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OK up to VMC
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QMC.org
@ -41,24 +41,20 @@
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computes a statistical estimate of the expectation value of the energy
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associated with a given wave function.
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Finally, we introduce the diffusion Monte Carlo (DMC) method which
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gives the exact energy of the $H_2$ molecule.
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gives the exact energy of the hydrogen atom and of the $H_2$ molecule.
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Code examples will be given in Python and Fortran. Whatever language
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can be chosen.
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Code examples will be given in Python and Fortran. You can use
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whatever language you prefer to write the program.
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We consider the stationary solution of the Schrödinger equation, so
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the wave functions considered here are real: for an $N$ electron
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system where the electrons move in the 3-dimensional space,
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$\Psi : \mathbb{R}^{3N} \rightarrow \mathbb{R}$. In addition, $\Psi$
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is defined everywhere, continuous and infinitely differentiable.
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*Note*
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#+begin_important
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In Fortran, when you use a double precision constant, don't forget
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to put ~d0~ as a suffix (for example ~2.0d0~), or it will be
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interpreted as a single precision value
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#+end_important
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All the quantities are expressed in /atomic units/ (energies,
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coordinates, etc).
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* Numerical evaluation of the energy
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@ -180,7 +176,7 @@ end function potential
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**** Python
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#+BEGIN_SRC python :results none
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#+BEGIN_SRC python :results none :tangle none
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def psi(a, r):
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# TODO
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#+END_SRC
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@ -192,7 +188,7 @@ def psi(a, r):
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#+END_SRC
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**** Fortran
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#+BEGIN_SRC f90
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#+BEGIN_SRC f90 :tangle none
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double precision function psi(a, r)
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implicit none
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double precision, intent(in) :: a, r(3)
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@ -247,7 +243,7 @@ end function psi
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$$
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**** Python
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#+BEGIN_SRC python :results none
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#+BEGIN_SRC python :results none :tangle none
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def kinetic(a,r):
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# TODO
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#+END_SRC
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@ -259,7 +255,7 @@ def kinetic(a,r):
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#+END_SRC
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**** Fortran
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#+BEGIN_SRC f90
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#+BEGIN_SRC f90 :tangle none
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double precision function kinetic(a,r)
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implicit none
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double precision, intent(in) :: a, r(3)
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@ -291,7 +287,7 @@ end function kinetic
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**** Python
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#+BEGIN_SRC python :results none
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#+BEGIN_SRC python :results none :tangle none
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def e_loc(a,r):
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#TODO
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#+END_SRC
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@ -303,7 +299,7 @@ def e_loc(a,r):
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#+END_SRC
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**** Fortran
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#+BEGIN_SRC f90
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#+BEGIN_SRC f90 :tangle none
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double precision function e_loc(a,r)
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implicit none
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double precision, intent(in) :: a, r(3)
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@ -337,7 +333,7 @@ end function e_loc
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#+end_exercise
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**** Python
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#+BEGIN_SRC python :results none
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#+BEGIN_SRC python :results none :tangle none
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import numpy as np
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import matplotlib.pyplot as plt
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@ -366,7 +362,7 @@ plt.figure(figsize=(10,5))
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for a in [0.1, 0.2, 0.5, 1., 1.5, 2.]:
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y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
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plt.plot(x,y,label=f"a={a}")
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plt.tight_layout()
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plt.legend()
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plt.savefig("plot_py.png")
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@ -377,7 +373,7 @@ plt.savefig("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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#+begin_src f90 :tangle none
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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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@ -393,18 +389,18 @@ program plot
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! TODO
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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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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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@ -414,10 +410,10 @@ 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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**** Fortran :solution:
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#+begin_src f90
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#+begin_src f90
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program plot
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implicit none
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double precision, external :: e_loc
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@ -446,20 +442,20 @@ 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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@ -469,12 +465,12 @@ 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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** TODO Numerical estimation 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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@ -505,8 +501,24 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
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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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**** Python
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#+BEGIN_SRC python :results none :tangle none
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import numpy as np
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from hydrogen import e_loc, psi
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interval = np.linspace(-5,5,num=50)
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delta = (interval[1]-interval[0])**3
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r = np.array([0.,0.,0.])
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for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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# TODO
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print(f"a = {a} \t E = {E}")
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#+end_src
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**** Python :solution:
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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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@ -518,32 +530,62 @@ r = np.array([0.,0.,0.])
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for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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E = 0.
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norm = 0.
