From 953942fba8b01da74e55574dfedf62acad402034 Mon Sep 17 00:00:00 2001 From: filippi-claudia Date: Sun, 31 Jan 2021 19:07:02 +0000 Subject: [PATCH] deploy: 47f2b41e9e71a9cc6e2a20bfaf5919efc3336452 --- index.html | 354 +++++++++++++++++++++++++++-------------------------- 1 file changed, 182 insertions(+), 172 deletions(-) diff --git a/index.html b/index.html index 2c435d2..3ae2a1c 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

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1 Introduction

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1 Introduction

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

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1.1 Energy and local energy

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+

1.1 Energy and local energy

For a given system with Hamiltonian \(\hat{H}\) and wave function \(\Psi\), we define the local energy as @@ -593,8 +593,8 @@ $$ E ≈ \frac{1}{N\rm MC} ∑i=1N\rm MC

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2 Numerical evaluation of the energy of the hydrogen atom

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2 Numerical evaluation of the energy of the hydrogen atom

In this section, we consider the hydrogen atom with the following @@ -623,8 +623,8 @@ To do that, we will compute the local energy and check whether it is constant.

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2.1 Local energy

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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. @@ -651,8 +651,8 @@ to catch the error.

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2.1.1 Exercise 1

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2.1.1 Exercise 1

@@ -696,8 +696,8 @@ and returns the potential.

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2.1.1.1 Solution   solution
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2.1.1.1 Solution   solution

Python @@ -737,8 +737,8 @@ and returns the potential.

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2.1.2 Exercise 2

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2.1.2 Exercise 2

@@ -773,8 +773,8 @@ input arguments, and returns a scalar.

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2.1.2.1 Solution   solution
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2.1.2.1 Solution   solution

Python @@ -801,8 +801,8 @@ input arguments, and returns a scalar.

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2.1.3 Exercise 3

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2.1.3 Exercise 3

@@ -883,8 +883,8 @@ Therefore, the local kinetic energy is

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2.1.3.1 Solution   solution
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2.1.3.1 Solution   solution

Python @@ -925,8 +925,8 @@ Therefore, the local kinetic energy is

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2.1.4 Exercise 4

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2.1.4 Exercise 4

@@ -969,8 +969,8 @@ local kinetic energy.

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2.1.4.1 Solution   solution
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2.1.4.1 Solution   solution

Python @@ -1000,8 +1000,8 @@ local kinetic energy.

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2.1.5 Exercise 5

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2.1.5 Exercise 5

@@ -1011,8 +1011,8 @@ Find the theoretical value of \(a\) for which \(\Psi\) is an eigenfunction of \(

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2.1.5.1 Solution   solution
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2.1.5.1 Solution   solution
\begin{eqnarray*} E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} - @@ -1032,8 +1032,8 @@ equal to -0.5 atomic units.
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2.2 Plot of the local energy along the \(x\) axis

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

@@ -1044,8 +1044,8 @@ choose a grid which does not contain the origin.

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2.2.1 Exercise

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2.2.1 Exercise

@@ -1128,8 +1128,8 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \

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2.2.1.1 Solution   solution
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2.2.1.1 Solution   solution

Python @@ -1204,8 +1204,8 @@ plt.savefig("plot_py.png")

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2.3 Numerical estimation of the energy

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2.3 Numerical estimation of the energy

If the space is discretized in small volume elements \(\mathbf{r}_i\) @@ -1235,8 +1235,8 @@ The energy is biased because:

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2.3.1 Exercise

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2.3.1 Exercise

@@ -1305,8 +1305,8 @@ To compile the Fortran and run it:

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2.3.1.1 Solution   solution
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2.3.1.1 Solution   solution

Python @@ -1421,8 +1421,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002

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2.4 Variance of the local energy

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2.4 Variance of the local energy

The variance of the local energy is a functional of \(\Psi\) @@ -1449,8 +1449,8 @@ energy can be used as a measure of the quality of a wave function.

