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QMC.org
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@ -1,9 +1,7 @@
#+TITLE: Quantum Monte Carlo
#+AUTHOR: Anthony Scemama, Claudia Filippi
# SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-readtheorg.setup
# SETUPFILE: https://fniessen.github.io/org-html-themes/org/theme-bigblow.setup
#+LANGUAGE: en
#+INFOJS_OPT: toc:t mouse:underline path:http://orgmode.org/org-info.js
#+INFOJS_OPT: toc:t mouse:underline path:org-info.js
#+STARTUP: latexpreview
#+LATEX_CLASS: report
#+LATEX_HEADER_EXTRA: \usepackage{minted}
@ -47,7 +45,7 @@
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 hydrogen atom and 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. You can use
whatever language you prefer to write the program.
@ -1156,17 +1154,17 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
initial position $\mathbf{r}_0$, we will realize a random walk as follows:
$$
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \mathbf{u}
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \mathbf{u}
$$
where $\tau$ is a fixed constant (the so-called /time-step/), and
where $\delta t$ is a fixed constant (the so-called /time-step/), and
$\mathbf{u}$ is a uniform random number in a 3-dimensional box
$(-1,-1,-1) \le \mathbf{u} \le (1,1,1)$. We will then add the
accept/reject step that guarantees that the distribution of the
$\mathbf{r}_n$ is $\Psi^2$:
1) Compute $\Psi$ at a new position $\mathbf{r'} = \mathbf{r}_n +
\tau \mathbf{u}$
\delta t\, \mathbf{u}$
2) Compute the ratio $R = \frac{\left[\Psi(\mathbf{r'})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}$
3) Draw a uniform random number $v \in [0,1]$
4) if $v \le R$, accept the move : set $\mathbf{r}_{n+1} = \mathbf{r'}$
@ -1213,15 +1211,15 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_uniform.f90 -o qmc_uniform
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,nmax,tau):
def MonteCarlo(a,nmax,dt):
# TODO
return energy/nmax, N_accep/nmax
# Run simulation
a = 0.9
nmax = 100000
tau = 1.3
X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
dt = 1.3
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]
# Energy
X = [ x for (x, _) in X0 ]
@ -1236,11 +1234,11 @@ print(f"A = {A} +/- {deltaA}")
*Fortran*
#+BEGIN_SRC f90 :tangle none
subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
subroutine metropolis_montecarlo(a,nmax,dt,energy,accep)
implicit none
double precision, intent(in) :: a
integer*8 , intent(in) :: nmax
double precision, intent(in) :: tau
double precision, intent(in) :: dt
double precision, intent(out) :: energy
double precision, intent(out) :: accep
@ -1257,7 +1255,7 @@ end subroutine metropolis_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9d0
double precision, parameter :: tau = 1.3d0
double precision, parameter :: dt = 1.3d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1266,7 +1264,7 @@ program qmc
double precision :: ave, err
do irun=1,nruns
call metropolis_montecarlo(a,nmax,tau,X(irun),Y(irun))
call metropolis_montecarlo(a,nmax,dt,X(irun),Y(irun))
enddo
call ave_error(X,nruns,ave,err)
@ -1288,13 +1286,13 @@ gfortran hydrogen.f90 qmc_stats.f90 qmc_metropolis.f90 -o qmc_metropolis
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,nmax,tau):
def MonteCarlo(a,nmax,dt):
energy = 0.
