id="text-2-4">
<p>
The variance of the local energy is a functional of \(\Psi\)
which measures the magnitude of the fluctuations of the local
energy associated with \(\Psi\) around the average:
</p>
<p>
\[
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
\]
which can be simplified as
</p>
<p>
\[ \sigma^2(E_L) = \langle E_L^2 \rangle - \langle E_L \rangle^2 \]
</p>
<p>
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.
</p>
</div>
<div id="outline-container-orgb444cc3" class="outline-4">
<h4 id="orgb444cc3"><span class="section-number-4">2.4.1</span> Exercise (optional)</h4>
<div class="outline-text-4" id="text-2-4-1">
<div class="exercise">
<p>
Prove that :
\[ \sigma^2(E_L) = \langle E^2 \rangle - \langle E \rangle^2 \]
</p>
</div>
</div>
</div>
<div id="outline-container-org11be305" class="outline-4">
<h4 id="org11be305"><span class="section-number-4">2.4.2</span> Exercise</h4>
<div class="outline-text-4" id="text-2-4-2">
<div class="exercise">
<p>
Add the calculation of the variance to the previous code, and
compute a numerical estimate of the variance of the local energy
in a grid of \(50\times50\times50\) points in the range
\((-5,-5,-5)
\le \mathbf{r} \le (5,5,5)\) for different values of \(a\).
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">import</span> numpy <span style="color: #a020f0;">as</span> np
<span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> e_loc, psi
<span style="color: #a0522d;">interval</span> = np.linspace(-5,5,num=50)
<span style="color: #a0522d;">delta</span> = (interval[1]-interval[0])**3
<span style="color: #a0522d;">r</span> = np.array([0.,0.,0.])
<span style="color: #a020f0;">for</span> a <span style="color: #a020f0;">in</span> [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
<span style="color: #a0522d;">E</span> = 0.
<span style="color: #a0522d;">E2</span> = 0.
<span style="color: #a0522d;">norm</span> = 0.
<span style="color: #a020f0;">for</span> x <span style="color: #a020f0;">in</span> interval:
<span style="color: #a0522d;">r</span>[0] = x
<span style="color: #a020f0;">for</span> y <span style="color: #a020f0;">in</span> interval:
<span style="color: #a0522d;">r</span>[1] = y
<span style="color: #a020f0;">for</span> z <span style="color: #a020f0;">in</span> interval:
<span style="color: #a0522d;">r</span>[2] = z
<span style="color: #a0522d;">w</span> = psi(a, r)
<span style="color: #a0522d;">w</span> = w * w * delta
<span style="color: #a0522d;">El</span> = e_loc(a, r)
<span style="color: #a0522d;">E</span> += w * El
<span style="color: #a0522d;">E2</span> += w * El*El
<span style="color: #a0522d;">norm</span> += w
<span style="color: #a0522d;">E</span> = E / norm
<span style="color: #a0522d;">E2</span> = E2 / norm
<span style="color: #a0522d;">s2</span> = E2 - E*E
<span style="color: #a020f0;">print</span>(f<span style="color: #8b2252;">"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}"</span>)
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">program</span> <span style="color: #0000ff;">variance_hydrogen</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> x(50), w, delta, energy, dx, r(3), a(6), norm, s2</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> i, k, l, j</span>
a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
dx = 10.d0/(<span style="color: #a020f0;">size</span>(x)-1)
<span style="color: #a020f0;">do</span> i=1,<span style="color: #a020f0;">size</span>(x)
x(i) = -5.d0 + (i-1)*dx
<span style="color: #a020f0;">end do</span>
delta = dx**3
r(:) = 0.d0
<span style="color: #a020f0;">do</span> j=1,<span style="color: #a020f0;">size</span>(a)
energy = 0.d0
norm = 0.d0
<span style="color: #a020f0;">do</span> i=1,<span style="color: #a020f0;">size</span>(x)
r(1) = x(i)
<span style="color: #a020f0;">do</span> k=1,<span style="color: #a020f0;">size</span>(x)
r(2) = x(k)
<span style="color: #a020f0;">do</span> l=1,<span style="color: #a020f0;">size</span>(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
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end do</span>
energy = energy / norm
s2 = 0.d0
norm = 0.d0
<span style="color: #a020f0;">do</span> i=1,<span style="color: #a020f0;">size</span>(x)
r(1) = x(i)
<span style="color: #a020f0;">do</span> k=1,<span style="color: #a020f0;">size</span>(x)
r(2) = x(k)
<span style="color: #a020f0;">do</span> l=1,<span style="color: #a020f0;">size</span>(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
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end do</span>
s2 = s2 / norm
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'a = '</span>, a(j), <span style="color: #8b2252;">' E = '</span>, energy, <span style="color: #8b2252;">' s2 = '</span>, s2
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">end program</span> <span style="color: #0000ff;">variance_hydrogen</span>
</pre>
</div>
<p>
To compile and run:
</p>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
./variance_hydrogen
</pre>
</div>
<pre class="example">
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
</pre>
</div>
</div>
</div>
</div>
<div id="outline-container-org478ecf8" class="outline-2">
<h2 id="org478ecf8"><span class="section-number-2">3</span> Variational Monte Carlo</h2>
<div class="outline-text-2" id="text-3">
<p>
Numerical integration with deterministic methods is very efficient
in low dimensions. When the number of dimensions becomes large,
instead of computing the average energy as a numerical integration
on a grid, it is usually more efficient to do a Monte Carlo sampling.