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2.4.1 Exercise (optional)

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2.4.1 Exercise (optional)

@@ -1461,8 +1461,8 @@ Prove that :

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2.4.1.1 Solution   solution
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2.4.1.1 Solution   solution

\(\bar{E} = \langle E \rangle\) is a constant, so \(\langle \bar{E} @@ -1481,8 +1481,8 @@ Prove that :

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2.4.2 Exercise

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2.4.2 Exercise

@@ -1556,8 +1556,8 @@ To compile and run:

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2.4.2.1 Solution   solution
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2.4.2.1 Solution   solution

Python @@ -1694,8 +1694,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814

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3 Variational Monte Carlo

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3 Variational Monte Carlo

Numerical integration with deterministic methods is very efficient @@ -1711,8 +1711,8 @@ interval.

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3.1 Computation of the statistical error

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3.1 Computation of the statistical error

To compute the statistical error, you need to perform \(M\) @@ -1752,8 +1752,8 @@ And the confidence interval is given by

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3.1.1 Exercise

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3.1.1 Exercise

@@ -1791,8 +1791,8 @@ input array.

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3.1.1.1 Solution   solution
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3.1.1.1 Solution   solution

Python @@ -1851,8 +1851,8 @@ input array.

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3.2 Uniform sampling in the box

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3.2 Uniform sampling in the box

We will now perform our first Monte Carlo calculation to compute the @@ -1913,8 +1913,8 @@ compute the statistical error.

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3.2.1 Exercise

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3.2.1 Exercise

@@ -2014,8 +2014,8 @@ well as the index of the current step.

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3.2.1.1 Solution   solution
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3.2.1.1 Solution   solution

Python @@ -2129,8 +2129,8 @@ E = -0.49518773675598715 +/- 5.2391494923686175E-004

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

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

We will now use the square of the wave function to sample random @@ -2269,8 +2269,8 @@ become clear later.

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3.3.1 Exercise

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3.3.1 Exercise

@@ -2377,8 +2377,8 @@ Can you observe a reduction in the statistical error?

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3.3.1.1 Solution   solution
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3.3.1.1 Solution   solution

Python @@ -2523,8 +2523,8 @@ A = 0.51695266666666673 +/- 4.0445505648997396E-004

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3.4 Gaussian random number generator

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3.4 Gaussian random number generator

To obtain Gaussian-distributed random numbers, you can apply the @@ -2587,8 +2587,8 @@ In Python, you can use the -

3.5 Generalized Metropolis algorithm

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3.5 Generalized Metropolis algorithm

One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability. @@ -2720,8 +2720,8 @@ Evaluate \(\Psi\) and \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\) at th

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3.5.1 Exercise 1

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3.5.1 Exercise 1

@@ -2755,8 +2755,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

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3.5.1.1 Solution   solution
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3.5.1.1 Solution   solution

Python @@ -2789,8 +2789,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P

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3.5.2 Exercise 2

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3.5.2 Exercise 2

@@ -2884,8 +2884,8 @@ Modify the previous program to introduce the drift-diffusion scheme.

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3.5.2.1 Solution   solution
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3.5.2.1 Solution   solution

Python @@ -3071,12 +3071,12 @@ A = 0.78839866666666658 +/- 3.2503783452043152E-004

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4 Diffusion Monte Carlo   solution

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4 Diffusion Monte Carlo   solution

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4.1 Schrödinger equation in imaginary time

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4.1 Schrödinger equation in imaginary time

Consider the time-dependent Schrödinger equation: @@ -3127,12 +3127,13 @@ Now, if we replace the time variable \(t\) by an imaginary time variable

where \(\psi(\mathbf{r},\tau) = \Psi(\mathbf{r},-i\,)\) and -\begin{eqnarray*} -ψ(\mathbf{r},τ) &=& ∑k ak exp( -Ek\, τ) φk(\mathbf{r})
- &=& exp(-(E0-ET)\, τ)∑k ak exp( -(Ek-E0)\, τ) φk(\mathbf{r})\,. -\begin{eqnarray*}

