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Working on VMC
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QMC.org
@ -90,9 +90,14 @@
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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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*** Exercise 1
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#+begin_exercise
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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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#+end_exercise
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$\mathbf{r}=\left( \begin{array}{c} x \\ y\\ z\end{array} \right)$, so
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$$
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@ -117,9 +122,12 @@ double precision function potential(r)
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end function potential
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#+END_SRC
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*** Write a function which computes the wave function at $\mathbf{r}$
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*** Exercise 2
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#+begin_exercise
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Write a function which computes the wave function at $\mathbf{r}$.
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The function accepts a scalar =a= and a 3-dimensional vector =r= as
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input arguments, and returns a scalar.
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#+end_exercise
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*Python*
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@ -137,9 +145,12 @@ double precision function psi(a, r)
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end function psi
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#+END_SRC
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*** Write a function which computes the local kinetic energy at $\mathbf{r}$
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*** Exercise 3
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#+begin_exercise
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Write a function which computes the local kinetic energy at $\mathbf{r}$.
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The function accepts =a= and =r= as input arguments and returns the
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local kinetic energy.
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#+end_exercise
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The local kinetic energy is defined as $$-\frac{1}{2}\frac{\Delta \Psi}{\Psi}.$$
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@ -187,9 +198,12 @@ double precision function kinetic(a,r)
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end function kinetic
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#+END_SRC
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*** Write a function which computes the local energy at $\mathbf{r}$
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*** Exercise 4
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#+begin_exercise
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Write a function which computes the local energy at $\mathbf{r}$.
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The function accepts =x,y,z= as input arguments and returns the
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local energy.
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#+end_exercise
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$$
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E_L(\mathbf{r}) = -\frac{1}{2} \frac{\Delta \Psi}{\Psi} (\mathbf{r}) + V(\mathbf{r})
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@ -218,11 +232,15 @@ end function e_loc
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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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*Python*
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#+BEGIN_SRC python :results none
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*** Exercise
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#+begin_exercise
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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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#+end_exercise
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*Python*
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#+BEGIN_SRC python :results none
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import numpy as np
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import matplotlib.pyplot as plt
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@ -244,16 +262,16 @@ 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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*Fortran*
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#+begin_src f90
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*Fortran*
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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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@ -282,20 +300,20 @@ 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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@ -305,12 +323,12 @@ 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 expectation value of the energy
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** Numerical estimation of the energy
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:PROPERTIES:
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:header-args:python: :tangle energy_hydrogen.py
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:header-args:f90: :tangle energy_hydrogen.f90
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@ -327,18 +345,22 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
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w_i = \left[\Psi(\mathbf{r}_i)\right]^2 \delta \mathbf{r}
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$$
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In this section, we will compute a numerical estimate of the
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energy in a grid of $50\times50\times50$ points in the range
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$(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
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#+begin_note
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The energy is biased because:
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- The volume elements are not infinitely small (discretization error)
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- The energy is evaluated only inside the box (incompleteness of the space)
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#+end_note
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*Python*
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#+BEGIN_SRC python :results none
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*** Exercise
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#+begin_exercise
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Compute a numerical estimate of the energy in a grid of
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$50\times50\times50$ points in the range $(-5,-5,-5) \le
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\mathbf{r} \le (5,5,5)$.
