Solving Quantum Mechanical Many-body Problems with Machine Learning Algorithms

What is this talk about?

More material

Why? Basic motivation

Overview

What is Machine Learning?

What are the Machine Learning calculations here based on?

A new world

Lots of room for creativity

Job market, a personal statement: A familiarity with ML is almost becoming a prerequisite for many of the most exciting employment opportunities. And add quantum computing and there you are!

Some members of the ML family

What are the basic ingredients?

Types of Machine Learning

References

Fitting Nuclear Masses

Let us now dive into nuclear physics and remind ourselves briefly about some basic features about binding energies. A basic quantity which can be measured for the ground states of nuclei is the atomic mass \( M(N, Z) \) of the neutral atom with atomic mass number \( A \) and charge \( Z \). The number of neutrons is \( N \). There are indeed several sophisticated experiments worldwide which allow us to measure this quantity to high precision (parts per million even).

Atomic masses are usually tabulated in terms of the mass excess defined by

 
$$ \Delta M(N, Z) = M(N, Z) - uA, $$

 
where \( u \) is the Atomic Mass Unit

 
$$ u = M(^{12}\mathrm{C})/12 = 931.4940954(57) \hspace{0.1cm} \mathrm{MeV}/c^2. $$

In the 2016 mass evaluation of by W.J.Huang, G.Audi, M.Wang, F.G.Kondev, S.Naimi and X.Xu there are data on masses and decays of 3437 nuclei.

The nuclear binding energy is defined as the energy required to break up a given nucleus into its constituent parts of \( N \) neutrons and \( Z \) protons. In terms of the atomic masses \( M(N, Z) \) the binding energy is defined by

 
$$ BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 , $$

 
where \( M_H \) is the mass of the hydrogen atom and \( m_n \) is the mass of the neutron. In terms of the mass excess the binding energy is given by

 
$$ BE(N, Z) = Z\Delta_H c^2 + N\Delta_n c^2 -\Delta(N, Z)c^2 , $$

 
where \( \Delta_H c^2 = 7.2890 \) MeV and \( \Delta_n c^2 = 8.0713 \) MeV.

A popular and physically intuitive model which can be used to parametrize the experimental binding energies as function of \( A \), is the so-called liquid drop model. The ansatz is based on the following expression

 
$$ BE(N,Z) = a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{(N-Z)^2}{A}, $$

where \( A \) stands for the number of nucleons and the $a_i$s are parameters which are determined by a fit to the experimental data.

Organizing our data

Let us start with reading and organizing our data. We start with the compilation of masses and binding energies from 2016. After having downloaded this file to our own computer, we are now ready to read the file and start structuring our data.

We start with preparing folders for storing our calculations and the data file over masses and binding energies. We import also various modules that we will find useful in order to present various Machine Learning methods. Here we focus mainly on the functionality of scikit-learn.

# Common imports
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import sklearn.linear_model as skl
from sklearn.model_selection import train_test_split
from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
import os

# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
DATA_ID = "DataFiles/"

if not os.path.exists(PROJECT_ROOT_DIR):
    os.mkdir(PROJECT_ROOT_DIR)

if not os.path.exists(FIGURE_ID):
    os.makedirs(FIGURE_ID)

if not os.path.exists(DATA_ID):
    os.makedirs(DATA_ID)

def image_path(fig_id):
    return os.path.join(FIGURE_ID, fig_id)

def data_path(dat_id):
    return os.path.join(DATA_ID, dat_id)

def save_fig(fig_id):
    plt.savefig(image_path(fig_id) + ".png", format='png')

infile = open(data_path("MassEval2016.dat"),'r')

Before we proceed, we define also a function for making our plots. You can obviously avoid this and simply set up various matplotlib commands every time you need them. You may however find it convenient to collect all such commands in one function and simply call this function.

from pylab import plt, mpl
plt.style.use('seaborn')
mpl.rcParams['font.family'] = 'serif'

def MakePlot(x,y, styles, labels, axlabels):
    plt.figure(figsize=(10,6))
    for i in range(len(x)):
        plt.plot(x[i], y[i], styles[i], label = labels[i])
        plt.xlabel(axlabels[0])
        plt.ylabel(axlabels[1])
    plt.legend(loc=0)

Our next step is to read the data on experimental binding energies and reorganize them as functions of the mass number \( A \), the number of protons \( Z \) and neutrons \( N \) using pandas. Before we do this it is always useful (unless you have a binary file or other types of compressed data) to actually open the file and simply take a look at it!

In particular, the program that outputs the final nuclear masses is written in Fortran with a specific format. It means that we need to figure out the format and which columns contain the data we are interested in. Pandas comes with a function that reads formatted output. After having admired the file, we are now ready to start massaging it with pandas. The file begins with some basic format information.

"""                                                                                                                         
This is taken from the data file of the mass 2016 evaluation.                                                               
All files are 3436 lines long with 124 character per line.                                                                  
       Headers are 39 lines long.                                                                                           
   col 1     :  Fortran character control: 1 = page feed  0 = line feed                                                     
   format    :  a1,i3,i5,i5,i5,1x,a3,a4,1x,f13.5,f11.5,f11.3,f9.3,1x,a2,f11.3,f9.3,1x,i3,1x,f12.5,f11.5                     
   These formats are reflected in the pandas widths variable below, see the statement                                       
   widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),                                                            
   Pandas has also a variable header, with length 39 in this case.                                                          
"""

The data we are interested in are in columns 2, 3, 4 and 11, giving us the number of neutrons, protons, mass numbers and binding energies, respectively. We add also for the sake of completeness the element name. The data are in fixed-width formatted lines and we will covert them into the pandas DataFrame structure.

