Overview

Topics

Reminder from last week and derivation of equations for the gradients of the energy

Gradient methods:
1. Semi-Newton methods (Broyden's algorithm) and gradient descent
2. Steepest descent and conjugate gradient descent
3. Discussion of codes for VMC calculations

Video and handwriten notes

Reading suggestions

Video of lecture

"Handwritten notes:"https://github.com/CompPhysics/ComputationalPhysics2/blob/gh-pages/doc/HandWrittenNotes/2025/NotesFebruary28.pdf"

Teaching Material, videos and written material

These lecture notes

Video on the Conjugate Gradient methods

Recommended background literature, Convex Optimization by Boyd and Vandenberghe. Their lecture slides are very useful (warning, these are some 300 pages).

Brief reminder on Newton-Raphson's method

Let us quickly remind ourselves on how we derive the above method.

Perhaps the most celebrated of all one-dimensional root-finding routines is Newton's method, also called the Newton-Raphson method. This method requires the evaluation of both the function $f$ and its derivative $f'$ at arbitrary points. If you can only calculate the derivative numerically and/or your function is not of the smooth type, we normally discourage the use of this method.

The equations

The Newton-Raphson formula consists geometrically of extending the tangent line at a current point until it crosses zero, then setting the next guess to the abscissa of that zero-crossing. The mathematics behind this method is rather simple. Employing a Taylor expansion for $x$ sufficiently close to the solution $s$ , we have

$f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots. \tag{1}$

For small enough values of the function and for well-behaved functions, the terms beyond linear are unimportant, hence we obtain

$f(x)+(s-x)f'(x)\approx 0,$

yielding

$s\approx x-\frac{f(x)}{f'(x)}.$

Having in mind an iterative procedure, it is natural to start iterating with

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$

Simple geometric interpretation

The above is Newton-Raphson's method. It has a simple geometric interpretation, namely $x_{n+1}$ is the point where the tangent from $(x_n,f(x_n))$ crosses the $x$ -axis. Close to the solution, Newton-Raphson converges fast to the desired result. However, if we are far from a root, where the higher-order terms in the series are important, the Newton-Raphson formula can give grossly inaccurate results. For instance, the initial guess for the root might be so far from the true root as to let the search interval include a local maximum or minimum of the function. If an iteration places a trial guess near such a local extremum, so that the first derivative nearly vanishes, then Newton-Raphson may fail totally

Extending to more than one variable

Newton's method can be generalized to systems of several non-linear equations and variables. Consider the case with two equations

$\begin{array}{cc} f_1(x_1,x_2) &=0\\ f_2(x_1,x_2) &=0,\end{array}$

which we Taylor expand to obtain

$\begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1 \partial f_1/\partial x_1+h_2 \partial f_1/\partial x_2+\dots\\ 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1 \partial f_2/\partial x_1+h_2 \partial f_2/\partial x_2+\dots \end{array}.$

Defining the Jacobian matrix $\hat{J}$ we have

$\hat{J}=\left( \begin{array}{cc} \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\ \partial f_2/\partial x_1 &\partial f_2/\partial x_2 \end{array} \right),$

we can rephrase Newton's method as

$\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)= \left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right),$

where we have defined

$\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)= -{\bf \hat{J}}^{-1} \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right).$

We need thus to compute the inverse of the Jacobian matrix and it is to understand that difficulties may arise in case $\hat{J}$ is nearly singular.

It is rather straightforward to extend the above scheme to systems of more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.

Steepest descent

The basic idea of gradient descent is that a function $F(\mathbf{x})$ , $\mathbf{x} \equiv (x_1,\cdots,x_n)$ , decreases fastest if one goes from $\bf {x}$ in the direction of the negative gradient $-\nabla F(\mathbf{x})$ .

It can be shown that if

$\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k),$

with $\gamma_k > 0$ .

For $\gamma_k$ small enough, then $F(\mathbf{x}_{k+1}) \leq F(\mathbf{x}_k)$ . This means that for a sufficiently small $\gamma_k$ we are always moving towards smaller function values, i.e a minimum.

