OK up to VMC

QMC.org

@@ -41,10 +41,10 @@
   computes a statistical estimate of the expectation value of the energy
   associated with a given wave function.
   Finally, we introduce the diffusion Monte Carlo (DMC) method which
-  gives the exact energy of the $H_2$ molecule. 
+  gives the exact energy of the hydrogen atom and of the $H_2$ molecule. 
 
-  Code examples will be given in Python and Fortran. Whatever language
+  Code examples will be given in Python and Fortran. You can use
-  can be chosen.
+  whatever language you prefer to write the program.
 
   We consider the stationary solution of the Schrödinger equation, so
   the wave functions considered here are real: for an $N$ electron
@@ -52,12 +52,8 @@
   $\Psi : \mathbb{R}^{3N} \rightarrow \mathbb{R}$. In addition, $\Psi$
   is defined everywhere, continuous and infinitely differentiable.
 
- *Note*
+  All the quantities are expressed in /atomic units/ (energies,
-  #+begin_important
+  coordinates, etc).
-  In Fortran, when  you use a double precision  constant, don't forget
-  to  put ~d0~ as  a  suffix (for example ~2.0d0~),  or  it will  be
-  interpreted as a single precision value
-  #+end_important
   
 
 * Numerical evaluation of the energy
@@ -180,7 +176,7 @@ end function potential
 
     
 **** Python
-     #+BEGIN_SRC python :results none
+     #+BEGIN_SRC python :results none  :tangle none
 def psi(a, r):
     # TODO
      #+END_SRC
@@ -192,7 +188,7 @@ def psi(a, r):
      #+END_SRC
 
 **** Fortran
-     #+BEGIN_SRC f90 
+     #+BEGIN_SRC f90  :tangle none
 double precision function psi(a, r)
   implicit none
   double precision, intent(in) :: a, r(3)
@@ -247,7 +243,7 @@ end function psi
     $$
      
 **** Python
-     #+BEGIN_SRC python :results none
+     #+BEGIN_SRC python :results none :tangle none
 def kinetic(a,r):
     # TODO
      #+END_SRC
@@ -259,7 +255,7 @@ def kinetic(a,r):
      #+END_SRC
 
 **** Fortran
-     #+BEGIN_SRC f90 
+     #+BEGIN_SRC f90  :tangle none
 double precision function kinetic(a,r)
   implicit none
   double precision, intent(in) :: a, r(3)
@@ -291,7 +287,7 @@ end function kinetic
 
     
 **** Python
-     #+BEGIN_SRC python :results none
+     #+BEGIN_SRC python :results none :tangle none
 def e_loc(a,r):
     #TODO
      #+END_SRC
@@ -303,7 +299,7 @@ def e_loc(a,r):
      #+END_SRC
 
 **** Fortran
-     #+BEGIN_SRC f90
+     #+BEGIN_SRC f90 :tangle none
 double precision function e_loc(a,r)
   implicit none
   double precision, intent(in) :: a, r(3)
@@ -337,7 +333,7 @@ end function e_loc
     #+end_exercise
 
 **** Python
-     #+BEGIN_SRC python :results none
+     #+BEGIN_SRC python :results none :tangle none
 import numpy as np
 import matplotlib.pyplot as plt
 
@@ -377,7 +373,7 @@ plt.savefig("plot_py.png")
      [[./plot_py.png]]
 
 **** Fortran
-        #+begin_src f90 
+     #+begin_src f90  :tangle none
 program plot
   implicit none
   double precision, external :: e_loc
@@ -402,7 +398,7 @@ gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
 ./plot_hydrogen > data
      #+end_src
 
-        To plot the data using gnuplot:
+     To plot the data using Gnuplot:
 
      #+begin_src gnuplot :file plot.png :exports both
 set grid
@@ -457,7 +453,7 @@ gfortran hydrogen.f90 plot_hydrogen.f90 -o plot_hydrogen
 
      #+RESULTS:
 
-        To plot the data using gnuplot:
+     To plot the data using Gnuplot:
 
      #+begin_src gnuplot :file plot.png :exports both
 set grid
@@ -474,7 +470,7 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
      #+RESULTS:
      [[file:plot.png]]
 
-** TODO Numerical estimation of the energy
+** Numerical estimation of the energy
    :PROPERTIES:
    :header-args:python: :tangle energy_hydrogen.py
    :header-args:f90: :tangle energy_hydrogen.f90
@@ -505,7 +501,23 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
     \mathbf{r} \le (5,5,5)$.
      #+end_exercise
 
