Computational Physics Lectures: Variational Monte Carlo methods

Given a hamiltonian $H$ and a trial wave function $\Psi_T$, the variational principle states that the expectation value of $\langle H \rangle$, defined through

 
$$E[H]= \langle H \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})H(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})},$$

 
is an upper bound to the ground state energy $E_0$ of the hamiltonian $H$, that is

 
$$E_0 \le \langle H \rangle .$$

 
In general, the integrals involved in the calculation of various expectation values are multi-dimensional ones. Traditional integration methods such as the Gauss-Legendre will not be adequate for say the computation of the energy of a many-body system.

The trial wave function can be expanded in the eigenstates of the hamiltonian since they form a complete set, viz.,

 
$$\Psi_T(\boldsymbol{R})=\sum_i a_i\Psi_i(\boldsymbol{R}),$$

 
and assuming the set of eigenfunctions to be normalized one obtains

 
$$\frac{\sum_{nm}a^*_ma_n \int d\boldsymbol{R}\Psi^{\ast}_m(\boldsymbol{R})H(\boldsymbol{R})\Psi_n(\boldsymbol{R})} {\sum_{nm}a^*_ma_n \int d\boldsymbol{R}\Psi^{\ast}_m(\boldsymbol{R})\Psi_n(\boldsymbol{R})} =\frac{\sum_{n}a^2_n E_n} {\sum_{n}a^2_n} \ge E_0,$$

 
where we used that $H(\boldsymbol{R})\Psi_n(\boldsymbol{R})=E_n\Psi_n(\boldsymbol{R})$. In general, the integrals involved in the calculation of various expectation values are multi-dimensional ones. The variational principle yields the lowest state of a given symmetry.

In most cases, a wave function has only small values in large parts of configuration space, and a straightforward procedure which uses homogenously distributed random points in configuration space will most likely lead to poor results. This may suggest that some kind of importance sampling combined with e.g., the Metropolis algorithm may be a more efficient way of obtaining the ground state energy. The hope is then that those regions of configurations space where the wave function assumes appreciable values are sampled more efficiently.

The tedious part in a VMC calculation is the search for the variational minimum. A good knowledge of the system is required in order to carry out reasonable VMC calculations. This is not always the case, and often VMC calculations serve rather as the starting point for so-called diffusion Monte Carlo calculations (DMC). DMC is a way of solving exactly the many-body Schroedinger equation by means of a stochastic procedure. A good guess on the binding energy and its wave function is however necessary. A carefully performed VMC calculation can aid in this context.

• Construct first a trial wave function $\psi_T(\boldsymbol{R},\boldsymbol{\alpha})$, for a many-body system consisting of $N$ particles located at positions

$\boldsymbol{R}=(\boldsymbol{R}_1,\dots ,\boldsymbol{R}_N)$. The trial wave function depends on $\alpha$ variational parameters $\boldsymbol{\alpha}=(\alpha_1,\dots ,\alpha_M)$.

• Then we evaluate the expectation value of the hamiltonian $H$

 
$$E[H]=\langle H \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}_{T}(\boldsymbol{R},\boldsymbol{\alpha})H(\boldsymbol{R})\Psi_{T}(\boldsymbol{R},\boldsymbol{\alpha})} {\int d\boldsymbol{R}\Psi^{\ast}_{T}(\boldsymbol{R},\boldsymbol{\alpha})\Psi_{T}(\boldsymbol{R},\boldsymbol{\alpha})}.$$

• Thereafter we vary $\alpha$ according to some minimization algorithm and return to the first step.

Choose a trial wave function $\psi_T(\boldsymbol{R})$.

 
$$P(\boldsymbol{R})= \frac{\left|\psi_T(\boldsymbol{R})\right|^2}{\int \left|\psi_T(\boldsymbol{R})\right|^2d\boldsymbol{R}}.$$

 
This is our new probability distribution function (PDF). The approximation to the expectation value of the Hamiltonian is now

 
$$E[H(\boldsymbol{\alpha})] = \frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R},\boldsymbol{\alpha})H(\boldsymbol{R})\Psi_T(\boldsymbol{R},\boldsymbol{\alpha})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R},\boldsymbol{\alpha})\Psi_T(\boldsymbol{R},\boldsymbol{\alpha})}.$$

