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Fokker-Planck
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@ -1,8 +1,13 @@
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#+TITLE: Quantum Monte Carlo
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#+AUTHOR: Anthony Scemama, Claudia Filippi
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#+SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup
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# SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup
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# SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-bigblow.setup
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#+STARTUP: latexpreview
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#+HTML_HEAD: <link rel="stylesheet" title="Standard" href="https://orgmode.org/worg/style/worg.css" type="text/css" />
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#+HTML_HEAD: <link rel="alternate stylesheet" title="Zenburn" href="https://orgmode.org/worg/style/worg-zenburn.css" type="text/css" />
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#+HTML_HEAD: <link rel="alternate stylesheet" title="Classic" href="https://orgmode.org/worg/style/worg-classic.css" type="text/css" />
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* Introduction
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@ -14,10 +19,17 @@
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computes a statistical estimate of the expectation value of the energy
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associated with a given wave function.
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Finally, we introduce the diffusion Monte Carlo (DMC) method which
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gives the exact energy of the H$_2$ molecule.
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gives the exact energy of the $H_2$ molecule.
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Code examples will be given in Python and Fortran. Whatever language
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can be chosen.
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We consider the stationary solution of the Schrödinger equation, so
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the wave functions considered here are real: for an $N$ electron
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system where the electrons move in the 3-dimensional space,
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$\Psi : \mathbb{R}^{3N} \rightarrow \mathbb{R}$. In addition, $\Psi$
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is defined everywhere, continuous and infinitely differentiable.
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** Python
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** Fortran
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@ -45,7 +57,7 @@
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$$
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when $a=1$, by checking that $\hat{H}\Psi(\mathbf{r}) = E\Psi(\mathbf{r})$ for
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all $\mathbf{r}$: we will check that the local energy, defined as
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all $\mathbf{r}$. We will check that the local energy, defined as
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$$
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E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
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@ -54,6 +66,30 @@
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is constant.
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The probabilistic /expected value/ of an arbitrary function $f(x)$
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with respect to a probability density function $p(x)$ is given by
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$$ \langle f \rangle_p = \int_{-\infty}^\infty p(x)\, f(x)\,dx $$.
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Recall that a probability density function $p(x)$ is non-negative
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and integrates to one:
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$$ \int_{-\infty}^\infty p(x)\,dx = 1 $$.
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The electronic energy of a system is the expectation value of the
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local energy $E(\mathbf{r})$ with respect to the $3N$-dimensional
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electron density given by the square of the wave function:
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\begin{eqnarray}
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E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} \\
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& = & \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\
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& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\
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& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, E_L(\mathbf{r})\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
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= \langle E_L \rangle_{\Psi^2}
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\end{eqnarray}
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** Local energy
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:PROPERTIES:
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:header-args:python: :tangle hydrogen.py
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@ -63,9 +99,9 @@
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The function accepts a 3-dimensional vector =r= as input arguments
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and returns the potential.
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$\mathbf{r}=\sqrt{x^2 + y^2 + z^2})$, so
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$\mathbf{r}=\sqrt{x^2 + y^2 + z^2}$, so
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$$
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V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$
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V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2}}
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$$
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#+BEGIN_SRC python :results none
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@ -175,7 +211,7 @@ double precision function e_loc(a,r)
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end function e_loc
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#+END_SRC
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** Plot the local energy along the x axis
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** Plot of the local energy along the $x$ axis
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:PROPERTIES:
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:header-args:python: :tangle plot_hydrogen.py
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:header-args:f90: :tangle plot_hydrogen.f90
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@ -184,7 +220,7 @@ end function e_loc
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For multiple values of $a$ (0.1, 0.2, 0.5, 1., 1.5, 2.), plot the
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local energy along the $x$ axis.
