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Gaussian sampling
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QMC.org
@ -6,185 +6,185 @@
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* Introduction
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We propose different exercises to understand quantum Monte Carlo (QMC)
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methods. In the first section, we propose to compute the energy of a
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hydrogen atom using numerical integration. The goal of this section is
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to introduce the /local energy/.
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Then we introduce the variational Monte Carlo (VMC) method which
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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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We propose different exercises to understand quantum Monte Carlo (QMC)
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methods. In the first section, we propose to compute the energy of a
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hydrogen atom using numerical integration. The goal of this section is
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to introduce the /local energy/.
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Then we introduce the variational Monte Carlo (VMC) method which
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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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Code examples will be given in Python and Fortran. Whatever language
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can be chosen.
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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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** Python
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** Fortran
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- 1.d0
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- external
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- r(:) = 0.d0
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- a = (/ 0.1, 0.2 /)
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- size(x)
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- 1.d0
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- external
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- r(:) = 0.d0
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- a = (/ 0.1, 0.2 /)
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- size(x)
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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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:END:
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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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:END:
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*** Write a function which computes the potential at $\mathbf{r}$
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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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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 :results none
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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 :results none
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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 :results none
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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 :results none
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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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: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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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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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 output
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import numpy as np
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import matplotlib.pyplot as plt
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@ -206,14 +206,14 @@ plt.tight_layout()
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plt.legend()
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plt.savefig("plot_py.png")
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#+end_src
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#+end_src
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#+RESULTS:
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#+RESULTS:
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[[./plot_py.png]]
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[[./plot_py.png]]
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#+begin_src f90
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#+begin_src f90
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program plot
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implicit none
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double precision, external :: e_loc
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@ -242,20 +242,20 @@ program plot
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end do
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end program plot
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#+end_src
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#+end_src
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To compile and run:
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To compile and run:
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#+begin_src sh :exports both
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#+begin_src sh :exports both
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gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
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./plot_hydrogen > data
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#+end_src
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#+end_src
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#+RESULTS:
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#+RESULTS:
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To plot the data using gnuplot"
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To plot the data using gnuplot"
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#+begin_src gnuplot :file plot.png :exports both
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#+begin_src gnuplot :file plot.png :exports both
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set grid
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set xrange [-5:5]
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set yrange [-2:1]
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@ -265,39 +265,39 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
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'./data' index 3 using 1:2 with lines title 'a=1.0', \
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'./data' index 4 using 1:2 with lines title 'a=1.5', \
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'./data' index 5 using 1:2 with lines title 'a=2.0'
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#+end_src
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#+end_src
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#+RESULTS:
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[[file:plot.png]]
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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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: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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: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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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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\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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\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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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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$$
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E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
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w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta \mathbf{r}
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$$
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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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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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@ -305,12 +305,12 @@ Note: the energy is biased because:
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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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@ -338,7 +338,7 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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: a = 2.0 E = -0.08086980667844901
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#+begin_src f90
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#+begin_src f90
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program energy_hydrogen
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implicit none
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double precision, external :: e_loc, psi
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@ -378,43 +378,43 @@ program energy_hydrogen
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end do
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end program energy_hydrogen
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#+end_src
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#+end_src
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To compile and run:
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To compile and run:
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#+begin_src sh :results output :exports both
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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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./energy_hydrogen
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#+end_src
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#+end_src
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#+RESULTS:
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: a = 0.10000000000000001 E = -0.24518438948809140
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: a = 0.20000000000000001 E = -0.26966057967803236
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: a = 0.50000000000000000 E = -0.38563576125173815
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: a = 1.0000000000000000 E = -0.50000000000000000
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: a = 1.5000000000000000 E = -0.39242967082602065
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: a = 2.0000000000000000 E = -8.0869806678448772E-002
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#+RESULTS:
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: a = 0.10000000000000001 E = -0.24518438948809140
|
||||
: a = 0.20000000000000001 E = -0.26966057967803236
|
||||
: a = 0.50000000000000000 E = -0.38563576125173815
|
||||
: a = 1.0000000000000000 E = -0.50000000000000000
|
||||
: a = 1.5000000000000000 E = -0.39242967082602065
|
||||
: a = 2.0000000000000000 E = -8.0869806678448772E-002
|
||||
|
||||
** Compute the variance of the local energy
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle variance_hydrogen.py
|
||||
:header-args:f90: :tangle variance_hydrogen.f90
|
||||
:END:
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle variance_hydrogen.py
|
||||
:header-args:f90: :tangle variance_hydrogen.f90
|
||||
:END:
|
||||
|
||||
The variance of the local energy measures the magnitude of the
|
||||
fluctuations of the local energy around the average. If the local
|
||||
energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$)
|
||||
the variance is zero.
