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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="#org929911c">1. Introduction</a>
<ul>
<li><a href="#orga2abb91">1.1. Energy and local energy</a></li>
</ul>
</li>
<li><a href="#org1fb2fcb">2. Numerical evaluation of the energy of the hydrogen atom</a>
<ul>
<li><a href="#org389532d">2.1. Local energy</a>
<ul>
<li><a href="#org0472b89">2.1.1. Exercise 1</a>
<ul>
<li><a href="#org3bafbca">2.1.1.1. Solution</a></li>
</ul>
</li>
<li><a href="#orgad48d97">2.1.2. Exercise 2</a>
<ul>
<li><a href="#orgf449f06">2.1.2.1. Solution</a></li>
</ul>
</li>
<li><a href="#org965213c">2.1.3. Exercise 3</a>
<ul>
<li><a href="#org0fe7a9a">2.1.3.1. Solution</a></li>
</ul>
</li>
<li><a href="#org8ed5b25">2.1.4. Exercise 4</a>
<ul>
<li><a href="#org1b60e83">2.1.4.1. Solution</a></li>
</ul>
</li>
<li><a href="#orgbfd6da2">2.1.5. Exercise 5</a>
<ul>
<li><a href="#org4320ea5">2.1.5.1. Solution</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org44b3c40">2.2. Plot of the local energy along the \(x\) axis</a>
<ul>
<li><a href="#org838197e">2.2.1. Exercise</a>
<ul>
<li><a href="#org9ce92e4">2.2.1.1. Solution</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org9586528">2.3. Numerical estimation of the energy</a>
<ul>
<li><a href="#org1ac15c2">2.3.1. Exercise</a>
<ul>
<li><a href="#orga479ac3">2.3.1.1. Solution</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#orgacf6fc5">2.4. Variance of the local energy</a>
<ul>
<li><a href="#org58436be">2.4.1. Exercise (optional)</a>
<ul>
<li><a href="#org0d03637">2.4.1.1. <span class="done DONE">DONE</span> Solution</a></li>
</ul>
</li>
<li><a href="#orged3345c">2.4.2. Exercise</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org95501a9">3. Variational Monte Carlo</a>
<ul>
<li><a href="#org1225c45">3.1. Computation of the statistical error</a>
<ul>
<li><a href="#org6dbe18d">3.1.1. Exercise</a></li>
</ul>
</li>
<li><a href="#orgf6d3040">3.2. Uniform sampling in the box</a>
<ul>
<li><a href="#org4cb0213">3.2.1. Exercise</a></li>
</ul>
</li>
<li><a href="#org9992a5c">3.3. Metropolis sampling with \(\Psi^2\)</a>
<ul>
<li><a href="#orga67f9ac">3.3.1. Optimal step size</a></li>
<li><a href="#org5ba7d14">3.3.2. Exercise</a></li>
</ul>
</li>
<li><a href="#org111a1e5">3.4. Generalized Metropolis algorithm</a>
<ul>
<li><a href="#org40cf0d9">3.4.1. Gaussian random number generator</a></li>
<li><a href="#org360cef1">3.4.2. Exercise 1</a></li>
<li><a href="#orga909007">3.4.3. Exercise 2</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#orgb2f6f52">4. Diffusion Monte Carlo</a>
<ul>
<li><a href="#org1cc4b9f">4.1. Schrödinger equation in imaginary time</a></li>
<li><a href="#org837b4d0">4.2. Relation to diffusion</a></li>
<li><a href="#org14723a7">4.3. Importance sampling</a>
<ul>
<li><a href="#orgdfdce4c">4.3.1. Appendix : Details of the Derivation</a></li>
</ul>
</li>
<li><a href="#org132ba01">4.4. Pure Diffusion Monte Carlo</a></li>
<li><a href="#orgefdaa89">4.5. Hydrogen atom</a>
<ul>
<li><a href="#org678058c">4.5.1. Exercise</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#orgb44f3af">5. Project</a></li>
<li><a href="#orgaa62257">6. Acknowledgments</a></li>
</ul>
</div>
</div>
<div id="outline-container-org929911c" class="outline-2">
<h2 id="org929911c"><span class="section-number-2">1</span> Introduction</h2>
<div class="outline-text-2" id="text-1">
<p>
This website contains the QMC tutorial of the 2021 LTTC winter school
<a href="https://www.irsamc.ups-tlse.fr/lttc/Luchon">Tutorials in Theoretical Chemistry</a>.
</p>
<p>
We propose different exercises to understand quantum Monte Carlo (QMC)
methods. In the first section, we start with the computation of the energy of a
hydrogen atom using numerical integration. The goal of this section is
to familarize yourself with the concept of <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, and apply this approach to the
hydrogen atom.
Finally, we present the diffusion Monte Carlo (DMC) method which
we use here to estimate the exact energy of the hydrogen atom and of the H<sub>2</sub> molecule,
starting from an approximate wave function.
</p>
<p>
Code examples will be given in Python3 and Fortran. You can use
whatever language you prefer to write the programs.
</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>
All the quantities are expressed in <i>atomic units</i> (energies,
coordinates, etc).
</p>
</div>
<div id="outline-container-orga2abb91" class="outline-3">
<h3 id="orga2abb91"><span class="section-number-3">1.1</span> Energy and local energy</h3>
<div class="outline-text-3" id="text-1-1">
<p>
For a given system with Hamiltonian \(\hat{H}\) and wave function \(\Psi\), we define the local energy as
</p>
<p>
\[
E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
\]
</p>
<p>
where \(\mathbf{r}\) denotes the 3N-dimensional electronic coordinates.
</p>
<p>
The electronic energy of a system, \(E\), can be rewritten in terms of the
local energy \(E_L(\mathbf{r})\) as
</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 |\Psi(\mathbf{r}) |^2 d\mathbf{r}} \\
& = & \frac{\int |\Psi(\mathbf{r})|^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}
= \frac{\int |\Psi(\mathbf{r})|^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}
\end{eqnarray*}
<p>
For few dimensions, one can easily compute \(E\) by evaluating the
integrals on a grid but, for a high number of dimensions, one can
resort to Monte Carlo techniques to compute \(E\).