More on Steepest descent

The previous observation is the basis of the method of steepest descent, which is also referred to as just gradient descent (GD). One starts with an initial guess $\mathbf{x}_0$ for a minimum of $F$ and computes new approximations according to

$\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0.$

The parameter $\gamma_k$ is often referred to as the step length or the learning rate within the context of Machine Learning.

The ideal

Ideally the sequence $\{\mathbf{x}_k \}_{k=0}$ converges to a global minimum of the function $F$ . In general we do not know if we are in a global or local minimum. In the special case when $F$ is a convex function, all local minima are also global minima, so in this case gradient descent can converge to the global solution. The advantage of this scheme is that it is conceptually simple and straightforward to implement. However the method in this form has some severe limitations:

In machine learing we are often faced with non-convex high dimensional cost functions with many local minima. Since GD is deterministic we will get stuck in a local minimum, if the method converges, unless we have a very good intial guess. This also implies that the scheme is sensitive to the chosen initial condition.

Note that the gradient is a function of $\mathbf{x} = (x_1,\cdots,x_n)$ which makes it expensive to compute numerically.

The sensitiveness of the gradient descent

The gradient descent method is sensitive to the choice of learning rate $\gamma_k$ . This is due to the fact that we are only guaranteed that $F(\mathbf{x}_{k+1}) \leq F(\mathbf{x}_k)$ for sufficiently small $\gamma_k$ . The problem is to determine an optimal learning rate. If the learning rate is chosen too small the method will take a long time to converge and if it is too large we can experience erratic behavior.

Many of these shortcomings can be alleviated by introducing randomness. One such method is that of Stochastic Gradient Descent (SGD), see below.

Convex functions

Ideally we want our cost/loss function to be convex(concave).

First we give the definition of a convex set: A set $C$ in $\mathbb{R}^n$ is said to be convex if, for all $x$ and $y$ in $C$ and all $t \in (0,1)$ , the point $(1 − t)x + ty$ also belongs to C. Geometrically this means that every point on the line segment connecting $x$ and $y$ is in $C$ as discussed below.

The convex subsets of $\mathbb{R}$ are the intervals of $\mathbb{R}$ . Examples of convex sets of $\mathbb{R}^2$ are the regular polygons (triangles, rectangles, pentagons, etc...).

Convex function

Convex function: Let $X \subset \mathbb{R}^n$ be a convex set. Assume that the function $f: X \rightarrow \mathbb{R}$ is continuous, then $f$ is said to be convex if

$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)$

for all $x_1, x_2 \in X$ and for all $t \in [0,1]$ . If $\leq$ is replaced with a strict inequaltiy in the definition, we demand $x_1 \neq x_2$ and $t\in(0,1)$ then $f$ is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting $f(x_1)$ and $f(x_2)$ , the value of the function on the interval $[x_1,x_2]$ is always below the line as illustrated below.

Conditions on convex functions

In the following we state first and second-order conditions which ensures convexity of a function $f$ . We write $D_f$ to denote the domain of $f$ , i.e the subset of $R^n$ where $f$ is defined. For more details and proofs we refer to: S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press.

First order condition

Suppose $f$ is differentiable (i.e $\nabla f(x)$ is well defined for all $x$ in the domain of $f$ ). Then $f$ is convex if and only if $D_f$ is a convex set and

$f(y) \geq f(x) + \nabla f(x)^T (y-x)$

holds for all $x,y \in D_f$ . This condition means that for a convex function the first order Taylor expansion (right hand side above) at any point a global under estimator of the function. To convince yourself you can make a drawing of $f(x) = x^2+1$ and draw the tangent line to $f(x)$ and note that it is always below the graph.

Second order condition

Assume that $f$ is twice differentiable, i.e the Hessian matrix exists at each point in $D_f$ . Then $f$ is convex if and only if $D_f$ is a convex set and its Hessian is positive semi-definite for all $x\in D_f$ . For a single-variable function this reduces to $f''(x) \geq 0$ . Geometrically this means that $f$ has nonnegative curvature everywhere.

This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.

More on convex functions

The next result is of great importance to us and the reason why we are going on about convex functions. In machine learning we frequently have to minimize a loss/cost function in order to find the best parameters for the model we are considering.