# Read the experimental data with Pandas
Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
              names=('N', 'Z', 'A', 'Element', 'Ebinding'),
              widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
              header=39,
              index_col=False)

# Extrapolated values are indicated by '#' in place of the decimal place, so
# the Ebinding column won't be numeric. Coerce to float and drop these entries.
Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
Masses = Masses.dropna()
# Convert from keV to MeV.
Masses['Ebinding'] /= 1000

# Group the DataFrame by nucleon number, A.
Masses = Masses.groupby('A')
# Find the rows of the grouped DataFrame with the maximum binding energy.
Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])

We have now read in the data, grouped them according to the variables we are interested in. We see how easy it is to reorganize the data using pandas. If we were to do these operations in C/C++ or Fortran, we would have had to write various functions/subroutines which perform the above reorganizations for us. Having reorganized the data, we can now start to make some simple fits using both the functionalities in numpy and Scikit-Learn afterwards.

Now we define five variables which contain the number of nucleons \( A \), the number of protons \( Z \) and the number of neutrons \( N \), the element name and finally the energies themselves.

A = Masses['A']
Z = Masses['Z']
N = Masses['N']
Element = Masses['Element']
Energies = Masses['Ebinding']
print(Masses)

The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). It has dimensionality \( p\times n \), where \( n \) is the number of data points and \( p \) are the so-called predictors. In our case here they are given by the number of polynomials in \( A \) we wish to include in the fit.

# Now we set up the design matrix X
X = np.zeros((len(A),5))
X[:,0] = 1
X[:,1] = A
X[:,2] = A**(2.0/3.0)
X[:,3] = A**(-1.0/3.0)
X[:,4] = A**(-1.0)

With scikitlearn we are now ready to use linear regression and fit our data.

clf = skl.LinearRegression().fit(X, Energies)
fity = clf.predict(X)

Pretty simple! Now we can print measures of how our fit is doing, the coefficients from the fits and plot the final fit together with our data.

# The mean squared error                               
print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
# Explained variance score: 1 is perfect prediction                                 
print('Variance score: %.2f' % r2_score(Energies, fity))
# Mean absolute error                                                           
print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
print(clf.coef_, clf.intercept_)

Masses['Eapprox']  = fity
# Generate a plot comparing the experimental with the fitted values values.
fig, ax = plt.subplots()
ax.set_xlabel(r'$A = N + Z$')
ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
            label='Ame2016')
ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
            label='Fit')
ax.legend()
save_fig("Masses2016")
plt.show()

Artificial neurons

A simple perceptron model

Neural network types

The system: two electrons in a harmonic oscillator trap in two dimensions

The Hamiltonian of the quantum dot is given by

 
$$ \hat{H} = \hat{H}_0 + \hat{V}, $$

 
where \( \hat{H}_0 \) is the many-body HO Hamiltonian, and \( \hat{V} \) is the inter-electron Coulomb interactions. In dimensionless units,

 
$$ \hat{V}= \sum_{i < j}^N \frac{1}{r_{ij}}, $$

 
with \( r_{ij}=\sqrt{\mathbf{r}_i^2 - \mathbf{r}_j^2} \).

This leads to the separable Hamiltonian, with the relative motion part given by (\( r_{ij}=r \))

 
$$ \hat{H}_r=-\nabla^2_r + \frac{1}{4}\omega^2r^2+ \frac{1}{r}, $$

 
plus a standard Harmonic Oscillator problem for the center-of-mass motion. This system has analytical solutions in two and three dimensions (M. Taut 1993 and 1994).

Quantum Monte Carlo Motivation

Quantum Monte Carlo Motivation

Quantum Monte Carlo Motivation

Quantum Monte Carlo

The trial wave function

The correlation part of the wave function

To find an ansatz for the correlated part of the wave function, it is useful to rewrite the two-particle local energy in terms of the relative and center-of-mass motion. Let us denote the distance between the two electrons as \( r_{12} \). We omit the center-of-mass motion since we are only interested in the case when \( r_{12} \rightarrow 0 \). The contribution from the center-of-mass (CoM) variable \( \boldsymbol{R}_{\mathrm{CoM}} \) gives only a finite contribution. We focus only on the terms that are relevant for \( r_{12} \) and for three dimensions. The relevant local energy operator becomes then (with \( l=0 \))

 
$$ \lim_{r_{12} \rightarrow 0}E_L(R)= \frac{1}{{\cal R}_T(r_{12})}\left(-2\frac{d^2}{dr_{ij}^2}-\frac{4}{r_{ij}}\frac{d}{dr_{ij}}+ \frac{2}{r_{ij}}\right){\cal R}_T(r_{12}). $$

 
In order to avoid divergencies when \( r_{12}\rightarrow 0 \) we obtain the so-called cusp condition

 
$$ \frac{d {\cal R}_T(r_{12})}{dr_{12}} = \frac{1}{2} {\cal R}_T(r_{12})\qquad r_{12}\to 0 $$

Resulting ansatz

 
$$ {\cal R}_T \propto \exp{(r_{ij}/2)}, $$

 
for anti-parallel spins and

 
$$ {\cal R}_T \propto \exp{(r_{ij}/4)}, $$

 
for anti-parallel spins. This is the so-called cusp condition for the relative motion, resulting in a minimal requirement for the correlation part of the wave fuction. For general systems containing more than say two electrons, we have this condition for each electron pair \( ij \).

