diff --git a/index.html b/index.html index 76dabc5..7eaf996 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

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-

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:

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-

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.

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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.

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2.1.2.1 Solution   solution
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+
2.1.2.1 Solution   solution

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

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2.1.3 Exercise 3

+
+

2.1.3 Exercise 3

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

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2.1.3.1 Solution   solution
+
+
2.1.3.1 Solution   solution

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

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-

2.1.4 Exercise 4

+
+

2.1.4 Exercise 4

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

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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 \(

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-
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', \

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2.2.1.1 Solution   solution
+
+
2.2.1.1 Solution   solution

Python @@ -1262,8 +1262,8 @@ plt.savefig("plot_py.png")

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-

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:

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2.3.1.1 Solution   solution
+
+
2.3.1.1 Solution   solution

Python @@ -1483,8 +1483,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002

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-

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.

-
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2.4.1 Exercise (optional)

+
+

2.4.1 Exercise (optional)

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

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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 :

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-

2.4.2 Exercise

+
+

2.4.2 Exercise

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

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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

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3 Variational Monte Carlo

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+

3 Variational Monte Carlo

Numerical integration with deterministic methods is very efficient @@ -1777,8 +1777,8 @@ interval.

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-

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

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-

3.1.1 Exercise

+
+

3.1.1 Exercise

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

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3.1.1.1 Solution   solution
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3.1.1.1 Solution   solution

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

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3.2.1 Exercise

+
+

3.2.1 Exercise

@@ -2086,8 +2086,8 @@ 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 @@ -2193,8 +2193,8 @@ E = -0.48084122147238995 +/- 2.4983775878329355E-003

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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 @@ -2313,8 +2313,8 @@ All samples should be kept, from both accepted and rejected moves.

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3.3.1 Optimal step size

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+

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.

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3.3.2 Exercise

+
+

3.3.2 Exercise

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

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3.3.2.1 Solution   solution
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3.3.2.1 Solution   solution

Python @@ -2607,8 +2607,8 @@ A = 0.50762633333333318 +/- 3.4601756760043725E-004

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3.4 Generalized Metropolis algorithm

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+

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:

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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

-
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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

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3.4.3 Exercise 2

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+

3.4.3 Exercise 2

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

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3.4.3.1 Solution   solution
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3.4.3.1 Solution   solution

Python @@ -3157,12 +3157,26 @@ A = 0.62037333333333333 +/- 4.8970160591451110E-004

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4 Diffusion Monte Carlo   solution

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+

4 Diffusion Monte Carlo   solution

+

+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). +

+ +

+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. +

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-

4.1 Schrödinger equation in imaginary time

+ +
+

4.1 Schrödinger equation in imaginary time

Consider the time-dependent Schrödinger equation: @@ -3230,59 +3244,38 @@ system.

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4.2 Diffusion and branching

+
+

4.2 Relation to diffusion

-The imaginary-time Schrödinger equation can be explicitly written in terms of the kinetic and -potential energies as +The diffusion equation of particles is given by

\[ - \frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,. + \frac{\partial \psi(\mathbf{r},t)}{\partial t} = D\, \Delta \psi(\mathbf{r},t) \]

-We can simulate this differential equation as a diffusion-branching process. -

- - -

-To see this, recall that the diffusion equation of particles is given by +where \(D\) is the diffusion coefficient. When the imaginary-time +Schrödinger equation is written in terms of the kinetic energy and +potential,

\[ - \frac{\partial \psi(\mathbf{r},t)}{\partial t} = D\, \Delta \psi(\mathbf{r},t). + \frac{\partial \psi(\mathbf{r}, \tau)}{\partial \tau} = + \left(\frac{1}{2}\Delta - [V(\mathbf{r}) -E_{\rm ref}]\right) \psi(\mathbf{r}, \tau)\,, \]

-Furthermore, the rate of reaction \(v\) is the speed at which a chemical reaction -takes place. In a solution, the rate is given as a function of the -concentration \([A]\) by -

- -

-\[ - v = \frac{d[A]}{dt}, - \] -

- -

-where the concentration \([A]\) is proportional to the number of particles. -

- -

-These two equations allow us to interpret the Schrödinger equation -in imaginary time as the combination of: +it can be identified as the combination of:

    -
  • a diffusion equation with a diffusion coefficient \(D=1/2\) for the -kinetic energy, and
    • -
  • a rate equation for the potential.
    • +
  • a diffusion equation (Laplacian)
    • +
  • an equation whose solution is an exponential (potential)

@@ -3294,16 +3287,12 @@ The diffusion equation can be simulated by a Brownian motion:

-where \(\chi\) is a Gaussian random variable, and the rate equation +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). +so-called branching process) or by simply considering it as a +cumulative multiplicative weight along the diffusion trajectory.

-

-In Diffusion Monte Carlo (DMC), one onbtains the ground state of a -system by simulating the Schrödinger equation in imaginary time via -the combination of a diffusion process and a branching process. -

We note that the ground-state wave function of a Fermionic system is @@ -3319,7 +3308,8 @@ For the systems you will study, this is not an issue:

  • Hydrogen atom: You only have one electron!
    • -
  • Two-electron system (\(H_2\) or He): The ground-wave function is antisymmetric in the spin variables but symmetric in the space ones.
    • +
  • Two-electron system (\(H_2\) or He): The ground-wave function is +antisymmetric in the spin variables but symmetric in the space ones.

@@ -3328,15 +3318,14 @@ 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 -and, in fact, diverges at the inter-particle coalescence points. Hence, when the -rate equation is simulated, it results in very large fluctuations -in the numbers of particles, making the calculations impossible in -practice. +and, in fact, diverges at the inter-particle coalescence points. Hence, +it results in very large fluctuations of the term associated with +the potental, making the calculations impossible in practice. Fortunately, if we multiply the Schrödinger equation by a chosen trial wave function \(\Psi_T(\mathbf{r})\) (Hartree-Fock, Kohn-Sham determinant, CI wave function, etc), one obtains @@ -3366,8 +3355,8 @@ Defining \(\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})\), (s 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 -changing the number of particles according to \(\exp\left[ -\delta t\, +when \(\Psi_T\) gets closer to the exact wave function. +This term can be simulated by t particles according to \(\exp\left[ -\delta t\, \left(E_L(\mathbf{r}) - E_{\rm ref}\right)\right]\) where \(E_{\rm ref}\) is the constant we had introduced above, which is adjusted to the running average energy to keep the number of particles @@ -3425,8 +3414,8 @@ energies computed with the trial wave function.

-
-

4.3.1 Appendix : Details of the Derivation

+
+

4.3.1 Appendix : Details of the Derivation

\[ @@ -3487,8 +3476,8 @@ 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 @@ -3564,13 +3553,13 @@ 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

@@ -3669,8 +3658,8 @@ energy of H for any value of \(a\).

-
-
4.5.1.1 Solution   solution
+
+
4.5.1.1 Solution   solution

Python @@ -3888,8 +3877,8 @@ 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 @@ -3910,8 +3899,8 @@ 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
    • @@ -3925,8 +3914,8 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.
-
-

6 Schedule

+
+

6 Schedule

@@ -3990,7 +3979,7 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.

Author: Anthony Scemama, Claudia Filippi

-

Created: 2021-02-02 Tue 16:05

+

Created: 2021-02-02 Tue 21:49

Validate