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Added drifted diffusion
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@ -988,125 +988,140 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
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: E = -0.49478505004797046 +/- 2.0493795299184956E-004
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: A = 0.51737800000000000 +/- 4.1827406733181444E-004
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** Sampling with $\Psi^2$
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We will now use the square of the wave function to make the sampling:
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\[
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P(\mathbf{r}) = \left[\Psi(\mathbf{r})\right]^2
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\]
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The expression for the energy will be simplified to the average of
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the local energies, each with a weight of 1.
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$$
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E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)
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$$
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** Gaussian random number generator
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*** Importance sampling
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To obtain Gaussian-distributed random numbers, you can apply the
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[[https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform][Box Muller transform]] to uniform random numbers:
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\begin{eqnarray*}
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z_1 &=& \sqrt{-2 \ln u_1} \cos(2 \pi u_2) \\
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z_2 &=& \sqrt{-2 \ln u_1} \sin(2 \pi u_2)
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\end{eqnarray*}
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Below is a Fortran implementation returning a Gaussian-distributed
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n-dimensional vector $\mathbf{z}$. This will be useful for the
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following sections.
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*Fortran*
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#+BEGIN_SRC f90 :tangle qmc_stats.f90
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subroutine random_gauss(z,n)
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implicit none
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integer, intent(in) :: n
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double precision, intent(out) :: z(n)
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double precision :: u(n+1)
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double precision, parameter :: two_pi = 2.d0*dacos(-1.d0)
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integer :: i
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call random_number(u)
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if (iand(n,1) == 0) then
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! n is even
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do i=1,n,2
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z(i) = dsqrt(-2.d0*dlog(u(i)))
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z(i+1) = z(i) * dsin( two_pi*u(i+1) )
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z(i) = z(i) * dcos( two_pi*u(i+1) )
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end do
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else
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! n is odd
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do i=1,n-1,2
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z(i) = dsqrt(-2.d0*dlog(u(i)))
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z(i+1) = z(i) * dsin( two_pi*u(i+1) )
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z(i) = z(i) * dcos( two_pi*u(i+1) )
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end do
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z(n) = dsqrt(-2.d0*dlog(u(n)))
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z(n) = z(n) * dcos( two_pi*u(n+1) )
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end if
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end subroutine random_gauss
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#+END_SRC
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** Generalized Metropolis algorithm
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:PROPERTIES:
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:header-args:python: :tangle vmc.py
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:header-args:f90: :tangle vmc.f90
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:header-args:python: :tangle vmc_metropolis.py
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:header-args:f90: :tangle vmc_metropolis.f90
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:END:
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To generate the probability density $\Psi^2$, we consider a
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diffusion process characterized by a time-dependent density
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$[\Psi(\mathbf{r},t)]^2$, which obeys the Fokker-Planck equation:
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One can use more efficient numerical schemes to move the electrons.
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But in that case, the Metropolis accepation step has to be adapted
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accordingly: the acceptance
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probability $A$ is chosen so that it is consistent with the
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probability of leaving $\mathbf{r}_n$ and the probability of
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entering $\mathbf{r}_{n+1}$:
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\[
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\frac{\partial \Psi^2}{\partial t} = \sum_i D
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\frac{\partial}{\partial \mathbf{r}_i} \left(
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\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
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[\Psi(\mathbf{r},t)]^2.
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\]
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\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1,
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\frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
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{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}
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\right)
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\]
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where $T(\mathbf{r}_n \rightarrow \mathbf{r}_{n+1})$ is the
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probability of transition from $\mathbf{r}_n$ to
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$\mathbf{r}_{n+1}$.
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In the previous example, we were using uniform random
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numbers. Hence, the transition probability was
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\[
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
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\text{constant}
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\]
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So the expression of $A$ was simplified to the ratios of the squared
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wave functions.