</p>
<p>
To this aim, recall that 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>
where a probability density function \(P(x)\) is non-negative
and integrates to one:
</p>
<p>
\[ \int_{-\infty}^\infty P(x)\,dx = 1. \]
</p>
<p>
Similarly, we can view the the energy of a system, \(E\), as the expected value of the local energy with respect to
a probability density \(P(\mathbf{r})\) defined in 3\(N\) dimensions:
</p>
<p>
\[ E = \int E_L(\mathbf{r}) P(\mathbf{r})\,d\mathbf{r} \equiv \langle E_L \rangle_{P}\,, \]
</p>
<p>
where the probability density is given by the square of the wave function:
</p>
<p>
\[ P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,. \]
</p>
<p>
If we can sample \(N_{\rm MC}\) configurations \(\{\mathbf{r}\}\)
distributed as \(P\), we can estimate \(E\) as the average of the local
energy computed over these configurations:
</p>
<p>
\[ E \approx \frac{1}{N_{\rm MC}} \sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i) \,. \]
</p>
</div>
</div>
</div>
<div id="outline-container-org1fb2fcb" class="outline-2">
<h2 id="org1fb2fcb"><span class="section-number-2">2</span> Numerical evaluation of the energy of the hydrogen atom</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, for a particular value of \(a\), \(\Psi\) is an
eigenfunction of the Hamiltonian
</p>
<p>
\[
\hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|}
\]
</p>
<p>
To do that, we will compute the local energy and check whether it is constant.
</p>
</div>
<div id="outline-container-org389532d" class="outline-3">
<h3 id="org389532d"><span class="section-number-3">2.1</span> Local energy</h3>
<div class="outline-text-3" id="text-2-1">
<p>
You will now program all quantities needed to compute the local energy of the H atom for the given wave function.
</p>
<p>
Write all the functions of this section in a single file :
<code>hydrogen.py</code> if you use Python, or <code>hydrogen.f90</code> is you use
Fortran.
</p>
<div class="note">
<ul class="org-ul">
<li>When computing a square root in \(\mathbb{R}\), <b>always</b> make sure
that the argument of the square root is non-negative.</li>
<li>When you divide, <b>always</b> make sure that you will not divide by zero</li>
</ul>
<p>
If a <i>floating-point exception</i> can occur, you should make a test
to catch the error.
</p>
</div>
</div>
<div id="outline-container-org0472b89" class="outline-4">
<h4 id="org0472b89"><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 argument
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: #b22222;">!/usr/bin/env python3</span>
<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: #b22222;">TODO</span>
</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>
! <span style="color: #b22222;">TODO</span>
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">potential</span>
</pre>
</div>
</div>
<div id="outline-container-org3bafbca" class="outline-5">
<h5 id="org3bafbca"><span class="section-number-5">2.1.1.1</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-1-1-1">
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python">#<span style="color: #b22222;">!/usr/bin/env python3</span>
<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: #a0522d;">distance</span> = np.sqrt(np.dot(r,r))
<span style="color: #a020f0;">assert</span> (distance &gt; 0)
<span style="color: #a020f0;">return</span> -1. / distance
</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>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> distance</span>
distance = dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )
<span style="color: #a020f0;">if</span> (distance &gt; 0.d0) <span style="color: #a020f0;">then</span>
potential = -1.d0 / distance
<span style="color: #a020f0;">else</span>
<span style="color: #a020f0;">stop</span> <span style="color: #8b2252;">'potential at r=0.d0 diverges'</span>
<span style="color: #a020f0;">end if</span>
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">potential</span>
</pre>
</div>
</div>
</div>
</div>
<div id="outline-container-orgad48d97" class="outline-4">
<h4 id="orgad48d97"><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: #b22222;">TODO</span>
</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>
! <span style="color: #b22222;">TODO</span>
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">psi</span>
</pre>
</div>
</div>
<div id="outline-container-orgf449f06" class="outline-5">
<h5 id="orgf449f06"><span class="section-number-5">2.1.2.1</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-1-2-1">
<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>
<div id="outline-container-org965213c" class="outline-4">
<h4 id="org965213c"><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 \[T_L(\mathbf{r}) = -\frac{1}{2}\frac{\Delta \Psi(\mathbf{r})}{\Psi(\mathbf{r})}.\]
</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>
Therefore, the local kinetic energy is
\[
T_L (\mathbf{r}) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right)
\]
</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: #b22222;">TODO</span>
</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>
! <span style="color: #b22222;">TODO</span>
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">kinetic</span>
</pre>
</div>
</div>
<div id="outline-container-org0fe7a9a" class="outline-5">
<h5 id="org0fe7a9a"><span class="section-number-5">2.1.3.1</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-1-3-1">
<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: #a0522d;">distance</span> = np.sqrt(np.dot(r,r))
<span style="color: #a020f0;">assert</span> (distance &gt; 0.)
<span style="color: #a020f0;">return</span> a * (1./distance - 0.5 * a)
</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>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> distance</span>
distance = dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )
<span style="color: #a020f0;">if</span> (distance &gt; 0.d0) <span style="color: #a020f0;">then</span>
kinetic = a * (1.d0 / distance - 0.5d0 * a)
<span style="color: #a020f0;">else</span>
<span style="color: #a020f0;">stop</span> <span style="color: #8b2252;">'kinetic energy diverges at r=0'</span>
<span style="color: #a020f0;">end if</span>
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">kinetic</span>
</pre>
</div>
</div>
</div>
</div>
<div id="outline-container-org8ed5b25" class="outline-4">
<h4 id="org8ed5b25"><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 \(\mathbf{r}\),
using the previously defined functions.
The function accepts <code>a</code> and <code>r</code> as input arguments and returns the
local kinetic 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: #b22222;">TODO</span>
</pre>
</div>
<p>
<b>Fortran</b>
</p>
<div class="note">
<p>
When you call a function in Fortran, you need to declare its
return type.
You might by accident choose a function name which is the
same as an internal function of Fortran. So it is recommended to
<b>always</b> use the keyword <code>external</code> to make sure the function you
are calling is yours.
</p>
</div>
<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</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> potential</span>
! <span style="color: #b22222;">TODO</span>
<span style="color: #a020f0;">end function</span> <span style="color: #0000ff;">e_loc</span>
</pre>
</div>
</div>
<div id="outline-container-org1b60e83" class="outline-5">
<h5 id="org1b60e83"><span class="section-number-5">2.1.4.1</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-1-4-1">
<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</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> 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-orgbfd6da2" class="outline-4">
<h4 id="orgbfd6da2"><span class="section-number-4">2.1.5</span> Exercise 5</h4>
<div class="outline-text-4" id="text-2-1-5">
<div class="exercise">
<p>
Find the theoretical value of \(a\) for which \(\Psi\) is an eigenfunction of \(\hat{H}\).