N_accep = 0
r_old = np.random.uniform(-tau, tau, (3))
r_old = np.random.uniform(-dt, dt, (3))
psi_old = psi(a,r_old)
for istep in range(nmax):
r_new = r_old + np.random.uniform(-tau,tau,(3))
r_new = r_old + np.random.uniform(-dt,dt,(3))
psi_new = psi(a,r_new)
ratio = (psi_new / psi_old)**2
v = np.random.uniform()
@ -1308,8 +1306,8 @@ def MonteCarlo(a,nmax,tau):
# Run simulation
a = 0.9
nmax = 100000
tau = 1.3
X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
dt = 1.3
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]
# Energy
X = [ x for (x, _) in X0 ]
@ -1328,11 +1326,11 @@ print(f"A = {A} +/- {deltaA}")
*Fortran*
#+BEGIN_SRC f90 :exports code
subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
subroutine metropolis_montecarlo(a,nmax,dt,energy,accep)
implicit none
double precision, intent(in) :: a
integer*8 , intent(in) :: nmax
double precision, intent(in) :: tau
double precision, intent(in) :: dt
double precision, intent(out) :: energy
double precision, intent(out) :: accep
@ -1346,11 +1344,11 @@ subroutine metropolis_montecarlo(a,nmax,tau,energy,accep)
energy = 0.d0
n_accep = 0_8
call random_number(r_old)
r_old(:) = tau * (2.d0*r_old(:) - 1.d0)
r_old(:) = dt * (2.d0*r_old(:) - 1.d0)
psi_old = psi(a,r_old)
do istep = 1,nmax
call random_number(r_new)
r_new(:) = r_old(:) + tau * (2.d0*r_new(:) - 1.d0)
r_new(:) = r_old(:) + dt * (2.d0*r_new(:) - 1.d0)
psi_new = psi(a,r_new)
ratio = (psi_new / psi_old)**2
call random_number(v)
@ -1368,7 +1366,7 @@ end subroutine metropolis_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9d0
double precision, parameter :: tau = 1.3d0
double precision, parameter :: dt = 1.3d0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1377,7 +1375,7 @@ program qmc
double precision :: ave, err
do irun=1,nruns
call metropolis_montecarlo(a,nmax,tau,X(irun),Y(irun))
call metropolis_montecarlo(a,nmax,dt,X(irun),Y(irun))
enddo
call ave_error(X,nruns,ave,err)
@ -1477,12 +1475,12 @@ end subroutine random_gauss
Now, if instead of drawing uniform random numbers we
choose to draw Gaussian random numbers with zero mean and variance
$\tau$, the transition probability becomes:
$\delta t$, the transition probability becomes:
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\tau} \right]
\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\delta t} \right]
\]
@ -1500,19 +1498,19 @@ end subroutine random_gauss
The numerical scheme becomes a drifted diffusion:
\[
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \tau \frac{\nabla
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
\]
where $\chi$ is a Gaussian random variable with zero mean and
variance $\tau$.
variance $\delta t$.
The transition probability becomes:
\[
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
\frac{1}{(2\pi\,\tau)^{3/2}} \exp \left[ - \frac{\left(
\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
\mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\tau} \right]
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]
\]
@ -1570,14 +1568,14 @@ end subroutine drift
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,nmax,tau):
def MonteCarlo(a,nmax,dt):
# TODO
# Run simulation
a = 0.9
nmax = 100000
tau = 1.3
X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
dt = 1.3
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]
# Energy
X = [ x for (x, _) in X0 ]
@ -1592,14 +1590,14 @@ print(f"A = {A} +/- {deltaA}")
*Fortran*
#+BEGIN_SRC f90 :tangle none
subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
subroutine variational_montecarlo(a,dt,nmax,energy,accep_rate)
implicit none
double precision, intent(in) :: a, tau
double precision, intent(in) :: a, dt
integer*8 , intent(in) :: nmax
double precision, intent(out) :: energy, accep_rate
integer*8 :: istep
double precision :: sq_tau, chi(3)
double precision :: sq_dt, chi(3)
double precision :: psi_old, psi_new
double precision :: r_old(3), r_new(3)
double precision :: d_old(3), d_new(3)
@ -1612,7 +1610,7 @@ end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: tau = 1.0
double precision, parameter :: dt = 1.0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1621,7 +1619,7 @@ program qmc
double precision :: ave, err
do irun=1,nruns
call variational_montecarlo(a,tau,nmax,X(irun),accep(irun))
call variational_montecarlo(a,dt,nmax,X(irun),accep(irun))
enddo
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
@ -1641,23 +1639,23 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
from hydrogen import *
from qmc_stats import *
def MonteCarlo(a,nmax,tau):
def MonteCarlo(a,nmax,dt):
energy = 0.
accep_rate = 0.
