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Introduced fixed-node error
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QMC.org
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QMC.org
@ -714,98 +714,83 @@ gfortran hydrogen.f90 energy_hydrogen.f90 -o energy_hydrogen
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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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**** TODO Solution :solution:
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**** Solution :solution:
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$\bar{E} = \langle E \rangle$ is a constant, so $\langle \bar{E}
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\rangle = \bar{E}$ .
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\begin{eqnarray*}
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\langle E - \bar{E} \rangle^2 & = &
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\langle E^2 - 2 E \bar{E} + \bar{E}^2 \rangle \\
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&=& \langle E^2 \rangle - 2 \langle E \bar{E} \rangle + \langle \bar{E}^2 \rangle \\
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&=& \langle E^2 \rangle - 2 \langle E \rangle \bar{E} + \bar{E}^2 \\
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&=& \langle E^2 \rangle - 2 \bar{E}^2 + \bar{E}^2 \\
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&=& \langle E^2 \rangle - \bar{E}^2 \\
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&=& \langle E^2 \rangle - \langle E \rangle^2 \\
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\end{eqnarray*}
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*** Exercise
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#+begin_exercise
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Add the calculation of the variance to the previous code, and
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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
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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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Add the calculation of the variance to the previous code, and
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compute a numerical estimate of the variance of the local energy in
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a grid of $50\times50\times50$ points in the range $(-5,-5,-5) \le
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\mathbf{r} \le (5,5,5)$ 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 :tangle none
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import numpy as np
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from hydrogen import e_loc, psi
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import numpy as np 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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interval = np.linspace(-5,5,num=50) delta =
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(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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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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*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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program variance_hydrogen implicit none double precision, external ::
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e_loc, psi double precision :: x(50), w, delta, energy, dx, r(3),
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a(6), norm, s2 double precision :: e, energy2 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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dx = 10.d0/(size(x)-1) do i=1,size(x) x(i) = -5.d0 + (i-1)*dx 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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do j=1,size(a) ! TODO print *, 'a = ', a(j), ' E = ', energy, ' s2 =
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', s2 end do
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end program variance_hydrogen
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#+end_src
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end program variance_hydrogen #+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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./variance_hydrogen #+end_src
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**** Solution :solution:
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*Python*
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#+begin_src python :results none :exports both
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import numpy as np
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from hydrogen import e_loc, psi
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import numpy as np 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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interval = np.linspace(-5,5,num=50) delta =
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(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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E = 0.
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E2 = 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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El = e_loc(a, r)
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E += w * El
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E2 += w * El*El
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norm += w
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E = E / norm
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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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for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]: E = 0. E2 = 0. norm = 0.
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for x in interval: r[0] = x for y in interval: r[1] = y for z in
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interval: r[2] = z w = psi(a, r) w = w * w * delta El = e_loc(a,
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r) E += w * El E2 += w * El*El norm += w E = E / norm E2 = E2 /
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norm s2 = E2 - E*E print(f"a = {a} \t E = {E:10.8f} \t \sigma^2 =
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{s2:10.8f}") #+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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@ -818,56 +803,31 @@ for a in [0.1, 0.2, 0.5, 0.9, 1., 1.5, 2.]:
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*Fortran*
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#+begin_src f90
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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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program variance_hydrogen implicit none double precision, external ::
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e_loc, psi double precision :: x(50), w, delta, energy, dx, r(3),
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a(6), norm, s2 double precision :: e, energy2 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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dx = 10.d0/(size(x)-1) do i=1,size(x) x(i) = -5.d0 + (i-1)*dx 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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energy = 0.d0
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energy2 = 0.d0
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norm = 0.d0
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do i=1,size(x)
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r(1) = x(i)
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do k=1,size(x)
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r(2) = x(k)
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do l=1,size(x)
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r(3) = x(l)
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w = psi(a(j),r)
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w = w * w * delta
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e = e_loc(a(j), r)
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energy = energy + w * e
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energy2 = energy2 + w * e * e
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norm = norm + w
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end do
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end do
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end do
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energy = energy / norm
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energy2 = energy2 / norm
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s2 = energy2 - energy*energy
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print *, 'a = ', a(j), ' E = ', energy, ' s2 = ', s2
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end do
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do j=1,size(a) energy = 0.d0 energy2 = 0.d0 norm = 0.d0 do
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i=1,size(x) r(1) = x(i) do k=1,size(x) r(2) = x(k) do l=1,size(x)
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r(3) = x(l) w = psi(a(j),r) w = w * w * delta e = e_loc(a(j), r)
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energy = energy + w * e energy2 = energy2 + w * e * e norm =
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norm + w end do end do end do energy = energy / norm energy2 =
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energy2 / norm s2 = energy2 - energy*energy print *, 'a = ',
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a(j), ' E = ', energy, ' s2 = ', s2 end do
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end program variance_hydrogen
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#+end_src
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end program variance_hydrogen #+end_src
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#+begin_src sh :results output :exports results
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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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./variance_hydrogen #+end_src
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#+RESULTS:
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: a = 0.10000000000000001 E = -0.24518438948809140 s2 = 2.6965218719722767E-002
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@ -1860,7 +1820,6 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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imaginary time will converge to the exact ground state of the
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system.
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** Diffusion and branching
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The [[https://en.wikipedia.org/wiki/Diffusion_equation][diffusion equation]] of particles is given by
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@ -1904,7 +1863,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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practice.
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Fortunately, if we multiply the Schrödinger equation by a chosen
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/trial wave function/ $\Psi_T(\mathbf{r})$ (Hartree-Fock, Kohn-Sham
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determinant, CI wave function, /etc/), one obtains
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determinant, CI wave function, /etc/), one obtains (see appendix
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for details)
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\[
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-\frac{\partial \psi(\mathbf{r},\tau)}{\partial \tau} \Psi_T(\mathbf{r}) =
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@ -1926,7 +1886,16 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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can be simulated by the drifted diffusion scheme presented in the
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previous section (VMC).
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This equation generates the /N/-electron density $\Pi$, which is the
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product of the ground state with the trial wave function. It
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introduces the constraint that $\Pi(\mathbf{r},\tau)=0$ where
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$\Psi_T(\mathbf{r})=0$. If the wave function has the same sign
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everywhere, as in the hydrogen atom or the H_2 molecule, this
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constraint is harmless and we can obtain the exact energy. But for
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systems where the wave function has nodes (systems with 3 or more
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electrons), this scheme introduces an error known as the /fixed node
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error/.
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*** Appendix : Details of the Derivation
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\[
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@ -1973,6 +1942,8 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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\right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau)
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\]
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** Pure Diffusion Monte Carlo
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** TODO Hydrogen atom
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*** Exercise
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