</p>
</div>
</div>
<div id="outline-container-org4320ea5" class="outline-5">
<h5 id="org4320ea5"><span class="section-number-5">2.1.5.1</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-1-5-1">
\begin{eqnarray*}
E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} -
\frac{1}{|\mathbf{r}|} \\
&=& -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right) -
\frac{1}{|\mathbf{r}|} \\
&=&
-\frac{1}{2} a^2 + \frac{a-1}{\mathbf{|r|}}
\end{eqnarray*}
<p>
\(a=1\) cancels the \(1/|r|\) term, and makes the energy constant and
equal to -0.5 atomic units.
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-org44b3c40" class="outline-3">
<h3 id="org44b3c40"><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">
<p>
The program you will write in this section will be written in
another file (<code>plot_hydrogen.py</code> or <code>plot_hydrogen.f90</code> for
example).
It will use the functions previously defined.
</p>
<p>
In Python, you should put at the beginning of the file
</p>
<div class="org-src-container">
<pre class="src src-python">#<span style="color: #b22222;">!/usr/bin/env python3</span>
<span style="color: #a020f0;">from</span> hydrogen <span style="color: #a020f0;">import</span> e_loc
</pre>
</div>
<p>
to be able to use the <code>e_loc</code> function of the <code>hydrogen.py</code> file.
</p>
<p>
In Fortran, you will need to compile all the source files together:
</p>
<div class="org-src-container">
<pre class="src src-sh">gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
</pre>
</div>
</div>
<div id="outline-container-org838197e" class="outline-4">
<h4 id="org838197e"><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. In Python, you can use matplotlib
for example. In Fortran, it is convenient to write in a text file
the values of \(x\) and \(E_L(\mathbf{r})\) for each point, and use
Gnuplot to plot the files. With Gnuplot, you will need 2 blank
lines to separate the data corresponding to different values of \(a\).
</p>
</div>
<div class="note">
<p>
The potential and the kinetic energy both diverge at \(r=0\), so we
choose a grid which does not contain the origin to avoid numerical issues.
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python">#<span style="color: #b22222;">!/usr/bin/env python3</span>
<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)
plt.figure(figsize=(10,5))
# <span style="color: #b22222;">TODO</span>
plt.tight_layout()
plt.legend()
plt.savefig(<span style="color: #8b2252;">"plot_py.png"</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;">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), dx</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> i, j</span>
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>
! <span style="color: #b22222;">TODO</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>
<div id="outline-container-org9ce92e4" class="outline-5">
<h5 id="org9ce92e4"><span class="section-number-5">2.2.1.1</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-2-1-1">
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python">#<span style="color: #b22222;">!/usr/bin/env python3</span>
<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)
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>=np.array([ e_loc(a, np.array([t,0.,0.]) ) <span style="color: #a020f0;">for</span> t <span style="color: #a020f0;">in</span> x])
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>
<div class="figure">
<p><img src="plot.png" alt="plot.png" />
</p>
</div>
</div>
</div>
</div>
</div>
<div id="outline-container-org9586528" class="outline-3">
<h3 id="org9586528"><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 wave function square 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-org1ac15c2" class="outline-4">
<h4 id="org1ac15c2"><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 using 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: #b22222;">!/usr/bin/env python3</span>
<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: #b22222;">TODO</span>
<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>
<span style="color: #a020f0;">do</span> j=1,<span style="color: #a020f0;">size</span>(a)
! <span style="color: #b22222;">TODO</span>
<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>
</div>
<div id="outline-container-orga479ac3" class="outline-5">
<h5 id="orga479ac3"><span class="section-number-5">2.3.1.1</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-3-1-1">
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python">#<span style="color: #b22222;">!/usr/bin/env python3</span>
<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>
<pre class="example">
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
</pre>
<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>
<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>
<div id="outline-container-orgacf6fc5" class="outline-3">
<h3 id="orgacf6fc5"><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 its average:
</p>
<p>
\[
\sigma^2(E_L) = \frac{\int |\Psi(\mathbf{r})|^2\, \left[
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}
\]
which can be simplified as
</p>
<p>
\[ \sigma^2(E_L) = \langle E_L^2 \rangle_{\Psi^2} - \langle E_L \rangle_{\Psi^2}^2.\]
</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-org58436be" class="outline-4">
<h4 id="org58436be"><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 :
\[\langle \left( E - \langle E \rangle_{\Psi^2} \right)^2\rangle_{\Psi^2} = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 \]
</p>
</div>
</div>
<div id="outline-container-org0d03637" class="outline-5">
<h5 id="org0d03637"><span class="section-number-5">2.4.1.1</span> <span class="done DONE">DONE</span> Solution&#xa0;&#xa0;&#xa0;<span class="tag"><span class="solution2">solution2</span></span></h5>
<div class="outline-text-5" id="text-2-4-1-1">
<p>
\(\bar{E} = \langle E \rangle\) is a constant, so \(\langle \bar{E}
\rangle = \bar{E}\) .
</p>
\begin{eqnarray*}
\langle (E - \bar{E})^2 \rangle & = &
\langle E^2 - 2 E \bar{E} + \bar{E}^2 \rangle \\
&=& \langle E^2 \rangle - 2 \langle E \bar{E} \rangle + \langle \bar{E}^2 \rangle \\
&=& \langle E^2 \rangle - 2 \langle E \rangle \bar{E} + \bar{E}^2 \\
&=& \langle E^2 \rangle - 2 \bar{E}^2 + \bar{E}^2 \\
&=& \langle E^2 \rangle - \bar{E}^2 \\
&=& \langle E^2 \rangle - \langle E \rangle^2 \\
\end{eqnarray*}
</div>
</div>
</div>
<div id="outline-container-orged3345c" class="outline-4">
<h4 id="orged3345c"><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 using
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: #b22222;">!/usr/bin/env python3</span>
<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: #b22222;">TODO</span>
<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: #a0522d;"> x(50), w, delta, energy, energy2</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> dx, r(3), a(6), norm, e_tmp, s2</span>
<span style="color: #228b22;">integer</span> ::<span style="color: #a0522d;"> i, k, l, j</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi</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>
<span style="color: #a020f0;">do</span> j=1,<span style="color: #a020f0;">size</span>(a)
! <span style="color: #b22222;">TODO</span>
<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;">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>
</div>
</div>
</div>
</div>
<div id="outline-container-org95501a9" class="outline-2">
<h2 id="org95501a9"><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 use Monte Carlo sampling.