sq_tau = np.sqrt(tau)
sq_dt = np.sqrt(dt)
r_old = np.random.normal(loc=0., scale=1.0, size=(3))
d_old = drift(a,r_old)
d2_old = np.dot(d_old,d_old)
psi_old = psi(a,r_old)
for istep in range(nmax):
chi = np.random.normal(loc=0., scale=1.0, size=(3))
r_new = r_old + tau * d_old + sq_tau * chi
r_new = r_old + dt * d_old + sq_dt * chi
d_new = drift(a,r_new)
d2_new = np.dot(d_new,d_new)
psi_new = psi(a,r_new)
# Metropolis
prod = np.dot((d_new + d_old), (r_new - r_old))
argexpo = 0.5 * (d2_new - d2_old)*tau + prod
argexpo = 0.5 * (d2_new - d2_old)*dt + prod
q = psi_new / psi_old
q = np.exp(-argexpo) * q*q
if np.random.uniform() < q:
@ -1673,8 +1671,8 @@ def MonteCarlo(a,nmax,tau):
# Run simulation
a = 0.9
nmax = 100000
tau = 1.3
X0 = [ MonteCarlo(a,nmax,tau) for i in range(30)]
dt = 1.3
X0 = [ MonteCarlo(a,nmax,dt) for i in range(30)]
# Energy
X = [ x for (x, _) in X0 ]
@ -1693,20 +1691,20 @@ print(f"A = {A} +/- {deltaA}")
*Fortran*
#+BEGIN_SRC f90
subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
subroutine variational_montecarlo(a,dt,nmax,energy,accep_rate)
implicit none
double precision, intent(in) :: a, tau
double precision, intent(in) :: a, dt
integer*8 , intent(in) :: nmax
double precision, intent(out) :: energy, accep_rate
integer*8 :: istep
double precision :: sq_tau, chi(3), d2_old, prod, u
double precision :: sq_dt, chi(3), d2_old, prod, u
double precision :: psi_old, psi_new, d2_new, argexpo, q
double precision :: r_old(3), r_new(3)
double precision :: d_old(3), d_new(3)
double precision, external :: e_loc, psi
sq_tau = dsqrt(tau)
sq_dt = dsqrt(dt)
! Initialization
energy = 0.d0
@ -1718,7 +1716,7 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
do istep = 1,nmax
call random_gauss(chi,3)
r_new(:) = r_old(:) + tau * d_old(:) + chi(:)*sq_tau
r_new(:) = r_old(:) + dt * d_old(:) + chi(:)*sq_dt
call drift(a,r_new,d_new)
d2_new = d_new(1)*d_new(1) + d_new(2)*d_new(2) + d_new(3)*d_new(3)
psi_new = psi(a,r_new)
@ -1726,7 +1724,7 @@ subroutine variational_montecarlo(a,tau,nmax,energy,accep_rate)
prod = (d_new(1) + d_old(1))*(r_new(1) - r_old(1)) + &
(d_new(2) + d_old(2))*(r_new(2) - r_old(2)) + &
(d_new(3) + d_old(3))*(r_new(3) - r_old(3))
argexpo = 0.5d0 * (d2_new - d2_old)*tau + prod
argexpo = 0.5d0 * (d2_new - d2_old)*dt + prod
q = psi_new / psi_old
q = dexp(-argexpo) * q*q
call random_number(u)
@ -1746,7 +1744,7 @@ end subroutine variational_montecarlo
program qmc
implicit none
double precision, parameter :: a = 0.9
double precision, parameter :: tau = 1.0
double precision, parameter :: dt = 1.0
integer*8 , parameter :: nmax = 100000
integer , parameter :: nruns = 30
@ -1755,7 +1753,7 @@ program qmc
double precision :: ave, err
do irun=1,nruns
call variational_montecarlo(a,tau,nmax,X(irun),accep(irun))
call variational_montecarlo(a,dt,nmax,X(irun),accep(irun))
enddo
call ave_error(X,nruns,ave,err)
print *, 'E = ', ave, '+/-', err
@ -1845,7 +1843,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
- a rate equation for the potential.
The diffusion equation can be simulated by a Brownian motion:
\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\tau}\, \chi \]
\[ \mathbf{r}_{n+1} = \mathbf{r}_{n} + \sqrt{\delta t}\, \chi \]
where $\chi$ is a Gaussian random variable, and the rate equation
can be simulated by creating or destroying particles over time (a
so-called branching process).
@ -1871,7 +1869,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
\left[ -\frac{1}{2} \Delta \psi(\mathbf{r},\tau) + V(\mathbf{r}) \psi(\mathbf{r},\tau) \right] \Psi_T(\mathbf{r})
\]
Defining $\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)
Defining $\Pi(\mathbf{r},\tau) = \psi(\mathbf{r},\tau)
\Psi_T(\mathbf{r})$,
\[
@ -1881,10 +1879,15 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
\right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau)
\]
The new "kinetic energy" can be simulated by the drifted diffusion
scheme presented in the previous section (VMC).
The new "potential" is the local energy, which has smaller fluctuations
as $\Psi_T$ tends to the exact wave function. The new "kinetic energy"
can be simulated by the drifted diffusion scheme presented in the
previous section (VMC).
when $\Psi_T$ gets closer to the exact wave function. It can be simulated by
changing the number of particles according to $\exp\left[ -\delta t\,
\left(E_L(\mathbf{r}) - E_\text{ref}\right)\right]$
where $E_{\text{ref}}$ is a constant introduced so that the average
of this term is close to one, keeping the number of particles rather
constant.
This equation generates the /N/-electron density $\Pi$, which is the
product of the ground state with the trial wave function. It
@ -1895,7 +1898,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
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
\[
@ -1942,15 +1945,87 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
\right] + E_L(\mathbf{r}) \Pi(\mathbf{r},\tau)
\]
** Fixed-node DMC energy
Now that we have a process to sample $\Pi(\mathbf{r},\tau) =
\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, we can compute the exact
energy of the system, within the fixed-node constraint, as:
\[
E = \lim_{\tau \to \infty} \frac{\int \Pi(\mathbf{r},\tau) E_L(\mathbf{r}) d\mathbf{r}}
{\int \Pi(\mathbf{r},\tau) d\mathbf{r}} = \lim_{\tau \to
\infty} E(\tau).