</p>
<p>
Moreover, a Monte Carlo sampling will alow us to remove the bias due
to the discretization of space, and compute a statistical confidence
interval.
</p>
</div>
<div id="outline-container-org279f738" class="outline-3">
<h3 id="org279f738"><span class="section-number-3">3.1</span> Computation of the statistical error</h3>
<div class="outline-text-3" id="text-3-1">
<p>
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.
</p>
<p>
The estimate of the energy is
</p>
<p>
\[
E = \frac{1}{M} \sum_{i=1}^M E_M
\]
</p>
<p>
The variance of the average energies can be computed as
</p>
<p>
\[
\sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_M - E)^2
\]
</p>
<p>
And the confidence interval is given by
</p>
<p>
\[
E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}}
\]
</p>
</div>
<div id="outline-container-org9fc9374" class="outline-4">
<h4 id="org9fc9374"><span class="section-number-4">3.1.1</span> Exercise</h4>
<div class="outline-text-4" id="text-3-1-1">
<div class="exercise">
<p>
Write a function returning the average and statistical error of an
input array.
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">from</span> math <span style="color: #a020f0;">import</span> sqrt
<span style="color: #a020f0;">def</span> <span style="color: #0000ff;">ave_error</span>(arr):
<span style="color: #a0522d;">M</span> = <span style="color: #483d8b;">len</span>(arr)
<span style="color: #a020f0;">assert</span> (M&gt;1)
<span style="color: #a0522d;">average</span> = <span style="color: #483d8b;">sum</span>(arr)/M
<span style="color: #a0522d;">variance</span> = 1./(M-1) * <span style="color: #483d8b;">sum</span>( [ (x - average)**2 <span style="color: #a020f0;">for</span> x <span style="color: #a020f0;">in</span> arr ] )
<span style="color: #a020f0;">return</span> (average, sqrt(variance/M))
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">subroutine</span> <span style="color: #0000ff;">ave_error</span>(x,n,ave,err)
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">integer</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> n </span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> x(n) </span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> ave, err</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> variance</span>
<span style="color: #a020f0;">if</span> (n == 1) <span style="color: #a020f0;">then</span>
ave = x(1)
err = 0.d0
<span style="color: #a020f0;">else</span>
ave = <span style="color: #a020f0;">sum</span>(x(:)) / <span style="color: #a020f0;">dble</span>(n)
variance = <span style="color: #a020f0;">sum</span>( (x(:) - ave)**2 ) / <span style="color: #a020f0;">dble</span>(n-1)
err = dsqrt(variance/<span style="color: #a020f0;">dble</span>(n))
<span style="color: #a020f0;">endif</span>
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">ave_error</span>
</pre>
</div>
</div>
</div>
</div>
<div id="outline-container-org00499df" class="outline-3">
<h3 id="org00499df"><span class="section-number-3">3.2</span> Uniform sampling in the box</h3>
<div class="outline-text-3" id="text-3-2">
<p>
We will now do our first Monte Carlo calculation to compute the
energy of the hydrogen atom.
</p>
<p>
At every Monte Carlo step:
</p>
<ul class="org-ul">
<li>Draw a random point \(\mathbf{r}_i\) in the box \((-5,-5,-5) \le
(x,y,z) \le (5,5,5)\)</li>
<li>Compute \([\Psi(\mathbf{r}_i)]^2\) and accumulate the result in a
variable <code>normalization</code></li>
<li>Compute \([\Psi(\mathbf{r}_i)]^2 \times E_L(\mathbf{r}_i)\), and accumulate the
result in a variable <code>energy</code></li>
</ul>
<p>
One Monte Carlo run will consist of \(N\) Monte Carlo steps. Once all the
steps have been computed, the run returns the average energy
\(\bar{E}_k\) over the \(N\) steps of the run.
</p>
<p>
To compute the statistical error, perform \(M\) runs. The final
estimate of the energy will be the average over the \(\bar{E}_k\),
and the variance of the \(\bar{E}_k\) will be used to compute the
statistical error.
</p>
</div>
<div id="outline-container-org0e626da" class="outline-4">
<h4 id="org0e626da"><span class="section-number-4">3.2.1</span> Exercise</h4>
<div class="outline-text-4" id="text-3-2-1">
<div class="exercise">
<p>
Parameterize the wave function with \(a=0.9\). Perform 30
independent Monte Carlo runs, each with 100 000 Monte Carlo
steps. Store the final energies of each run and use this array to
compute the average energy and the associated error bar.
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">from</span> qmc_stats <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">def</span> <span style="color: #0000ff;">MonteCarlo</span>(a, nmax):
<span style="color: #a0522d;">E</span> = 0.
<span style="color: #a0522d;">N</span> = 0.