|
||||
The variance of the local energy measures the magnitude of the
|
||||
fluctuations of the local energy around the average. If the local
|
||||
energy is constant (i.e. $\Psi$ is an eigenfunction of $\hat{H}$)
|
||||
the variance is zero.
|
||||
|
||||
$$
|
||||
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
|
||||
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
|
||||
$$
|
||||
$$
|
||||
\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
|
||||
E_L(\mathbf{r}) - E \right]^2 \, d\mathbf{r}}{\int \left[\Psi(\mathbf{r}) \right]^2 d\mathbf{r}}
|
||||
$$
|
||||
|
||||
Compute a numerical estimate of the variance of the local energy
|
||||
in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
|
||||
Compute a numerical estimate of the variance of the local energy
|
||||
in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
|
||||
|
||||
#+BEGIN_SRC python :results output :exports both
|
||||
#+BEGIN_SRC python :results output :exports both
|
||||
import numpy as np
|
||||
from hydrogen import e_loc, psi
|
||||
|
||||
@ -452,18 +452,18 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
s2 = s2 / norm
|
||||
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
|
||||
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
#+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
|
||||
#+begin_src f90
|
||||
program variance_hydrogen
|
||||
implicit none
|
||||
double precision, external :: e_loc, psi
|
||||
@ -521,69 +521,68 @@ program variance_hydrogen
|
||||
end do
|
||||
|
||||
end program variance_hydrogen
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
To compile and run:
|
||||
To compile and run:
|
||||
|
||||
#+begin_src sh :results output :exports both
|
||||
#+begin_src sh :results output :exports both
|
||||
gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
|
||||
./variance_hydrogen
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
#+RESULTS:
|
||||
: a = 0.10000000000000001 E = -0.24518438948809140 s2 = 2.6965218719733813E-002
|
||||
: a = 0.20000000000000001 E = -0.26966057967803236 s2 = 3.7197072370217653E-002
|
||||
: a = 0.50000000000000000 E = -0.38563576125173815 s2 = 5.3185967578488862E-002
|
||||
: a = 1.0000000000000000 E = -0.50000000000000000 s2 = 0.0000000000000000
|
||||
: a = 1.5000000000000000 E = -0.39242967082602065 s2 = 0.31449670909180444
|
||||
: a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814270851303
|
||||
#+RESULTS:
|
||||
: a = 0.10000000000000001 E = -0.24518438948809140 s2 = 2.6965218719733813E-002
|
||||
: a = 0.20000000000000001 E = -0.26966057967803236 s2 = 3.7197072370217653E-002
|
||||
: a = 0.50000000000000000 E = -0.38563576125173815 s2 = 5.3185967578488862E-002
|
||||
: a = 1.0000000000000000 E = -0.50000000000000000 s2 = 0.0000000000000000
|
||||
: a = 1.5000000000000000 E = -0.39242967082602065 s2 = 0.31449670909180444
|
||||
: a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814270851303
|
||||
|
||||
|
||||
* 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 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.
|
||||
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.
