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Uniform sampling
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@ -1,9 +1,7 @@
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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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#+STARTUP: latexpreview
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#+STARTUP: indent
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* Introduction
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@ -33,162 +31,160 @@ can be chosen.
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* Numerical evaluation of the energy
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In this section we consider the Hydrogen atom with the following
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wave function:
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In this section we consider the Hydrogen atom with the following
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wave function:
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$$
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\Psi(\mathbf{r}) = \exp(-a |\mathbf{r}|)
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$$
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$$
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\Psi(\mathbf{r}) = \exp(-a |\mathbf{r}|)
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$$
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We will first verify that $\Psi$ is an eigenfunction of the Hamiltonian
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We will first verify that $\Psi$ is an eigenfunction of the Hamiltonian
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$$
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\hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|}
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$$
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$$
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\hat{H} = \hat{T} + \hat{V} = - \frac{1}{2} \Delta - \frac{1}{|\mathbf{r}|}
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$$
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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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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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$$
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E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
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$$
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$$
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E_L(\mathbf{r}) = \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})},
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$$
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is constant.
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is constant.
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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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:header-args:f90: :tangle hydrogen.f90
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:header-args:f90: :tangle hydrogen.f90
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:END:
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*** Write a function which computes the potential at $\mathbf{r}$
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The function accepts q 3-dimensional vector =r= as input arguments
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and returns the potential.
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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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$$
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V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$
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$$
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$\mathbf{r}=\sqrt{x^2 + y^2 + z^2})$, so
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$$
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V(x,y,z) = -\frac{1}{\sqrt{x^2 + y^2 + z^2})$
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$$
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#+BEGIN_SRC python
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#+BEGIN_SRC python :results none
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import numpy as np
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def potential(r):
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return -1. / np.sqrt(np.dot(r,r))
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#+END_SRC
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#+END_SRC
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#+BEGIN_SRC f90
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#+BEGIN_SRC f90
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double precision function potential(r)
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implicit none
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double precision, intent(in) :: r(3)
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potential = -1.d0 / dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) )
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end function potential
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#+END_SRC
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#+END_SRC
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*** Write a function which computes the wave function at $\mathbf{r}$
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The function accepts a scalar =a= and a 3-dimensional vector =r= as
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input arguments, and returns a scalar.
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The function accepts a scalar =a= and a 3-dimensional vector =r= as
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input arguments, and returns a scalar.
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#+BEGIN_SRC python
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#+BEGIN_SRC python :results none
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def psi(a, r):
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return np.exp(-a*np.sqrt(np.dot(r,r)))
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#+END_SRC
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#+END_SRC
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#+BEGIN_SRC f90
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#+BEGIN_SRC f90
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double precision function psi(a, r)
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implicit none
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double precision, intent(in) :: a, r(3)
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psi = dexp(-a * dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
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end function psi
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#+END_SRC
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#+END_SRC
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*** Write a function which computes the local kinetic energy at $\mathbf{r}$
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The function accepts =a= and =r= as input arguments and returns the
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local kinetic energy.
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The function accepts =a= and =r= as input arguments and returns the
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local kinetic energy.
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The local kinetic energy is defined as $$-\frac{1}{2}\frac{\Delta \Psi}{\Psi}$$.
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The local kinetic energy is defined as $$-\frac{1}{2}\frac{\Delta \Psi}{\Psi}$$.
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$$
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\Psi(x,y,z) = \exp(-a\,\sqrt{x^2 + y^2 + z^2}).
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$$
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$$
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\Psi(x,y,z) = \exp(-a\,\sqrt{x^2 + y^2 + z^2}).
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$$
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We differentiate $\Psi$ with respect to $x$:
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We differentiate $\Psi$ with respect to $x$:
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$$
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\frac{\partial \Psi}{\partial x}
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= \frac{\partial \Psi}{\partial r} \frac{\partial r}{\partial x}
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= - \frac{a\,x}{|\mathbf{r}|} \Psi(x,y,z)
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$$
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$$
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\frac{\partial \Psi}{\partial x}
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= \frac{\partial \Psi}{\partial r} \frac{\partial r}{\partial x}
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= - \frac{a\,x}{|\mathbf{r}|} \Psi(x,y,z)
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$$
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and we differentiate a second time:
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and we differentiate a second time:
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$$
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\frac{\partial^2 \Psi}{\partial x^2} =
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\left( \frac{a^2\,x^2}{|\mathbf{r}|^2} - \frac{a(y^2+z^2)}{|\mathbf{r}|^{3}} \right) \Psi(x,y,z).