The VMC code

# Importing various packages
from math import exp, sqrt
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys

#Trial wave function for quantum dots in two dims
def WaveFunction(r,alpha,beta):
    r1 = r[0,0]**2 + r[0,1]**2
    r2 = r[1,0]**2 + r[1,1]**2
    r12 = sqrt((r[0,0]-r[1,0])**2 + (r[0,1]-r[1,1])**2)
    deno = r12/(1+beta*r12)
    return exp(-0.5*alpha*(r1+r2)+deno)

#Local energy  for quantum dots in two dims, using analytical local energy
def LocalEnergy(r,alpha,beta):
    
    r1 = (r[0,0]**2 + r[0,1]**2)
    r2 = (r[1,0]**2 + r[1,1]**2)
    r12 = sqrt((r[0,0]-r[1,0])**2 + (r[0,1]-r[1,1])**2)
    deno = 1.0/(1+beta*r12)
    deno2 = deno*deno
    return 0.5*(1-alpha*alpha)*(r1 + r2) +2.0*alpha + 1.0/r12+deno2*(alpha*r12-deno2+2*beta*deno-1.0/r12)

# The Monte Carlo sampling with the Metropolis algo
def MonteCarloSampling():

    NumberMCcycles= 100000
    StepSize = 1.0
    # positions
    PositionOld = np.zeros((NumberParticles,Dimension), np.double)
    PositionNew = np.zeros((NumberParticles,Dimension), np.double)
    # seed for rng generator
    seed()
    # start variational parameter
    alpha = 0.9
    for ia in range(MaxVariations):
        alpha += .025
        AlphaValues[ia] = alpha
        beta = 0.2 
        for jb in range(MaxVariations):
            beta += .01
            BetaValues[jb] = beta
            energy = energy2 = 0.0
            DeltaE = 0.0
            #Initial position
            for i in range(NumberParticles):
                for j in range(Dimension):
                    PositionOld[i,j] = StepSize * (random() - .5)
            wfold = WaveFunction(PositionOld,alpha,beta)

            #Loop over MC MCcycles
            for MCcycle in range(NumberMCcycles):
                #Trial position
                for i in range(NumberParticles):
                    for j in range(Dimension):
                        PositionNew[i,j] = PositionOld[i,j] + StepSize * (random() - .5)
                wfnew = WaveFunction(PositionNew,alpha,beta)

                #Metropolis test to see whether we accept the move
                if random() < wfnew**2 / wfold**2:
                   PositionOld = PositionNew.copy()
                   wfold = wfnew
                   DeltaE = LocalEnergy(PositionOld,alpha,beta)
                energy += DeltaE
                energy2 += DeltaE**2

            #We calculate mean, variance and error ...
            energy /= NumberMCcycles
            energy2 /= NumberMCcycles
            variance = energy2 - energy**2
            error = sqrt(variance/NumberMCcycles)
            Energies[ia,jb] = energy    
    return Energies, AlphaValues, BetaValues


#Here starts the main program with variable declarations
NumberParticles = 2
Dimension = 2
MaxVariations = 10
Energies = np.zeros((MaxVariations,MaxVariations))
AlphaValues = np.zeros(MaxVariations)
BetaValues = np.zeros(MaxVariations)
(Energies, AlphaValues, BetaValues) = MonteCarloSampling()

# Prepare for plots
fig = plt.figure()
ax = fig.gca(projection='3d')
# Plot the surface.
X, Y = np.meshgrid(AlphaValues, BetaValues)
surf = ax.plot_surface(X, Y, Energies,cmap=cm.coolwarm,linewidth=0, antialiased=False)
# Customize the z axis.
zmin = np.matrix(Energies).min()
zmax = np.matrix(Energies).max()
ax.set_zlim(zmin, zmax)
ax.set_xlabel(r'$\alpha$')
ax.set_ylabel(r'$\beta$')
ax.set_zlabel(r'$\langle E \rangle$')
ax.zaxis.set_major_locator(LinearLocator(10))
ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f'))
# Add a color bar which maps values to colors.
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.show()

Technical aspect, improvements and how to define the cost function

Energy derivatives

Derivatives of the local energy

How do we define our cost function?

Meet the variance and its derivatives

The variance defines the cost function

The code for two electrons in two dims with no Coulomb interaction

We present here the code (including importance sampling) for finding the optimal parameter \( \alpha \) using gradient descent with a given learning rate \( \eta \). In principle we should run calculations for various learning rates.

Again, we start first with set up of various files.

# 2-electron VMC code for 2dim quantum dot with importance sampling
# No Coulomb interaction
# Using gaussian rng for new positions and Metropolis- Hastings 
# Energy minimization using standard gradient descent 

# Common imports
from math import exp, sqrt
from random import random, seed, normalvariate
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys
from numba import jit
from scipy.optimize import minimize


# Trial wave function for the 2-electron quantum dot in two dims
def WaveFunction(r,alpha):
    r1 = r[0,0]**2 + r[0,1]**2
    r2 = r[1,0]**2 + r[1,1]**2
    return exp(-0.5*alpha*(r1+r2))

# Local energy  for the 2-electron quantum dot in two dims, using analytical local energy
def LocalEnergy(r,alpha):
    
    r1 = (r[0,0]**2 + r[0,1]**2)
    r2 = (r[1,0]**2 + r[1,1]**2)
    return 0.5*(1-alpha*alpha)*(r1 + r2) +2.0*alpha

# Derivate of wave function ansatz as function of variational parameters
def DerivativeWFansatz(r,alpha):
    
    r1 = (r[0,0]**2 + r[0,1]**2)
    r2 = (r[1,0]**2 + r[1,1]**2)
    WfDer = -0.5*(r1+r2)
    return  WfDer

# Setting up the quantum force for the two-electron quantum dot, recall that it is a vector
def QuantumForce(r,alpha):

    qforce = np.zeros((NumberParticles,Dimension), np.double)
    qforce[0,:] = -2*r[0,:]*alpha
    qforce[1,:] = -2*r[1,:]*alpha
    return qforce

Then comes our Monte Carlo sampling.