</p>
<p>
Moreover, Monte Carlo sampling will allow us to remove the bias due
to the discretization of space, and compute a statistical confidence
interval.
</p>
</div>
<div id="outline-container-org1225c45" class="outline-3">
<h3 id="org1225c45"><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 for large \(M\), according to the <a href="https://en.wikipedia.org/wiki/Central_limit_theorem">Central Limit Theorem</a>.
</p>
<p>
The estimate of the energy is
</p>
<p>
\[
E = \frac{1}{M} \sum_{i=1}^M E_i
\]
</p>
<p>
The variance of the average energies can be computed as
</p>
<p>
\[
\sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_i - E)^2
\]
</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-org6dbe18d" class="outline-4">
<h4 id="org6dbe18d"><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: #b22222;">!/usr/bin/env python3</span>
<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: #b22222;">TODO</span>
<span style="color: #a020f0;">return</span> (average, error)
</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: #b22222;">TODO</span>
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">ave_error</span>
</pre>
</div>
</div>
</div>
</div>
<div id="outline-container-orgf6d3040" class="outline-3">
<h3 id="orgf6d3040"><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 perform our first Monte Carlo calculation to compute the
energy of the hydrogen atom.
</p>
<p>
Consider again the expression of the energy
</p>
\begin{eqnarray*}
E & = & \frac{\int E_L(\mathbf{r})|\Psi(\mathbf{r})|^2\,d\mathbf{r}}{\int |\Psi(\mathbf{r}) |^2 d\mathbf{r}}\,.
\end{eqnarray*}
<p>
Clearly, the square of the wave function is a good choice of probability density to sample but we will start with something simpler and rewrite the energy as
</p>
\begin{eqnarray*}
E & = & \frac{\int E_L(\mathbf{r})\frac{|\Psi(\mathbf{r})|^2}{P(\mathbf{r})}P(\mathbf{r})\, \,d\mathbf{r}}{\int \frac{|\Psi(\mathbf{r})|^2 }{P(\mathbf{r})}P(\mathbf{r})d\mathbf{r}}\,.
\end{eqnarray*}
<p>
Here, we will sample a uniform probability \(P(\mathbf{r})\) in a cube of volume \(L^3\) centered at the origin:
</p>
<p>
\[ P(\mathbf{r}) = \frac{1}{L^3}\,, \]
</p>
<p>
and zero outside the cube.
</p>
<p>
One Monte Carlo run will consist of \(N_{\rm MC}\) Monte Carlo iterations. At every Monte Carlo iteration:
</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>
Once all the iterations have been computed, the run returns the average energy
\(\bar{E}_k\) over the \(N_{\rm MC}\) iterations of the run.
</p>
<p>
To compute the statistical error, perform \(M\) independent 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-org4cb0213" class="outline-4">
<h4 id="org4cb0213"><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 \(a=1.2\). 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="note">
<p>
To draw a uniform random number in Python, you can use
the <a href="https://numpy.org/doc/stable/reference/random/generated/numpy.random.uniform.html"><code>random.uniform</code></a> function of Numpy.
</p>
</div>
<div class="org-src-container">
<pre class="src src-python">#<span style="color: #b22222;">!/usr/bin/env python3</span>
<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: #b22222;">TODO</span>
<span style="color: #a0522d;">a</span> = 1.2
<span style="color: #a0522d;">nmax</span> = 100000
#<span style="color: #b22222;">TODO</span>
<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>
To draw a uniform random number in Fortran, you can use
the <a href="https://gcc.gnu.org/onlinedocs/gfortran/RANDOM_005fNUMBER.html"><code>RANDOM_NUMBER</code></a> subroutine.
</p>
</div>
<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>
! <span style="color: #b22222;">TODO</span>
<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 = 1.2d0</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: #b22222;">TODO</span>
<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>
</div>
</div>
</div>
<div id="outline-container-org9992a5c" class="outline-3">
<h3 id="org9992a5c"><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
\[
P(\mathbf{r}) = \frac{|\Psi(\mathbf{r})|^2}{\int |\Psi(\mathbf{r})|^2 d\mathbf{r}}\,.
\]
</p>
<p>
The expression of the average energy is now simplified as the average of
the local energies, since the weights are taken care of by the
sampling:
</p>
<p>
\[
E \approx \frac{1}{N_{\rm MC}}\sum_{i=1}^{N_{\rm MC}} E_L(\mathbf{r}_i)\,.
\]
</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:
</p>
<p>
\[ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \rightarrow \mathbf{r}_{N_{\rm MC}}\,, \]
</p>
<p>
according to the following algorithm.
</p>
<p>
At every step, we propose a new move according to a transition probability \(T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})\) of our choice.
</p>
<p>
For simplicity, we will move the electron in a 3-dimensional box of side \(2\delta L\) centered at the current position
of the electron:
</p>
<p>
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta L \, \mathbf{u}
\]
</p>
<p>
where \(\delta L\) is a fixed constant, and
\(\mathbf{u}\) is a uniform random number in a 3-dimensional box
\((-1,-1,-1) \le \mathbf{u} \le (1,1,1)\).
</p>
<p>
After having moved the electron, we add the
accept/reject step that guarantees that the distribution of the
\(\mathbf{r}_n\) is \(\Psi^2\). This amounts to accepting the move with
probability
</p>
<p>
\[
A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{T(\mathbf{r}_{n+1}\rightarrow\mathbf{r}_{n}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1})P(\mathbf{r}_{n})}\right)\,,
\]
</p>
<p>
which, for our choice of transition probability, becomes
</p>
<p>
\[
A(\mathbf{r}_{n}\rightarrow\mathbf{r}_{n+1}) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{|\Psi(\mathbf{r}_{n+1})|^2}{|\Psi(\mathbf{r}_{n})|^2}\right)\,.
\]
</p>
<div class="exercise">
<p>
Explain why the transition probability cancels out in the
expression of \(A\).
</p>
</div>
<p>
Also note that we do not need to compute the norm of the wave function!
</p>
<p>
The algorithm is summarized as follows:
</p>
<ol class="org-ol">
<li>Evaluate the local energy at \(\mathbf{r}_n\) and accumulate it</li>
<li>Compute a new position \(\mathbf{r'} = \mathbf{r}_n + \delta L\, \mathbf{u}\)</li>
<li>Evaluate \(\Psi(\mathbf{r}')\) at the new position</li>
<li>Compute the ratio \(A = \frac{\left|\Psi(\mathbf{r'})\right|^2}{\left|\Psi(\mathbf{r}_{n})\right|^2}\)</li>
<li>Draw a uniform random number \(v \in [0,1]\)</li>
<li>if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)</li>
<li>else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)</li>
</ol>
<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 <i>and</i> rejected moves.
