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Introduced fixed-node error

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Anthony Scemama 2021-01-27 15:21:44 +01:00
parent d49102cf2e
commit f9354e4401

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@ -714,23 +714,34 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
$$\left( \langle E - \langle E \rangle_{\Psi^2} \rangle_{\Psi^2} \right)^2 = \langle E^2 \rangle_{\Psi^2} - \langle E \rangle_{\Psi^2}^2 $$
#+end_exercise
**** TODO Solution :solution:
**** Solution :solution:
$\bar{E} = \langle E \rangle$ is a constant, so $\langle \bar{E}
\rangle = \bar{E}$ .
\begin{eqnarray*}
\langle E - \bar{E} \rangle^2 & = &
\langle E^2 - 2 E \bar{E} + \bar{E}^2 \rangle \\
&=& \langle E^2 \rangle - 2 \langle E \bar{E} \rangle + \langle \bar{E}^2 \rangle \\
&=& \langle E^2 \rangle - 2 \langle E \rangle \bar{E} + \bar{E}^2 \\
&=& \langle E^2 \rangle - 2 \bar{E}^2 + \bar{E}^2 \\
&=& \langle E^2 \rangle - \bar{E}^2 \\
&=& \langle E^2 \rangle - \langle E \rangle^2 \\
\end{eqnarray*}
*** Exercise
#+begin_exercise
Add the calculation of the variance to the previous code, and
compute a numerical estimate of the variance of the local energy
in a grid of $50\times50\times50$ points in the range
$(-5,-5,-5)
\le \mathbf{r} \le (5,5,5)$ for different values of $a$.
compute a numerical estimate of the variance of the local energy in
a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le
\mathbf{r} \le (5,5,5)$ for different values of $a$.
#+end_exercise
*Python*
#+begin_src python :results none :tangle none
import numpy as np
from hydrogen import e_loc, psi
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.])
@ -741,71 +752,45 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
*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
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
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
do j=1,size(a) ! TODO print *, 'a = ', a(j), ' E = ', energy, ' s2 =
', s2 end do
end program variance_hydrogen
#+end_src
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
./variance_hydrogen #+end_src
**** Solution :solution:
*Python*
#+begin_src python :results none :exports both
import numpy as np
from hydrogen import e_loc, psi
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.])
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
E = 0.
E2 = 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
El = e_loc(a, r)
E += w * El
E2 += w * El*El
norm += w
E = E / norm
E2 = E2 / norm
s2 = E2 - E*E
print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 = {s2:10.8f}")
#+end_src
for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]: E = 0. E2 = 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 El = e_loc(a,
r) E += w * El E2 += w * El*El norm += w E = E / norm E2 = E2 /
norm s2 = E2 - E*E print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 =
{s2:10.8f}") #+end_src
#+RESULTS:
: a = 0.1 E = -0.24518439 \sigma^2 = 0.02696522
@ -818,56 +803,31 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
*Fortran*
#+begin_src f90
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
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
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)
energy = 0.d0
energy2 = 0.d0
norm = 0.d0
do i=1,size(x)
r(1) = x(i)
do k=1,size(x)
r(2) = x(k)
do l=1,size(x)
r(3) = x(l)
w = psi(a(j),r)
w = w * w * delta
e = e_loc(a(j), r)
energy = energy + w * e
energy2 = energy2 + w * e * e
norm = norm + w
end do
end do
end do
energy = energy / norm
energy2 = energy2 / norm
s2 = energy2 - energy*energy
print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
end do
do j=1,size(a) energy = 0.d0 energy2 = 0.d0 norm = 0.d0 do
i=1,size(x) r(1) = x(i) do k=1,size(x) r(2) = x(k) do l=1,size(x)
r(3) = x(l) w = psi(a(j),r) w = w * w * delta e = e_loc(a(j), r)
energy = energy + w * e energy2 = energy2 + w * e * e norm =
norm + w end do end do end do energy = energy / norm energy2 =
energy2 / norm s2 = energy2 - energy*energy print *, 'a = ',
a(j), ' E = ', energy, ' s2 = ', s2 end do
end program variance_hydrogen
#+end_src
end program variance_hydrogen #+end_src
#+begin_src sh :results output :exports results
gfortran hydrogen.f90 variance_hydrogen.f90 -o variance_hydrogen
./variance_hydrogen
#+end_src
./variance_hydrogen #+end_src
#+RESULTS:
: a = 0.10000000000000001 E = -0.24518438948809140 s2 = 2.6965218719722767E-002
@ -1860,7 +1820,6 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
imaginary time will converge to the exact ground state of the
system.
** Diffusion and branching
The [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by
@ -1904,7 +1863,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
practice.
Fortunately, if we multiply the Schrödinger equation by a chosen
/trial wave function/ $\Psi_T(\mathbf{r})$ (Hartree-Fock, Kohn-Sham
determinant, CI wave function, /etc/), one obtains
determinant, CI wave function, /etc/), one obtains (see appendix
for details)
\[
-\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) =
@ -1926,6 +1886,15 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
can be simulated by the drifted diffusion scheme presented in the
previous section (VMC).
This equation generates the /N/-electron density $\Pi$, which is the
product of the ground state with the trial wave function. It
introduces the constraint that $\Pi(\mathbf{r},\tau)=0$ where
$\Psi_T(\mathbf{r})=0$. If the wave function has the same sign
everywhere, as in the hydrogen atom or the H_2 molecule, this
constraint is harmless and we can obtain the exact energy. But for
systems where the wave function has nodes (systems with 3 or more
electrons), this scheme introduces an error known as the /fixed node
error/.
*** Appendix : Details of the Derivation
@ -1973,6 +1942,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
\right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau)
\]
** Pure Diffusion Monte Carlo
** TODO Hydrogen atom
*** Exercise