|
||||
|
||||
** Computation of the statistical error
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle qmc_stats.py
|
||||
:header-args:f90: :tangle qmc_stats.f90
|
||||
:END:
|
||||
: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
|
||||
estimates of the energy, which are expected to have a Gaussian
|
||||
distribution by the central limit theorem.
|
||||
To compute the statistical error, you need to perform $M$
|
||||
independent Monte Carlo calculations. You will obtain $M$ different
|
||||
estimates of the energy, which are expected to have a Gaussian
|
||||
distribution by the central limit theorem.
|
||||
|
||||
The estimate of the energy is
|
||||
The estimate of the energy is
|
||||
|
||||
$$
|
||||
E = \frac{1}{M} \sum_{i=1}^M E_M
|
||||
$$
|
||||
$$
|
||||
E = \frac{1}{M} \sum_{i=1}^M E_M
|
||||
$$
|
||||
|
||||
The variance of the average energies can be computed as
|
||||
The variance of the average energies can be computed as
|
||||
|
||||
$$
|
||||
\sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_M - E)^2
|
||||
$$
|
||||
$$
|
||||
\sigma^2 = \frac{1}{M-1} \sum_{i=1}^{M} (E_M - E)^2
|
||||
$$
|
||||
|
||||
And the confidence interval is given by
|
||||
And the confidence interval is given by
|
||||
|
||||
$$
|
||||
E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}}
|
||||
$$
|
||||
$$
|
||||
E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}}
|
||||
$$
|
||||
|
||||
Write a function returning the average and statistical error of an
|
||||
input array.
|
||||
Write a function returning the average and statistical error of an
|
||||
input array.
|
||||
|
||||
#+BEGIN_SRC python
|
||||
#+BEGIN_SRC python
|
||||
from math import sqrt
|
||||
def ave_error(arr):
|
||||
M = len(arr)
|
||||
@ -591,9 +590,9 @@ def ave_error(arr):
|
||||
average = sum(arr)/M
|
||||
variance = 1./(M-1) * sum( [ (x - average)**2 for x in arr ] )
|
||||
return (average, sqrt(variance/M))
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+BEGIN_SRC f90
|
||||
#+BEGIN_SRC f90
|
||||
subroutine ave_error(x,n,ave,err)
|
||||
implicit none
|
||||
integer, intent(in) :: n
|
||||
@ -609,23 +608,23 @@ subroutine ave_error(x,n,ave,err)
|
||||
err = dsqrt(variance/dble(n))
|
||||
endif
|
||||
end subroutine ave_error
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
** Uniform sampling in the box
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle qmc_uniform.py
|
||||
:header-args:f90: :tangle qmc_uniform.f90
|
||||
:END:
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle qmc_uniform.py
|
||||
:header-args:f90: :tangle qmc_uniform.f90
|
||||
:END:
|
||||
|
||||
In this section we 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
|
||||
- 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
|
||||
- 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
|
||||
- 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.
|
||||
@ -635,12 +634,11 @@ At every Monte Carlo step:
|
||||
|
||||
Compute the energy of the wave function with $a=0.9$.
|
||||
|
||||
|
||||
#+BEGIN_SRC python :results output
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
|
||||
def MonteCarlo(a, nmax):
|
||||
def MonteCarlo(a, nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
for istep in range(nmax):
|
||||
@ -651,17 +649,17 @@ def MonteCarlo(a, nmax):
|
||||
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}")
|
||||
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
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.4956255109300764 +/- 0.0007082875482711226
|
||||
|
||||
#+BEGIN_SRC f90
|
||||
#+BEGIN_SRC f90
|
||||
subroutine uniform_montecarlo(a,nmax,energy)
|
||||
implicit none
|
||||
double precision, intent(in) :: a
|
||||
@ -703,50 +701,93 @@ program qmc
|
||||
call ave_error(X,nruns,ave,err)
|
||||
print *, 'E = ', ave, '+/-', err
|
||||
end program qmc
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+begin_src sh :results output :exports both
|
||||
#+begin_src sh :results output :exports both
|
||||
gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
|
||||
./qmc_uniform
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.49588321986667677 +/- 7.1758863546737969E-004
|
||||
#+RESULTS:
|
||||
: E = -0.49588321986667677 +/- 7.1758863546737969E-004
|
||||
|
||||
** Gaussian sampling
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle qmc_gaussian.py
|
||||
:header-args:f90: :tangle qmc_gaussian.f90
|
||||
:END:
|
||||
|
||||
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.