+\begin{eqnarray*} +\psi(\mathbf{r},\tau) &=& \sum_k a_k \exp( -E_k\, \tau) \phi_k(\mathbf{r})\\ + &=& \exp(-(E_0-E_T)\, \tau)\sum_k a_k \exp( -(E_k-E_0)\, \tau) \phi_k(\mathbf{r})\,. +\end{eqnarray*} +

For large positive values of \(\tau\), \(\psi\) is dominated by the \(k=0\) term, namely, the lowest eigenstate. If we adjust \(E_T\) to the running estimate of \(E_0\), @@ -3143,8 +3144,8 @@ system.

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4.2 Diffusion and branching

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4.2 Diffusion and branching

The imaginary-time Schrödinger equation can be explicitly written in terms of the kinetic and @@ -3163,7 +3164,7 @@ We can simulate this differential equation as a diffusion-branching process.

-To this this, recall that the diffusion equation of particles is given by +To see this, recall that the diffusion equation of particles is given by

@@ -3217,11 +3218,20 @@ so-called branching process). system by simulating the Schrödinger equation in imaginary time, by the combination of a diffusion process and a branching process.

+ +

+We note here that the ground-state wave function of a Fermionic system is +antisymmetric and changes sign. +

+ +

+I AM HERE +

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4.3 Importance sampling

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4.3 Importance sampling

In a molecular system, the potential is far from being constant @@ -3250,20 +3260,20 @@ Defining \(\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})\), (s -\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) + \right] + (E_L(\mathbf{r})-E_T)\Pi(\mathbf{r},\tau) \]

-The new "kinetic energy" can be simulated by the drifted diffusion +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. It can be simulated by changing the number of particles according to \(\exp\left[ -\delta t\, - \left(E_L(\mathbf{r}) - E_\text{ref}\right)\right]\) -where \(E_{\text{ref}}\) is a constant introduced so that the average -of this term is close to one, keeping the number of particles rather -constant. + \left(E_L(\mathbf{r}) - E_T\right)\right]\) +where \(E_T\) is the constant we had introduced above, which is adjusted to +the running average energy and is introduced to keep the number of particles +reasonably constant.

@@ -3278,8 +3288,8 @@ error known as the fixed node error.

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4.3.1 Appendix : Details of the Derivation

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4.3.1 Appendix : Details of the Derivation

\[ @@ -3341,8 +3351,8 @@ Defining \(\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)

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4.4 Fixed-node DMC energy

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4.4 Fixed-node DMC energy

Now that we have a process to sample \(\Pi(\mathbf{r},\tau) = @@ -3394,8 +3404,8 @@ energies computed with the trial wave function.

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4.5 Pure Diffusion Monte Carlo (PDMC)

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4.5 Pure Diffusion Monte Carlo (PDMC)

Instead of having a variable number of particles to simulate the @@ -3447,13 +3457,13 @@ code, so this is what we will do in the next section.

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4.6 Hydrogen atom

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4.6 Hydrogen atom

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4.6.1 Exercise

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4.6.1 Exercise

@@ -3552,8 +3562,8 @@ energy of H for any value of \(a\).

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4.6.1.1 Solution   solution
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4.6.1.1 Solution   solution

Python @@ -3769,8 +3779,8 @@ A = 0.98788066666666663 +/- 7.2889356133441110E-005

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4.7 TODO H2

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4.7 TODO H2

We will now consider the H2 molecule in a minimal basis composed of the @@ -3791,8 +3801,8 @@ the nuclei.

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5 TODO [0/3] Last things to do

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5 TODO [0/3] Last things to do

  • [ ] Give some hints of how much time is required for each section
    • @@ -3808,7 +3818,7 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.

Author: Anthony Scemama, Claudia Filippi

-

Created: 2021-01-31 Sun 18:17

+

Created: 2021-01-31 Sun 19:07

Validate