<span style="color: #a020f0;">for</span> istep <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(nmax):
<span style="color: #a0522d;">r</span> = np.random.uniform(-5., 5., (3))
<span style="color: #a0522d;">w</span> = psi(a,r)
<span style="color: #a0522d;">w</span> = w*w
<span style="color: #a0522d;">N</span> += w
<span style="color: #a0522d;">E</span> += w * e_loc(a,r)
<span style="color: #a020f0;">return</span> E/N
<span style="color: #a0522d;">a</span> = 0.9
<span style="color: #a0522d;">nmax</span> = 100000
<span style="color: #a0522d;">X</span> = [MonteCarlo(a,nmax) <span style="color: #a020f0;">for</span> i <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(30)]
<span style="color: #a0522d;">E</span>, <span style="color: #a0522d;">deltaE</span> = ave_error(X)
<span style="color: #a020f0;">print</span>(f<span style="color: #8b2252;">"E = {E} +/- {deltaE}"</span>)
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="note">
<p>
When running Monte Carlo calculations, the number of steps is
usually very large. We expect <code>nmax</code> to be possibly larger than 2
billion, so we use 8-byte integers (<code>integer*8</code>) to represent it, as
well as the index of the current step.
</p>
</div>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">subroutine</span> <span style="color: #0000ff;">uniform_montecarlo</span>(a,nmax,energy)
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> nmax </span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> energy</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> istep</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> norm, r(3), w</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi</span>
energy = 0.d0
norm = 0.d0
<span style="color: #a020f0;">do</span> istep = 1,nmax
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_number</span>(r)
r(:) = -5.d0 + 10.d0*r(:)
w = psi(a,r)
w = w*w
norm = norm + w
energy = energy + w * e_loc(a,r)
<span style="color: #a020f0;">end do</span>
energy = energy / norm
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">uniform_montecarlo</span>
<span style="color: #a020f0;">program</span> <span style="color: #0000ff;">qmc</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> a = 0.9</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nmax = 100000</span>
<span style="color: #228b22;">integer</span> , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nruns = 30</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> irun</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> X(nruns)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> ave, err</span>
<span style="color: #a020f0;">do</span> irun=1,nruns
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">uniform_montecarlo</span>(a,nmax,X(irun))
<span style="color: #a020f0;">enddo</span>
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">ave_error</span>(X,nruns,ave,err)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'E = '</span>, ave, <span style="color: #8b2252;">'+/-'</span>, err
<span style="color: #a020f0;">end program</span> <span style="color: #0000ff;">qmc</span>
</pre>
</div>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
./qmc_uniform
</pre>
</div>
<pre class="example">
E = -0.49588321986667677 +/- 7.1758863546737969E-004
</pre>
</div>
</div>
</div>
<div id="outline-container-org3c75811" class="outline-3">
<h3 id="org3c75811"><span class="section-number-3">3.3</span> Metropolis sampling with \(\Psi^2\)</h3>
<div class="outline-text-3" id="text-3-3">
<p>
We will now use the square of the wave function to sample random
points distributed with the probability density
\[
P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
\]
</p>
<p>
The expression of the average energy is now simplified to the average of
the local energies, since the weights are taken care of by the
sampling :
</p>
<p>
\[
E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)
\]
</p>
<p>
To sample a chosen probability density, an efficient method is the
<a href="https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm">Metropolis-Hastings sampling algorithm</a>. Starting from a random
initial position \(\mathbf{r}_0\), we will realize a random walk as follows:
</p>
<p>
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \mathbf{u}
\]
</p>
<p>
where \(\tau\) is a fixed constant (the so-called <i>time-step</i>), and
\(\mathbf{u}\) is a uniform random number in a 3-dimensional box
\((-1,-1,-1) \le \mathbf{u} \le (1,1,1)\). We will then add the
accept/reject step that will guarantee that the distribution of the
\(\mathbf{r}_n\) is \(\Psi^2\):
</p>
<ul class="org-ul">
<li>Compute a new position \(\mathbf{r}_{n+1}\)</li>
<li>Draw a uniform random number \(v \in [0,1]\)</li>
<li>Compute the ratio \(R = \frac{\left[\Psi(\mathbf{r}_{n+1})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}\)</li>
<li>if \(v \le R\), accept the move (do nothing)</li>
<li>else, reject the move (set \(\mathbf{r}_{n+1} = \mathbf{r}_n\))</li>
<li>evaluate the local energy at \(\mathbf{r}_{n+1}\)</li>
</ul>
<div class="note">
<p>
A common error is to remove the rejected samples from the
calculation of the average. <b>Don't do it!</b>
</p>
<p>
All samples should be kept, from both accepted and rejected moves.
</p>
</div>
<p>
If the time step is infinitely small, the ratio will be very close
to one and all the steps will be accepted. But the trajectory will
be infinitely too short to have statistical significance.
</p>
<p>
On the other hand, as the time step increases, the number of
accepted steps will decrease because the ratios might become
small. If the number of accepted steps is close to zero, then the
space is not well sampled either.
</p>
<p>
The time step should be adjusted so that it is as large as
possible, keeping the number of accepted steps not too small. To
achieve that we define the acceptance rate as the number of
accepted steps over the total number of steps. Adjusting the time
step such that the acceptance rate is close to 0.5 is a good compromise.