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$$
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$$
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\frac{\partial^2 \Psi}{\partial x^2} =
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\left( \frac{a^2\,x^2}{|\mathbf{r}|^2} - \frac{a(y^2+z^2)}{|\mathbf{r}|^{3}} \right) \Psi(x,y,z).
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$$
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The Laplacian operator $\Delta = \frac{\partial^2}{\partial x^2} +
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\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$
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applied to the wave function gives:
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The Laplacian operator $\Delta = \frac{\partial^2}{\partial x^2} +
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\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$
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applied to the wave function gives:
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$$
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\Delta \Psi (x,y,z) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(x,y,z)
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$$
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$$
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\Delta \Psi (x,y,z) = \left(a^2 - \frac{2a}{\mathbf{|r|}} \right) \Psi(x,y,z)
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$$
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So the local kinetic energy is
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$$
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-\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right)
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$$
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So the local kinetic energy is
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$$
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-\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) = -\frac{1}{2}\left(a^2 - \frac{2a}{\mathbf{|r|}} \right)
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$$
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#+BEGIN_SRC python
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#+BEGIN_SRC python :results none
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def kinetic(a,r):
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return -0.5 * (a**2 - (2.*a)/np.sqrt(np.dot(r,r)))
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#+END_SRC
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#+END_SRC
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#+BEGIN_SRC f90
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#+BEGIN_SRC f90
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double precision function kinetic(a,r)
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implicit none
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double precision, intent(in) :: a, r(3)
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kinetic = -0.5d0 * (a*a - (2.d0*a) / &
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dsqrt( r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ) )
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end function kinetic
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#+END_SRC
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#+END_SRC
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*** Write a function which computes the local energy at $\mathbf{r}$
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The function accepts =x,y,z= as input arguments and returns the
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local energy.
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The function accepts =x,y,z= as input arguments and returns the
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local energy.
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$$
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E_L(x,y,z) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) + V(x,y,z)
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$$
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$$
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E_L(x,y,z) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (x,y,z) + V(x,y,z)
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$$
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#+BEGIN_SRC python
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#+BEGIN_SRC python :results none
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def e_loc(a,r):
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return kinetic(a,r) + potential(r)
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#+END_SRC
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#+END_SRC
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#+BEGIN_SRC f90
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#+BEGIN_SRC f90
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double precision function e_loc(a,r)
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implicit none
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double precision, intent(in) :: a, r(3)
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double precision, external :: kinetic, potential
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e_loc = kinetic(a,r) + potential(r)
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end function e_loc
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#+END_SRC
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#+END_SRC
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** Plot 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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:END:
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:LOGBOOK:
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CLOCK: [2021-01-03 Sun 17:48]
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:header-args:f90: :tangle plot_hydrogen.f90
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:END:
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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
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#+begin_src python :results output
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import numpy as np
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import matplotlib.pyplot as plt
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@ -212,6 +208,8 @@ plt.legend()
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plt.savefig("plot_py.png")
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#+end_src
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#+RESULTS:
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[[./plot_py.png]]
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@ -275,36 +273,35 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
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** Compute numerically the average 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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:header-args:f90: :tangle energy_hydrogen.f90
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:END:
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We want to compute
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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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\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
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$\delta x\, \delta y\, \delta z$, this last equation corresponds
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to a weighted average of the local energy, where the weights are
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the values of the square of the wave function at $(x,y,z)$
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multiplied by the volume element:
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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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$$
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E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
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w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta x\, \delta y\, \delta z
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$$
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$$
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E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
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w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta \mathbf{r}
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$$
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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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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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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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#+BEGIN_SRC python :results output :exports both
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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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#+BEGIN_SRC python :results output :exports both
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import numpy as np
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from hydrogen import e_loc, psi
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@ -314,31 +311,32 @@ delta = (interval[1]-interval[0])**3
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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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E = 0.