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for x in interval:
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r[0] = x
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for y in interval:
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r[1] = y
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for z in interval:
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r[2] = z
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w = psi(a,r)
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w = w * w * delta
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E += w * e_loc(a,r)
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norm += w
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for x in interval:
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r[0] = x
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for y in interval:
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r[1] = y
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for z in interval:
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r[2] = z
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w = psi(a,r)
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w = w * w * delta
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E += w * e_loc(a,r)
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norm += w
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E = E / norm
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print(f"a = {a} \t E = {E}")
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#+end_src
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#+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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double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm
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integer :: i, k, l, j
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a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
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dx = 10.d0/(size(x)-1)
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do i=1,size(x)
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x(i) = -5.d0 + (i-1)*dx
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end do
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do j=1,size(a)
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! TODO
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print *, 'a = ', a(j), ' E = ', energy
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end do
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end program energy_hydrogen
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#+end_src
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To compile the Fortran and run it:
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#+begin_src sh :results output :exports both
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gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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./energy_hydrogen
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#+end_src
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**** Fortran :solution:
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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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@ -582,24 +624,24 @@ 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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** TODO 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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@ -607,7 +649,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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The variance of the local energy is a functional of $\Psi$
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which measures the magnitude of the fluctuations of the local
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energy associated with $\Psi$ around the average:
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energy associated with $\Psi$ around its average:
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$$
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\sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
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@ -615,7 +657,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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$$
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which can be simplified as
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$$ \sigma^2(E_L) = \langle E_L^2 \rangle - \langle E_L \rangle^2 $$
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$$ \sigma^2(E_L) = \langle E_L^2 \rangle_{\Psi^2} - \langle E_L \rangle_{\Psi^2}^2.$$
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If the local energy is constant (i.e. $\Psi$ is an eigenfunction of
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$\hat{H}$) the variance is zero, so the variance of the local
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@ -624,7 +666,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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*** Exercise (optional)
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#+begin_exercise
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Prove that :
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$$\langle E - \langle E \rangle \rangle^2 = \langle E^2 \rangle - \langle E \rangle^2 $$
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$$\left( \langle E - \langle E \rangle_{\Psi^2} \rangle_{\Psi^2} \right)^2 = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$
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#+end_exercise
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*** Exercise
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@ -636,8 +678,23 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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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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**** Python
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#+begin_src python :results none :tangle none
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import numpy as np
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from hydrogen import e_loc, psi
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interval = np.linspace(-5,5,num=50)
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delta = (interval[1]-interval[0])**3
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r = np.array([0.,0.,0.])
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for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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# TODO
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print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
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#+end_src
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**** Python :solution:
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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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@ -666,19 +723,54 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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E2 = E2 / norm
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s2 = E2 - E*E
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print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
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#+end_src
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#+end_src
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#+RESULTS:
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: a = 0.1 E = -0.24518439 \sigma^2 = 0.02696522
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: a = 0.2 E = -0.26966058 \sigma^2 = 0.03719707
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: a = 0.5 E = -0.38563576 \sigma^2 = 0.05318597
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: a = 0.9 E = -0.49435710 \sigma^2 = 0.00577812
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: a = 1.0 E = -0.50000000 \sigma^2 = 0.00000000
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: a = 1.5 E = -0.39242967 \sigma^2 = 0.31449671
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: a = 2.0 E = -0.08086981 \sigma^2 = 1.80688143
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#+RESULTS:
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: a = 0.1 E = -0.24518439 \sigma^2 = 0.02696522
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: a = 0.2 E = -0.26966058 \sigma^2 = 0.03719707
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: a = 0.5 E = -0.38563576 \sigma^2 = 0.05318597
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: a = 0.9 E = -0.49435710 \sigma^2 = 0.00577812
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: a = 1.0 E = -0.50000000 \sigma^2 = 0.00000000
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: a = 1.5 E = -0.39242967 \sigma^2 = 0.31449671
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: a = 2.0 E = -0.08086981 \sigma^2 = 1.80688143
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**** Fortran
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#+begin_src f90 :tangle none
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program variance_hydrogen
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implicit none
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double precision, external :: e_loc, psi
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double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm, s2
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double precision :: e, energy2
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integer :: i, k, l, j
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*Fortran*
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||||
#+begin_src f90
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a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
|
||||
|
||||
dx = 10.d0/(size(x)-1)
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||||
do i=1,size(x)
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||||
x(i) = -5.d0 + (i-1)*dx
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||||
end do
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||||
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||||
delta = dx**3
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||||
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||||
r(:) = 0.d0
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||||
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||||
do j=1,size(a)
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! TODO
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||||
print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
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||||
end do
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||||
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||||
end program variance_hydrogen
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#+end_src
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||||
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To compile and run:
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||||
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#+begin_src sh :results output :exports both
|
||||
gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
|
||||
./variance_hydrogen
|
||||
#+end_src
|
||||
|
||||
**** Fortran :solution:
|
||||
#+begin_src f90
|
||||
program variance_hydrogen
|
||||
implicit none
|
||||
double precision, external :: e_loc, psi
|
||||
@ -723,22 +815,22 @@ program variance_hydrogen
|
||||
end do
|
||||
|
||||
end program variance_hydrogen
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
To compile and run:
|
||||
To compile and run:
|
||||
|
||||
#+begin_src sh :results output :exports both
|
||||
#+begin_src sh :results output :exports both
|
||||
gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
|
||||
./variance_hydrogen
|
||||
#+end_src
|
||||
#+end_src
|
||||
|
||||
#+RESULTS:
|
||||
: a = 0.10000000000000001 E = -0.24518438948809140 s2 = 2.6965218719722767E-002
|
||||
: a = 0.20000000000000001 E = -0.26966057967803236 s2 = 3.7197072370201284E-002
|
||||
: a = 0.50000000000000000 E = -0.38563576125173815 s2 = 5.3185967578480653E-002
|
||||
: a = 1.0000000000000000 E = -0.50000000000000000 s2 = 0.0000000000000000
|
||||
: a = 1.5000000000000000 E = -0.39242967082602065 s2 = 0.31449670909172917
|
||||
: a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814270846534
|
||||
#+RESULTS:
|
||||
: a = 0.10000000000000001 E = -0.24518438948809140 s2 = 2.6965218719722767E-002
|
||||
: a = 0.20000000000000001 E = -0.26966057967803236 s2 = 3.7197072370201284E-002
|
||||
: a = 0.50000000000000000 E = -0.38563576125173815 s2 = 5.3185967578480653E-002
|
||||
: a = 1.0000000000000000 E = -0.50000000000000000 s2 = 0.0000000000000000
|
||||
: a = 1.5000000000000000 E = -0.39242967082602065 s2 = 0.31449670909172917
|
||||
: a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814270846534
|
||||
|
||||
|
||||
* TODO Variational Monte Carlo
|
||||
@ -1897,3 +1989,12 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
|
||||
#+RESULTS:
|
||||
: E = -0.48584030499187431 +/- 1.0411743995438257E-004
|
||||
|
||||
|
||||
* TODO [0/1] Last things to do
|
||||
|
||||
- [ ] Prepare 4 questions for the exam: multiple-choice questions
|
||||
with 4 possible answers. Questions should be independent because
|
||||
they will be asked in a random order.
|
||||
- [ ] Propose a project for the students to continue the
|
||||
programs. Idea: Modify the program to compute the exact energy of
|
||||
the H$_2$ molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.
|
||||
|
||||
@ -1,3 +1,23 @@
|
||||
program energy_hydrogen
|
||||
implicit none
|
||||
double precision, external :: e_loc, psi
|
||||
double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm
|
||||
integer :: i, k, l, j
|
||||
|
||||
a = (/ 0.1d0, 0.2d0, 0.5d0, 1.d0, 1.5d0, 2.d0 /)
|
||||
|
||||
dx = 10.d0/(size(x)-1)
|
||||
do i=1,size(x)
|
||||
x(i) = -5.d0 + (i-1)*dx
|
||||
end do
|
||||
|
||||
do j=1,size(a)
|
||||
! TODO
|
||||
print *, 'a = ', a(j), ' E = ', energy
|
||||
end do
|
||||
|
||||
end program energy_hydrogen
|
||||
|
||||
program energy_hydrogen
|
||||
implicit none
|
||||
double precision, external :: e_loc, psi
|
||||
|
||||
@ -9,15 +9,15 @@ 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
|
||||
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}")
|
||||
|
||||
@ -4,18 +4,12 @@ import matplotlib.pyplot as plt
|
||||
from hydrogen import e_loc
|
||||
|
||||
x=np.linspace(-5,5)
|
||||
|
||||
def make_array(a):
|
||||
y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
|
||||
return y
|
||||
|
||||
plt.figure(figsize=(10,5))
|
||||
|
||||
for a in [0.1, 0.2, 0.5, 1., 1.5, 2.]:
|
||||
y = make_array(a)
|
||||
y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
|
||||
plt.plot(x,y,label=f"a={a}")
|
||||
|
||||
|
||||
plt.tight_layout()
|
||||
|
||||
plt.legend()
|
||||
|
||||
plt.savefig("plot_py.png")
|
||||
|
||||
Loading…
Reference in New Issue
Block a user