- *Python*
+**** Python
+     #+BEGIN_SRC python :results none :tangle none
+import numpy as np
+from hydrogen import e_loc, psi
+
+interval = np.linspace(-5,5,num=50)
+delta = (interval[1]-interval[0])**3
+
+r = np.array([0.,0.,0.])
+
+for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
+    # TODO
+    print(f"a = {a} \t E = {E}")                
+
+     #+end_src
+
+**** Python                                                        :solution:
      #+BEGIN_SRC python :results none
 import numpy as np
 from hydrogen import e_loc, psi
@@ -542,7 +554,37 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
      : a = 1.5 	 E = -0.39242967082602226
      : a = 2.0 	 E = -0.08086980667844901
 
- *Fortran*
+**** Fortran
+     #+begin_src f90 
+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
+     #+end_src
+
+     To compile the Fortran and run it:
+
+     #+begin_src sh :results output :exports both
+gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
+./energy_hydrogen 
+     #+end_src
+
+**** Fortran                                                       :solution:
      #+begin_src f90 
 program energy_hydrogen
   implicit none
@@ -599,7 +641,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
      :  a =    1.5000000000000000          E =  -0.39242967082602065     
      :  a =    2.0000000000000000          E =   -8.0869806678448772E-002
 
-** TODO Variance of the local energy
+** Variance of the local energy
    :PROPERTIES:
    :header-args:python: :tangle variance_hydrogen.py
    :header-args:f90: :tangle variance_hydrogen.f90
@@ -607,7 +649,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
 
    The variance of the local energy is a functional of $\Psi$
    which measures the magnitude of the fluctuations of the local
-   energy associated with $\Psi$ around the average:
+   energy associated with $\Psi$ around its average:
 
    $$
    \sigma^2(E_L) = \frac{\int \left[\Psi(\mathbf{r})\right]^2\, \left[
@@ -615,7 +657,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
    $$
    which can be simplified as
    
-   $$ \sigma^2(E_L) = \langle E_L^2 \rangle - \langle E_L \rangle^2 $$
+   $$ \sigma^2(E_L) = \langle E_L^2 \rangle_{\Psi^2} - \langle E_L \rangle_{\Psi^2}^2.$$
 
    If the local energy is constant (i.e. $\Psi$ is an eigenfunction of
    $\hat{H}$) the variance is zero, so the variance of the local
@@ -624,7 +666,7 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
 *** Exercise (optional)
    #+begin_exercise
    Prove that :
-   $$\langle E - \langle E \rangle \rangle^2 = \langle E^2 \rangle - \langle E \rangle^2 $$
+   $$\left( \langle E - \langle E \rangle_{\Psi^2} \rangle_{\Psi^2} \right)^2 = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$
    #+end_exercise
    
 *** Exercise
@@ -636,7 +678,22 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
    \le \mathbf{r} \le (5,5,5)$ for different values of $a$.
    #+end_exercise
      
- *Python*
+**** Python
+     #+begin_src python :results none :tangle none
+import numpy as np
+from hydrogen import e_loc, psi
+
+interval = np.linspace(-5,5,num=50)
+delta = (interval[1]-interval[0])**3
+
+r = np.array([0.,0.,0.])
+
+for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
+    # TODO
+    print(f"a = {a} \t E = {E:10.8f}  \t  \sigma^2 = {s2:10.8f}")
+     #+end_src
+
+**** Python                                                        :solution:
      #+begin_src python :results none
 import numpy as np
 from hydrogen import e_loc, psi
@@ -677,7 +734,42 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
      : a = 1.5 	 E = -0.39242967  	  \sigma^2 = 0.31449671
      : a = 2.0 	 E = -0.08086981  	  \sigma^2 = 1.80688143
      
- *Fortran*
+**** Fortran
+     #+begin_src f90 :tangle none
+program variance_hydrogen
+  implicit none
+  double precision, external :: e_loc, psi
+  double precision :: x(50), w, delta, energy, dx, r(3), a(6), norm, s2
+  double precision :: e, energy2
+  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
+  
+  delta = dx**3
+  
+  r(:) = 0.d0
+  
+  do j=1,size(a)
+     ! TODO
+     print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
+  end do
+
+end program variance_hydrogen
+     #+end_src
+
+     To compile and run:
+
+     #+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
@@ -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.

energy_hydrogen.f90

@@ -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

plot_hydrogen.py

@@ -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")