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norm = 0.
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for x in interval:
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r[0] = x
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for y in interval:
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r[1] = y
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for z in interval:
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r[2] = z
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w = psi(a,r)
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w = w * w * delta
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E += w * e_loc(a,r)
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norm += w
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E = E / norm
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print(f"a = {a} \t E = {E}")
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#+end_src
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#+RESULTS:
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: a = 0.1 E = -0.24518438948809218
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: a = 0.2 E = -0.26966057967803525
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: a = 0.5 E = -0.3856357612517407
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: a = 0.9 E = -0.49435709786716214
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: a = 1.0 E = -0.5
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: a = 1.5 E = -0.39242967082602226
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: a = 2.0 E = -0.08086980667844901
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E = 0.
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norm = 0.
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for x in interval:
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r[0] = x
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for y in interval:
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r[1] = y
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for z in interval:
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r[2] = z
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w = psi(a,r)
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w = w * w * delta
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E += w * e_loc(a,r)
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norm += w
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E = E / norm
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print(f"a = {a} \t E = {E}")
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#+end_src
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#+RESULTS:
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: a = 0.1 E = -0.24518438948809218
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: a = 0.2 E = -0.26966057967803525
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: a = 0.5 E = -0.3856357612517407
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: a = 0.9 E = -0.49435709786716214
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: a = 1.0 E = -0.5
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: a = 1.5 E = -0.39242967082602226
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: a = 2.0 E = -0.08086980667844901
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#+begin_src f90
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program energy_hydrogen
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@ -400,23 +398,23 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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** Compute the variance of the local energy
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:PROPERTIES:
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:header-args:python: :tangle variance_hydrogen.py
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:header-args:f90: :tangle variance_hydrogen.f90
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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 intensity 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 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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$$
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\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
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E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
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$$
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$$
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\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
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E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
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$$
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Compute an 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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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 output :exports both
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import numpy as np
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from hydrogen import e_loc, psi
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@ -450,20 +448,20 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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w = psi(a, r)
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w = w * w * delta
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El = e_loc(a, r)
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s2 += w * (El - E)**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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#+end_src
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#+RESULTS:
|
||||
: a = 0.1 E = -0.24518439 \sigma^2 = 0.02696522
|
||||
: a = 0.2 E = -0.26966058 \sigma^2 = 0.03719707
|
||||
: a = 0.5 E = -0.38563576 \sigma^2 = 0.05318597
|
||||
: a = 0.9 E = -0.49435710 \sigma^2 = 0.00577812
|
||||
: a = 1.0 E = -0.50000000 \sigma^2 = 0.00000000
|
||||
: a = 1.5 E = -0.39242967 \sigma^2 = 0.31449671
|
||||
: a = 2.0 E = -0.08086981 \sigma^2 = 1.80688143
|
||||
#+RESULTS:
|
||||
: a = 0.1 E = -0.24518439 \sigma^2 = 0.02696522
|
||||
: a = 0.2 E = -0.26966058 \sigma^2 = 0.03719707
|
||||
: a = 0.5 E = -0.38563576 \sigma^2 = 0.05318597
|
||||
: a = 0.9 E = -0.49435710 \sigma^2 = 0.00577812
|
||||
: a = 1.0 E = -0.50000000 \sigma^2 = 0.00000000
|
||||
: a = 1.5 E = -0.39242967 \sigma^2 = 0.31449671
|
||||
: a = 2.0 E = -0.08086981 \sigma^2 = 1.80688143
|
||||
|
||||
#+begin_src f90
|
||||
program variance_hydrogen
|
||||
@ -521,7 +519,7 @@ program variance_hydrogen
|
||||
s2 = s2 / norm
|
||||
print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
|
||||
end do
|
||||
|
||||
|
||||
end program variance_hydrogen
|
||||
#+end_src
|
||||
|
||||
@ -543,15 +541,21 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
|
||||
|
||||
* Variational Monte Carlo
|
||||
|
||||
Instead of computing the average energy as a numerical integration
|
||||
on a grid, we will do a Monte Carlo sampling, which is an extremely
|
||||
efficient method to compute integrals in large dimensions.
