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OK up to VMC
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@ -41,10 +41,10 @@
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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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@ -52,12 +52,8 @@
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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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@ -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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@ -402,7 +398,7 @@ 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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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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set grid
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@ -457,7 +453,7 @@ gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
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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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set grid
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@ -474,7 +470,7 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
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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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@ -505,7 +501,23 @@ 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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**** 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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@ -542,7 +554,37 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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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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**** 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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@ -599,7 +641,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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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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@ -636,7 +678,22 @@ 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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**** 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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@ -677,7 +734,42 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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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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**** 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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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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delta = dx**3
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r(:) = 0.d0
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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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end program variance_hydrogen
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#+end_src
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To compile and run:
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#+begin_src sh :results output :exports both
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gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
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./variance_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 variance_hydrogen
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implicit none
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@ -1897,3 +1989,12 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc.f90 -o vmc
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#+RESULTS:
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: E = -0.48584030499187431 +/- 1.0411743995438257E-004
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* TODO [0/1] Last things to do
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- [ ] Prepare 4 questions for the exam: multiple-choice questions
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with 4 possible answers. Questions should be independent because
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they will be asked in a random order.
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- [ ] Propose a project for the students to continue the
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programs. Idea: Modify the program to compute the exact energy of
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the H$_2$ molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.
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@ -1,3 +1,23 @@
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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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program energy_hydrogen
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implicit none
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double precision, external :: e_loc, psi
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@ -4,18 +4,12 @@ import matplotlib.pyplot as plt
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from hydrogen import e_loc
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x=np.linspace(-5,5)
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def make_array(a):
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y=np.array([ e_loc(a, np.array([t,0.,0.]) ) for t in x])
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return y
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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 = make_array(a)
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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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