</p>
</div>
<div id="outline-container-org41c65bb" class="outline-4">
<h4 id="org41c65bb"><span class="section-number-4">3.3.1</span> Exercise</h4>
<div class="outline-text-4" id="text-3-3-1">
<div class="exercise">
<p>
Modify the program of the previous section to compute the energy, sampling with
\(Psi^2\).
Compute also the acceptance rate, so that you can adapt the time
step in order to have an acceptance rate close to 0.5 .
Can you observe a reduction in the statistical error?
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">from</span> qmc_stats <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">def</span> <span style="color: #0000ff;">MonteCarlo</span>(a,nmax,tau):
<span style="color: #a0522d;">E</span> = 0.
<span style="color: #a0522d;">N</span> = 0.
<span style="color: #a0522d;">N_accep</span> = 0.
<span style="color: #a0522d;">r_old</span> = np.random.uniform(-tau, tau, (3))
<span style="color: #a0522d;">psi_old</span> = psi(a,r_old)
<span style="color: #a020f0;">for</span> istep <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(nmax):
<span style="color: #a0522d;">r_new</span> = r_old + np.random.uniform(-tau,tau,(3))
<span style="color: #a0522d;">psi_new</span> = psi(a,r_new)
<span style="color: #a0522d;">ratio</span> = (psi_new / psi_old)**2
<span style="color: #a0522d;">v</span> = np.random.uniform(0,1,(1))
<span style="color: #a020f0;">if</span> v &lt; ratio:
<span style="color: #a0522d;">N_accep</span> += 1.
<span style="color: #a0522d;">r_old</span> = r_new
<span style="color: #a0522d;">psi_old</span> = psi_new
<span style="color: #a0522d;">N</span> += 1.
<span style="color: #a0522d;">E</span> += e_loc(a,r_old)
<span style="color: #a020f0;">return</span> E/N, N_accep/N
<span style="color: #a0522d;">a</span> = 0.9
<span style="color: #a0522d;">nmax</span> = 100000
<span style="color: #a0522d;">tau</span> = 1.3
<span style="color: #a0522d;">X0</span> = [ MonteCarlo(a,nmax,tau) <span style="color: #a020f0;">for</span> i <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(30)]
<span style="color: #a0522d;">X</span> = [ x <span style="color: #a020f0;">for</span> x, _ <span style="color: #a020f0;">in</span> X0 ]
<span style="color: #a0522d;">A</span> = [ x <span style="color: #a020f0;">for</span> _, x <span style="color: #a020f0;">in</span> X0 ]
<span style="color: #a0522d;">E</span>, <span style="color: #a0522d;">deltaE</span> = ave_error(X)
<span style="color: #a0522d;">A</span>, <span style="color: #a0522d;">deltaA</span> = ave_error(A)
<span style="color: #a020f0;">print</span>(f<span style="color: #8b2252;">"E = {E} +/- {deltaE}"</span>)
<span style="color: #a020f0;">print</span>(f<span style="color: #8b2252;">"A = {A} +/- {deltaA}"</span>)
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">subroutine</span> <span style="color: #0000ff;">metropolis_montecarlo</span>(a,nmax,tau,energy,accep)
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> nmax </span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> tau</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> energy</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> accep</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> istep</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> norm, r_old(3), r_new(3), psi_old, psi_new</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> v, ratio, n_accep</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi, gaussian</span>
energy = 0.d0
norm = 0.d0
n_accep = 0.d0
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_number</span>(r_old)
r_old(:) = tau * (2.d0*r_old(:) - 1.d0)
psi_old = psi(a,r_old)
<span style="color: #a020f0;">do</span> istep = 1,nmax
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_number</span>(r_new)
r_new(:) = r_old(:) + tau * (2.d0*r_new(:) - 1.d0)
psi_new = psi(a,r_new)
ratio = (psi_new / psi_old)**2
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_number</span>(v)
<span style="color: #a020f0;">if</span> (v &lt; ratio) <span style="color: #a020f0;">then</span>
r_old(:) = r_new(:)
psi_old = psi_new
n_accep = n_accep + 1.d0
<span style="color: #a020f0;">endif</span>
norm = norm + 1.d0
energy = energy + e_loc(a,r_old)
<span style="color: #a020f0;">end do</span>
energy = energy / norm
accep = n_accep / norm
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">metropolis_montecarlo</span>
<span style="color: #a020f0;">program</span> <span style="color: #0000ff;">qmc</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> a = 0.9d0</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> tau = 1.3d0</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nmax = 100000</span>
<span style="color: #228b22;">integer</span> , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nruns = 30</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> irun</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> X(nruns), Y(nruns)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> ave, err</span>
<span style="color: #a020f0;">do</span> irun=1,nruns
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">metropolis_montecarlo</span>(a,nmax,tau,X(irun),Y(irun))
<span style="color: #a020f0;">enddo</span>
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">ave_error</span>(X,nruns,ave,err)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'E = '</span>, ave, <span style="color: #8b2252;">'+/-'</span>, err
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">ave_error</span>(Y,nruns,ave,err)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'A = '</span>, ave, <span style="color: #8b2252;">'+/-'</span>, err
<span style="color: #a020f0;">end program</span> <span style="color: #0000ff;">qmc</span>
</pre>
</div>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
./qmc_metropolis
</pre>
</div>
<pre class="example">
E = -0.49478505004797046 +/- 2.0493795299184956E-004
A = 0.51737800000000000 +/- 4.1827406733181444E-004
</pre>
</div>
</div>
</div>
<div id="outline-container-orgd11f044" class="outline-3">
<h3 id="orgd11f044"><span class="section-number-3">3.4</span> Gaussian random number generator</h3>
<div class="outline-text-3" id="text-3-4">
<p>
To obtain Gaussian-distributed random numbers, you can apply the
<a href="https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform">Box Muller transform</a> to uniform random numbers:
</p>
\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*}
<p>
Below is a Fortran implementation returning a Gaussian-distributed
n-dimensional vector \(\mathbf{z}\). This will be useful for the
following sections.