|
||||
Instead of computing the average energy as a numerical integration
|
||||
on a grid, we will do a Monte Carlo sampling, which is an extremely
|
||||
efficient method to compute integrals when the number of dimensions is
|
||||
large.
|
||||
|
||||
Moreover, a Monte Carlo sampling will alow us to remove the bias due
|
||||
to the discretization of space, and compute a statistical confidence
|
||||
interval.
|
||||
Moreover, a Monte Carlo sampling will alow us to remove the bias due
|
||||
to the discretization of space, and compute a statistical confidence
|
||||
interval.
|
||||
|
||||
|
||||
** Computation of the statistical error
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle qmc_stats.py
|
||||
:header-args:f90: :tangle qmc_stats.f90
|
||||
:END:
|
||||
|
||||
To compute the statistical error, you need to perform $M$
|
||||
independent Monte Carlo calculations. You will obtain $M$ different
|
||||
@ -579,7 +583,8 @@ $$
|
||||
Write a function returning the average and statistical error of an
|
||||
input array.
|
||||
|
||||
#+BEGIN_SRC python :results output
|
||||
#+BEGIN_SRC python
|
||||
from math import sqrt
|
||||
def ave_error(arr):
|
||||
M = len(arr)
|
||||
assert (M>1)
|
||||
@ -588,28 +593,53 @@ def ave_error(arr):
|
||||
return (average, sqrt(variance/M))
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
#+BEGIN_SRC f90
|
||||
subroutine ave_error(x,n,ave,err)
|
||||
implicit none
|
||||
integer, intent(in) :: n
|
||||
double precision, intent(in) :: x(n)
|
||||
double precision, intent(out) :: ave, err
|
||||
double precision :: variance
|
||||
if (n == 1) then
|
||||
ave = x(1)
|
||||
err = 0.d0
|
||||
else
|
||||
ave = sum(x(:)) / dble(n)
|
||||
variance = sum( (x(:) - ave)**2 ) / dble(n-1)
|
||||
err = dsqrt(variance/dble(n))
|
||||
endif
|
||||
end subroutine ave_error
|
||||
#+END_SRC
|
||||
|
||||
** Uniform sampling in the box
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle qmc_uniform.py
|
||||
:header-args:f90: :tangle qmc_uniform.f90
|
||||
:END:
|
||||
|
||||
Write a function to perform a Monte Carlo calculation of the
|
||||
average energy. At every Monte Carlo step,
|
||||
In this section we write a function to perform a Monte Carlo
|
||||
calculation of the average energy.
|
||||
At every Monte Carlo step:
|
||||
|
||||
- Draw 3 uniform random numbers in the interval $(-5,-5,-5) \le
|
||||
(x,y,z) \le (5,5,5)$
|
||||
- Compute $\Psi^2 \times E_L$ at this point and accumulate the
|
||||
result in E
|
||||
- Compute $\Psi^2$ at this point and accumulate the result in N
|
||||
- Draw 3 uniform random numbers in the interval $(-5,-5,-5) \le
|
||||
(x,y,z) \le (5,5,5)$
|
||||
- Compute $\Psi^2 \times E_L$ at this point and accumulate the
|
||||
result in E
|
||||
- Compute $\Psi^2$ at this point and accumulate the result in N
|
||||
|
||||
Once all the steps have been computed, return the average energy
|
||||
computed on the Monte Carlo calculation.
|
||||
Once all the steps have been computed, return the average energy
|
||||
computed on the Monte Carlo calculation.
|
||||
|
||||
Then, write a loop to perform 30 Monte Carlo runs, and compute the
|
||||
average energy and the associated statistical error.
|
||||
In the main program, write a loop to perform 30 Monte Carlo runs,
|
||||
and compute the average energy and the associated statistical error.
|
||||
|
||||
Compute the energy of the wave function with $a=0.9$.
|
||||
Compute the energy of the wave function with $a=0.9$.