Ideally we want the global minimum (for high-dimensional models it is hard to know if we have local or global minimum). However, if the cost/loss function is convex the following result provides invaluable information:

Any minimum is global for convex functions

Consider the problem of finding $x \in \mathbb{R}^n$ such that $f(x)$ is minimal, where $f$ is convex and differentiable. Then, any point $x^*$ that satisfies $\nabla f(x^*) = 0$ is a global minimum.

This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.

Some simple problems

Show that $f(x)=x^2$ is convex for $x \in \mathbb{R}$ using the definition of convexity. Hint: If you re-write the definition, $f$ is convex if the following holds for all $x,y \in D_f$ and any $\lambda \in [0,1]$ $\lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0$.

Using the second order condition show that the following functions are convex on the specified domain.

$f(x) = e^x$ is convex for $x \in \mathbb{R}$ .

$g(x) = -\ln(x)$ is convex for $x \in (0,\infty)$ .

Let $f(x) = x^2$ and $g(x) = e^x$ . Show that $f(g(x))$ and $g(f(x))$ is convex for $x \in \mathbb{R}$ . Also show that if $f(x)$ is any convex function than $h(x) = e^{f(x)}$ is convex.

A norm is any function that satisfy the following properties

$f(\alpha x) = |\alpha| f(x)$ for all $\alpha \in \mathbb{R}$ .

$f(x+y) \leq f(x) + f(y)$

$f(x) \leq 0$ for all $x \in \mathbb{R}^n$ with equality if and only if $x = 0$

Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).

Standard steepest descent

Before we proceed, we would like to discuss the approach called the standard Steepest descent, which again leads to us having to be able to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).

The success of the CG method

for finding solutions of non-linear problems is based on the theory of conjugate gradients for linear systems of equations. It belongs to the class of iterative methods for solving problems from linear algebra of the type

$\begin{equation*} \hat{A}\hat{x} = \hat{b}. \end{equation*}$

In the iterative process we end up with a problem like

$\begin{equation*} \hat{r}= \hat{b}-\hat{A}\hat{x}, \end{equation*}$

where $\hat{r}$ is the so-called residual or error in the iterative process.

When we have found the exact solution, $\hat{r}=0$ .

Gradient method

The residual is zero when we reach the minimum of the quadratic equation

$\begin{equation*} P(\hat{x})=\frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T\hat{b}, \end{equation*}$

with the constraint that the matrix $\hat{A}$ is positive definite and symmetric. This defines also the Hessian and we want it to be positive definite.

Steepest descent method

We denote the initial guess for $\hat{x}$ as $\hat{x}_0$ . We can assume without loss of generality that

$\begin{equation*} \hat{x}_0=0, \end{equation*}$

or consider the system

$\begin{equation*} \hat{A}\hat{z} = \hat{b}-\hat{A}\hat{x}_0, \end{equation*}$

instead.

Steepest descent method

One can show that the solution $\hat{x}$ is also the unique minimizer of the quadratic form

$\begin{equation*} f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n. \end{equation*}$

This suggests taking the first basis vector $\hat{r}_1$ (see below for definition) to be the gradient of $f$ at $\hat{x}=\hat{x}_0$ , which equals

$\begin{equation*} \hat{A}\hat{x}_0-\hat{b}, \end{equation*}$

and $\hat{x}_0=0$ it is equal $-\hat{b}$ .

Final expressions

We can compute the residual iteratively as

$\begin{equation*} \hat{r}_{k+1}=\hat{b}-\hat{A}\hat{x}_{k+1}, \end{equation*}$

which equals

$\begin{equation*} \hat{b}-\hat{A}(\hat{x}_k+\alpha_k\hat{r}_k), \end{equation*}$

or

$\begin{equation*} (\hat{b}-\hat{A}\hat{x}_k)-\alpha_k\hat{A}\hat{r}_k, \end{equation*}$

which gives

$\alpha_k = \frac{\hat{r}_k^T\hat{r}_k}{\hat{r}_k^T\hat{A}\hat{r}_k}$

leading to the iterative scheme

$\begin{equation*} \hat{x}_{k+1}=\hat{x}_k-\alpha_k\hat{r}_{k}, \end{equation*}$

Our simple $2\times 2$ example

Last week we introduced the simple two-dimensional function

$f(x_1,x_2)=x_1^2+x_1x_2+10x_2^2-5x_1-3x_2,$

which is of the form (in terms of vectors and matrices)