Define a new quantity

 
$$E_L(\boldsymbol{R},\boldsymbol{\alpha})=\frac{1}{\psi_T(\boldsymbol{R},\boldsymbol{\alpha})}H\psi_T(\boldsymbol{R},\boldsymbol{\alpha}), \tag{1}$$

 
called the local energy, which, together with our trial PDF yields

 
$$E[H(\boldsymbol{\alpha})]=\int P(\boldsymbol{R})E_L(\boldsymbol{R}) d\boldsymbol{R}\approx \frac{1}{N}\sum_{i=1}^NP(\boldsymbol{R_i},\boldsymbol{\alpha})E_L(\boldsymbol{R_i},\boldsymbol{\alpha}) \tag{2}$$

 
with $N$ being the number of Monte Carlo samples.

The Algorithm for performing a variational Monte Carlo calculations runs thus as this

• Initialisation: Fix the number of Monte Carlo steps. Choose an initial $\boldsymbol{R}$ and variational parameters $\alpha$ and calculate $\left|\psi_T^{\alpha}(\boldsymbol{R})\right|^2$.
• Initialise the energy and the variance and start the Monte Carlo calculation.

• Calculate a trial position $\boldsymbol{R}_p=\boldsymbol{R}+r*step$ where $r$ is a random variable $r \in [0,1]$.
• Metropolis algorithm to accept or reject this move $w = P(\boldsymbol{R}_p)/P(\boldsymbol{R})$.
• If the step is accepted, then we set $\boldsymbol{R}=\boldsymbol{R}_p$.
• Update averages
• Finish and compute final averages.

Observe that the jumping in space is governed by the variable step. This is Called brute-force sampling. Need importance sampling to get more relevant sampling, see lectures below.

The radial Schroedinger equation for the hydrogen atom can be written as

 
$$-\frac{\hbar^2}{2m}\frac{\partial^2 u(r)}{\partial r^2}- \left(\frac{ke^2}{r}-\frac{\hbar^2l(l+1)}{2mr^2}\right)u(r)=Eu(r),$$

 
or with dimensionless variables

 
$$-\frac{1}{2}\frac{\partial^2 u(\rho)}{\partial \rho^2}- \frac{u(\rho)}{\rho}+\frac{l(l+1)}{2\rho^2}u(\rho)-\lambda u(\rho)=0, \tag{3}$$

 
with the hamiltonian

 
$$H=-\frac{1}{2}\frac{\partial^2 }{\partial \rho^2}- \frac{1}{\rho}+\frac{l(l+1)}{2\rho^2}.$$

 
Use variational parameter $\alpha$ in the trial wave function

 
$$u_T^{\alpha}(\rho)=\alpha\rho e^{-\alpha\rho}. \tag{4}$$

Inserting this wave function into the expression for the local energy $E_L$ gives

 
$$E_L(\rho)=-\frac{1}{\rho}- \frac{\alpha}{2}\left(\alpha-\frac{2}{\rho}\right).$$

 
A simple variational Monte Carlo calculation results in

$\alpha$	$\langle H \rangle$	$\sigma^2$	$\sigma/\sqrt{N}$
7.00000E-01	-4.57759E-01	4.51201E-02	6.71715E-04
8.00000E-01	-4.81461E-01	3.05736E-02	5.52934E-04
9.00000E-01	-4.95899E-01	8.20497E-03	2.86443E-04
1.00000E-00	-5.00000E-01	0.00000E+00	0.00000E+00
1.10000E+00	-4.93738E-01	1.16989E-02	3.42036E-04
1.20000E+00	-4.75563E-01	8.85899E-02	9.41222E-04
1.30000E+00	-4.54341E-01	1.45171E-01	1.20487E-03

We note that at $\alpha=1$ we obtain the exact result, and the variance is zero, as it should. The reason is that we then have the exact wave function, and the action of the hamiltionan on the wave function

 
$$H\psi = \mathrm{constant}\times \psi,$$

 
yields just a constant. The integral which defines various expectation values involving moments of the hamiltonian becomes then

 
$$\langle H^n \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})H^n(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})}= \mathrm{constant}\times\frac{\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}_T(\boldsymbol{R})\Psi_T(\boldsymbol{R})}=\mathrm{constant}.$$

 
This gives an important information: the exact wave function leads to zero variance! Variation is then performed by minimizing both the energy and the variance.