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#+begin_src python :results output
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#+BEGIN_SRC python :results none
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import numpy as np
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import matplotlib.pyplot as plt
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@ -270,47 +306,41 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
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#+RESULTS:
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[[file:plot.png]]
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** Compute numerically the average energy
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** Compute numerically the expectation value of the energy
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:PROPERTIES:
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:header-args:python: :tangle energy_hydrogen.py
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:header-args:f90: :tangle energy_hydrogen.f90
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:END:
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We want to compute
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\begin{eqnarray}
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E & = & \frac{\langle \Psi| \hat{H} | \Psi\rangle}{\langle \Psi |\Psi \rangle} \\
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& = & \frac{\int \Psi(\mathbf{r})\, \hat{H} \Psi(\mathbf{r})\, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}} \\
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& = & \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
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\end{eqnarray}
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If the space is discretized in small volume elements $\delta
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\mathbf{r}$, this last equation corresponds to a weighted average of
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the local energy, where the weights are the values of the square of
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the wave function at $\mathbf{r}$ multiplied by the volume element:
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\mathbf{r}$, the expression of \langle E_L \rangle_{\Psi^2}$ becomes
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a weighted average of the local energy, where the weights are the
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values of the probability density at $\mathbf{r}$ multiplied
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by the volume element:
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$$
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E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
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\langle E \rangle_{\Psi^2} \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
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w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta \mathbf{r}
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$$
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We now compute an numerical estimate of the energy in a grid of
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$50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
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In this section, we will compute a numerical estimate of the
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energy in a grid of $50\times50\times50$ points in the range
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$(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
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Note: the energy is biased because:
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- The energy is evaluated only inside the box
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- The volume elements are not infinitely small
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- The volume elements are not infinitely small (discretization error)
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- The energy is evaluated only inside the box (incompleteness of the space)
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#+BEGIN_SRC python :results output :exports both
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#+BEGIN_SRC python :results none
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import numpy as np
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from hydrogen import e_loc, psi
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interval = np.linspace(-5,5,num=50)
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delta = (interval[1]-interval[0])**3
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interval = np.linspace(-5,5,num=50)
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delta = (interval[1]-interval[0])**3
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r = np.array([0.,0.,0.])
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r = np.array([0.,0.,0.])
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for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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E = 0.
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norm = 0.
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for x in interval:
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@ -367,7 +397,6 @@ program energy_hydrogen
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r(3) = x(l)
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w = psi(a(j),r)
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w = w * w * delta
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energy = energy + w * e_loc(a(j), r)
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norm = norm + w
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end do
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@ -380,7 +409,7 @@ program energy_hydrogen
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end program energy_hydrogen
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#+end_src
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To compile and run:
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To compile the Fortran and run it:
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#+begin_src sh :results output :exports both
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gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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@ -401,20 +430,23 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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:header-args:f90: :tangle variance_hydrogen.f90
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:END:
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The variance of the local energy measures the magnitude of the
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fluctuations of the local energy around the average. If the local
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energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$)
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the variance is zero.
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The variance of the local energy is a functional of $\Psi$
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which measures the magnitude of the fluctuations of the local
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energy associated with $\Psi$ around the average:
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$$
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\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
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E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
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$$
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If the local energy is constant (i.e. $\Psi$ is an eigenfunction of
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$\hat{H}$) the variance is zero, so the variance of the local
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energy can be used as a measure of the quality of a wave function.
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Compute a numerical estimate of the variance of the local energy
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in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
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#+BEGIN_SRC python :results output :exports both
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#+begin_src python :results none
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import numpy as np
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from hydrogen import e_loc, psi
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@ -451,7 +483,6 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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s2 += w * (El - E)**2
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s2 = s2 / norm
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print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
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#+end_src
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#+RESULTS:
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@ -541,6 +572,8 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
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* Variational Monte Carlo
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Numerical integration with deterministic methods is very efficient
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in low dimensions. When the number of dimensions becomes larger than
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Instead of computing the average energy as a numerical integration
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on a grid, we will do a Monte Carlo sampling, which is an extremely
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efficient method to compute integrals when the number of dimensions is
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@ -582,7 +615,7 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
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Write a function returning the average and statistical error of an
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input array.