</p>
</div>
</div>
<div id="outline-container-orga67f9ac" class="outline-4">
<h4 id="orga67f9ac"><span class="section-number-4">3.3.1</span> Optimal step size</h4>
<div class="outline-text-4" id="text-3-3-1">
<p>
If the box is infinitely small, the ratio will be very close
to one and all the steps will be accepted. However, the moves will be
very correlated and you will visit the configurational space very slowly.
</p>
<p>
On the other hand, if you propose too large moves, 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 size of the move 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 for the current problem.
</p>
<div class="note">
<p>
Below, we use the symbol \(\delta t\) to denote \(\delta L\) since we will use
the same variable later on to store a time step.
</p>
</div>
</div>
</div>
<div id="outline-container-org5ba7d14" class="outline-4">
<h4 id="org5ba7d14"><span class="section-number-4">3.3.2</span> Exercise</h4>
<div class="outline-text-4" id="text-3-3-2">
<div class="exercise">
<p>
Modify the program of the previous section to compute the energy,
sampled with \(\Psi^2\).
</p>
<p>
Compute also the acceptance rate, so that you can adapt the time
step in order to have an acceptance rate close to 0.5.
</p>
<p>
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: #b22222;">!/usr/bin/env python3</span>
<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,dt):
# <span style="color: #b22222;">TODO</span>
<span style="color: #a020f0;">return</span> energy/nmax, N_accep/nmax
# <span style="color: #b22222;">Run simulation</span>
<span style="color: #a0522d;">a</span> = 1.2
<span style="color: #a0522d;">nmax</span> = 100000
<span style="color: #a0522d;">dt</span> = #<span style="color: #b22222;">TODO</span>
<span style="color: #a0522d;">X0</span> = [ MonteCarlo(a,nmax,dt) <span style="color: #a020f0;">for</span> i <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(30)]
# <span style="color: #b22222;">Energy</span>
<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;">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>)
# <span style="color: #b22222;">Acceptance rate</span>
<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>, <span style="color: #a0522d;">deltaA</span> = ave_error(X)
<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,dt,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;"> dt </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;">integer</span>*8 ::<span style="color: #a0522d;"> n_accep</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> r_old(3), r_new(3), psi_old, psi_new</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> v, ratio</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">external</span> ::<span style="color: #a0522d;"> e_loc, psi, gaussian</span>
! <span style="color: #b22222;">TODO</span>
<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 = 1.2d0</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> dt = </span>! <span style="color: #b22222;">TODO</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,dt,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>
</div>
</div>
</div>
<div id="outline-container-org111a1e5" class="outline-3">
<h3 id="org111a1e5"><span class="section-number-3">3.4</span> Generalized Metropolis algorithm</h3>
<div class="outline-text-3" id="text-3-4">
<p>
One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability.
</p>
<p>
The Metropolis acceptance step has to be adapted accordingly to ensure that the detailed balance condition is satisfied. This means that
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 sampling in a box centered
at the current position. Hence, the transition probability was symmetric
</p>
<p>
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n})
= \text{constant}\,,
\]
</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, we
choose to draw Gaussian random numbers with zero mean and variance
\(\delta t\), the transition probability becomes:
</p>
<p>
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\delta t} \right]\,.
\]
</p>
<p>
Furthermore, to sample the density even better, we can "push" the electrons
into in the regions of high probability, and "pull" them away from
the low-probability regions. This will increase the
acceptance ratios and improve the sampling.
</p>
<p>
To do this, we can use the gradient of the probability density
</p>
<p>
\[
\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,,
\]
</p>
<p>
and add the so-called drift vector, \(\frac{\nabla \Psi}{\Psi}\), so that the numerical scheme becomes a
drifted diffusion with transition probability:
</p>
<p>
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} - \delta t\frac{\nabla
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,.
\]
</p>
<p>
The corresponding move is proposed as
</p>
<p>
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi \,,
\]
</p>
<p>
where \(\chi\) is a Gaussian random variable with zero mean and
variance \(\delta t\).
</p>
<p>
The algorithm of the previous exercise is only slighlty modified as:
</p>
<ol class="org-ol">
<li>Evaluate the local energy at \(\mathbf{r}_{n}\) and accumulate it</li>
<li>Compute a new position \(\mathbf{r'} = \mathbf{r}_n +
\delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)</li>
<li>Evaluate \(\Psi(\mathbf{r}')\) and \(\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}\) at the new position</li>
<li>Compute the ratio \(A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}\)</li>
<li>Draw a uniform random number \(v \in [0,1]\)</li>
<li>if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)</li>
<li>else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)</li>
</ol>
</div>
<div id="outline-container-org40cf0d9" class="outline-4">
<h4 id="org40cf0d9"><span class="section-number-4">3.4.1</span> Gaussian random number generator</h4>
<div class="outline-text-4" id="text-3-4-1">
<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>
<p>
In Python, you can use the <a href="https://numpy.org/doc/stable/reference/random/generated/numpy.random.normal.html"><code>random.normal</code></a> function of Numpy.
</p>
</div>
</div>
<div id="outline-container-org360cef1" class="outline-4">
<h4 id="org360cef1"><span class="section-number-4">3.4.2</span> Exercise 1</h4>
<div class="outline-text-4" id="text-3-4-2">
<div class="exercise">
<p>
If you use Fortran, copy/paste the <code>random_gauss</code> function in
a Fortran file.
</p>
</div>
<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: #b22222;">TODO</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;">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: #b22222;">TODO</span>
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">drift</span>
</pre>
</div>
</div>
</div>
<div id="outline-container-orga909007" class="outline-4">
<h4 id="orga909007"><span class="section-number-4">3.4.3</span> Exercise 2</h4>
<div class="outline-text-4" id="text-3-4-3">
<div class="exercise">
<p>
Modify the previous program to introduce the drift-diffusion scheme.
(This is a necessary step for the next section on diffusion Monte Carlo).