\]
Indeed, this is equivalent to
\[
E(\tau) = \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}}
= \frac{\int \psi(\mathbf{r},\tau) \hat{H} \Psi_T(\mathbf{r}) d\mathbf{r}}
{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}}
= \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle}
{\langle \psi_\tau | \Psi_T \rangle}
\]
As $\hat{H}$ is Hermitian,
\[
E(\tau) = \frac{\langle \psi_\tau | \hat{H} | \Psi_T \rangle}
{\langle \psi_\tau | \Psi_T \rangle}
= \frac{\langle \Psi_T | \hat{H} | \psi_\tau \rangle}
{\langle \Psi_T | \psi_\tau \rangle}
= E[\psi_\tau] \frac{\langle \Psi_T | \psi_\tau \rangle}
{\langle \Psi_T | \psi_\tau \rangle}
= E[\psi_\tau]
\]
So computing the energy within DMC consists in generating the
density $\Pi$, and simply averaging the local energies computed
with the trial wave function.
** Pure Diffusion Monte Carlo
Instead of having a variable number of particles to simulate the
branching process, one can choose to sample $[\Psi_T(\mathbf{r})]^2$ instead of
$\psi(\mathbf{r},\tau) \Psi_T(\mathbf{r})$, and consider the term
$\exp \left( -\delta t\,( E_L(\mathbf{r}) - E_{\text{ref}} \right)$ as a
cumulative product of branching weights:
\[
W(\mathbf{r}_n, \tau)
= \exp \left( \int_0^\tau - (E_L(\mathbf{r}_t) - E_{\text{ref}}) dt \right)
\approx \prod_{i=1}^{n} \exp \left( -\delta t\, (E_L(\mathbf{r}_i) - E_{\text{ref}}) \right).
\]
The wave function becomes
\[
\psi(\mathbf{r},\tau) = \Psi_T(\mathbf{r}) W(\mathbf{r},\tau)
\]
and the expression of the fixed-node DMC energy is
\begin{eqnarray*}
E(\tau) & = & \frac{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) E_L(\mathbf{r}) d\mathbf{r}}
{\int \psi(\mathbf{r},\tau) \Psi_T(\mathbf{r}) d\mathbf{r}} \\
& = & \frac{\int \left[ \Psi_T(\mathbf{r}) \right]^2 W(\mathbf{r},\tau) E_L(\mathbf{r}) d\mathbf{r}}
{\int \left[ \Psi_T(\mathbf{r}) \right]^2 W(\mathbf{r},\tau) d\mathbf{r}} \\
\end{eqnarray*}
This algorithm is less stable than the branching algorithm: it
requires to have a value of $E_\text{ref}$ which is close to the
fixed-node energy, and a good trial wave function. Its big
advantage is that it is very easy to program starting from a VMC code.
** TODO Hydrogen atom
*** Exercise
#+begin_exercise
Modify the Metropolis VMC program to introduce the PDMC weight.
In the limit $\tau \rightarrow 0$, you should recover the exact
In the limit $\delta t \rightarrow 0$, you should recover the exact
energy of H for any value of $a$.
#+end_exercise
@ -2094,7 +2169,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
: A = 0.78861366666666655 +/- 3.5096729498002445E-004
** TODO Dihydrogen
** TODO H_2
We will now consider the H_2 molecule in a minimal basis composed of the
$1s$ orbitals of the hydrogen atoms:
@ -2108,7 +2183,7 @@ gfortran hydrogen.f90 qmc_stats.f90 vmc_metropolis.f90 -o vmc_metropolis
the nuclei.
* Appendix :noexport:
* Old sections to be removed :noexport:
** Gaussian sampling :noexport:
:PROPERTIES:

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--b){if(this.HOOKS[a][b]==c){this.HOOKS[a].splice(b,1);}}}}};function OrgHtmlManagerKeyEvent(b){var d;
if(!b){b=window.event;}if(b.which){d=b.which;}else{if(b.keyCode){d=b.keyCode;}}if(b.ctrlKey){return;
}var a=String.fromCharCode(d);if(b.shiftKey){org_html_manager.CI.value=org_html_manager.CI.value+a;
}else{org_html_manager.CI.value=org_html_manager.CI.value+a.toLowerCase();}org_html_manager.getKey();
}function OrgHtmlManagerLoadCheck(){org_html_manager.init();}