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#+BEGIN_SRC python
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#+BEGIN_SRC python :results none
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from math import sqrt
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def ave_error(arr):
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M = len(arr)
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@ -638,7 +671,7 @@ end subroutine ave_error
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from hydrogen import *
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from qmc_stats import *
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def MonteCarlo(a, nmax):
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def MonteCarlo(a, nmax):
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E = 0.
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N = 0.
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for istep in range(nmax):
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@ -649,11 +682,11 @@ from qmc_stats import *
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E += w * e_loc(a,r)
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return E/N
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a = 0.9
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nmax = 100000
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X = [MonteCarlo(a,nmax) for i in range(30)]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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a = 0.9
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nmax = 100000
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X = [MonteCarlo(a,nmax) for i in range(30)]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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#+END_SRC
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#+RESULTS:
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@ -868,7 +901,12 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
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#+RESULTS:
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: E = -0.49606057056767766 +/- 1.3918807547836872E-004
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** Sampling with $\Psi^2$
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:PROPERTIES:
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:header-args:python: :tangle vmc.py
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:header-args:f90: :tangle vmc.f90
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:END:
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We will now use the square of the wave function to make the sampling:
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@ -876,19 +914,79 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
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P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
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\]
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Now, the expression for the energy will be simplified to the
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average of the local energies, each with a weight of 1.
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The expression for the energy will be simplified to the average of
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the local energies, each with a weight of 1.
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$$
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E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)}
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E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)
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$$
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To generate the probability density $\Psi^2$, we can use a drifted
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diffusion scheme:
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To generate the probability density $\Psi^2$, we consider a
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diffusion process characterized by a time-dependent density
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$[\Psi(\mathbf{r},t)]^2$, which obeys the Fokker-Planck equation:
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\[
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\frac{\partial \Psi^2}{\partial t} = \sum_i D
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\frac{\partial}{\partial \mathbf{r}_i} \left(
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\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
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[\Psi(\mathbf{r},t)]^2.
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\]
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$D$ is the diffusion constant and $F_i$ is the i-th component of a
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drift velocity caused by an external potential. For a stationary
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density, \( \frac{\partial \Psi^2}{\partial t} = 0 \), so
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\begin{eqnarray*}
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0 & = & \sum_i D
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\frac{\partial}{\partial \mathbf{r}_i} \left(
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\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
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[\Psi(\mathbf{r})]^2 \\
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0 & = & \sum_i D
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\frac{\partial}{\partial \mathbf{r}_i} \left(
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\frac{\partial [\Psi(\mathbf{r})]^2}{\partial \mathbf{r}_i} -
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F_i(\mathbf{r})\,[\Psi(\mathbf{r})]^2 \right) \\
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0 & = &
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\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} -
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\frac{\partial F_i }{\partial \mathbf{r}_i}[\Psi(\mathbf{r})]^2 -
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\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
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\end{eqnarray*}
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we search for a drift function which satisfies
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\[
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\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
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[\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
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\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
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\]
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to obtain a second derivative on the left, we need the drift to be
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of the form
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\[
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F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
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\]
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\[
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\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
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[\Psi(\mathbf{r})]^2 \frac{\partial
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g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} +
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[\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2
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\Psi^2}{\partial \mathbf{r}_i^2} +
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\frac{\partial \Psi^2}{\partial \mathbf{r}_i}
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g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
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\]
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$g = 1 / \Psi^2$ satisfies this equation, so
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\[
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F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
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\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
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\]
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following drifted diffusion scheme:
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\[
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\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
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\Psi(r)}{\Psi(r)} + \eta \sqrt{\tau}
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\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \eta \sqrt{\tau}
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\]
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where $\eta$ is a normally-distributed Gaussian random number.