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#+end_exercise
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*Python*
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#+BEGIN_SRC python :results none
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import numpy as np
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from hydrogen import e_loc, psi
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@ -363,19 +385,19 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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E = E / norm
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print(f"a = {a} \t E = {E}")
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#+end_src
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#+end_src
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#+RESULTS:
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: a = 0.1 E = -0.24518438948809218
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: a = 0.2 E = -0.26966057967803525
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: a = 0.5 E = -0.3856357612517407
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: a = 0.9 E = -0.49435709786716214
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: a = 1.0 E = -0.5
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: a = 1.5 E = -0.39242967082602226
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: a = 2.0 E = -0.08086980667844901
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#+RESULTS:
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: a = 0.1 E = -0.24518438948809218
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: a = 0.2 E = -0.26966057967803525
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: a = 0.5 E = -0.3856357612517407
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: a = 0.9 E = -0.49435709786716214
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: a = 1.0 E = -0.5
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: a = 1.5 E = -0.39242967082602226
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: a = 2.0 E = -0.08086980667844901
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*Fortran*
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#+begin_src f90
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*Fortran*
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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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@ -414,24 +436,24 @@ 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 the Fortran and run it:
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To compile the Fortran and run it:
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#+begin_src sh :results output :exports both
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#+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
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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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** Compute the variance of the local energy
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** Variance of the local energy
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:PROPERTIES:
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:header-args:python: :tangle variance_hydrogen.py
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:header-args:f90: :tangle variance_hydrogen.f90
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@ -450,8 +472,13 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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$\hat{H}$) the variance is zero, so the variance of the local
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energy can be used as a measure of the quality of a wave function.
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*** Exercise
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#+begin_exercise
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Compute a numerical estimate of the variance of the local energy
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in a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le \mathbf{r} \le (5,5,5)$.
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in a grid of $50\times50\times50$ points in the range
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$(-5,-5,-5)
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\le \mathbf{r} \le (5,5,5)$ for different values of $a$.
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#+end_exercise
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*Python*
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#+begin_src python :results none
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@ -617,8 +644,11 @@ gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
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E \pm \delta E, \text{ where } \delta E = \frac{\sigma}{\sqrt{M}}
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$$
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*** Exercise
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#+begin_exercise
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Write a function returning the average and statistical error of an
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input array.
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#+end_exercise
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*Python*
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#+BEGIN_SRC python :results none
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@ -656,26 +686,38 @@ end subroutine ave_error
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:header-args:f90: :tangle qmc_uniform.f90
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:END:
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In this section we write a function to perform a Monte Carlo
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calculation of the average energy.
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We will now do our first Monte Carlo calculation to compute the
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energy of the hydrogen atom.
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At every Monte Carlo step:
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- Draw 3 uniform random numbers in the interval $(-5,-5,-5) \le
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- Draw a random point $\mathbf{r}_i$ in the box $(-5,-5,-5) \le
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(x,y,z) \le (5,5,5)$
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- Compute $\Psi^2 \times E_L$ at this point and accumulate the
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result in E
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- Compute $\Psi^2$ at this point and accumulate the result in N
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- Compute $[\Psi(\mathbf{r}_i)]^2$ and accumulate the result in a
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variable =normalization=
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- Compute $[\Psi(\mathbf{r}_i)]^2 \times E_L(\mathbf{r}_i)$, and accumulate the
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result in a variable =energy=
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Once all the steps have been computed, return the average energy
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computed on the Monte Carlo calculation.
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One Monte Carlo run will consist of $N$ Monte Carlo steps. Once all the
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steps have been computed, the run returns the average energy
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$\bar{E}_k$ over the $N$ steps of the run.
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In the main program, write a loop to perform 30 Monte Carlo runs,
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and compute the average energy and the associated statistical error.
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To compute the statistical error, perform $M$ runs. The final
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estimate of the energy will be the average over the $\bar{E}_k$,
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and the variance of the $\bar{E}_k$ will be used to compute the
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statistical error.
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Compute the energy of the wave function with $a=0.9$.
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*** Exercise
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*Python*
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#+BEGIN_SRC python :results output
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#+begin_exercise
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Parameterize the wave function with $a=0.9$. Perform 30
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independent Monte Carlo runs, each with 100 000 Monte Carlo
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steps. Store the final energies of each run and use this array to
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compute the average energy and the associated error bar.