|
||||
|
||||
\[
|
||||
E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}}
|
||||
\]
|
||||
with
|
||||
\[
|
||||
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right)
|
||||
\]
|
||||
To obtain Gaussian-distributed random numbers, you can apply the
|
||||
[[https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform][Box Muller transform]] to uniform random numbers:
|
||||
|
||||
As the coordinates are drawn with probability $P(\mathbf{r})$, the
|
||||
average energy can be computed as
|
||||
\begin{eqnarray*}
|
||||
z_1 &=& \sqrt{-2 \ln u_1} \cos(2 \pi u_2) \\
|
||||
z_2 &=& \sqrt{-2 \ln u_1} \sin(2 \pi u_2)
|
||||
\end{eqnarray*}
|
||||
|
||||
$$
|
||||
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
|
||||
w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r}_i)} \delta \mathbf{r}
|
||||
$$
|
||||
#+BEGIN_SRC f90 :tangle qmc_stats.f90
|
||||
subroutine random_gauss(z,n)
|
||||
implicit none
|
||||
integer, intent(in) :: n
|
||||
double precision, intent(out) :: z(n)
|
||||
double precision :: u(n+1)
|
||||
double precision, parameter :: two_pi = 2.d0*dacos(-1.d0)
|
||||
integer :: i
|
||||
|
||||
call random_number(u)
|
||||
if (iand(n,1) == 0) then
|
||||
! n is even
|
||||
do i=1,n,2
|
||||
z(i) = dsqrt(-2.d0*dlog(u(i)))
|
||||
z(i+1) = z(i) + dsin( two_pi*u(i+1) )
|
||||
z(i) = z(i) + dcos( two_pi*u(i+1) )
|
||||
end do
|
||||
else
|
||||
! n is odd
|
||||
do i=1,n-1,2
|
||||
z(i) = dsqrt(-2.d0*dlog(u(i)))
|
||||
z(i+1) = z(i) + dsin( two_pi*u(i+1) )
|
||||
z(i) = z(i) + dcos( two_pi*u(i+1) )
|
||||
end do
|
||||
z(n) = dsqrt(-2.d0*dlog(u(n)))
|
||||
z(n) = z(n) + dcos( two_pi*u(n+1) )
|
||||
end if
|
||||
end subroutine random_gauss
|
||||
#+END_SRC
|
||||
|
||||
|
||||
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
|
||||
\[
|
||||
P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right)
|
||||
\]
|
||||
|
||||
As the coordinates are drawn with probability $P(\mathbf{r})$, the
|
||||
average energy can be computed as
|
||||
|
||||
$$
|
||||
E \approx \frac{\sum_i w_i E_L(\mathbf{r}_i)}{\sum_i w_i}, \;\;
|
||||
w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r}_i)} \delta \mathbf{r}
|
||||
$$
|
||||
|
||||
#+BEGIN_SRC python :results output
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
|
||||
#+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
|
||||
|
||||
#+RESULTS:
|
||||
|
||||
#+BEGIN_SRC python
|
||||
def MonteCarlo(a,nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
@ -757,58 +798,113 @@ def MonteCarlo(a,nmax):
|
||||
N += w
|
||||
E += w * e_loc(a,r)
|
||||
return E/N
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
|
||||
#+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 = -0.49507506093129827 +/- 0.00014164037765553668
|
||||
|
||||
|
||||
#+BEGIN_SRC f90
|
||||
double precision function gaussian(r)
|
||||
implicit none
|
||||
double precision, intent(in) :: r(3)
|
||||
double precision, parameter :: norm_gauss = 1.d0/(2.d0*dacos(-1.d0))**(1.5d0)
