diff --git a/index.html b/index.html index db29ff8..efe6ecc 100644 --- a/index.html +++ b/index.html @@ -3,7 +3,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> - + Quantum Monte Carlo @@ -329,153 +329,153 @@ 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 @@ -515,8 +515,8 @@ 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 @@ -599,8 +599,8 @@ 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 @@ -629,8 +629,8 @@ 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. @@ -657,8 +657,8 @@ to catch the error.

-
-

2.1.1 Exercise 1

+
+

2.1.1 Exercise 1

@@ -703,8 +703,8 @@ and returns the potential.

-
-
2.1.1.1 Solution   solution
+
+
2.1.1.1 Solution   solution

Python @@ -745,8 +745,8 @@ and returns the potential.

-
-

2.1.2 Exercise 2

+
+

2.1.2 Exercise 2

@@ -781,8 +781,8 @@ input arguments, and returns a scalar.

-
-
2.1.2.1 Solution   solution
+
+
2.1.2.1 Solution   solution

Python @@ -809,8 +809,8 @@ input arguments, and returns a scalar.

-
-

2.1.3 Exercise 3

+
+

2.1.3 Exercise 3

@@ -891,8 +891,8 @@ Therefore, the local kinetic energy is

-
-
2.1.3.1 Solution   solution
+
+
2.1.3.1 Solution   solution

Python @@ -933,8 +933,8 @@ Therefore, the local kinetic energy is

-
-

2.1.4 Exercise 4

+
+

2.1.4 Exercise 4

@@ -993,8 +993,8 @@ are calling is yours.

-
-
2.1.4.1 Solution   solution
+
+
2.1.4.1 Solution   solution

Python @@ -1025,8 +1025,8 @@ are calling is yours.

-
-

2.1.5 Exercise 5

+
+

2.1.5 Exercise 5

@@ -1036,8 +1036,8 @@ 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} - @@ -1057,8 +1057,8 @@ 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 @@ -1089,8 +1089,8 @@ In Fortran, you will need to compile all the source files together:

-
-

2.2.1 Exercise

+
+

2.2.1 Exercise

@@ -1184,8 +1184,8 @@ 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 @@ -1262,8 +1262,8 @@ 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 \(\mathbf{r}_i\) @@ -1293,8 +1293,8 @@ The energy is biased because:

-
-

2.3.1 Exercise

+
+

2.3.1 Exercise

@@ -1365,8 +1365,8 @@ To compile the Fortran and run it:

-
-
2.3.1.1 Solution   solution
+
+
2.3.1.1 Solution   solution

Python @@ -1483,8 +1483,8 @@ 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 \(\Psi\) @@ -1511,8 +1511,8 @@ energy can be used as a measure of the quality of a wave function.

-
-

2.4.1 Exercise (optional)

+
+

2.4.1 Exercise (optional)

@@ -1523,8 +1523,8 @@ 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} @@ -1543,8 +1543,8 @@ Prove that :

-
-

2.4.2 Exercise

+
+

2.4.2 Exercise

@@ -1620,8 +1620,8 @@ To compile and run:

-
-
2.4.2.1 Solution   solution
+
+
2.4.2.1 Solution   solution

Python @@ -1760,8 +1760,8 @@ 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 @@ -1777,8 +1777,8 @@ interval.

-
-

3.1 Computation of the statistical error

+
+

3.1 Computation of the statistical error

To compute the statistical error, you need to perform \(M\) @@ -1818,8 +1818,8 @@ And the confidence interval is given by

-
-

3.1.1 Exercise

+
+

3.1.1 Exercise

@@ -1859,8 +1859,8 @@ input array.

-
-
3.1.1.1 Solution   solution
+
+
3.1.1.1 Solution   solution

Python @@ -1921,8 +1921,8 @@ 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 @@ -1983,8 +1983,8 @@ compute the statistical error.

-
-

3.2.1 Exercise

+
+

3.2.1 Exercise

@@ -2086,8 +2086,8 @@ well as the index of the current step.

-
-
3.2.1.1 Solution   solution
+
+
3.2.1.1 Solution   solution

Python @@ -2193,8 +2193,8 @@ 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 @@ -2313,8 +2313,8 @@ 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 @@ -2349,8 +2349,8 @@ the same variable later on to store a time step.

-
-

3.3.2 Exercise

+
+

3.3.2 Exercise

@@ -2459,8 +2459,8 @@ Can you observe a reduction in the statistical error?

-
-
3.3.2.1 Solution   solution
+
+
3.3.2.1 Solution   solution

Python @@ -2607,8 +2607,8 @@ 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. @@ -2729,8 +2729,8 @@ 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 @@ -2794,8 +2794,8 @@ In Python, you can use the -

3.4.2 Exercise 1

+
+

3.4.2 Exercise 1

@@ -2837,8 +2837,8 @@ 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 @@ -2871,8 +2871,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

-
-

3.4.3 Exercise 2

+
+

3.4.3 Exercise 2

@@ -2968,8 +2968,8 @@ Modify the previous program to introduce the drift-diffusion scheme.

-
-
3.4.3.1 Solution   solution
+
+
3.4.3.1 Solution   solution

Python @@ -3151,12 +3151,12 @@ A = 0.62037333333333333 +/- 4.8970160591451110E-004

-
-

4 Diffusion Monte Carlo   solution

+
+

4 Diffusion Monte Carlo   solution

-
-

4.1 Schrödinger equation in imaginary time

+
+

4.1 Schrödinger equation in imaginary time

Consider the time-dependent Schrödinger equation: @@ -3224,8 +3224,8 @@ system.

-
-

4.2 Diffusion and branching

+
+

4.2 Diffusion and branching

The imaginary-time Schrödinger equation can be explicitly written in terms of the kinetic and @@ -3322,8 +3322,8 @@ Therefore, in both cases, you are dealing with a "Bosonic" ground state.

-
-

4.3 Importance sampling

+
+

4.3 Importance sampling

In a molecular system, the potential is far from being constant @@ -3419,8 +3419,8 @@ energies computed with the trial wave function.

-
-

4.3.1 Appendix : Details of the Derivation

+
+

4.3.1 Appendix : Details of the Derivation

\[ @@ -3481,8 +3481,8 @@ Defining \(\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)

-
-

4.4 Pure Diffusion Monte Carlo (PDMC)

+
+

4.4 Pure Diffusion Monte Carlo (PDMC)

Instead of having a variable number of particles to simulate the @@ -3563,13 +3563,13 @@ code, so this is what we will do in the next section.

-
-

4.5 Hydrogen atom

+
+

4.5 Hydrogen atom

-
-

4.5.1 Exercise

+
+

4.5.1 Exercise

@@ -3668,8 +3668,8 @@ energy of H for any value of \(a\).

-
-
4.5.1.1 Solution   solution
+
+
4.5.1.1 Solution   solution

Python @@ -3887,8 +3887,8 @@ A = 0.98788066666666663 +/- 7.2889356133441110E-005

-
-

4.6 TODO H2

+
+

4.6 TODO H2

We will now consider the H2 molecule in a minimal basis composed of the @@ -3909,8 +3909,8 @@ the nuclei.

-
-

5 TODO [0/3] Last things to do

+
+

5 TODO [0/3] Last things to do

  • [ ] Give some hints of how much time is required for each section
    • @@ -3924,8 +3924,8 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.
-
-

6 Schedule

+
+

6 Schedule

@@ -3989,7 +3989,7 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.

Author: Anthony Scemama, Claudia Filippi

-

Created: 2021-02-02 Tue 13:25

+

Created: 2021-02-02 Tue 15:42

Validate