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Now, if instead of drawing uniform random numbers
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choose to draw Gaussian random numbers with mean 0 and variance
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$\tau$, the transition probability becomes:
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\[
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
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\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
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\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\tau} \right]
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\]
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To sample even better the density, we can "push" the electrons
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into in the regions of high probability, and "pull" them away from
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the low-probability regions. This will mechanically increase the
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acceptance ratios and improve the sampling.
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To do this, we can add the drift vector
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\[
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\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}
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\].
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The numerical scheme becomes a drifted diffusion:
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\[
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\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
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\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
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\]
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where $\chi$ is a Gaussian random variable with zero mean and
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variance $\tau$.
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The transition probability becomes:
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\[
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
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\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
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\mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla
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\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\tau} \right]
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\]
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*** Exercise 1
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#+begin_exercise
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Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
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#+end_exercise
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$D$ is the diffusion constant and $F_i$ is the i-th component of a
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drift velocity caused by an external potential. For a stationary
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density, \( \frac{\partial \Psi^2}{\partial t} = 0 \), so
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\begin{eqnarray*}
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0 & = & \sum_i D
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\frac{\partial}{\partial \mathbf{r}_i} \left(
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\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
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[\Psi(\mathbf{r})]^2 \\
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0 & = & \sum_i D
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\frac{\partial}{\partial \mathbf{r}_i} \left(
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\frac{\partial [\Psi(\mathbf{r})]^2}{\partial \mathbf{r}_i} -
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F_i(\mathbf{r})\,[\Psi(\mathbf{r})]^2 \right) \\
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0 & = &
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\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} -
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\frac{\partial F_i }{\partial \mathbf{r}_i}[\Psi(\mathbf{r})]^2 -
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\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
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\end{eqnarray*}
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we search for a drift function which satisfies
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\[
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\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
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[\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
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\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
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\]
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to obtain a second derivative on the left, we need the drift to be
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of the form
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\[
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F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
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\]
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\[
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\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
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[\Psi(\mathbf{r})]^2 \frac{\partial
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g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} +
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[\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2
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\Psi^2}{\partial \mathbf{r}_i^2} +
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\frac{\partial \Psi^2}{\partial \mathbf{r}_i}
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g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
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\]
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$g = 1 / \Psi^2$ satisfies this equation, so
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\[
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F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
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\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
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\]
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In statistical mechanics, Fokker-Planck trajectories are generated
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by a Langevin equation:
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\[
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\frac{\partial \mathbf{r}(t)}{\partial t} = 2D \frac{\nabla
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\Psi(\mathbf{r}(t))}{\Psi} + \eta
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\]
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where $\eta$ is a normally-distributed fluctuating random force.
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Discretizing this differential equation gives the following drifted
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diffusion scheme:
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\[
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\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau\, 2D \frac{\nabla
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\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
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\]
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where $\chi$ is a Gaussian random variable with zero mean and
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variance $\tau\,2D$.
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**** Exercise 1
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#+begin_exercise
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Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
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#+end_exercise
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*Python*
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#+BEGIN_SRC python :tangle hydrogen.py
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*Python*
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#+BEGIN_SRC python :tangle hydrogen.py
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def drift(a,r):
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ar_inv = -a/np.sqrt(np.dot(r,r))
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return r * ar_inv
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#+END_SRC
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#+END_SRC
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*Fortran*
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#+BEGIN_SRC f90 :tangle hydrogen.f90
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*Fortran*
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#+BEGIN_SRC f90 :tangle hydrogen.f90
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subroutine drift(a,r,b)
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implicit none
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double precision, intent(in) :: a, r(3)
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@ -1115,167 +1130,17 @@ subroutine drift(a,r,b)
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ar_inv = -a / dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3))
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b(:) = r(:) * ar_inv
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end subroutine drift
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#+END_SRC
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#+END_SRC
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**** TODO Exercise 2
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*** Exercise 2
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#+begin_exercise
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Sample $\Psi^2$ approximately using the drifted diffusion scheme,
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with a diffusion constant $D=1/2$. You can use a time step of
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0.001 a.u.