</p>
</div>
<p>
<b>Python</b>
</p>
<div class="org-src-container">
<pre class="src src-python">#<span style="color: #b22222;">!/usr/bin/env python3</span>
<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,dt):
# <span style="color: #b22222;">TODO</span>
# <span style="color: #b22222;">Run simulation</span>
<span style="color: #a0522d;">a</span> = 1.2
<span style="color: #a0522d;">nmax</span> = 100000
<span style="color: #a0522d;">dt</span> = # <span style="color: #b22222;">TODO</span>
<span style="color: #a0522d;">X0</span> = [ MonteCarlo(a,nmax,dt) <span style="color: #a020f0;">for</span> i <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(30)]
# <span style="color: #b22222;">Energy</span>
<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;">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>)
# <span style="color: #b22222;">Acceptance rate</span>
<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>, <span style="color: #a0522d;">deltaA</span> = ave_error(X)
<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;">variational_montecarlo</span>(a,dt,nmax,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, dt</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</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> istep</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> n_accep</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> sq_dt, chi(3)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> psi_old, psi_new</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>
! <span style="color: #b22222;">TODO</span>
<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 = 1.2d0</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> dt = </span>! <span style="color: #b22222;">TODO</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,dt,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>
</div>
</div>
</div>
</div>
<div id="outline-container-orgb2f6f52" class="outline-2">
<h2 id="orgb2f6f52"><span class="section-number-2">4</span> Diffusion Monte Carlo</h2>
<div class="outline-text-2" id="text-4">
<p>
As we have seen, Variational Monte Carlo is a powerful method to
compute integrals in large dimensions. It is often used in cases
where the expression of the wave function is such that the integrals
can't be evaluated (multi-centered Slater-type orbitals, correlation
factors, etc).
</p>
<p>
Diffusion Monte Carlo is different. It goes beyond the computation
of the integrals associated with an input wave function, and aims at
finding a near-exact numerical solution to the Schrödinger equation.
</p>
</div>
<div id="outline-container-org1cc4b9f" class="outline-3">
<h3 id="org1cc4b9f"><span class="section-number-3">4.1</span> Schrödinger equation in imaginary time</h3>
<div class="outline-text-3" id="text-4-1">
<p>
Consider the time-dependent Schrödinger equation:
</p>
<p>
\[
i\frac{\partial \Psi(\mathbf{r},t)}{\partial t} = (\hat{H} -E_{\rm ref}) \Psi(\mathbf{r},t)\,.
\]
</p>
<p>
where we introduced a shift in the energy, \(E_{\rm ref}\), for reasons which will become apparent below.
</p>
<p>
We can expand a given starting wave function, \(\Psi(\mathbf{r},0)\), in the basis of the eigenstates
of the time-independent Hamiltonian, \(\Phi_k\), with energies \(E_k\):
</p>
<p>
\[
\Psi(\mathbf{r},0) = \sum_k a_k\, \Phi_k(\mathbf{r}).
\]
</p>
<p>
The solution of the Schrödinger equation at time \(t\) is
</p>
<p>
\[
\Psi(\mathbf{r},t) = \sum_k a_k \exp \left( -i\, (E_k-E_{\rm ref})\, t \right) \Phi_k(\mathbf{r}).
\]
</p>
<p>
Now, if we replace the time variable \(t\) by an imaginary time variable
\(\tau=i\,t\), we obtain
</p>
<p>
\[
-\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = (\hat{H} -E_{\rm ref}) \psi(\mathbf{r}, \tau)
\]
</p>
<p>
where \(\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,\tau)\)
and
</p>
\begin{eqnarray*}
\psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -(E_k-E_{\rm ref})\, \tau) \Phi_k(\mathbf{r})\\
&=& \exp(-(E_0-E_{\rm ref})\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \Phi_k(\mathbf{r})\,.
\end{eqnarray*}
<p>
For large positive values of \(\tau\), \(\psi\) is dominated by the
\(k=0\) term, namely, the lowest eigenstate. If we adjust \(E_{\rm ref}\) to the running estimate of \(E_0\),
we can expect that simulating the differetial equation in
imaginary time will converge to the exact ground state of the
system.
</p>
</div>
</div>
<div id="outline-container-org837b4d0" class="outline-3">
<h3 id="org837b4d0"><span class="section-number-3">4.2</span> Relation to diffusion</h3>
<div class="outline-text-3" id="text-4-2">
<p>
The <a href="https://en.wikipedia.org/wiki/Diffusion_equation">diffusion equation</a> of particles is given by
</p>
<p>
\[
\frac{\partial \psi(\mathbf{r},t)}{\partial t} = D\, \Delta \psi(\mathbf{r},t)
\]
</p>
<p>
where \(D\) is the diffusion coefficient. When the imaginary-time
Schrödinger equation is written in terms of the kinetic energy and
potential,
</p>
<p>
\[
\frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} =
\left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,,
\]
</p>
<p>
it can be identified as the combination of:
</p>
<ul class="org-ul">
<li>a diffusion equation (Laplacian)</li>
<li>an equation whose solution is an exponential (potential)</li>
</ul>
<p>
The diffusion equation can be simulated by a Brownian motion:
</p>
<p>
\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]
</p>
<p>
where \(\chi\) is a Gaussian random variable, and the potential term
can be simulated by creating or destroying particles over time (a
so-called branching process) or by simply considering it as a
cumulative multiplicative weight along the diffusion trajectory
(pure Diffusion Monte Carlo):
</p>
<p>
\[
\prod_i \exp \left( - (V(\mathbf{r}_i) - E_{\text{ref}}) \delta t \right).
\]
</p>
<p>
We note that the ground-state wave function of a Fermionic system is
antisymmetric and changes sign. Therefore, its interpretation as a probability
distribution is somewhat problematic. In fact, mathematically, since
the Bosonic ground state is lower in energy than the Fermionic one, for
large \(\tau\), the system will evolve towards the Bosonic solution.
</p>
<p>
For the systems you will study, this is not an issue:
</p>
<ul class="org-ul">
<li>Hydrogen atom: You only have one electron!</li>
<li>Two-electron system (\(H_2\) or He): The ground-wave function is
antisymmetric in the spin variables but symmetric in the space ones.</li>
</ul>
<p>
Therefore, in both cases, you are dealing with a "Bosonic" ground state.
</p>
</div>
</div>
<div id="outline-container-org14723a7" class="outline-3">
<h3 id="org14723a7"><span class="section-number-3">4.3</span> Importance sampling</h3>
<div class="outline-text-3" id="text-4-3">
<p>
In a molecular system, the potential is far from being constant
and, in fact, diverges at the inter-particle coalescence points. Hence,
it results in very large fluctuations of the erm weight associated with
the potental, making the calculations impossible in practice.