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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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@ -695,13 +737,13 @@ nmax = 100000
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X = [MonteCarlo(a,nmax) for i in range(30)]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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#+END_SRC
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#+END_SRC
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#+RESULTS:
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: E = -0.4956255109300764 +/- 0.0007082875482711226
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#+RESULTS:
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: E = -0.4956255109300764 +/- 0.0007082875482711226
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*Fortran*
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#+BEGIN_SRC f90
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*Fortran*
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#+BEGIN_SRC f90
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subroutine uniform_montecarlo(a,nmax,energy)
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implicit none
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double precision, intent(in) :: a
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@ -743,15 +785,15 @@ program qmc
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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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#+END_SRC
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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 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
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./qmc_uniform
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#+end_src
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#+end_src
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#+RESULTS:
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: E = -0.49588321986667677 +/- 7.1758863546737969E-004
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#+RESULTS:
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: E = -0.49588321986667677 +/- 7.1758863546737969E-004
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** Gaussian sampling
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:PROPERTIES:
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@ -773,6 +815,9 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
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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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Here is a Fortran implementation returning a Gaussian-distributed
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n-dimensional vector $\mathbf{z}$;
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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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@ -805,14 +850,16 @@ end subroutine random_gauss
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#+END_SRC
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Now the equation for the energy is changed into
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Now the sampling probability can be inserted into the equation of the energy:
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\[
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E = \frac{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r})\right]^2}{P(\mathbf{r})}\, \frac{\hat{H} \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\,d\mathbf{r}}{\int P(\mathbf{r}) \frac{\left[\Psi(\mathbf{r}) \right]^2}{P(\mathbf{r})} d\mathbf{r}}
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\]
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with
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with the Gaussian probability
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\[
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P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right)
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P(\mathbf{r}) = \frac{1}{(2 \pi)^{3/2}}\exp\left( -\frac{\mathbf{r}^2}{2} \right).
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\]
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As the coordinates are drawn with probability $P(\mathbf{r})$, the
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@ -823,8 +870,17 @@ end subroutine random_gauss
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w_i = \frac{\left[\Psi(\mathbf{r}_i)\right]^2}{P(\mathbf{r}_i)} \delta \mathbf{r}
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$$
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*Python*
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#+BEGIN_SRC python :results output
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*** Exercise
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#+begin_exercise
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Modify the exercise of the previous section to sample with
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Gaussian-distributed random numbers. Can you see an reduction in
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the statistical error?
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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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@ -848,14 +904,14 @@ nmax = 100000
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X = [MonteCarlo(a,nmax) for i in range(30)]
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E, deltaE = ave_error(X)
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print(f"E = {E} +/- {deltaE}")
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#+END_SRC
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#+END_SRC
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#+RESULTS:
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: E = -0.49507506093129827 +/- 0.00014164037765553668
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#+RESULTS:
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: E = -0.49507506093129827 +/- 0.00014164037765553668
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*Fortran*
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#+BEGIN_SRC f90
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*Fortran*
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#+BEGIN_SRC f90
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double precision function gaussian(r)
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implicit none
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double precision, intent(in) :: r(3)
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@ -904,15 +960,15 @@ program qmc
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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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#+END_SRC
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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 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
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./qmc_gaussian
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#+end_src
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#+end_src
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#+RESULTS:
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: E = -0.49606057056767766 +/- 1.3918807547836872E-004
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#+RESULTS:
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: E = -0.49606057056767766 +/- 1.3918807547836872E-004
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** Sampling with $\Psi^2$
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:PROPERTIES:
|
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@ -933,88 +989,105 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_gaussian.f90 -o qmc_gaussian
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E \approx \frac{1}{M}\sum_{i=1}^M E_L(\mathbf{r}_i)
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$$
|
||||
|
||||
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:
|
||||
|
||||
\[
|
||||
\frac{\partial \Psi^2}{\partial t} = \sum_i D
|
||||
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
||||
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
|
||||
[\Psi(\mathbf{r},t)]^2.
|
||||
\]
|
||||
*** Importance sampling
|
||||
|
||||
$D$ 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
|
||||
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:
|
||||
|
||||
\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*}
|
||||
\[
|
||||
\frac{\partial \Psi^2}{\partial t} = \sum_i D
|
||||
\frac{\partial}{\partial \mathbf{r}_i} \left(
|
||||
\frac{\partial}{\partial \mathbf{r}_i} - F_i(\mathbf{r}) \right)
|
||||
[\Psi(\mathbf{r},t)]^2.