# Computing the derivative of the energy and the energy 
# jit decorator tells Numba to compile this function.
# The argument types will be inferred by Numba when function is called.
@jit
def EnergyMinimization(alpha):

    NumberMCcycles= 1000
    # Parameters in the Fokker-Planck simulation of the quantum force
    D = 0.5
    TimeStep = 0.05
    # positions
    PositionOld = np.zeros((NumberParticles,Dimension), np.double)
    PositionNew = np.zeros((NumberParticles,Dimension), np.double)
    # Quantum force
    QuantumForceOld = np.zeros((NumberParticles,Dimension), np.double)
    QuantumForceNew = np.zeros((NumberParticles,Dimension), np.double)

    # seed for rng generator 
    seed()
    energy = 0.0
    DeltaE = 0.0
    EnergyDer = 0.0
    DeltaPsi = 0.0
    DerivativePsiE = 0.0
    #Initial position
    for i in range(NumberParticles):
        for j in range(Dimension):
            PositionOld[i,j] = normalvariate(0.0,1.0)*sqrt(TimeStep)
    wfold = WaveFunction(PositionOld,alpha)
    QuantumForceOld = QuantumForce(PositionOld,alpha)

    #Loop over MC MCcycles
    for MCcycle in range(NumberMCcycles):
        #Trial position moving one particle at the time
        for i in range(NumberParticles):
            for j in range(Dimension):
                PositionNew[i,j] = PositionOld[i,j]+normalvariate(0.0,1.0)*sqrt(TimeStep)+\
                                       QuantumForceOld[i,j]*TimeStep*D
            wfnew = WaveFunction(PositionNew,alpha)
            QuantumForceNew = QuantumForce(PositionNew,alpha)
            GreensFunction = 0.0
            for j in range(Dimension):
                GreensFunction += 0.5*(QuantumForceOld[i,j]+QuantumForceNew[i,j])*\
	                              (D*TimeStep*0.5*(QuantumForceOld[i,j]-QuantumForceNew[i,j])-\
                                      PositionNew[i,j]+PositionOld[i,j])
      
            GreensFunction = 1.0#exp(GreensFunction)
            ProbabilityRatio = GreensFunction*wfnew**2/wfold**2
            #Metropolis-Hastings test to see whether we accept the move
            if random() <= ProbabilityRatio:
                for j in range(Dimension):
                    PositionOld[i,j] = PositionNew[i,j]
                    QuantumForceOld[i,j] = QuantumForceNew[i,j]
                wfold = wfnew
        DeltaE = LocalEnergy(PositionOld,alpha)
        DerPsi = DerivativeWFansatz(PositionOld,alpha)
        DeltaPsi +=DerPsi
        energy += DeltaE
        DerivativePsiE += DerPsi*DeltaE
            
    # We calculate mean values
    energy /= NumberMCcycles
    DerivativePsiE /= NumberMCcycles
    DeltaPsi /= NumberMCcycles
    EnergyDer  = 2*(DerivativePsiE-DeltaPsi*energy)
    return energy, EnergyDer

Finally, here we use the gradient descent method with a fixed learning rate and a fixed number of iterations. This code is meant for illustrative purposes only. We could for example add a test which stops the number of terations when the derivative has reached a certain by us fixed minimal value.

#Here starts the main program with variable declarations
NumberParticles = 2
Dimension = 2
# guess for variational parameters
x0 = 0.5
# Set up iteration using stochastic gradient method
Energy =0 ; EnergyDer = 0
Energy, EnergyDer = EnergyMinimization(x0)

# No adaptive search for a minimum
eta = 0.5
Niterations = 50

Energies = np.zeros(Niterations)
EnergyDerivatives = np.zeros(Niterations)
AlphaValues = np.zeros(Niterations)
Totiterations = np.zeros(Niterations)

for iter in range(Niterations):
    gradients = EnergyDer
    x0 -= eta*gradients
    Energy, EnergyDer = EnergyMinimization(x0)
    Energies[iter] = Energy
    EnergyDerivatives[iter] = EnergyDer
    AlphaValues[iter] = x0
    Totiterations[iter] = iter

plt.subplot(2, 1, 1)
plt.plot(Totiterations, Energies, 'o-')
plt.title('Energy and energy derivatives')
plt.ylabel('Dimensionless energy')
plt.subplot(2, 1, 2)
plt.plot(Totiterations, EnergyDerivatives, '.-')
plt.xlabel(r'$\mathrm{Iterations}$', fontsize=15)
plt.ylabel('Energy derivative')
plt.show()
#nice printout with Pandas
import pandas as pd
from pandas import DataFrame
data ={'Alpha':AlphaValues, 'Energy':Energies,'Derivative':EnergyDerivatives}

frame = pd.DataFrame(data)
print(frame)

We see that the first derivative becomes smaller and smaller and after some forty iterations, it is for all practical purposes almost vanishing. The exact energy is \( 2.0 \) and the optimal variational parameter is \( 1.0 \), as it should.

Why Boltzmann machines?

What is known as restricted Boltzmann Machines (RMB) have received a lot of attention lately. One of the major reasons is that they can be stacked layer-wise to build deep neural networks that capture complicated statistics.

The original RBMs had just one visible layer and a hidden layer, but recently so-called Gaussian-binary RBMs have gained quite some popularity in imaging since they are capable of modeling continuous data that are common to natural images.

Furthermore, they have been used to solve complicated quantum mechanical many-particle problems or classical statistical physics problems like the Ising and Potts classes of models.

Boltzmann Machines

Why use a generative model rather than the more well known discriminative deep neural networks (DNN)?

• Discriminitave methods have several limitations: They are mainly supervised learning methods, thus requiring labeled data. And there are tasks they cannot accomplish, like drawing new examples from an unknown probability distribution.
• A generative model can learn to represent and sample from a probability distribution. The core idea is to learn a parametric model of the probability distribution from which the training data was drawn. As an example

  1. A model for images could learn to draw new examples of cats and dogs, given a training dataset of images of cats and dogs.
  2. Generate a sample of an ordered or disordered Ising model phase, having been given samples of such phases.
  3. Model the trial function for Monte Carlo calculations

Some similarities and differences from DNNs

1. Both use gradient-descent based learning procedures for minimizing cost functions
2. Energy based models don't use backpropagation and automatic differentiation for computing gradients, instead turning to Markov Chain Monte Carlo methods.
3. DNNs often have several hidden layers. A restricted Boltzmann machine has normally only one hidden layer, however several RBMs can be stacked to make up Deep Belief Networks, of which they constitute the building blocks.

History: The RBM was developed by amongst others Geoffrey Hinton, called by some the "Godfather of Deep Learning", working with the University of Toronto and Google.