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#+end_exercise
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#+begin_exercise
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Modify the previous program to introduce the drifted diffusion scheme.
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(This is a necessary step for the next section).
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#+end_exercise
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*Python*
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#+BEGIN_SRC python :results output
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from hydrogen import *
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from qmc_stats import *
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def MonteCarlo(a,tau,nmax):
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sq_tau = np.sqrt(tau)
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# Initialization
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E = 0.
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N = 0.
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r_old = np.random.normal(loc=0., scale=1.0, size=(3))
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for istep in range(nmax):
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d_old = drift(a,r_old)
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chi = np.random.normal(loc=0., scale=1.0, size=(3))
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r_new = r_old + tau * d_old + chi*sq_tau
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N += 1.
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E += e_loc(a,r_new)
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r_old = r_new
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return E/N
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a = 0.9
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nmax = 100000
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tau = 0.2
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X = [MonteCarlo(a,tau,nmax) for i in range(30)]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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#+END_SRC
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#+RESULTS:
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: E = -0.4858534479298907 +/- 0.00010203236131158794
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*Fortran*
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#+BEGIN_SRC f90
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subroutine variational_montecarlo(a,tau,nmax,energy)
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implicit none
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double precision, intent(in) :: a, tau
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integer*8 , intent(in) :: nmax
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double precision, intent(out) :: energy
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integer*8 :: istep
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double precision :: norm, r_old(3), r_new(3), d_old(3), sq_tau, chi(3)
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double precision, external :: e_loc
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sq_tau = dsqrt(tau)
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! Initialization
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energy = 0.d0
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norm = 0.d0
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call random_gauss(r_old,3)
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do istep = 1,nmax
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call drift(a,r_old,d_old)
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call random_gauss(chi,3)
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r_new(:) = r_old(:) + tau * d_old(:) + chi(:)*sq_tau
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norm = norm + 1.d0
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energy = energy + e_loc(a,r_new)
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r_old(:) = r_new(:)
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end do
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energy = energy / norm
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end subroutine variational_montecarlo
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program qmc
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implicit none
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double precision, parameter :: a = 0.9
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double precision, parameter :: tau = 0.2
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integer*8 , parameter :: nmax = 100000
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integer , parameter :: nruns = 30
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integer :: irun
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double precision :: X(nruns)
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double precision :: ave, err
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do irun=1,nruns
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call variational_montecarlo(a,tau,nmax,X(irun))
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enddo
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call ave_error(X,nruns,ave,err)
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print *, 'E = ', ave, '+/-', err
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end program qmc
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#+END_SRC
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#+begin_src sh :results output :exports both
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gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
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./vmc
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#+end_src
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#+RESULTS:
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: E = -0.48584030499187431 +/- 1.0411743995438257E-004
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*** Generalized Metropolis algorithm
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:PROPERTIES:
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:header-args:python: :tangle vmc_metropolis.py
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:header-args:f90: :tangle vmc_metropolis.f90
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:END:
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Discretizing the differential equation to generate the desired
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probability density will suffer from a discretization error
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leading to biases in the averages. The [[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][Metropolis-Hastings
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sampling algorithm]] removes exactly the discretization errors, so
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large time steps can be employed.
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After the new position $\mathbf{r}_{n+1}$ has been computed, an
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additional accept/reject step is performed. The acceptance
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probability $A$ is chosen so that it is consistent with the
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probability of leaving $\mathbf{r}_n$ and the probability of
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entering $\mathbf{r}_{n+1}$:
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\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1,
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\frac{T(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
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{T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}
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\right)
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\]
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where $T(\mathbf{r}_n \rightarrow \mathbf{r}_{n+1})$ is the
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probability of transition from $\mathbf{r}_n$ to $\mathbf{r}_{n+1})$.