|
||||
|
||||
#+BEGIN_SRC python
|
||||
|
||||
#+BEGIN_SRC python :results output
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
|
||||
def MonteCarlo(a, nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
@ -620,53 +650,103 @@ def MonteCarlo(a, nmax):
|
||||
N += w
|
||||
E += w * e_loc(a,r)
|
||||
return E/N
|
||||
#+END_SRC
|
||||
|
||||
#+BEGIN_SRC python
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
X = [MonteCarlo(a,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.4952626284319677 +/- 0.0006877988969872546
|
||||
#+RESULTS:
|
||||
: E = -0.4956255109300764 +/- 0.0007082875482711226
|
||||
|
||||
#+BEGIN_SRC f90
|
||||
subroutine uniform_montecarlo(a,nmax,energy)
|
||||
implicit none
|
||||
double precision, intent(in) :: a
|
||||
integer , intent(in) :: nmax
|
||||
double precision, intent(out) :: energy
|
||||
|
||||
integer*8 :: istep
|
||||
|
||||
double precision :: norm, r(3), w
|
||||
|
||||
double precision, external :: e_loc, psi
|
||||
|
||||
energy = 0.d0
|
||||
norm = 0.d0
|
||||
do istep = 1,nmax
|
||||
call random_number(r)
|
||||
r(:) = -5.d0 + 10.d0*r(:)
|
||||
w = psi(a,r)
|
||||
w = w*w
|
||||
norm = norm + w
|
||||
energy = energy + w * e_loc(a,r)
|
||||
end do
|
||||
energy = energy / norm
|
||||
end subroutine uniform_montecarlo
|
||||
|
||||
program qmc
|
||||
implicit none
|
||||
double precision, parameter :: a = 0.9
|
||||
integer , parameter :: nmax = 100000
|
||||
integer , parameter :: nruns = 30
|
||||
|
||||
integer :: irun
|
||||
double precision :: X(nruns)
|
||||
double precision :: ave, err
|
||||
|
||||
do irun=1,nruns
|
||||
call uniform_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 qmc_uniform.f90 -o qmc_uniform
|
||||
./qmc_uniform
|
||||
#+end_src
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.49588321986667677 +/- 7.1758863546737969E-004
|
||||
|
||||
** Gaussian sampling
|
||||
|
||||
We will now improve the sampling and allow to sample in the whole
|
||||
3D space, correcting the bias related to the sampling in the box.
|
||||
We will now improve the sampling and allow to sample in the whole
|
||||
3D space, correcting the bias related to the sampling in the box.
|
||||
|
||||
Instead of drawing uniform random numbers, we will draw Gaussian
|
||||
random numbers centered on 0 and with a variance of 1. Now the
|
||||
equation for the energy is changed into
|
||||
Instead of drawing uniform random numbers, we will draw Gaussian
|
||||
random numbers centered on 0 and with a variance of 1. Now the
|
||||
equation for the energy is changed into
|
||||
|
||||
\[
|
||||
E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}}
|
||||
\]
|
||||
with
|
||||
\[
|
||||
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right)
|
||||
\]
|
||||
\[
|
||||
E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}}
|
||||
\]
|
||||
with
|
||||
\[
|
||||
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right)
|
||||
\]
|
||||
|
||||
As the coordinates are drawn with probability $P(\mathbf{r})$, the
|
||||
average energy can be computed as
|
||||
As the coordinates are drawn with probability $P(\mathbf{r})$, the
|
||||
average energy can be computed as
|
||||
|
||||
$$
|
||||
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
|
||||
w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r})} \delta x\, \delta y\, \delta z
|
||||
$$
|
||||
$$
|
||||
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
|
||||
w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r}_i)} \delta \mathbf{r}
|
||||
$$
|
||||
|
||||
#+BEGIN_SRC python
|
||||
#+BEGIN_SRC python
|
||||
norm_gauss = 1./(2.*np.pi)**(1.5)
|
||||
def gaussian(r):
|
||||
return norm_gauss * np.exp(-np.dot(r,r)*0.5)
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
#+RESULTS:
|
||||
|
||||
#+BEGIN_SRC python
|
||||
#+BEGIN_SRC python
|
||||
def MonteCarlo(a,nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
@ -677,58 +757,58 @@ def MonteCarlo(a,nmax):
|
||||
N += w
|
||||
E += w * e_loc(a,r)
|
||||
return E/N
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
#+RESULTS:
|
||||
|
||||
#+BEGIN_SRC python :results output
|
||||
#+BEGIN_SRC python :results output
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
X = [MonteCarlo(a,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.4952488228427792 +/- 0.00011913174676540714
|
||||
#+RESULTS:
|
||||
: E = -0.4952488228427792 +/- 0.00011913174676540714
|
||||
|
||||
** Sampling with $\Psi^2$
|
||||
|
||||
We will now use the square of the wave function to make the sampling:
|
||||
We will now use the square of the wave function to make the sampling:
|
||||
|
||||
\[
|
||||
P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
|
||||
\]
|
||||
\[
|
||||
P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
|
||||
\]
|
||||
|
||||
Now, the expression for the energy will be simplified to the
|
||||
average of the local energies, each with a weight of 1.