Boltzmann machines (BM)

A standard BM setup

However, they are often hard to train. This leads to the introduction of so-called restricted BMs, or RBMS. Here we take away all lateral connections between nodes in the visible layer as well as connections between nodes in the hidden layer. The network is illustrated in the figure below.

The structure of the RBM network

The network

The network layers:

1. A function \( \mathbf{x} \) that represents the visible layer, a vector of \( M \) elements (nodes). This layer represents both what the RBM might be given as training input, and what we want it to be able to reconstruct. This might for example be the pixels of an image, the spin values of the Ising model, or coefficients representing speech.
2. The function \( \mathbf{h} \) represents the hidden, or latent, layer. A vector of \( N \) elements (nodes). Also called "feature detectors".

Goals

The goal of the hidden layer is to increase the model's expressive power. We encode complex interactions between visible variables by introducing additional, hidden variables that interact with visible degrees of freedom in a simple manner, yet still reproduce the complex correlations between visible degrees in the data once marginalized over (integrated out).

Examples of this trick being employed in physics:

1. The Hubbard-Stratonovich transformation
2. The introduction of ghost fields in gauge theory
3. Shadow wave functions in Quantum Monte Carlo simulations

The network parameters, to be optimized/learned:

1. \( \mathbf{a} \) represents the visible bias, a vector of same length as \( \mathbf{x} \).
2. \( \mathbf{b} \) represents the hidden bias, a vector of same lenght as \( \mathbf{h} \).
3. \( W \) represents the interaction weights, a matrix of size \( M\times N \).

Joint distribution

The restricted Boltzmann machine is described by a Boltzmann distribution

 
$$ \begin{align} P_{rbm}(\mathbf{x},\mathbf{h}) = \frac{1}{Z} e^{-\frac{1}{T_0}E(\mathbf{x},\mathbf{h})}, \tag{3} \end{align} $$

 
where \( Z \) is the normalization constant or partition function, defined as

 
$$ \begin{align} Z = \int \int e^{-\frac{1}{T_0}E(\mathbf{x},\mathbf{h})} d\mathbf{x} d\mathbf{h}. \tag{4} \end{align} $$

 
It is common to ignore \( T_0 \) by setting it to one.

Network Elements, the energy function

The function \( E(\mathbf{x},\mathbf{h}) \) gives the energy of a configuration (pair of vectors) \( (\mathbf{x}, \mathbf{h}) \). The lower the energy of a configuration, the higher the probability of it. This function also depends on the parameters \( \mathbf{a} \), \( \mathbf{b} \) and \( W \). Thus, when we adjust them during the learning procedure, we are adjusting the energy function to best fit our problem.

An expression for the energy function is

 
$$ E(\hat{x},\hat{h}) = -\sum_{ia}^{NA}b_i^a \alpha_i^a(x_i)-\sum_{jd}^{MD}c_j^d \beta_j^d(h_j)-\sum_{ijad}^{NAMD}b_i^a \alpha_i^a(x_i)c_j^d \beta_j^d(h_j)w_{ij}^{ad}. $$

Here \( \beta_j^d(h_j) \) and \( \alpha_i^a(x_j) \) are so-called transfer functions that map a given input value to a desired feature value. The labels \( a \) and \( d \) denote that there can be multiple transfer functions per variable. The first sum depends only on the visible units. The second on the hidden ones. Note that there is no connection between nodes in a layer.

The quantities \( b \) and \( c \) can be interpreted as the visible and hidden biases, respectively.

The connection between the nodes in the two layers is given by the weights \( w_{ij} \).

Defining different types of RBMs

There are different variants of RBMs, and the differences lie in the types of visible and hidden units we choose as well as in the implementation of the energy function \( E(\mathbf{x},\mathbf{h}) \).

Sampling: Metropolis sampling

In order to sample from the RBM probability distribution it is common to use Markov Chain Monte Carlo (MCMC) algorithms such as Metropolis-Hastings or Gibbs sampling.

RBMs for the quantum many body problem

The idea of applying RBMs to quantum many body problems was presented by G. Carleo and M. Troyer, working with ETH Zurich and Microsoft Research.

Some of their motivation included

• The wave function \( \Psi \) is a monolithic mathematical quantity that contains all the information on a quantum state, be it a single particle or a complex molecule. In principle, an exponential amount of information is needed to fully encode a generic many-body quantum state.
• There are still interesting open problems, including fundamental questions ranging from the dynamical properties of high-dimensional systems to the exact ground-state properties of strongly interacting fermions.
• The difficulty lies in finding a general strategy to reduce the exponential complexity of the full many-body wave function down to its most essential features. That is

  1. \( \rightarrow \) Dimensional reduction
  2. \( \rightarrow \) Feature extraction

• Among the most successful techniques to attack these challenges, artifical neural networks play a prominent role.
• Want to understand whether an artifical neural network may adapt to describe a quantum system.

Choose the right RBM

Carleo and Troyer applied the RBM to the quantum mechanical spin lattice systems of the Ising model and Heisenberg model, with encouraging results. Our goal is to test the method on systems of moving particles. For the spin lattice systems it was natural to use a binary-binary RBM, with the nodes taking values of 1 and -1. For moving particles, on the other hand, we want the visible nodes to be continuous, representing position coordinates. Thus, we start by choosing a Gaussian-binary RBM, where the visible nodes are continuous and hidden nodes take on values of 0 or 1. If eventually we would like the hidden nodes to be continuous as well the rectified linear units seem like the most relevant choice.