Fortunately, if we multiply the Schrödinger equation by a chosen
<i>trial wave function</i> \(\Psi_T(\mathbf{r})\) (Hartree-Fock, Kohn-Sham
determinant, CI wave function, <i>etc</i>), one obtains
</p>
<p>
\[
-\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) =
\left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r})
\]
</p>
<p>
Defining \(\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})\), (see appendix for details)
</p>
<p>
\[
-\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau}
= -\frac{1}{2} \Delta \Pi(\mathbf{r},\tau) +
\nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})}
\right] + (E_L(\mathbf{r})-E_{\rm ref})\Pi(\mathbf{r},\tau)
\]
</p>
<p>
The new "kinetic energy" can be simulated by the drift-diffusion
scheme presented in the previous section (VMC).
The new "potential" is the local energy, which has smaller fluctuations
when \(\Psi_T\) gets closer to the exact wave function.
This term can be simulated by
\[
\prod_i \exp \left( - (E_L(\mathbf{r}_i) - E_{\text{ref}}) \delta t \right).
\]
where \(E_{\rm ref}\) is the constant we had introduced above, which is adjusted to
an estimate of the average energy to keep the weights close to one.
</p>
<p>
This equation generates the <i>N</i>-electron density \(\Pi\), which is the
product of the ground state solution with the trial wave
function. You may then ask: how can we compute the total energy of
the system?
</p>
<p>
To this aim, we use the <i>mixed estimator</i> of the energy:
</p>
\begin{eqnarray*}
E(\tau) &=& \frac{\langle \psi(\tau) | \hat{H} | \Psi_T \rangle}{\langle \psi(\tau) | \Psi_T \rangle}\\
&=& \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}}
{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \\
&=& \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \,.
\end{eqnarray*}
<p>
For large \(\tau\), we have that
</p>
<p>
\[
\Pi(\mathbf{r},\tau) =\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \rightarrow \Phi_0(\mathbf{r}) \Psi_T(\mathbf{r})\,,
\]
</p>
<p>
and, using that \(\hat{H}\) is Hermitian and that \(\Phi_0\) is an
eigenstate of the Hamiltonian, we obtain for large \(\tau\)
</p>
<p>
\[
E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle}
{\langle \psi_\tau | \Psi_T \rangle}
= \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle}
{\langle \Psi_T | \psi_\tau \rangle}
\rightarrow E_0 \frac{\langle \Psi_T | \Phi_0 \rangle}
{\langle \Psi_T | \Phi_0 \rangle}
= E_0
\]
</p>
<p>
Therefore, we can compute the energy within DMC by generating the
density \(\Pi\) with random walks, and simply averaging the local
energies computed with the trial wave function.
</p>
</div>
<div id="outline-container-orgdfdce4c" class="outline-4">
<h4 id="orgdfdce4c"><span class="section-number-4">4.3.1</span> Appendix : Details of the Derivation</h4>
<div class="outline-text-4" id="text-4-3-1">
<p>
\[
-\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) =
\left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r})
\]
</p>
<p>
\[
-\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau}
= -\frac{1}{2} \Big( \Delta \big[
\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] -
\psi(\mathbf{r},\tau) \Delta \Psi_T(\mathbf{r}) - 2
\nabla \psi(\mathbf{r},\tau) \nabla \Psi_T(\mathbf{r}) \Big) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})
\]
</p>
<p>
\[
-\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau}
= -\frac{1}{2} \Delta \big[\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] +
\frac{1}{2} \psi(\mathbf{r},\tau) \Delta \Psi_T(\mathbf{r}) +
\Psi_T(\mathbf{r})\nabla \psi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} + V(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})
\]
</p>
<p>
\[
-\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau}
= -\frac{1}{2} \Delta \big[\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] +
\psi(\mathbf{r},\tau) \Delta \Psi_T(\mathbf{r}) +
\Psi_T(\mathbf{r})\nabla \psi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})} + E_L(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})
\]
\[
-\frac{\partial \big[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big]}{\partial \tau}
= -\frac{1}{2} \Delta \big[\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) \big] +
\nabla \left[ \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})
\frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})}
\right] + E_L(\mathbf{r}) \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})
\]
</p>
<p>
Defining \(\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)
\Psi_T(\mathbf{r})\),
</p>
<p>
\[
-\frac{\partial \Pi(\mathbf{r},\tau)}{\partial \tau}
= -\frac{1}{2} \Delta \Pi(\mathbf{r},\tau) +
\nabla \left[ \Pi(\mathbf{r},\tau) \frac{\nabla \Psi_T(\mathbf{r})}{\Psi_T(\mathbf{r})}
\right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau)
\]
</p>
</div>
</div>
</div>
<div id="outline-container-org132ba01" class="outline-3">
<h3 id="org132ba01"><span class="section-number-3">4.4</span> Pure Diffusion Monte Carlo</h3>
<div class="outline-text-3" id="text-4-4">
<p>
Instead of having a variable number of particles to simulate the
branching process as in the <i>Diffusion Monte Carlo</i> (DMC) algorithm, we
use variant called <i>pure Diffusion Monte Carlo</i> (PDMC) where
the potential term is considered as a cumulative product of weights:
</p>
\begin{eqnarray*}
W(\mathbf{r}_n, \tau) = \prod_{i=1}^{n} \exp \left( -\delta t\,
(E_L(\mathbf{r}_i) - E_{\text{ref}}) \right) =
\prod_{i=1}^{n} w(\mathbf{r}_i)
\end{eqnarray*}
<p>
where \(\mathbf{r}_i\) are the coordinates along the trajectory and
we introduced a <i>time-step variable</i> \(\delta t\) to discretize the
integral.
</p>
<p>
The PDMC algorithm is less stable than the DMC algorithm: it
requires to have a value of \(E_\text{ref}\) which is close to the
fixed-node energy, and a good trial wave function. Moreover, we
can't let \(\tau\) become too large because the weight whether
explode or vanish: we need to have a fixed value of \(\tau\)
(projection time).