</p>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">subroutine</span> <span style="color: #0000ff;">random_gauss</span>(z,n)
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">integer</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> n</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> z(n)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> u(n+1)</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> two_pi = 2.d0*dacos(-1.d0)</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> i</span>
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_number</span>(u)
<span style="color: #a020f0;">if</span> (<span style="color: #a020f0;">iand</span>(n,1) == 0) <span style="color: #a020f0;">then</span>
! <span style="color: #b22222;">n is even</span>
<span style="color: #a020f0;">do</span> 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) )
<span style="color: #a020f0;">end do</span>
<span style="color: #a020f0;">else</span>
! <span style="color: #b22222;">n is odd</span>
<span style="color: #a020f0;">do</span> 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) )
<span style="color: #a020f0;">end do</span>
z(n) = dsqrt(-2.d0*dlog(u(n)))
z(n) = z(n) * dcos( two_pi*u(n+1) )
<span style="color: #a020f0;">end if</span>
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">random_gauss</span>
</pre>
</div>
</div>
</div>
<div id="outline-container-orgd59955a" class="outline-3">
<h3 id="orgd59955a"><span class="section-number-3">3.5</span> Generalized Metropolis algorithm</h3>
<div class="outline-text-3" id="text-3-5">
<p>
One can use more efficient numerical schemes to move the electrons.
But in that case, the Metropolis accepation step has to be adapted
accordingly: the acceptance
probability \(A\) is chosen so that it is consistent with the
probability of leaving \(\mathbf{r}_n\) and the probability of
entering \(\mathbf{r}_{n+1}\):
</p>
<p>
\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1,
\frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}
\right)
\]
where \(T(\mathbf{r}_n \rightarrow \mathbf{r}_{n+1})\) is the
probability of transition from \(\mathbf{r}_n\) to
\(\mathbf{r}_{n+1}\).
</p>
<p>
In the previous example, we were using uniform random
numbers. Hence, the transition probability was
</p>
<p>
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
\text{constant}
\]
</p>
<p>
So the expression of \(A\) was simplified to the ratios of the squared
wave functions.
</p>
<p>
Now, if instead of drawing uniform random numbers
choose to draw Gaussian random numbers with mean 0 and variance
\(\tau\), the transition probability becomes:
</p>
<p>
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\tau} \right]
\]
</p>
<p>
To sample even better the density, we can "push" the electrons
into in the regions of high probability, and "pull" them away from
the low-probability regions. This will mechanically increase the
acceptance ratios and improve the sampling.
</p>
<p>
To do this, we can add the drift vector
</p>
<p>
\[
\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}
\].
</p>
<p>
The numerical scheme becomes a drifted diffusion:
</p>
<p>
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
\]
</p>
<p>
where \(\chi\) is a Gaussian random variable with zero mean and
variance \(\tau\).
The transition probability becomes:
</p>
<p>
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\tau} \right]
\]
</p>
</div>
<div id="outline-container-orgae8d2ad" class="outline-4">
<h4 id="orgae8d2ad"><span class="section-number-4">3.5.1</span> Exercise 1</h4>
<div class="outline-text-4" id="text-3-5-1">
<div class="exercise">
<p>
Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\).
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">def</span> <span style="color: #0000ff;">drift</span>(a,r):
<span style="color: #a0522d;">ar_inv</span> = -a/np.sqrt(np.dot(r,r))
<span style="color: #a020f0;">return</span> r * ar_inv
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">subroutine</span> <span style="color: #0000ff;">drift</span>(a,r,b)
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a, r(3)</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> b(3)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> ar_inv</span>
ar_inv = -a / dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3))
b(:) = r(:) * ar_inv
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">drift</span>
</pre>
</div>
</div>
</div>
<div id="outline-container-orgbb7142f" class="outline-4">
<h4 id="orgbb7142f"><span class="section-number-4">3.5.2</span> Exercise 2</h4>
<div class="outline-text-4" id="text-3-5-2">
<div class="exercise">
<p>
Modify the previous program to introduce the drifted diffusion scheme.
(This is a necessary step for the next section).
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">from</span> qmc_stats <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">def</span> <span style="color: #0000ff;">MonteCarlo</span>(a,tau,nmax):
<span style="color: #a0522d;">E</span> = 0.
