From 63ebb51d0a12b3e1e2f2a0506823661901f1c067 Mon Sep 17 00:00:00 2001 From: scemama Date: Wed, 9 Jun 2021 14:50:21 +0000 Subject: [PATCH] deploy: 784af90fa0415bf6f01f65933c03d985c6d01028 --- index.html | 322 ++++++++++++++++++++++++++--------------------------- 1 file changed, 161 insertions(+), 161 deletions(-) diff --git a/index.html b/index.html index e538905..adb1ed2 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> - + Quantum Monte Carlo @@ -329,152 +329,152 @@ for the JavaScript code in this tag.

Table of Contents

-
-

1 Introduction

+
+

1 Introduction

This website contains the QMC tutorial of the 2021 LTTC winter school @@ -514,8 +514,8 @@ coordinates, etc).

-
-

1.1 Energy and local energy

+
+

1.1 Energy and local energy

For a given system with Hamiltonian \(\hat{H}\) and wave function \(\Psi\), we define the local energy as @@ -598,8 +598,8 @@ energy computed over these configurations:

-
-

2 Numerical evaluation of the energy of the hydrogen atom

+
+

2 Numerical evaluation of the energy of the hydrogen atom

In this section, we consider the hydrogen atom with the following @@ -628,8 +628,8 @@ To do that, we will compute the local energy and check whether it is constant.

-
-

2.1 Local energy

+
+

2.1 Local energy

You will now program all quantities needed to compute the local energy of the H atom for the given wave function. @@ -656,8 +656,8 @@ to catch the error.

-
-

2.1.1 Exercise 1

+
+

2.1.1 Exercise 1

@@ -702,8 +702,8 @@ and returns the potential.

-
-
2.1.1.1 Solution   solution
+
+
2.1.1.1 Solution   solution

Python @@ -744,8 +744,8 @@ and returns the potential.

-
-

2.1.2 Exercise 2

+
+

2.1.2 Exercise 2

@@ -780,8 +780,8 @@ input arguments, and returns a scalar.

-
-
2.1.2.1 Solution   solution
+
+
2.1.2.1 Solution   solution

Python @@ -808,8 +808,8 @@ input arguments, and returns a scalar.

-
-

2.1.3 Exercise 3

+
+

2.1.3 Exercise 3

@@ -890,8 +890,8 @@ Therefore, the local kinetic energy is

-
-
2.1.3.1 Solution   solution
+
+
2.1.3.1 Solution   solution

Python @@ -932,8 +932,8 @@ Therefore, the local kinetic energy is

-
-

2.1.4 Exercise 4

+
+

2.1.4 Exercise 4

@@ -992,8 +992,8 @@ are calling is yours.

-
-
2.1.4.1 Solution   solution
+
+
2.1.4.1 Solution   solution

Python @@ -1024,8 +1024,8 @@ are calling is yours.

-
-

2.1.5 Exercise 5

+
+

2.1.5 Exercise 5

@@ -1035,8 +1035,8 @@ Find the theoretical value of \(a\) for which \(\Psi\) is an eigenfunction of \(

-
-
2.1.5.1 Solution   solution
+
+
2.1.5.1 Solution   solution
\begin{eqnarray*} E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} - @@ -1056,8 +1056,8 @@ equal to -0.5 atomic units.
-
-

2.2 Plot of the local energy along the \(x\) axis

+
+

2.2 Plot of the local energy along the \(x\) axis

The program you will write in this section will be written in @@ -1088,8 +1088,8 @@ In Fortran, you will need to compile all the source files together:

-
-

2.2.1 Exercise

+
+

2.2.1 Exercise

@@ -1183,8 +1183,8 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \

-
-
2.2.1.1 Solution   solution
+
+
2.2.1.1 Solution   solution

Python @@ -1261,8 +1261,8 @@ plt.savefig("plot_py.png")

-
-

2.3 Numerical estimation of the energy

+
+

2.3 Numerical estimation of the energy

If the space is discretized in small volume elements \(\mathbf{r}_i\) @@ -1292,8 +1292,8 @@ The energy is biased because:

-
-

2.3.1 Exercise

+
+

2.3.1 Exercise

@@ -1364,8 +1364,8 @@ To compile the Fortran and run it:

-
-
2.3.1.1 Solution   solution
+
+
2.3.1.1 Solution   solution

Python @@ -1482,8 +1482,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002

-
-

2.4 Variance of the local energy

+
+

2.4 Variance of the local energy

The variance of the local energy is a functional of \(\Psi\) @@ -1510,8 +1510,8 @@ energy can be used as a measure of the quality of a wave function.