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In our Hydrogen atom example, $P$ is $\Psi^2$ and $g$ is a
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solution of the discretized Fokker-Planck equation:
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\begin{eqnarray*}
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P(r_{n}) &=& \Psi^2(\mathbf{r}_n) \\
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T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
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\frac{1}{(4\pi\,D\,\tau)^{3/2}} \exp \left[ - \frac{\left(
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\mathbf{r}_{n+1} - \mathbf{r}_{n} - 2D \frac{\nabla
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\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{4D\,\tau} \right]
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\end{eqnarray*}
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The accept/reject step is the following:
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- Compute $A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$.
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- Draw a uniform random number $u$
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- if $u \le A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, accept
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the move
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- if $u>A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, reject
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the move: set $\mathbf{r}_{n+1} = \mathbf{r}_{n}$, but *don't remove the sample from the average!*
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The /acceptance rate/ is the ratio of the number of accepted step
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over the total number of steps. The time step should be adapted so
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that the acceptance rate is around 0.5 for a good efficiency of
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the simulation.
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**** Exercise
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#+begin_exercise
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Modify the previous program to introduce the accept/reject step.
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You should recover the unbiased result.
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Adjust the time-step so that the acceptance rate is 0.5.
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#+end_exercise
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*Python*
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#+BEGIN_SRC python :results output
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*Python*
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#+BEGIN_SRC python :results output
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from hydrogen import *
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from qmc_stats import *
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@ -1305,8 +1170,8 @@ def MonteCarlo(a,tau,nmax):
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d_old = d_new
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d2_old = d2_new
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psi_old = psi_new
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N += 1.
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E += e_loc(a,r_old)
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N += 1.
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E += e_loc(a,r_old)
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return E/N, accep_rate/N
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@ -1317,14 +1182,14 @@ X = [MonteCarlo(a,tau,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error([x[0] for x in X])
|
||||
A, deltaA = ave_error([x[1] for x in X])
|
||||
print(f"E = {E} +/- {deltaE}\nA = {A} +/- {deltaA}")
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.4949730317138491 +/- 0.00012478601801760644
|
||||
: A = 0.7887163333333334 +/- 0.00026834549840347617
|
||||
#+RESULTS:
|
||||
: E = -0.4949730317138491 +/- 0.00012478601801760644
|
||||
: A = 0.7887163333333334 +/- 0.00026834549840347617
|
||||
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90
|
||||
subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
|
||||
implicit none
|
||||
double precision, intent(in) :: a, tau
|
||||
@ -1339,7 +1204,7 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
|
||||
double precision, external :: e_loc, psi
|
||||
|
||||
sq_tau = dsqrt(tau)
|
||||
|
||||
|
||||
! Initialization
|
||||
energy = 0.d0
|
||||
norm = 0.d0
|
||||
@ -1357,8 +1222,8 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
|
||||
psi_new = psi(a,r_new)
|
||||
! Metropolis
|
||||
prod = (d_new(1) + d_old(1))*(r_new(1) - r_old(1)) + &
|
||||
(d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + &
|
||||
(d_new(3) + d_old(3))*(r_new(3) - r_old(3))
|
||||
(d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + &
|
||||
(d_new(3) + d_old(3))*(r_new(3) - r_old(3))
|
||||
argexpo = 0.5d0 * (d2_new - d2_old)*tau + prod
|
||||
q = psi_new / psi_old
|
||||
q = dexp(-argexpo) * q*q
|
||||
@ -1396,18 +1261,17 @@ program qmc
|
||||
call ave_error(accep,nruns,ave,err)
|
||||
print *, 'A = ', 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 vmc_metropolis.f90 -o vmc_metropolis
|
||||
./vmc_metropolis
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.49499990423528023 +/- 1.5958250761863871E-004
|
||||
: A = 0.78861366666666655 +/- 3.5096729498002445E-004
|
||||
#+RESULTS:
|
||||
: E = -0.49499990423528023 +/- 1.5958250761863871E-004
|
||||
: A = 0.78861366666666655 +/- 3.5096729498002445E-004
|
||||
|
||||
|
||||
* TODO Diffusion Monte Carlo
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle dmc.py
|
||||
@ -1576,12 +1440,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
|
||||
coordinates and $\mathbf{R}_A$ and $\mathbf{R}_B$ the coordinates of
|
||||
the nuclei.