The helium atom consists of two electrons and a nucleus with charge $Z=2$. The contribution to the potential energy due to the attraction from the nucleus is

 
$$-\frac{2ke^2}{r_1}-\frac{2ke^2}{r_2},$$

 
and if we add the repulsion arising from the two interacting electrons, we obtain the potential energy

 
$$V(r_1, r_2)=-\frac{2ke^2}{r_1}-\frac{2ke^2}{r_2}+ \frac{ke^2}{r_{12}},$$

 
with the electrons separated at a distance $r_{12}=|\boldsymbol{r}_1-\boldsymbol{r}_2|$.

The hamiltonian becomes then

 
$$\hat{H}=-\frac{\hbar^2\nabla_1^2}{2m}-\frac{\hbar^2\nabla_2^2}{2m} -\frac{2ke^2}{r_1}-\frac{2ke^2}{r_2}+ \frac{ke^2}{r_{12}},$$

 
and Schroedingers equation reads

 
$$\hat{H}\psi=E\psi.$$

 
All observables are evaluated with respect to the probability distribution

 
$$P(\boldsymbol{R})= \frac{\left|\psi_T(\boldsymbol{R})\right|^2}{\int \left|\psi_T(\boldsymbol{R})\right|^2d\boldsymbol{R}}.$$

 
generated by the trial wave function. The trial wave function must approximate an exact eigenstate in order that accurate results are to be obtained.

Choice of trial wave function for Helium: Assume $r_1 \rightarrow 0$.

 
$$E_L(\boldsymbol{R})=\frac{1}{\psi_T(\boldsymbol{R})}H\psi_T(\boldsymbol{R})= \frac{1}{\psi_T(\boldsymbol{R})}\left(-\frac{1}{2}\nabla^2_1 -\frac{Z}{r_1}\right)\psi_T(\boldsymbol{R}) + \mathrm{finite \hspace{0.1cm}terms}.$$

 
$$E_L(R)= \frac{1}{{\cal R}_T(r_1)}\left(-\frac{1}{2}\frac{d^2}{dr_1^2}- \frac{1}{r_1}\frac{d}{dr_1} -\frac{Z}{r_1}\right){\cal R}_T(r_1) + \mathrm{finite\hspace{0.1cm} terms}$$

 
For small values of $r_1$, the terms which dominate are

 
$$\lim_{r_1 \rightarrow 0}E_L(R)= \frac{1}{{\cal R}_T(r_1)}\left(- \frac{1}{r_1}\frac{d}{dr_1} -\frac{Z}{r_1}\right){\cal R}_T(r_1),$$

 
since the second derivative does not diverge due to the finiteness of $\Psi$ at the origin.

This results in

 
$$\frac{1}{{\cal R}_T(r_1)}\frac{d {\cal R}_T(r_1)}{dr_1}=-Z,$$

 
and

 
$${\cal R}_T(r_1)\propto e^{-Zr_1}.$$

 
A similar condition applies to electron 2 as well. For orbital momenta $l > 0$ we have

 
$$\frac{1}{{\cal R}_T(r)}\frac{d {\cal R}_T(r)}{dr}=-\frac{Z}{l+1}.$$

 
Similarly, studying the case $r_{12}\rightarrow 0$ we can write a possible trial wave function as

 
$$\psi_T(\boldsymbol{R})=e^{-\alpha(r_1+r_2)}e^{\beta r_{12}}. \tag{5}$$

 
The last equation can be generalized to

 
$$\psi_T(\boldsymbol{R})=\phi(\boldsymbol{r}_1)\phi(\boldsymbol{r}_2)\dots\phi(\boldsymbol{r}_N) \prod_{i < j}f(r_{ij}),$$

 
for a system with $N$ electrons or particles.

During the development of our code we need to make several checks. It is also very instructive to compute a closed form expression for the local energy. Since our wave function is rather simple it is straightforward to find an analytic expressions. Consider first the case of the simple helium function

 
$$\Psi_T(\boldsymbol{r}_1,\boldsymbol{r}_2) = e^{-\alpha(r_1+r_2)}$$

 
The local energy is for this case

 
$$E_{L1} = \left(\alpha-Z\right)\left(\frac{1}{r_1}+\frac{1}{r_2}\right)+\frac{1}{r_{12}}-\alpha^2$$

 
which gives an expectation value for the local energy given by

 
$$\langle E_{L1} \rangle = \alpha^2-2\alpha\left(Z-\frac{5}{16}\right)$$

With closed form formulae we can speed up the computation of the correlation. In our case we write it as

 
$$\Psi_C= \exp{\left\{\sum_{i < j}\frac{ar_{ij}}{1+\beta r_{ij}}\right\}},$$

 
which means that the gradient needed for the so-called quantum force and local energy can be calculated analytically. This will speed up your code since the computation of the correlation part and the Slater determinant are the most time consuming parts in your code.