Representing the wave function

The wavefunction should be a probability amplitude depending on \( \boldsymbol{x} \). The RBM model is given by the joint distribution of \( \boldsymbol{x} \) and \( \boldsymbol{h} \)

 
$$ \begin{align} F_{rbm}(\mathbf{x},\mathbf{h}) = \frac{1}{Z} e^{-\frac{1}{T_0}E(\mathbf{x},\mathbf{h})}. \tag{7} \end{align} $$

 
To find the marginal distribution of \( \boldsymbol{x} \) we set:

 
$$ \begin{align} F_{rbm}(\mathbf{x}) &= \sum_\mathbf{h} F_{rbm}(\mathbf{x}, \mathbf{h}) \tag{8}\\ &= \frac{1}{Z}\sum_\mathbf{h} e^{-E(\mathbf{x}, \mathbf{h})}. \tag{9} \end{align} $$

 
Now this is what we use to represent the wave function, calling it a neural-network quantum state (NQS)

 
$$ \begin{align} \Psi (\mathbf{x}) &= F_{rbm}(\mathbf{x}) \tag{10}\\ &= \frac{1}{Z}\sum_{\boldsymbol{h}} e^{-E(\mathbf{x}, \mathbf{h})} \tag{11}\\ &= \frac{1}{Z} \sum_{\{h_j\}} e^{-\sum_i^M \frac{(x_i - a_i)^2}{2\sigma^2} + \sum_j^N b_j h_j + \sum_{i,j}^{M,N} \frac{x_i w_{ij} h_j}{\sigma^2}} \tag{12}\\ &= \frac{1}{Z} e^{-\sum_i^M \frac{(x_i - a_i)^2}{2\sigma^2}} \prod_j^N (1 + e^{b_j + \sum_i^M \frac{x_i w_{ij}}{\sigma^2}}). \tag{13}\\ \tag{14} \end{align} $$

Choose the cost/loss function

Now we don't necessarily have training data (unless we generate it by using some other method). However, what we do have is the variational principle which allows us to obtain the ground state wave function by minimizing the expectation value of the energy of a trial wavefunction (corresponding to the untrained NQS). Similarly to the traditional variational Monte Carlo method then, it is the local energy we wish to minimize. The gradient to use for the stochastic gradient descent procedure is

 
$$ \begin{align} C_i = \frac{\partial \langle E_L \rangle}{\partial \theta_i} = 2(\langle E_L \frac{1}{\Psi}\frac{\partial \Psi}{\partial \theta_i} \rangle - \langle E_L \rangle \langle \frac{1}{\Psi}\frac{\partial \Psi}{\partial \theta_i} \rangle ), \tag{15} \end{align} $$

 
where the local energy is given by

 
$$ \begin{align} E_L = \frac{1}{\Psi} \hat{\mathbf{H}} \Psi. \tag{16} \end{align} $$

Running the codes

Energy as function of iterations, \( N=2 \) electrons

Energy as function of iterations, no Physics info \( N=2 \) electrons

Energy as function of iterations, \( N=6 \) electrons

Wave function analysis, onebody densities \( N=2 \)

Onebody densities \( N=6 \), \( \hbar\omega=1.0 \) a.u.

Onebody densities \( N=6 \), \( \hbar\omega=0.1 \) a.u.

Onebody densities \( N=30 \), \( \hbar\omega=1.0 \) a.u.

Onebody densities \( N=30 \), \( \hbar\omega=0.1 \) a.u.

Conclusions and where do we stand

Appendix: Mathematical details

The original RBM had binary visible and hidden nodes. They were showne to be universal approximators of discrete distributions. It was also shown that adding hidden units yields strictly improved modelling power. The common choice of binary values are 0 and 1. However, in some physics applications, -1 and 1 might be a more natural choice. We will here use 0 and 1.

Here we look at some of the relevant equations for a binary-binary RBM.

 
$$ \begin{align} E_{BB}(\boldsymbol{x}, \mathbf{h}) = - \sum_i^M x_i a_i- \sum_j^N b_j h_j - \sum_{i,j}^{M,N} x_i w_{ij} h_j. \tag{17} \end{align} $$

 
$$ \begin{align} p_{BB}(\boldsymbol{x}, \boldsymbol{h}) =& \frac{1}{Z_{BB}} e^{\sum_i^M a_i x_i + \sum_j^N b_j h_j + \sum_{ij}^{M,N} x_i w_{ij} h_j} \tag{18}\\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} \tag{19} \end{align} $$

with the partition function

 
$$ \begin{align} Z_{BB} = \sum_{\boldsymbol{x}, \boldsymbol{h}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} . \tag{20} \end{align} $$

Marginal Probability Density Functions

In order to find the probability of any configuration of the visible units we derive the marginal probability density function.

 
$$ \begin{align} p_{BB} (\boldsymbol{x}) =& \sum_{\boldsymbol{h}} p_{BB} (\boldsymbol{x}, \boldsymbol{h}) \tag{21}\\ =& \frac{1}{Z_{BB}} \sum_{\boldsymbol{h}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \sum_{\boldsymbol{h}} e^{\sum_j^N (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \sum_{\boldsymbol{h}} \prod_j^N e^{ (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \bigg ( \sum_{h_1} e^{(b_1 + \boldsymbol{x}^T \boldsymbol{w}_{\ast 1})h_1} \times \sum_{h_2} e^{(b_2 + \boldsymbol{x}^T \boldsymbol{w}_{\ast 2})h_2} \times \nonumber \\ & ... \times \sum_{h_2} e^{(b_N + \boldsymbol{x}^T \boldsymbol{w}_{\ast N})h_N} \bigg ) \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N \sum_{h_j} e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}) h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N (1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}) . \tag{22} \end{align} $$

A similar derivation yields the marginal probability of the hidden units

 
$$ \begin{align} p_{BB} (\boldsymbol{h}) = \frac{1}{Z_{BB}} e^{\boldsymbol{b}^T \boldsymbol{h}} \prod_i^M (1 + e^{a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h}}) . \tag{23} \end{align} $$