The big advantage of PDMC is that it is rather simple to implement
starting from a VMC code:
</p>
<ol class="org-ol">
<li>Start with \(W(\mathbf{r}_0)=1, \tau_0 = 0\)</li>
<li>Evaluate the local energy at \(\mathbf{r}_{n}\)</li>
<li>Compute the contribution to the weight \(w(\mathbf{r}_n) =
\exp(-\delta t(E_L(\mathbf{r}_n)-E_\text{ref}))\)</li>
<li>Update \(W(\mathbf{r}_{n}) = W(\mathbf{r}_{n-1}) \times w(\mathbf{r}_n)\)</li>
<li>Accumulate the weighted energy \(W(\mathbf{r}_n) \times
E_L(\mathbf{r}_n)\),
and the weight \(W(\mathbf{r}_n)\) for the normalization</li>
<li>Update \(\tau_n = \tau_{n-1} + \delta t\)</li>
<li>If \(\tau_{n} > \tau_\text{max}\), the long projection time has
been reached and we can start an new trajectory from the current
position: reset \(W(r_n) = 1\) and \(\tau_n
= 0\)</li>
<li>Compute a new position \(\mathbf{r'} = \mathbf{r}_n +
\delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)</li>
<li>Evaluate \(\Psi(\mathbf{r}')\) and \(\frac{\nabla \Psi(\mathbf{r'})}{\Psi(\mathbf{r'})}\) at the new position</li>
<li>Compute the ratio \(A = \frac{T(\mathbf{r}' \rightarrow \mathbf{r}_{n}) P(\mathbf{r}')}{T(\mathbf{r}_{n} \rightarrow \mathbf{r}') P(\mathbf{r}_{n})}\)</li>
</ol>
<ol class="org-ol">
<li>Draw a uniform random number \(v \in [0,1]\)</li>
<li>if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)</li>
<li>else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)</li>
</ol>
<p>
Some comments are needed:
</p>
<ul class="org-ul">
<li><p>
You estimate the energy as
</p>
\begin{eqnarray*}
E = \frac{\sum_{k=1}^{N_{\rm MC}} E_L(\mathbf{r}_k) W(\mathbf{r}_k, k\delta t)}{\sum_{k=1}^{N_{\rm MC}} W(\mathbf{r}_k, k\delta t)}
\end{eqnarray*}</li>
<li><p>
The result will be affected by a time-step error
(the finite size of \(\delta t\)) due to the discretization of the
integral, and one has in principle to extrapolate to the limit
\(\delta t \rightarrow 0\). This amounts to fitting the energy
computed for multiple values of \(\delta t\).
</p>
<p>
Here, you will be using a small enough time-step and you should not worry about the extrapolation.
</p></li>
<li>The accept/reject step (steps 9-12 in the algorithm) is in principle not needed for the correctness of
the DMC algorithm. However, its use reduces significantly the time-step error.</li>
</ul>
</div>
</div>
<div id="outline-container-orgefdaa89" class="outline-3">
<h3 id="orgefdaa89"><span class="section-number-3">4.5</span> Hydrogen atom</h3>
<div class="outline-text-3" id="text-4-5">
</div>
<div id="outline-container-org678058c" class="outline-4">
<h4 id="org678058c"><span class="section-number-4">4.5.1</span> Exercise</h4>
<div class="outline-text-4" id="text-4-5-1">
<div class="exercise">
<p>
Modify the Metropolis VMC program into a PDMC program.
In the limit \(\delta t \rightarrow 0\), you should recover the exact
energy of H for any value of \(a\), as long as the simulation is stable.
We choose here a time step of 0.05 a.u. and a fixed projection
time \(\tau\) =100 a.u.
</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, dt, Eref):
# <span style="color: #b22222;">TODO</span>
# <span style="color: #b22222;">Run simulation</span>
<span style="color: #a0522d;">a</span> = 1.2
<span style="color: #a0522d;">nmax</span> = 100000
<span style="color: #a0522d;">dt</span> = 0.05
<span style="color: #a0522d;">tau</span> = 100.
<span style="color: #a0522d;">E_ref</span> = -0.5
<span style="color: #a0522d;">X0</span> = [ MonteCarlo(a, nmax, dt, E_ref) <span style="color: #a020f0;">for</span> i <span style="color: #a020f0;">in</span> <span style="color: #483d8b;">range</span>(30)]
# <span style="color: #b22222;">Energy</span>
<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;">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>)
# <span style="color: #b22222;">Acceptance rate</span>
<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>, <span style="color: #a0522d;">deltaA</span> = ave_error(X)
<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;">pdmc</span>(a, dt, nmax, energy, accep, tau, E_ref)
<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, dt, 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</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">intent</span>(in) ::<span style="color: #a0522d;"> E_ref</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> istep</span>
<span style="color: #228b22;">integer</span>*8 ::<span style="color: #a0522d;"> n_accep</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> sq_dt, chi(3)</span>
<span style="color: #228b22;">double precision</span> ::<span style="color: #a0522d;"> psi_old, psi_new</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>
! <span style="color: #b22222;">TODO</span>
<span style="color: #a020f0;">end subroutine</span> <span style="color: #0000ff;">pdmc</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 = 1.2d0</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> dt = 0.05d0</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> E_ref = -0.5d0</span>
<span style="color: #228b22;">double precision</span>, <span style="color: #a020f0;">parameter</span> ::<span style="color: #a0522d;"> tau = 100.d0</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;">pdmc</span>(a, dt, nmax, X(irun), accep(irun), tau, E_ref)
<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 pdmc.f90 -o pdmc
./pdmc
</pre>
</div>
</div>
</div>
</div>
</div>
<div id="outline-container-orgb44f3af" class="outline-2">
<h2 id="orgb44f3af"><span class="section-number-2">5</span> Project</h2>
<div class="outline-text-2" id="text-5">
<p>
Change your PDMC code for one of the following:
</p>
<ul class="org-ul">
<li>the Helium atom, or</li>
<li>the H<sub>2</sub> molecule at \(R\) =1.401 bohr.</li>
</ul>
<p>
And compute the ground state energy.
</p>
</div>
</div>
<div id="outline-container-orgaa62257" class="outline-2">
<h2 id="orgaa62257"><span class="section-number-2">6</span> Acknowledgments</h2>
<div class="outline-text-2" id="text-6">
<div class="figure">
<p><img src="https://trex-coe.eu/sites/default/files/inline-images/euflag.jpg" alt="euflag.jpg" />
</p>
</div>
<p>
<a href="https://trex-coe.eu">TREX</a> : Targeting Real Chemical Accuracy at the Exascale project
has received funding from the European Union’s Horizon 2020 - Research and
Innovation program - under grant agreement no. 952165. The content of this
document does not represent the opinion of the European Union, and the European
Union is not responsible for any use that might be made of such content.
</p>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Author: Anthony Scemama, Claudia Filippi</p>
<p class="date">Created: 2021-02-04 Thu 14:21</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
</div>
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