We will refer to this correlation function as $\Psi_C$ or the linear Pade-Jastrow.

We can test this by computing the local energy for our helium wave function

 
$$\psi_{T}(\boldsymbol{r}_1,\boldsymbol{r}_2) = \exp{\left(-\alpha(r_1+r_2)\right)} \exp{\left(\frac{r_{12}}{2(1+\beta r_{12})}\right)},$$

 
with $\alpha$ and $\beta$ as variational parameters.

The local energy is for this case

 
$$E_{L2} = E_{L1}+\frac{1}{2(1+\beta r_{12})^2}\left\{\frac{\alpha(r_1+r_2)}{r_{12}}(1-\frac{\boldsymbol{r}_1\boldsymbol{r}_2}{r_1r_2})-\frac{1}{2(1+\beta r_{12})^2}-\frac{2}{r_{12}}+\frac{2\beta}{1+\beta r_{12}}\right\}$$

 
It is very useful to test your code against these expressions. It means also that you don't need to compute a derivative numerically as discussed in the code example below.

For the computation of various derivatives with different types of wave functions, you will find it useful to use python with symbolic python, that is sympy, see online manual. Using sympy allows you autogenerate both Latex code as well c++, python or Fortran codes. Here you will find some simple examples. We choose the $2s$ hydrogen-orbital (not normalized) as an example

 
$$\phi_{2s}(\boldsymbol{r}) = (Zr - 2)\exp{-(\frac{1}{2}Zr)},$$

 
with $ r^2 = x^2 + y^2 + z^2$.

from sympy import symbols, diff, exp, sqrt
x, y, z, Z = symbols('x y z Z')
r = sqrt(x*x + y*y + z*z)
r
phi = (Z*r - 2)*exp(-Z*r/2)
phi
diff(phi, x)

This doesn't look very nice, but sympy provides several functions that allow for improving and simplifying the output.

We can improve our output by factorizing and substituting expressions

from sympy import symbols, diff, exp, sqrt, factor, Symbol, printing
x, y, z, Z = symbols('x y z Z')
r = sqrt(x*x + y*y + z*z)
phi = (Z*r - 2)*exp(-Z*r/2)
R = Symbol('r') #Creates a symbolic equivalent of r
#print latex and c++ code
print printing.latex(diff(phi, x).factor().subs(r, R))
print printing.ccode(diff(phi, x).factor().subs(r, R))

We can in turn look at second derivatives

from sympy import symbols, diff, exp, sqrt, factor, Symbol, printing
x, y, z, Z = symbols('x y z Z')
r = sqrt(x*x + y*y + z*z)
phi = (Z*r - 2)*exp(-Z*r/2)
R = Symbol('r') #Creates a symbolic equivalent of r
(diff(diff(phi, x), x) + diff(diff(phi, y), y) + diff(diff(phi, z), z)).factor().subs(r, R)
# Collect the Z values
(diff(diff(phi, x), x) + diff(diff(phi, y), y) +diff(diff(phi, z), z)).factor().collect(Z).subs(r, R)
# Factorize also the r**2 terms
(diff(diff(phi, x), x) + diff(diff(phi, y), y) + diff(diff(phi, z), z)).factor().collect(Z).subs(r, R).subs(r**2, R**2).factor()
print printing.ccode((diff(diff(phi, x), x) + diff(diff(phi, y), y) + diff(diff(phi, z), z)).factor().collect(Z).subs(r, R).subs(r**2, R**2).factor())

With some practice this allows one to be able to check one's own calculation and translate automatically into code lines.

The Metropolis algorithm , see the original article (see also the FYS3150 lectures) was invented by Metropolis et. al and is often simply called the Metropolis algorithm. It is a method to sample a normalized probability distribution by a stochastic process. We define ${\cal P}_i^{(n)}$ to be the probability for finding the system in the state $i$ at step $n$. The algorithm is then

• Sample a possible new state $j$ with some probability $T_{i\rightarrow j}$.
• Accept the new state $j$ with probability $A_{i \rightarrow j}$ and use it as the next sample. With probability $1-A_{i\rightarrow j}$ the move is rejected and the original state $i$ is used again as a sample.