Conditional Probability Density Functions

We derive the probability of the hidden units given the visible units using Bayes' rule

 
$$ \begin{align} p_{BB} (\boldsymbol{h}|\boldsymbol{x}) =& \frac{p_{BB} (\boldsymbol{x}, \boldsymbol{h})}{p_{BB} (\boldsymbol{x})} \nonumber \\ =& \frac{ \frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} } {\frac{1}{Z_{BB}} e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N (1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}})} \nonumber \\ =& \frac{ e^{\boldsymbol{x}^T \boldsymbol{a}} e^{ \sum_j^N (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} ) h_j} } { e^{\boldsymbol{x}^T \boldsymbol{a}} \prod_j^N (1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}})} \nonumber \\ =& \prod_j^N \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} ) h_j} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \nonumber \\ =& \prod_j^N p_{BB} (h_j| \boldsymbol{x}) . \tag{24} \end{align} $$

From this we find the probability of a hidden unit being "on" or "off":

 
$$ \begin{align} p_{BB} (h_j=1 | \boldsymbol{x}) =& \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} ) h_j} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \tag{25}\\ =& \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} )} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \tag{26}\\ =& \frac{ 1 }{1 + e^{-(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})} } , \tag{27} \end{align} $$

 
and

 
$$ \begin{align} p_{BB} (h_j=0 | \boldsymbol{x}) =\frac{ 1 }{1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}} } . \tag{28} \end{align} $$

Similarly we have that the conditional probability of the visible units given the hidden ones are

 
$$ \begin{align} p_{BB} (\boldsymbol{x}|\boldsymbol{h}) =& \prod_i^M \frac{ e^{ (a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h}) x_i} }{ 1 + e^{a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h}} } \tag{29}\\ &= \prod_i^M p_{BB} (x_i | \boldsymbol{h}) . \tag{30} \end{align} $$

 
$$ \begin{align} p_{BB} (x_i=1 | \boldsymbol{h}) =& \frac{1}{1 + e^{-(a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h} )}} \tag{31}\\ p_{BB} (x_i=0 | \boldsymbol{h}) =& \frac{1}{1 + e^{a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h} }} . \tag{32} \end{align} $$

Python version for the two non-interacting particles

# 2-electron VMC code for 2dim quantum dot with importance sampling
# Using gaussian rng for new positions and Metropolis- Hastings 
# Added restricted boltzmann machine method for dealing with the wavefunction
# RBM code based heavily off of:
# https://github.com/CompPhysics/ComputationalPhysics2/tree/gh-pages/doc/Programs/BoltzmannMachines/MLcpp/src/CppCode/ob
from math import exp, sqrt
from random import random, seed, normalvariate
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys



# Trial wave function for the 2-electron quantum dot in two dims
def WaveFunction(r,a,b,w):
    sigma=1.0
    sig2 = sigma**2
    Psi1 = 0.0
    Psi2 = 1.0
    Q = Qfac(r,b,w)
    
    for iq in range(NumberParticles):
        for ix in range(Dimension):
            Psi1 += (r[iq,ix]-a[iq,ix])**2
            
    for ih in range(NumberHidden):
        Psi2 *= (1.0 + np.exp(Q[ih]))
        
    Psi1 = np.exp(-Psi1/(2*sig2))

    return Psi1*Psi2

# Local energy  for the 2-electron quantum dot in two dims, using analytical local energy
def LocalEnergy(r,a,b,w):
    sigma=1.0
    sig2 = sigma**2
    locenergy = 0.0
    
    Q = Qfac(r,b,w)

    for iq in range(NumberParticles):
        for ix in range(Dimension):
            sum1 = 0.0
            sum2 = 0.0
            for ih in range(NumberHidden):
                sum1 += w[iq,ix,ih]/(1+np.exp(-Q[ih]))
                sum2 += w[iq,ix,ih]**2 * np.exp(Q[ih]) / (1.0 + np.exp(Q[ih]))**2
    
            dlnpsi1 = -(r[iq,ix] - a[iq,ix]) /sig2 + sum1/sig2
            dlnpsi2 = -1/sig2 + sum2/sig2**2
            locenergy += 0.5*(-dlnpsi1*dlnpsi1 - dlnpsi2 + r[iq,ix]**2)
            
    if(interaction==True):
        for iq1 in range(NumberParticles):
            for iq2 in range(iq1):
                distance = 0.0
                for ix in range(Dimension):
                    distance += (r[iq1,ix] - r[iq2,ix])**2
                    
                locenergy += 1/sqrt(distance)
                
    return locenergy

# Derivate of wave function ansatz as function of variational parameters
def DerivativeWFansatz(r,a,b,w):
    
    sigma=1.0
    sig2 = sigma**2
    
    Q = Qfac(r,b,w)
    
    WfDer = np.empty((3,),dtype=object)
    WfDer = [np.copy(a),np.copy(b),np.copy(w)]
    
    WfDer[0] = (r-a)/sig2
    WfDer[1] = 1 / (1 + np.exp(-Q))
    
    for ih in range(NumberHidden):
        WfDer[2][:,:,ih] = w[:,:,ih] / (sig2*(1+np.exp(-Q[ih])))
            
    return  WfDer

# Setting up the quantum force for the two-electron quantum dot, recall that it is a vector
def QuantumForce(r,a,b,w):

    sigma=1.0
    sig2 = sigma**2
    
    qforce = np.zeros((NumberParticles,Dimension), np.double)
    sum1 = np.zeros((NumberParticles,Dimension), np.double)
    
    Q = Qfac(r,b,w)
    
    for ih in range(NumberHidden):
        sum1 += w[:,:,ih]/(1+np.exp(-Q[ih]))
    
    qforce = 2*(-(r-a)/sig2 + sum1/sig2)
    
    return qforce
    
def Qfac(r,b,w):
    Q = np.zeros((NumberHidden), np.double)
    temp = np.zeros((NumberHidden), np.double)
    
    for ih in range(NumberHidden):
        temp[ih] = (r*w[:,:,ih]).sum()
        