|
||||
Now, the expression for the energy will be simplified to the
|
||||
average of the local energies, each with a weight of 1.
|
||||
|
||||
$$
|
||||
E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)}
|
||||
$$
|
||||
$$
|
||||
E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)}
|
||||
$$
|
||||
|
||||
To generate the probability density $\Psi^2$, we can use a drifted
|
||||
diffusion scheme:
|
||||
To generate the probability density $\Psi^2$, we can use a drifted
|
||||
diffusion scheme:
|
||||
|
||||
\[
|
||||
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
|
||||
\Psi(r)}{\Psi(r)} + \eta \sqrt{\tau}
|
||||
\]
|
||||
\[
|
||||
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
|
||||
\Psi(r)}{\Psi(r)} + \eta \sqrt{\tau}
|
||||
\]
|
||||
|
||||
where $\eta$ is a normally-distributed Gaussian random number.
|
||||
where $\eta$ is a normally-distributed Gaussian random number.
|
||||
|
||||
|
||||
First, write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
|
||||
First, write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
|
||||
|
||||
#+BEGIN_SRC python
|
||||
#+BEGIN_SRC python
|
||||
def drift(a,r):
|
||||
ar_inv = -a/np.sqrt(np.dot(r,r))
|
||||
return r * ar_inv
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
#+RESULTS:
|
||||
|
||||
#+BEGIN_SRC python
|
||||
#+BEGIN_SRC python
|
||||
def MonteCarlo(a,tau,nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
@ -753,36 +833,36 @@ def MonteCarlo(a,tau,nmax):
|
||||
d_old = d_new
|
||||
d2_old = d2_new
|
||||
psi_old = psi_new
|
||||
N += 1.
|
||||
E += e_loc(a,r_old)
|
||||
N += 1.
|
||||
E += e_loc(a,r_old)
|
||||
return E/N
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
#+RESULTS:
|
||||
|
||||
#+BEGIN_SRC python :results output
|
||||
#+BEGIN_SRC python :results output
|
||||
nmax = 100000
|
||||
tau = 0.1
|
||||
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.4951783346213532 +/- 0.00022067316984271938
|
||||
#+RESULTS:
|
||||
: E = -0.4951783346213532 +/- 0.00022067316984271938
|
||||
|
||||
|
||||
* 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.
|
||||
|
||||
|
||||
|
||||
@ -7,17 +7,17 @@ delta = (interval[1]-interval[0])**3
|
||||
r = np.array([0.,0.,0.])
|
||||
|
||||
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
E = 0.
|
||||
norm = 0.
|
||||
for x in interval:
|
||||
r[0] = x
|
||||
for y in interval:
|
||||
r[1] = y
|
||||
for z in interval:
|
||||
r[2] = z
|
||||
w = psi(a,r)
|
||||
w = w * w * delta
|
||||
E += w * e_loc(a,r)
|
||||
norm += w
|
||||
E = E / norm
|
||||
print(f"a = {a} \t E = {E}")
|
||||
E = 0.
|
||||
norm = 0.