We wish to derive the required properties of $T$ and $A$ such that ${\cal P}_i^{(n\rightarrow \infty)} \rightarrow p_i$ so that starting from any distribution, the method converges to the correct distribution. Note that the description here is for a discrete probability distribution. Replacing probabilities $p_i$ with expressions like $p(x_i)dx_i$ will take all of these over to the corresponding continuum expressions.

The dynamical equation for ${\cal P}_i^{(n)}$ can be written directly from the description above. The probability of being in the state $i$ at step $n$ is given by the probability of being in any state $j$ at the previous step, and making an accepted transition to $i$ added to the probability of being in the state $i$, making a transition to any state $j$ and rejecting the move:

 
$${\cal P}^{(n)}_i = \sum_j \left [ {\cal P}^{(n-1)}_jT_{j\rightarrow i} A_{j\rightarrow i} +{\cal P}^{(n-1)}_iT_{i\rightarrow j}\left ( 1- A_{i\rightarrow j} \right) \right ] \,.$$

 
Since the probability of making some transition must be 1, $\sum_j T_{i\rightarrow j} = 1$, and the above equation becomes

 
$${\cal P}^{(n)}_i = {\cal P}^{(n-1)}_i + \sum_j \left [ {\cal P}^{(n-1)}_jT_{j\rightarrow i} A_{j\rightarrow i} -{\cal P}^{(n-1)}_iT_{i\rightarrow j}A_{i\rightarrow j} \right ] \,.$$

For large $n$ we require that ${\cal P}^{(n\rightarrow \infty)}_i = p_i$, the desired probability distribution. Taking this limit, gives the balance requirement

 
$$\sum_j \left [ p_jT_{j\rightarrow i} A_{j\rightarrow i} -p_iT_{i\rightarrow j}A_{i\rightarrow j} \right ] = 0 \,.$$

 
The balance requirement is very weak. Typically the much stronger detailed balance requirement is enforced, that is rather than the sum being set to zero, we set each term separately to zero and use this to determine the acceptance probabilities. Rearranging, the result is

 
$$\frac{ A_{j\rightarrow i}}{A_{i\rightarrow j}} = \frac{p_iT_{i\rightarrow j}}{ p_jT_{j\rightarrow i}} \,.$$

The Metropolis choice is to maximize the $A$ values, that is

 
$$A_{j \rightarrow i} = \min \left ( 1, \frac{p_iT_{i\rightarrow j}}{ p_jT_{j\rightarrow i}}\right ).$$

 
Other choices are possible, but they all correspond to multilplying $A_{i\rightarrow j}$ and $A_{j\rightarrow i}$ by the same constant smaller than unity.\footnote{The penalty function method uses just such a factor to compensate for $p_i$ that are evaluated stochastically and are therefore noisy.}

Having chosen the acceptance probabilities, we have guaranteed that if the ${\cal P}_i^{(n)}$ has equilibrated, that is if it is equal to $p_i$, it will remain equilibrated. Next we need to find the circumstances for convergence to equilibrium.

The dynamical equation can be written as

 
$${\cal P}^{(n)}_i = \sum_j M_{ij}{\cal P}^{(n-1)}_j$$

 
with the matrix $M$ given by

 
$$M_{ij} = \delta_{ij}\left [ 1 -\sum_k T_{i\rightarrow k} A_{i \rightarrow k} \right ] + T_{j\rightarrow i} A_{j\rightarrow i} \,.$$

 
Summing over $i$ shows that $\sum_i M_{ij} = 1$, and since $\sum_k T_{i\rightarrow k} = 1$, and $A_{i \rightarrow k} \leq 1$, the elements of the matrix satisfy $M_{ij} \geq 0$. The matrix $M$ is therefore a stochastic matrix.

The Metropolis method is simply the power method for computing the right eigenvector of $M$ with the largest magnitude eigenvalue. By construction, the correct probability distribution is a right eigenvector with eigenvalue 1. Therefore, for the Metropolis method to converge to this result, we must show that $M$ has only one eigenvalue with this magnitude, and all other eigenvalues are smaller.