|
||||
\]
|
||||
|
||||
we search for a drift function which satisfies
|
||||
$D$ 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
|
||||
|
||||
\[
|
||||
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
|
||||
[\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
|
||||
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
|
||||
\]
|
||||
\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*}
|
||||
|
||||
to obtain a second derivative on the left, we need the drift to be
|
||||
of the form
|
||||
\[
|
||||
F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
\]
|
||||
we search for a drift function which satisfies
|
||||
|
||||
\[
|
||||
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
|
||||
[\Psi(\mathbf{r})]^2 \frac{\partial
|
||||
g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} +
|
||||
[\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2
|
||||
\Psi^2}{\partial \mathbf{r}_i^2} +
|
||||
\frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
\]
|
||||
\[
|
||||
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
|
||||
[\Psi(\mathbf{r})]^2 \frac{\partial F_i }{\partial \mathbf{r}_i} +
|
||||
\frac{\partial \Psi^2}{\partial \mathbf{r}_i} F_i(\mathbf{r})
|
||||
\]
|
||||
|
||||
$g = 1 / \Psi^2$ satisfies this equation, so
|
||||
to obtain a second derivative on the left, we need the drift to be
|
||||
of the form
|
||||
\[
|
||||
F_i(\mathbf{r}) = g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
\]
|
||||
|
||||
\[
|
||||
F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
|
||||
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
|
||||
\]
|
||||
\[
|
||||
\frac{\partial^2 \Psi^2}{\partial \mathbf{r}_i^2} =
|
||||
[\Psi(\mathbf{r})]^2 \frac{\partial
|
||||
g(\mathbf{r})}{\partial \mathbf{r}_i}\frac{\partial \Psi^2}{\partial \mathbf{r}_i} +
|
||||
[\Psi(\mathbf{r})]^2 g(\mathbf{r}) \frac{\partial^2
|
||||
\Psi^2}{\partial \mathbf{r}_i^2} +
|
||||
\frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
g(\mathbf{r}) \frac{\partial \Psi^2}{\partial \mathbf{r}_i}
|
||||
\]
|
||||
|
||||
following drifted diffusion scheme:
|
||||
$g = 1 / \Psi^2$ satisfies this equation, so
|
||||
|
||||
\[
|
||||
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
|
||||
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \eta \sqrt{\tau}
|
||||
\]
|
||||
\[
|
||||
F(\mathbf{r}) = \frac{\nabla [\Psi(\mathbf{r})]^2}{[\Psi(\mathbf{r})]^2} = 2 \frac{\nabla
|
||||
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} = 2 \nabla \left( \log \Psi(\mathbf{r}) \right)
|
||||
\]
|
||||
|
||||
where $\eta$ is a normally-distributed Gaussian random number.
|
||||
In statistical mechanics, Fokker-Planck trajectories are generated
|
||||
by a Langevin equation:
|
||||
|
||||
\[
|
||||
\frac{\partial \mathbf{r}(t)}{\partial t} = 2D \frac{\nabla
|
||||
\Psi(\mathbf{r}(t))}{\Psi} + \eta
|
||||
\]
|
||||
|
||||
First, write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
|
||||
where $\eta$ is a normally-distributed fluctuating random force.
|
||||
|
||||
*Python*
|
||||
#+BEGIN_SRC python
|
||||
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 1
|
||||
|
||||
#+begin_exercise
|
||||
Write a function to compute the drift vector $\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}$.