<span style="color: #a0522d;">N</span> = 0.
<span style="color: #a0522d;">accep_rate</span> = 0.
<span style="color: #a0522d;">sq_tau</span> = np.sqrt(tau)
<span style="color: #a0522d;">r_old</span> = np.random.normal(loc=0., scale=1.0, size=(3))
<span style="color: #a0522d;">d_old</span> = drift(a,r_old)
<span style="color: #a0522d;">d2_old</span> = np.dot(d_old,d_old)
<span style="color: #a0522d;">psi_old</span> = psi(a,r_old)
<span style="color: #a020f0;">for</span> istep <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(nmax):
<span style="color: #a0522d;">chi</span> = np.random.normal(loc=0., scale=1.0, size=(3))
<span style="color: #a0522d;">r_new</span> = r_old + tau * d_old + sq_tau * chi
<span style="color: #a0522d;">d_new</span> = drift(a,r_new)
<span style="color: #a0522d;">d2_new</span> = np.dot(d_new,d_new)
<span style="color: #a0522d;">psi_new</span> = psi(a,r_new)
# <span style="color: #b22222;">Metropolis</span>
<span style="color: #a0522d;">prod</span> = np.dot((d_new + d_old), (r_new - r_old))
<span style="color: #a0522d;">argexpo</span> = 0.5 * (d2_new - d2_old)*tau + prod
<span style="color: #a0522d;">q</span> = psi_new / psi_old
<span style="color: #a0522d;">q</span> = np.exp(-argexpo) * q*q
<span style="color: #a020f0;">if</span> np.random.uniform() &lt; q:
<span style="color: #a0522d;">accep_rate</span> += 1.
<span style="color: #a0522d;">r_old</span> = r_new
<span style="color: #a0522d;">d_old</span> = d_new
<span style="color: #a0522d;">d2_old</span> = d2_new
<span style="color: #a0522d;">psi_old</span> = psi_new
<span style="color: #a0522d;">N</span> += 1.
<span style="color: #a0522d;">E</span> += e_loc(a,r_old)
<span style="color: #a020f0;">return</span> E/N, accep_rate/N
<span style="color: #a0522d;">a</span> = 0.9
<span style="color: #a0522d;">nmax</span> = 100000
<span style="color: #a0522d;">tau</span> = 1.0
<span style="color: #a0522d;">X</span> = [MonteCarlo(a,tau,nmax) <span style="color: #a020f0;">for</span> i <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(30)]
<span style="color: #a0522d;">E</span>, <span style="color: #a0522d;">deltaE</span> = ave_error([x[0] <span style="color: #a020f0;">for</span> x <span style="color: #a020f0;">in</span> X])
<span style="color: #a0522d;">A</span>, <span style="color: #a0522d;">deltaA</span> = ave_error([x[1] <span style="color: #a020f0;">for</span> x <span style="color: #a020f0;">in</span> X])
<span style="color: #a020f0;">print</span>(f<span style="color: #8b2252;">"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}"</span>)
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">subroutine</span> <span style="color: #0000ff;">variational_montecarlo</span>(a,tau,nmax,energy,accep_rate)
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a, tau</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> nmax </span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> energy, accep_rate</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> istep</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> norm, sq_tau, chi(3), d2_old, prod, u</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> psi_old, psi_new, d2_new, argexpo, q</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> r_old(3), r_new(3)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> d_old(3), d_new(3)</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi</span>
sq_tau = dsqrt(tau)
! <span style="color: #b22222;">Initialization</span>
energy = 0.d0
norm = 0.d0
accep_rate = 0.d0
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_gauss</span>(r_old,3)
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">drift</span>(a,r_old,d_old)
d2_old = d_old(1)*d_old(1) + d_old(2)*d_old(2) + d_old(3)*d_old(3)
psi_old = psi(a,r_old)
<span style="color: #a020f0;">do</span> istep = 1,nmax
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_gauss</span>(chi,3)
r_new(:) = r_old(:) + tau * d_old(:) + chi(:)*sq_tau
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">drift</span>(a,r_new,d_new)
d2_new = d_new(1)*d_new(1) + d_new(2)*d_new(2) + d_new(3)*d_new(3)
psi_new = psi(a,r_new)
! <span style="color: #b22222;">Metropolis</span>
prod = (d_new(1) + d_old(1))*(r_new(1) - r_old(1)) + <span style="color: #a020f0;">&amp;</span>
(d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + <span style="color: #a020f0;">&amp;</span>
(d_new(3) + d_old(3))*(r_new(3) - r_old(3))
argexpo = 0.5d0 * (d2_new - d2_old)*tau + prod
q = psi_new / psi_old
q = dexp(-argexpo) * q*q
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_number</span>(u)
<span style="color: #a020f0;">if</span> (u&lt;q) <span style="color: #a020f0;">then</span>
accep_rate = accep_rate + 1.d0
r_old(:) = r_new(:)
d_old(:) = d_new(:)
d2_old = d2_new
psi_old = psi_new
<span style="color: #a020f0;">end if</span>
norm = norm + 1.d0
energy = energy + e_loc(a,r_old)
<span style="color: #a020f0;">end do</span>
energy = energy / norm
accep_rate = accep_rate / norm
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">variational_montecarlo</span>