|
||||
for x in interval:
|
||||
r[0] = x
|
||||
for y in interval:
|
||||
r[1] = y
|
||||
for z in interval:
|
||||
r[2] = z
|
||||
w = psi(a,r)
|
||||
w = w * w * delta
|
||||
E += w * e_loc(a,r)
|
||||
norm += w
|
||||
E = E / norm
|
||||
print(f"a = {a} \t E = {E}")
|
||||
|
||||
15
qmc_stats.f90
Normal file
15
qmc_stats.f90
Normal file
@ -0,0 +1,15 @@
|
||||
subroutine ave_error(x,n,ave,err)
|
||||
implicit none
|
||||
integer, intent(in) :: n
|
||||
double precision, intent(in) :: x(n)
|
||||
double precision, intent(out) :: ave, err
|
||||
double precision :: variance
|
||||
if (n == 1) then
|
||||
ave = x(1)
|
||||
err = 0.d0
|
||||
else
|
||||
ave = sum(x(:)) / dble(n)
|
||||
variance = sum( (x(:) - ave)**2 ) / dble(n-1)
|
||||
err = dsqrt(variance/dble(n))
|
||||
endif
|
||||
end subroutine ave_error
|
||||
7
qmc_stats.py
Normal file
7
qmc_stats.py
Normal file
@ -0,0 +1,7 @@
|
||||
from math import sqrt
|
||||
def ave_error(arr):
|
||||
M = len(arr)
|
||||
assert (M>1)
|
||||
average = sum(arr)/M
|
||||
variance = 1./(M-1) * sum( [ (x - average)**2 for x in arr ] )
|
||||
return (average, sqrt(variance/M))
|
||||
41
qmc_uniform.f90
Normal file
41
qmc_uniform.f90
Normal file
@ -0,0 +1,41 @@
|
||||
subroutine uniform_montecarlo(a,nmax,energy)
|
||||
implicit none
|
||||
double precision, intent(in) :: a
|
||||
integer , intent(in) :: nmax
|
||||
double precision, intent(out) :: energy
|
||||
|
||||
integer*8 :: istep
|
||||
|
||||
double precision :: norm, r(3), w
|
||||
|
||||
double precision, external :: e_loc, psi
|
||||
|
||||
energy = 0.d0
|
||||
norm = 0.d0
|
||||
do istep = 1,nmax
|
||||
call random_number(r)
|
||||
r(:) = -5.d0 + 10.d0*r(:)
|
||||
w = psi(a,r)
|
||||
w = w*w
|
||||
norm = norm + w
|
||||
energy = energy + w * e_loc(a,r)
|
||||
end do
|
||||
energy = energy / norm
|
||||
end subroutine uniform_montecarlo
|
||||
|
||||
program qmc
|
||||
implicit none
|
||||
double precision, parameter :: a = 0.9
|
||||
integer , parameter :: nmax = 100000
|
||||
integer , parameter :: nruns = 30
|
||||
|
||||
integer :: irun
|
||||
double precision :: X(nruns)
|
||||
double precision :: ave, err
|
||||
|
||||
do irun=1,nruns
|
||||
call uniform_montecarlo(a,nmax,X(irun))
|
||||
enddo
|
||||
call ave_error(X,nruns,ave,err)
|
||||
print *, 'E = ', ave, '+/-', err
|
||||
end program qmc
|
||||
19
qmc_uniform.py
Normal file
19
qmc_uniform.py
Normal file
@ -0,0 +1,19 @@
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
|
||||
def MonteCarlo(a, nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
for istep in range(nmax):
|
||||
r = np.random.uniform(-5., 5., (3))
|
||||
w = psi(a,r)
|
||||
w = w*w
|
||||
N += w
|
||||
E += w * e_loc(a,r)
|
||||
return E/N
|
||||
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
X = [MonteCarlo(a,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
@ -53,5 +53,5 @@ program variance_hydrogen
|
||||
s2 = s2 / norm
|
||||
print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
|
||||
end do
|
||||
|
||||
|
||||
end program variance_hydrogen
|
||||
|
||||
@ -31,6 +31,6 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
w = psi(a, r)
|
||||
w = w * w * delta
|
||||
El = e_loc(a, r)
|
||||
s2 += w * (El - E)**2
|
||||
s2 += w * (El - E)**2
|
||||
s2 = s2 / norm
|
||||
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
|
||||
|
||||
Loading…
Reference in New Issue
Block a user