|
||||
#+end_exercise
|
||||
|
||||
*Python*
|
||||
#+BEGIN_SRC python :tangle hydrogen.py
|
||||
def drift(a,r):
|
||||
ar_inv = -a/np.sqrt(np.dot(r,r))
|
||||
return r * ar_inv
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90 :tangle hydrogen.f90
|
||||
subroutine drift(a,r,b)
|
||||
implicit none
|
||||
double precision, intent(in) :: a, r(3)
|
||||
@ -1023,54 +1096,50 @@ subroutine drift(a,r,b)
|
||||
ar_inv = -a / dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3))
|
||||
b(:) = r(:) * ar_inv
|
||||
end subroutine drift
|
||||
#+END_SRC
|
||||
#+END_SRC
|
||||
|
||||
**** TODO Exercise 2
|
||||
|
||||
Now we can write the Monte Carlo sampling:
|
||||
#+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
|
||||
*Python*
|
||||
#+BEGIN_SRC python :results output :tangle vmc.py
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
def MonteCarlo(a,tau,nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
sq_tau = sqrt(tau)
|
||||
sq_tau = np.sqrt(tau)
|
||||
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
|
||||
d_old = drift(a,r_old)
|
||||
d2_old = np.dot(d_old,d_old)
|
||||
psi_old = psi(a,r_old)
|
||||
for istep in range(nmax):
|
||||
eta = np.random.normal(loc=0., scale=1.0, size=(3))
|
||||
r_new = r_old + tau * d_old + sq_tau * eta
|
||||
d_new = drift(a,r_new)
|
||||
d2_new = np.dot(d_new,d_new)
|
||||
psi_new = psi(a,r_new)
|
||||
# Metropolis
|
||||
prod = np.dot((d_new + d_old), (r_new - r_old))
|
||||
argexpo = 0.5 * (d2_new - d2_old)*tau + prod
|
||||
q = psi_new / psi_old
|
||||
q = np.exp(-argexpo) * q*q
|
||||
if np.random.uniform() < q:
|
||||
r_old = r_new
|
||||
d_old = d_new
|
||||
d2_old = d2_new
|
||||
psi_old = psi_new
|
||||
N += 1.
|
||||
E += e_loc(a,r_old)
|
||||
chi = np.random.normal(loc=0., scale=1.0, size=(3))
|
||||
r_new = r_old + tau * d_old + sq_tau * chi
|
||||
N += 1.
|
||||
E += e_loc(a,r_new)
|
||||
r_old = r_new
|
||||
return E/N
|
||||
|
||||
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
tau = 0.1
|
||||
tau = 0.001
|
||||
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 = -0.4112049153828464 +/- 0.00027934927432953063
|
||||
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90
|
||||
subroutine variational_montecarlo(a,nmax,energy)
|
||||
implicit none
|
||||
double precision, intent(in) :: a
|
||||
@ -1111,15 +1180,153 @@ 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 vmc.f90 -o vmc
|
||||
./vmc
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
*** Metropolis algorithm
|
||||
|
||||
Discretizing the differential equation to generate the desired
|
||||
probability density will suffer from a discretization error
|
||||
leading to biases in the averages. The [[https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm][Metropolis-Hastings
|
||||
sampling algorithm]] removes exactly the discretization errors, so
|
||||
large time steps can be employed.
|
||||
|
||||
After the new position $\mathbf{r}_{n+1}$ has been computed, an
|
||||
additional accept/reject step is performed. The acceptance
|
||||
probability $A$ is chosen so that it is consistent with the
|
||||
probability of leaving $\mathbf{r}_n$ and the probability of
|
||||
entering $\mathbf{r}_{n+1}$:
|
||||
|
||||
\[ A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) = \min \left( 1,
|
||||
\frac{g(\mathbf{r}_{n+1} \rightarrow \mathbf{r}_{n}) P(\mathbf{r}_{n+1})}
|
||||
{g(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) P(\mathbf{r}_{n})}
|
||||
\right)
|
||||
\]
|
||||
|
||||
In our Hydrogen atom example, $P$ is $\Psi^2$ and $g$ is a
|
||||
solution of the discretized Fokker-Planck equation:
|
||||
|
||||
\begin{eqnarray*}
|
||||
P(r_{n}) &=& \Psi^2(\mathbf{r}_n) \\
|
||||
g(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) & = &
|
||||
\frac{1}{(4\pi\,D\,\tau)^{3/2}} \exp \left[ - \frac{\left(
|
||||
\mathbf{r}_{n+1} - \mathbf{r}_{n} - 2D \frac{\nabla
|
||||
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{4D\,\tau} \right]
|
||||
\end{eqnarray*}
|
||||
|
||||
The accept/reject step is the following:
|
||||
- Compute $A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$.