|
||||
gaussian = norm_gauss * dexp( -0.5d0 * dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3) ))
|
||||
end function gaussian
|
||||
|
||||
|
||||
subroutine gaussian_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, gaussian
|
||||
|
||||
energy = 0.d0
|
||||
norm = 0.d0
|
||||
do istep = 1,nmax
|
||||
call random_gauss(r,3)
|
||||
w = psi(a,r)
|
||||
w = w*w / gaussian(r)
|
||||
norm = norm + w
|
||||
energy = energy + w * e_loc(a,r)
|
||||
end do
|
||||
energy = energy / norm
|
||||
end subroutine gaussian_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 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 qmc_gaussian.f90 -o qmc_gaussian
|
||||
./qmc_gaussian
|
||||
#+end_src
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.49606057056767766 +/- 1.3918807547836872E-004
|
||||
** 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.
|
||||
@ -836,20 +932,20 @@ def MonteCarlo(a,tau,nmax):
|
||||
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
|
||||
|
||||
@ -1,12 +1,12 @@
|
||||
import numpy as np
|
||||
from hydrogen import e_loc, psi
|
||||
|
||||
interval = np.linspace(-5,5,num=50)
|
||||
delta = (interval[1]-interval[0])**3
|
||||
interval = np.linspace(-5,5,num=50)
|
||||
delta = (interval[1]-interval[0])**3
|
||||
|
||||
r = np.array([0.,0.,0.])
|
||||
r = np.array([0.,0.,0.])
|
||||
|
||||
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
E = 0.
|
||||
norm = 0.
|
||||
for x in interval:
|
||||
|
||||
@ -13,3 +13,31 @@ subroutine ave_error(x,n,ave,err)
|
||||
err = dsqrt(variance/dble(n))
|
||||
endif
|
||||
end subroutine ave_error
|
||||
|
||||
subroutine random_gauss(z,n)
|
||||
implicit none
|
||||
integer, intent(in) :: n
|
||||
double precision, intent(out) :: z(n)
|
||||
double precision :: u(n+1)
|
||||
double precision, parameter :: two_pi = 2.d0*dacos(-1.d0)
|
||||
integer :: i
|
||||
|
||||
call random_number(u)
|
||||
if (iand(n,1) == 0) then
|
||||
! n is even
|
||||
do i=1,n,2
|
||||
z(i) = dsqrt(-2.d0*dlog(u(i)))
|
||||
z(i+1) = z(i) + dsin( two_pi*u(i+1) )
|
||||
z(i) = z(i) + dcos( two_pi*u(i+1) )
|
||||
end do
|
||||
else
|
||||
! n is odd
|
||||
do i=1,n-1,2
|
||||
z(i) = dsqrt(-2.d0*dlog(u(i)))
|
||||
z(i+1) = z(i) + dsin( two_pi*u(i+1) )
|
||||
z(i) = z(i) + dcos( two_pi*u(i+1) )
|
||||
end do
|
||||
z(n) = dsqrt(-2.d0*dlog(u(n)))
|
||||
z(n) = z(n) + dcos( two_pi*u(n+1) )
|
||||
end if
|
||||
end subroutine random_gauss
|
||||
|
||||
@ -1,7 +1,7 @@
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
|
||||
def MonteCarlo(a, nmax):
|
||||
def MonteCarlo(a, nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
for istep in range(nmax):
|
||||
@ -12,8 +12,8 @@ def MonteCarlo(a, nmax):
|
||||
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}")
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
X = [MonteCarlo(a,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
|
||||
Loading…
Reference in New Issue
Block a user