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@ -902,7 +1000,66 @@ def drift(a,r):
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return r * ar_inv
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#+END_SRC
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#+RESULTS:
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#+BEGIN_SRC f90
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subroutine drift(a,r,b)
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implicit none
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double precision, intent(in) :: a, r(3)
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double precision, intent(out) :: b(3)
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double precision :: ar_inv
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ar_inv = -a / dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3))
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b(:) = r(:) * ar_inv
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end subroutine drift
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#+END_SRC
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Now we can write the Monte Carlo sampling
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#+BEGIN_SRC f90
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subroutine variational_montecarlo(a,nmax,energy)
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implicit none
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double precision, intent(in) :: a
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integer , intent(in) :: nmax
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double precision, intent(out) :: energy
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integer*8 :: istep
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double precision :: norm, r(3), w
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double precision, external :: e_loc, psi, gaussian
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energy = 0.d0
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norm = 0.d0
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do istep = 1,nmax
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call random_gauss(r,3)
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w = psi(a,r)
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w = w*w / gaussian(r)
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norm = norm + w
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energy = energy + w * e_loc(a,r)
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end do
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energy = energy / norm
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end subroutine variational_montecarlo
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program qmc
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implicit none
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double precision, parameter :: a = 0.9
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integer , parameter :: nmax = 100000
|
||||
integer , parameter :: nruns = 30
|
||||
|
||||
integer :: irun
|
||||
double precision :: X(nruns)
|
||||
double precision :: ave, err
|
||||
|
||||
do irun=1,nruns
|
||||
call gaussian_montecarlo(a,nmax,X(irun))
|
||||
enddo
|
||||
call ave_error(X,nruns,ave,err)
|
||||
print *, 'E = ', ave, '+/-', err
|
||||
end program qmc
|
||||
#+END_SRC
|
||||
|
||||
#+begin_src sh :results output :exports both
|
||||
gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
|
||||
./vmc
|
||||
#+end_src
|
||||
|
||||
#+BEGIN_SRC python
|
||||
def MonteCarlo(a,tau,nmax):
|
||||
@ -932,11 +1089,8 @@ def MonteCarlo(a,tau,nmax):
|
||||
N += 1.
|
||||
E += e_loc(a,r_old)
|
||||
return E/N
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
|
||||
#+BEGIN_SRC python :results output
|
||||
nmax = 100000
|
||||
tau = 0.1
|
||||
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
|
||||
@ -950,15 +1104,15 @@ print(f"E = {E} +/- {deltaE}")
|
||||
|
||||
* Diffusion Monte Carlo
|
||||
|
||||
We will now consider the H_2 molecule in a minimal basis composed of the
|
||||
$1s$ orbitals of the hydrogen atoms:
|
||||
We will now consider the H_2 molecule in a minimal basis composed of the
|
||||
$1s$ orbitals of the hydrogen atoms:
|
||||
|
||||
$$
|
||||
\Psi(\mathbf{r}_1, \mathbf{r}_2) =
|
||||
\exp(-(\mathbf{r}_1 - \mathbf{R}_A)) +
|
||||
$$
|
||||
where $\mathbf{r}_1$ and $\mathbf{r}_2$ denote the electron
|
||||
coordinates and \mathbf{R}_A$ and $\mathbf{R}_B$ the coordinates of
|
||||
the nuclei.
|
||||
$$
|
||||
\Psi(\mathbf{r}_1, \mathbf{r}_2) =
|
||||
\exp(-(\mathbf{r}_1 - \mathbf{R}_A)) +
|
||||
$$
|
||||
where $\mathbf{r}_1$ and $\mathbf{r}_2$ denote the electron
|
||||
coordinates and $\mathbf{R}_A$ and $\mathbf{R}_B$ the coordinates of
|
||||
the nuclei.
|
||||
|
||||
|
||||
|
||||
Loading…
Reference in New Issue
Block a user