    Q = b + temp
    
    return Q
    
# Computing the derivative of the energy and the energy 
def EnergyMinimization(a,b,w):

    NumberMCcycles= 10000
    # Parameters in the Fokker-Planck simulation of the quantum force
    D = 0.5
    TimeStep = 0.05
    # positions
    PositionOld = np.zeros((NumberParticles,Dimension), np.double)
    PositionNew = np.zeros((NumberParticles,Dimension), np.double)
    # Quantum force
    QuantumForceOld = np.zeros((NumberParticles,Dimension), np.double)
    QuantumForceNew = np.zeros((NumberParticles,Dimension), np.double)

    # seed for rng generator 
    seed()
    energy = 0.0
    DeltaE = 0.0

    EnergyDer = np.empty((3,),dtype=object)
    DeltaPsi = np.empty((3,),dtype=object)
    DerivativePsiE = np.empty((3,),dtype=object)
    EnergyDer = [np.copy(a),np.copy(b),np.copy(w)]
    DeltaPsi = [np.copy(a),np.copy(b),np.copy(w)]
    DerivativePsiE = [np.copy(a),np.copy(b),np.copy(w)]
    for i in range(3): EnergyDer[i].fill(0.0)
    for i in range(3): DeltaPsi[i].fill(0.0)
    for i in range(3): DerivativePsiE[i].fill(0.0)

    
    #Initial position
    for i in range(NumberParticles):
        for j in range(Dimension):
            PositionOld[i,j] = normalvariate(0.0,1.0)*sqrt(TimeStep)
    wfold = WaveFunction(PositionOld,a,b,w)
    QuantumForceOld = QuantumForce(PositionOld,a,b,w)

    #Loop over MC MCcycles
    for MCcycle in range(NumberMCcycles):
        #Trial position moving one particle at the time
        for i in range(NumberParticles):
            for j in range(Dimension):
                PositionNew[i,j] = PositionOld[i,j]+normalvariate(0.0,1.0)*sqrt(TimeStep)+\
                                       QuantumForceOld[i,j]*TimeStep*D
            wfnew = WaveFunction(PositionNew,a,b,w)
            QuantumForceNew = QuantumForce(PositionNew,a,b,w)
            
            GreensFunction = 0.0
            for j in range(Dimension):
                GreensFunction += 0.5*(QuantumForceOld[i,j]+QuantumForceNew[i,j])*\
                                      (D*TimeStep*0.5*(QuantumForceOld[i,j]-QuantumForceNew[i,j])-\
                                      PositionNew[i,j]+PositionOld[i,j])
      
            GreensFunction = exp(GreensFunction)
            ProbabilityRatio = GreensFunction*wfnew**2/wfold**2
            #Metropolis-Hastings test to see whether we accept the move
            if random() <= ProbabilityRatio:
                for j in range(Dimension):
                    PositionOld[i,j] = PositionNew[i,j]
                    QuantumForceOld[i,j] = QuantumForceNew[i,j]
                wfold = wfnew
        #print("wf new:        ", wfnew)
        #print("force on 1 new:", QuantumForceNew[0,:])
        #print("pos of 1 new:  ", PositionNew[0,:])
        #print("force on 2 new:", QuantumForceNew[1,:])
        #print("pos of 2 new:  ", PositionNew[1,:])
        DeltaE = LocalEnergy(PositionOld,a,b,w)
        DerPsi = DerivativeWFansatz(PositionOld,a,b,w)
        
        DeltaPsi[0] += DerPsi[0]
        DeltaPsi[1] += DerPsi[1]
        DeltaPsi[2] += DerPsi[2]
        
        energy += DeltaE

        DerivativePsiE[0] += DerPsi[0]*DeltaE
        DerivativePsiE[1] += DerPsi[1]*DeltaE
        DerivativePsiE[2] += DerPsi[2]*DeltaE
            
    # We calculate mean values
    energy /= NumberMCcycles
    DerivativePsiE[0] /= NumberMCcycles
    DerivativePsiE[1] /= NumberMCcycles
    DerivativePsiE[2] /= NumberMCcycles
    DeltaPsi[0] /= NumberMCcycles
    DeltaPsi[1] /= NumberMCcycles
    DeltaPsi[2] /= NumberMCcycles
    EnergyDer[0]  = 2*(DerivativePsiE[0]-DeltaPsi[0]*energy)
    EnergyDer[1]  = 2*(DerivativePsiE[1]-DeltaPsi[1]*energy)
    EnergyDer[2]  = 2*(DerivativePsiE[2]-DeltaPsi[2]*energy)
    return energy, EnergyDer


#Here starts the main program with variable declarations
NumberParticles = 2
Dimension = 2
NumberHidden = 2

interaction=False

# guess for parameters
a=np.random.normal(loc=0.0, scale=0.001, size=(NumberParticles,Dimension))
b=np.random.normal(loc=0.0, scale=0.001, size=(NumberHidden))
w=np.random.normal(loc=0.0, scale=0.001, size=(NumberParticles,Dimension,NumberHidden))
# Set up iteration using stochastic gradient method
Energy = 0
EDerivative = np.empty((3,),dtype=object)
EDerivative = [np.copy(a),np.copy(b),np.copy(w)]
# Learning rate eta, max iterations, need to change to adaptive learning rate
eta = 0.001
MaxIterations = 50
iter = 0
np.seterr(invalid='raise')
Energies = np.zeros(MaxIterations)
EnergyDerivatives1 = np.zeros(MaxIterations)
EnergyDerivatives2 = np.zeros(MaxIterations)

while iter < MaxIterations:
    Energy, EDerivative = EnergyMinimization(a,b,w)
    agradient = EDerivative[0]
    bgradient = EDerivative[1]
    wgradient = EDerivative[2]
    a -= eta*agradient
    b -= eta*bgradient 
    w -= eta*wgradient 
    Energies[iter] = Energy
    print("Energy:",Energy)

    iter += 1

#nice printout with Pandas
import pandas as pd
from pandas import DataFrame
pd.set_option('max_columns', 6)
data ={'Energy':Energies}#,'A Derivative':EnergyDerivatives1,'B Derivative':EnergyDerivatives2,'Weights Derivative':EnergyDerivatives3}

frame = pd.DataFrame(data)
print(frame)