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
* Appendix
|
||||
* Appendix :noexport:
|
||||
|
||||
** Gaussian sampling :noexport:
|
||||
:PROPERTIES:
|
||||
@ -1594,49 +1454,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
|
||||
|
||||
Instead of drawing uniform random numbers, we will draw Gaussian
|
||||
random numbers centered on 0 and with a variance of 1.
|
||||
|
||||
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:
|
||||
|
||||
\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*}
|
||||
|
||||
Here is a Fortran implementation returning a Gaussian-distributed
|
||||
n-dimensional vector $\mathbf{z}$;
|
||||
|
||||
*Fortran*
|
||||
#+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 sampling probability can be inserted into the equation of the energy:
|
||||
|
||||
\[
|
||||
@ -1755,3 +1573,192 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.49517104619091717 +/- 1.0685523607878961E-004
|
||||
|
||||
** Improved sampling with $\Psi^2$ :noexport:
|
||||
|
||||
*** Importance sampling
|
||||
:PROPERTIES:
|
||||
:header-args:python: :tangle vmc.py
|
||||
:header-args:f90: :tangle vmc.f90
|
||||
:END:
|
||||
|
||||
To generate the probability density $\Psi^2$, we consider a
|
||||
diffusion process characterized by a time-dependent density
|
||||
$[\Psi(\mathbf{r},t)]^2$, which obeys the Fokker-Planck equation:
|
||||
|
||||
\[
|
||||
\frac{\partial \Psi^2}{\partial t} = \sum_i D
|
||||
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
||||
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
|
||||
[\Psi(\mathbf{r},t)]^2.
|
||||
\]
|
||||
|
||||
$D$ is the diffusion constant and $F_i$ is the i-th component of a
|
||||
drift velocity caused by an external potential. For a stationary
|
||||
density, \( \frac{\partial \Psi^2}{\partial t} = 0 \), so
|
||||
|
||||
\begin{eqnarray*}
|
||||
0 & = & \sum_i D
|
||||
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
||||
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
|
||||
[\Psi(\mathbf{r})]^2 \\
|
||||
0 & = & \sum_i D
|
||||
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
||||
\frac{\partial [\Psi(\mathbf{r})]^2}{\partial \mathbf{r}_i} -
|
||||
F_i(\mathbf{r})\,[\Psi(\mathbf{r})]^2 \right) \\
|
||||
0 & = &
|
||||
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} -
|
||||
\frac{\partial F_i }{\partial \mathbf{r}_i}[\Psi(\mathbf{r})]^2 -
|
||||
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
|
||||
\end{eqnarray*}
|
||||
|
||||
we search for a drift function which satisfies
|
||||
|
||||
\[
|
||||
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
|
||||
[\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
|
||||
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
|
||||
\]
|
||||
|
||||
to obtain a second derivative on the left, we need the drift to be
|
||||
of the form
|
||||
\[
|
||||
F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
\]
|
||||
|
||||
\[
|
||||
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
|
||||
[\Psi(\mathbf{r})]^2 \frac{\partial
|
||||
g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} +
|
||||
[\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2
|
||||
\Psi^2}{\partial \mathbf{r}_i^2} +
|
||||
\frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
\]
|
||||
|
||||
$g = 1 / \Psi^2$ satisfies this equation, so
|
||||
|
||||
\[
|
||||
F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
|
||||
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
|
||||
\]
|
||||
|
||||
In statistical mechanics, Fokker-Planck trajectories are generated
|
||||
by a Langevin equation:
|
||||
|
||||
\[
|
||||
\frac{\partial \mathbf{r}(t)}{\partial t} = 2D \frac{\nabla
|
||||
\Psi(\mathbf{r}(t))}{\Psi} + \eta
|
||||
\]
|
||||
|
||||
where $\eta$ is a normally-distributed fluctuating random force.