-
-

2.4.1 Exercise (optional)

+
+

2.4.1 Exercise (optional)

@@ -1522,8 +1522,8 @@ Prove that :

-
-
2.4.1.1 Solution   solution
+
+
2.4.1.1 Solution   solution

\(\bar{E} = \langle E \rangle\) is a constant, so \(\langle \bar{E} @@ -1542,8 +1542,8 @@ Prove that :

-
-

2.4.2 Exercise

+
+

2.4.2 Exercise

@@ -1619,8 +1619,8 @@ To compile and run:

-
-
2.4.2.1 Solution   solution
+
+
2.4.2.1 Solution   solution

Python @@ -1759,8 +1759,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814

-
-

3 Variational Monte Carlo

+
+

3 Variational Monte Carlo

Numerical integration with deterministic methods is very efficient @@ -1776,8 +1776,8 @@ interval.

-
-

3.1 Computation of the statistical error

+
+

3.1 Computation of the statistical error

To compute the statistical error, you need to perform \(M\) @@ -1817,8 +1817,8 @@ And the confidence interval is given by

-
-

3.1.1 Exercise

+
+

3.1.1 Exercise

@@ -1858,8 +1858,8 @@ input array.

-
-
3.1.1.1 Solution   solution
+
+
3.1.1.1 Solution   solution

Python @@ -1920,8 +1920,8 @@ input array.

-
-

3.2 Uniform sampling in the box

+
+

3.2 Uniform sampling in the box

We will now perform our first Monte Carlo calculation to compute the @@ -1982,8 +1982,8 @@ compute the statistical error.

-
-

3.2.1 Exercise

+
+

3.2.1 Exercise

@@ -2085,8 +2085,8 @@ well as the index of the current step.

-
-
3.2.1.1 Solution   solution
+
+
3.2.1.1 Solution   solution

Python @@ -2192,8 +2192,8 @@ E = -0.48084122147238995 +/- 2.4983775878329355E-003

-
-

3.3 Metropolis sampling with \(\Psi^2\)

+
+

3.3 Metropolis sampling with \(\Psi^2\)

We will now use the square of the wave function to sample random @@ -2312,8 +2312,8 @@ All samples should be kept, from both accepted and rejected moves.

-
-

3.3.1 Optimal step size

+
+

3.3.1 Optimal step size

If the box is infinitely small, the ratio will be very close @@ -2348,8 +2348,8 @@ the same variable later on to store a time step.

-
-

3.3.2 Exercise

+
+

3.3.2 Exercise

@@ -2458,8 +2458,8 @@ Can you observe a reduction in the statistical error?

-
-
3.3.2.1 Solution   solution
+
+
3.3.2.1 Solution   solution

Python @@ -2606,8 +2606,8 @@ A = 0.50762633333333318 +/- 3.4601756760043725E-004

-
-

3.4 Generalized Metropolis algorithm

+
+

3.4 Generalized Metropolis algorithm

One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability. @@ -2728,8 +2728,8 @@ The algorithm of the previous exercise is only slighlty modified as:

-
-

3.4.1 Gaussian random number generator

+
+

3.4.1 Gaussian random number generator

To obtain Gaussian-distributed random numbers, you can apply the @@ -2793,8 +2793,8 @@ In Python, you can use the -

3.4.2 Exercise 1

+
+

3.4.2 Exercise 1

@@ -2836,8 +2836,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

-
-
3.4.2.1 Solution   solution
+
+
3.4.2.1 Solution   solution

Python @@ -2870,8 +2870,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

-
-

3.4.3 Exercise 2

+
+

3.4.3 Exercise 2

@@ -2967,8 +2967,8 @@ Modify the previous program to introduce the drift-diffusion scheme.

-
-
3.4.3.1 Solution   solution
+
+
3.4.3.1 Solution   solution

Python @@ -3156,8 +3156,8 @@ A = 0.62037333333333333 +/- 4.8970160591451110E-004

-
-

4 Diffusion Monte Carlo

+
+

4 Diffusion Monte Carlo

As we have seen, Variational Monte Carlo is a powerful method to @@ -3174,8 +3174,8 @@ finding a near-exact numerical solution to the Schrödinger equation.

-
-

4.1 Schrödinger equation in imaginary time

+
+

4.1 Schrödinger equation in imaginary time

Consider the time-dependent Schrödinger equation: @@ -3243,8 +3243,8 @@ system.

-