|
||||
- Draw a uniform random number $u$
|
||||
- if $u \le A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, accept
|
||||
the move
|
||||
- if $u>A(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1})$, reject
|
||||
the move: set $\mathbf{r}_{n+1} = \mathbf{r}_{n}$, but *don't
|
||||
remove the sample from the average!*
|
||||
|
||||
|
||||
* Diffusion Monte Carlo
|
||||
**** TODO Exercise
|
||||
|
||||
#+begin_exercise
|
||||
Modify the previous program to introduce the accept/reject step.
|
||||
You should recover the unbiased result.
|
||||
#+end_exercise
|
||||
|
||||
*Python*
|
||||
#+BEGIN_SRC python
|
||||
def MonteCarlo(a,tau,nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
sq_tau = np.sqrt(tau)
|
||||
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
|
||||
d_old = drift(a,r_old)
|
||||
d2_old = np.dot(d_old,d_old)
|
||||
psi_old = psi(a,r_old)
|
||||
for istep in range(nmax):
|
||||
eta = np.random.normal(loc=0., scale=1.0, size=(3))
|
||||
r_new = r_old + tau * d_old + sq_tau * eta
|
||||
d_new = drift(a,r_new)
|
||||
d2_new = np.dot(d_new,d_new)
|
||||
psi_new = psi(a,r_new)
|
||||
# Metropolis
|
||||
prod = np.dot((d_new + d_old), (r_new - r_old))
|
||||
argexpo = 0.5 * (d2_new - d2_old)*tau + prod
|
||||
q = psi_new / psi_old
|
||||
q = np.exp(-argexpo) * q*q
|
||||
if np.random.uniform() < q:
|
||||
r_old = r_new
|
||||
d_old = d_new
|
||||
d2_old = d2_new
|
||||
psi_old = psi_new
|
||||
N += 1.
|
||||
E += e_loc(a,r_old)
|
||||
return E/N
|
||||
|
||||
|
||||
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
|
||||
|
||||
#+RESULTS:
|
||||
: E = -0.4951783346213532 +/- 0.00022067316984271938
|
||||
|
||||
*Fortran*
|
||||
#+BEGIN_SRC f90
|
||||
subroutine variational_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 variational_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 vmc.f90 -o vmc
|
||||
./vmc
|
||||
#+end_src
|
||||
|
||||
|
||||
* TODO Diffusion Monte Carlo
|
||||
|
||||
We will now consider the H_2 molecule in a minimal basis composed of the
|
||||
$1s$ orbitals of the hydrogen atoms:
|
||||
@ -1133,3 +1340,4 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
|
||||
the nuclei.
|
||||
|
||||
|
||||
|
||||
|
||||
@ -26,7 +26,6 @@ program energy_hydrogen
|
||||
r(3) = x(l)
|
||||
w = psi(a(j),r)
|
||||
w = w * w * delta
|
||||
|
||||
energy = energy + w * e_loc(a(j), r)
|
||||
norm = norm + w
|
||||
end do
|
||||
|
||||
@ -1,23 +1,23 @@
|
||||
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.]:
|
||||
E = 0.
|
||||
norm = 0.
|
||||
for x in interval:
|
||||
r[0] = x
|
||||
for y in interval:
|
||||
r[1] = y
|
||||
for z in interval:
|
||||
r[2] = z
|
||||
w = psi(a,r)
|
||||
w = w * w * delta
|
||||
E += w * e_loc(a,r)
|
||||
norm += w
|
||||
E = E / norm
|
||||
print(f"a = {a} \t E = {E}")
|
||||
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
E = 0.
|
||||
norm = 0.