|
||||
|
||||
Discretizing this differential equation gives the following drifted
|
||||
diffusion scheme:
|
||||
|
||||
\[
|
||||
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau\, 2D \frac{\nabla
|
||||
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
|
||||
\]
|
||||
where $\chi$ is a Gaussian random variable with zero mean and
|
||||
variance $\tau\,2D$.
|
||||
|
||||
**** Exercise 2
|
||||
|
||||
#+begin_exercise
|
||||
Sample $\Psi^2$ approximately using the drifted diffusion scheme,
|
||||
with a diffusion constant $D=1/2$. You can use a time step of
|
||||
0.001 a.u.
|
||||
#+end_exercise
|
||||
|
||||
*Python*
|
||||
#+BEGIN_SRC python :results output
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
|
||||
def MonteCarlo(a,tau,nmax):
|
||||
sq_tau = np.sqrt(tau)
|
||||
|
||||
# Initialization
|
||||
E = 0.
|
||||
N = 0.
|
||||
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
|
||||
|
||||
for istep in range(nmax):
|
||||
d_old = drift(a,r_old)
|
||||
chi = np.random.normal(loc=0., scale=1.0, size=(3))
|
||||
r_new = r_old + tau * d_old + chi*sq_tau
|
||||
N += 1.
|
||||
E += e_loc(a,r_new)
|
||||
r_old = r_new
|
||||
return E/N
|
||||
|
||||
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
tau = 0.2
|
||||
X = [MonteCarlo(a,tau,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
#+END_SRC
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.4858534479298907 +/- 0.00010203236131158794
|
||||
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90
|
||||
subroutine variational_montecarlo(a,tau,nmax,energy)
|
||||
implicit none
|
||||
double precision, intent(in) :: a, tau
|
||||
integer*8 , intent(in) :: nmax
|
||||
double precision, intent(out) :: energy
|
||||
|
||||
integer*8 :: istep
|
||||
double precision :: norm, r_old(3), r_new(3), d_old(3), sq_tau, chi(3)
|
||||
double precision, external :: e_loc
|
||||
|
||||
sq_tau = dsqrt(tau)
|
||||
|
||||
! Initialization
|
||||
energy = 0.d0
|
||||
norm = 0.d0
|
||||
call random_gauss(r_old,3)
|
||||
|
||||
do istep = 1,nmax
|
||||
call drift(a,r_old,d_old)
|
||||
call random_gauss(chi,3)
|
||||
r_new(:) = r_old(:) + tau * d_old(:) + chi(:)*sq_tau
|
||||
norm = norm + 1.d0
|
||||
energy = energy + e_loc(a,r_new)
|
||||
r_old(:) = r_new(:)
|
||||
end do
|
||||
energy = energy / norm
|
||||
end subroutine variational_montecarlo
|
||||
|
||||
program qmc
|
||||
implicit none
|
||||
double precision, parameter :: a = 0.9
|
||||
double precision, parameter :: tau = 0.2
|
||||
integer*8 , parameter :: nmax = 100000
|
||||
integer , parameter :: nruns = 30
|
||||
|
||||
integer :: irun
|
||||
double precision :: X(nruns)
|
||||
double precision :: ave, err
|
||||
|
||||
do irun=1,nruns
|
||||
call variational_montecarlo(a,tau,nmax,X(irun))
|
||||
enddo
|
||||
call ave_error(X,nruns,ave,err)
|
||||
print *, 'E = ', ave, '+/-', err
|
||||
end program qmc
|
||||
#+END_SRC
|
||||
|
||||
#+begin_src sh :results output :exports both
|
||||
gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
|
||||
./vmc
|
||||
#+end_src
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.48584030499187431 +/- 1.0411743995438257E-004
|
||||
|
||||
|
||||
Loading…
Reference in New Issue
Block a user