<span style="color: #a020f0;">program</span> <span style="color: #0000ff;">qmc</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> a = 0.9</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> tau = 1.0</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nmax = 100000</span>
<span style="color: #228b22;">integer</span> , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nruns = 30</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> irun</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> X(nruns), accep(nruns)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> ave, err</span>
<span style="color: #a020f0;">do</span> irun=1,nruns
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">variational_montecarlo</span>(a,tau,nmax,X(irun),accep(irun))
<span style="color: #a020f0;">enddo</span>
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">ave_error</span>(X,nruns,ave,err)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'E = '</span>, ave, <span style="color: #8b2252;">'+/-'</span>, err
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">ave_error</span>(accep,nruns,ave,err)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'A = '</span>, ave, <span style="color: #8b2252;">'+/-'</span>, err
<span style="color: #a020f0;">end program</span> <span style="color: #0000ff;">qmc</span>
</pre>
</div>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
./vmc_metropolis
</pre>
</div>
<pre class="example">
E = -0.49499990423528023 +/- 1.5958250761863871E-004
A = 0.78861366666666655 +/- 3.5096729498002445E-004
</pre>
</div>
</div>
</div>
</div>
<div id="outline-container-org2042bc2" class="outline-2">
<h2 id="org2042bc2"><span class="section-number-2">4</span> <span class="todo TODO">TODO</span> Diffusion Monte Carlo</h2>
<div class="outline-text-2" id="text-4">
</div>
<div id="outline-container-org964c160" class="outline-3">
<h3 id="org964c160"><span class="section-number-3">4.1</span> Hydrogen atom</h3>
<div class="outline-text-3" id="text-4-1">
</div>
<ol class="org-ol">
<li><a id="orge2cfd4b"></a>Exercise<br />
<div class="outline-text-5" id="text-4-1-0-1">
<div class="exercise">
<p>
Modify the Metropolis VMC program to introduce the PDMC weight.
In the limit \(\tau \rightarrow 0\), you should recover the exact
energy of H for any value of \(a\).
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">from</span> qmc_stats <span style="color: #a020f0;">import</span> *
<span style="color: #a020f0;">def</span> <span style="color: #0000ff;">MonteCarlo</span>(a,tau,nmax,Eref):
<span style="color: #a0522d;">E</span> = 0.
<span style="color: #a0522d;">N</span> = 0.
<span style="color: #a0522d;">accep_rate</span> = 0.
<span style="color: #a0522d;">sq_tau</span> = np.sqrt(tau)
<span style="color: #a0522d;">r_old</span> = np.random.normal(loc=0., scale=1.0, size=(3))
<span style="color: #a0522d;">d_old</span> = drift(a,r_old)
<span style="color: #a0522d;">d2_old</span> = np.dot(d_old,d_old)
<span style="color: #a0522d;">psi_old</span> = psi(a,r_old)
<span style="color: #a0522d;">w</span> = 1.0
<span style="color: #a020f0;">for</span> istep <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(nmax):
<span style="color: #a0522d;">chi</span> = np.random.normal(loc=0., scale=1.0, size=(3))
<span style="color: #a0522d;">el</span> = e_loc(a,r_old)
<span style="color: #a0522d;">w</span> *= np.exp(-tau*(el - Eref))
<span style="color: #a0522d;">N</span> += w
<span style="color: #a0522d;">E</span> += w * el
<span style="color: #a0522d;">r_new</span> = r_old + tau * d_old + sq_tau * chi
<span style="color: #a0522d;">d_new</span> = drift(a,r_new)
<span style="color: #a0522d;">d2_new</span> = np.dot(d_new,d_new)
<span style="color: #a0522d;">psi_new</span> = psi(a,r_new)
# <span style="color: #b22222;">Metropolis</span>
<span style="color: #a0522d;">prod</span> = np.dot((d_new + d_old), (r_new - r_old))
<span style="color: #a0522d;">argexpo</span> = 0.5 * (d2_new - d2_old)*tau + prod
<span style="color: #a0522d;">q</span> = psi_new / psi_old
<span style="color: #a0522d;">q</span> = np.exp(-argexpo) * q*q
# <span style="color: #b22222;">PDMC weight</span>
<span style="color: #a020f0;">if</span> np.random.uniform() &lt; q:
<span style="color: #a0522d;">accep_rate</span> += w
<span style="color: #a0522d;">r_old</span> = r_new
<span style="color: #a0522d;">d_old</span> = d_new
<span style="color: #a0522d;">d2_old</span> = d2_new
<span style="color: #a0522d;">psi_old</span> = psi_new
<span style="color: #a020f0;">return</span> E/N, accep_rate/N
<span style="color: #a0522d;">a</span> = 0.9
<span style="color: #a0522d;">nmax</span> = 10000
<span style="color: #a0522d;">tau</span> = .1
<span style="color: #a0522d;">X</span> = [MonteCarlo(a,tau,nmax,-0.5) <span style="color: #a020f0;">for</span> i <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(30)]
<span style="color: #a0522d;">E</span>, <span style="color: #a0522d;">deltaE</span> = ave_error([x[0] <span style="color: #a020f0;">for</span> x <span style="color: #a020f0;">in</span> X])
<span style="color: #a0522d;">A</span>, <span style="color: #a0522d;">deltaA</span> = ave_error([x[1] <span style="color: #a020f0;">for</span> x <span style="color: #a020f0;">in</span> X])