|
||||
for x in interval:
|
||||
r[0] = x
|
||||
for y in interval:
|
||||
r[1] = y
|
||||
for z in interval:
|
||||
r[2] = z
|
||||
w = psi(a,r)
|
||||
w = w * w * delta
|
||||
E += w * e_loc(a,r)
|
||||
norm += w
|
||||
E = E / norm
|
||||
print(f"a = {a} \t E = {E}")
|
||||
|
||||
@ -23,3 +23,12 @@ double precision function e_loc(a,r)
|
||||
double precision, external :: kinetic, potential
|
||||
e_loc = kinetic(a,r) + potential(r)
|
||||
end function e_loc
|
||||
|
||||
subroutine drift(a,r,b)
|
||||
implicit none
|
||||
double precision, intent(in) :: a, r(3)
|
||||
double precision, intent(out) :: b(3)
|
||||
double precision :: ar_inv
|
||||
ar_inv = -a / dsqrt(r(1)*r(1) + r(2)*r(2) + r(3)*r(3))
|
||||
b(:) = r(:) * ar_inv
|
||||
end subroutine drift
|
||||
|
||||
@ -11,3 +11,7 @@ def kinetic(a,r):
|
||||
|
||||
def e_loc(a,r):
|
||||
return kinetic(a,r) + potential(r)
|
||||
|
||||
def drift(a,r):
|
||||
ar_inv = -a/np.sqrt(np.dot(r,r))
|
||||
return r * ar_inv
|
||||
|
||||
@ -1,19 +1,19 @@
|
||||
from hydrogen import *
|
||||
from qmc_stats import *
|
||||
|
||||
def MonteCarlo(a, nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
for istep in range(nmax):
|
||||
r = np.random.uniform(-5., 5., (3))
|
||||
w = psi(a,r)
|
||||
w = w*w
|
||||
N += w
|
||||
E += w * e_loc(a,r)
|
||||
return E/N
|
||||
def MonteCarlo(a, nmax):
|
||||
E = 0.
|
||||
N = 0.
|
||||
for istep in range(nmax):
|
||||
r = np.random.uniform(-5., 5., (3))
|
||||
w = psi(a,r)
|
||||
w = w*w
|
||||
N += w
|
||||
E += w * e_loc(a,r)
|
||||
return E/N
|
||||
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
X = [MonteCarlo(a,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
a = 0.9
|
||||
nmax = 100000
|
||||
X = [MonteCarlo(a,nmax) for i in range(30)]
|
||||
E, deltaE = ave_error(X)
|
||||
print(f"E = {E} +/- {deltaE}")
|
||||
|
||||
@ -26,7 +26,6 @@ program variance_hydrogen
|
||||
r(3) = x(l)
|
||||
w = psi(a(j),r)
|
||||
w = w * w * delta
|
||||
|
||||
energy = energy + w * e_loc(a(j), r)
|
||||
norm = norm + w
|
||||
end do
|
||||
@ -44,7 +43,6 @@ program variance_hydrogen
|
||||
r(3) = x(l)
|
||||
w = psi(a(j),r)
|
||||
w = w * w * delta
|
||||
|
||||
s2 = s2 + w * ( e_loc(a(j), r) - energy )**2
|
||||
norm = norm + w
|
||||
end do
|
||||
|
||||
@ -20,8 +20,8 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
El = e_loc(a, r)
|
||||
E += w * El
|
||||
norm += w
|
||||
E = E / norm
|
||||
s2 = 0.
|
||||
E = E / norm
|
||||
s2 = 0.
|
||||
for x in interval:
|
||||
r[0] = x
|
||||
for y in interval:
|
||||
@ -32,5 +32,5 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
|
||||
w = w * w * delta
|
||||
El = e_loc(a, r)
|
||||
s2 += w * (El - E)**2
|
||||
s2 = s2 / norm
|
||||
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
|
||||
s2 = s2 / norm
|
||||
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
|
||||
|
||||
Loading…
Reference in New Issue
Block a user