<span style="color: #a020f0;">print</span>(f<span style="color: #8b2252;">"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}"</span>)
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="org-src-container">
<pre class="src src-f90"><span style="color: #a020f0;">subroutine</span> <span style="color: #0000ff;">variational_montecarlo</span>(a,tau,nmax,energy,accep_rate)
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> a, tau</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> nmax </span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(out) ::<span style="color: #a0522d;"> energy, accep_rate</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> istep</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> norm, sq_tau, chi(3), d2_old, prod, u</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> psi_old, psi_new, d2_new, argexpo, q</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> r_old(3), r_new(3)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> d_old(3), d_new(3)</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi</span>
sq_tau = dsqrt(tau)
! <span style="color: #b22222;">Initialization</span>
energy = 0.d0
norm = 0.d0
accep_rate = 0.d0
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_gauss</span>(r_old,3)
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">drift</span>(a,r_old,d_old)
d2_old = d_old(1)*d_old(1) + d_old(2)*d_old(2) + d_old(3)*d_old(3)
psi_old = psi(a,r_old)
<span style="color: #a020f0;">do</span> istep = 1,nmax
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_gauss</span>(chi,3)
r_new(:) = r_old(:) + tau * d_old(:) + chi(:)*sq_tau
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">drift</span>(a,r_new,d_new)
d2_new = d_new(1)*d_new(1) + d_new(2)*d_new(2) + d_new(3)*d_new(3)
psi_new = psi(a,r_new)
! <span style="color: #b22222;">Metropolis</span>
prod = (d_new(1) + d_old(1))*(r_new(1) - r_old(1)) + <span style="color: #a020f0;">&amp;</span>
(d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + <span style="color: #a020f0;">&amp;</span>
(d_new(3) + d_old(3))*(r_new(3) - r_old(3))
argexpo = 0.5d0 * (d2_new - d2_old)*tau + prod
q = psi_new / psi_old
q = dexp(-argexpo) * q*q
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">random_number</span>(u)
<span style="color: #a020f0;">if</span> (u&lt;q) <span style="color: #a020f0;">then</span>
accep_rate = accep_rate + 1.d0
r_old(:) = r_new(:)
d_old(:) = d_new(:)
d2_old = d2_new
psi_old = psi_new
<span style="color: #a020f0;">end if</span>
norm = norm + 1.d0
energy = energy + e_loc(a,r_old)
<span style="color: #a020f0;">end do</span>
energy = energy / norm
accep_rate = accep_rate / norm
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">variational_montecarlo</span>
<span style="color: #a020f0;">program</span> <span style="color: #0000ff;">qmc</span>
<span style="color: #a020f0;">implicit</span> <span style="color: #228b22;">none</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> a = 0.9</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> tau = 1.0</span>
<span style="color: #228b22;">integer</span>*8 , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nmax = 100000</span>
<span style="color: #228b22;">integer</span> , <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> nruns = 30</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> irun</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> X(nruns), accep(nruns)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> ave, err</span>
<span style="color: #a020f0;">do</span> irun=1,nruns
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">variational_montecarlo</span>(a,tau,nmax,X(irun),accep(irun))
<span style="color: #a020f0;">enddo</span>
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">ave_error</span>(X,nruns,ave,err)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'E = '</span>, ave, <span style="color: #8b2252;">'+/-'</span>, err
<span style="color: #a020f0;">call</span> <span style="color: #0000ff;">ave_error</span>(accep,nruns,ave,err)
<span style="color: #a020f0;">print</span> *, <span style="color: #8b2252;">'A = '</span>, ave, <span style="color: #8b2252;">'+/-'</span>, err
<span style="color: #a020f0;">end program</span> <span style="color: #0000ff;">qmc</span>
</pre>
</div>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
./vmc_metropolis
</pre>
</div>
<pre class="example">
E = -0.49499990423528023 +/- 1.5958250761863871E-004
A = 0.78861366666666655 +/- 3.5096729498002445E-004
</pre>
</div>
</li>
</ol>
</div>
<div id="outline-container-orgf75f8aa" class="outline-3">
<h3 id="orgf75f8aa"><span class="section-number-3">4.2</span> Dihydrogen</h3>
<div class="outline-text-3" id="text-4-2">
<p>
We will now consider the H<sub>2</sub> molecule in a minimal basis composed of the
\(1s\) orbitals of the hydrogen atoms:
</p>
<p>
\[
\Psi(\mathbf{r}_1, \mathbf{r}_2) =
\exp(-(\mathbf{r}_1 - \mathbf{R}_A)) +
\]
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.
</p>
</div>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Author: Anthony Scemama, Claudia Filippi</p>
<p class="date">Created: 2021-01-20 Wed 18:13</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
</div>
</body>
</html>