$f(\boldsymbol{x})=\frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x}-\boldsymbol{b}^T\boldsymbol{x},$

where we have

$\boldsymbol{x}=\begin{bmatrix} x_1 \\ x_2\end{bmatrix},$

$\boldsymbol{b}=\begin{bmatrix} 5 \\ 3\end{bmatrix},$

and

$\boldsymbol{A}=\begin{bmatrix} 2 & 1\\ 1& 20\end{bmatrix}.$

Derivatives and more

Optimizing the above equation, that is

$\nabla f = 0 = \boldsymbol{A}\boldsymbol{x}-\boldsymbol{b},$

which leads to a simple matrix-inversion problem

$\boldsymbol{x}=\boldsymbol{A}^{-1}\boldsymbol{b}.$

This problem is easy to solve since we can calculate the inverse. Alternatively, we can solve the two coupled equations with two unknowns

$\frac{\partial f}{\partial x_1}=2x_1+x_2-5=0,$

and

$\frac{\partial f}{\partial x_2}=x_1+20x_2-3=0,$

with solutions $x_1=97/39$ and $x_2=1/39$ .

First a simple gradient descent solution

from random import random, seed
import numpy as np
# The function we want to optimize
def f(x):
    return x[0]**2 + 10.0*x[1]**2+x[0]*x[1]-5.0*x[0]-3*x[1]

def df(x):
    return np.array([2*x[0]+x[1]-5.0, x[0]+20*x[1]-3.0])

# Matrix A
A = np.array([[2,1],[1,20]])
# Get the eigenvalues
EigValues, EigVectors = np.linalg.eig(A)
print(f"Eigenvalues of Matrix A:{EigValues}")
# Vector b
b = np.array([[5,3]])
# Exact solution
exact = np.linalg.inv(A) @ b.T
print("Exact solution:",exact)

# start simple gradient descent with random guess
x = np.random.randn(2,1)

# finding optimal learning rate
eta = 1.0/np.max(EigValues)
Niterations = 100

# a brute force implementation of gradient descent
for iter in range(Niterations):
    gradient = df(x)
#    print(gradient)
    x -= eta*gradient
    
print("Absolute error for Gradient descent solution",np.abs(x-exact))

Implementing the steepest descent

# start Steepest descent

IterMax = 30
tolerance = 1.0e-14
# start simple gradient descent with random guess again
x = np.random.randn(2,1)

# The residual
r = A @ x-b.T 
i = 0
# a more intelligent iterative process
while (i <= IterMax):
    z = A @ r
    c = r.T @ r
    deno = r.T @ z
    alpha = c/deno
    x -= alpha*r
    r =  (A @ x)-b.T;
    if(np.sqrt(r.T @ r) < tolerance):
        break
    i += 1

print("Absolute error for Steepest descent solution",np.abs(x-exact))

Simple codes for steepest descent and conjugate gradient using a $2\times 2$ matrix, in c++

#include <cmath>
#include <iostream>
#include <fstream>
#include <iomanip>
#include "vectormatrixclass.h"
using namespace  std;
//   Main function begins here
int main(int  argc, char * argv[]){
  int dim = 2;
  Vector x(dim),xsd(dim), b(dim),x0(dim);
  Matrix A(dim,dim);

  // Set our initial guess
  x0(0) = x0(1) = 0;
  // Set the matrix
  A(0,0) =  2;    A(1,0) =  1;   A(0,1) =  1;   A(1,1) =  20;
  b(0) = 5; b(1) = 3;
  cout << "The Matrix A that we are using: " << endl;
  A.Print();
  cout << endl;
  xsd = SteepestDescent(A,b,x0);
  cout << "The approximate solution using Steepest Descent is: " << endl;
  xsd.Print();
  cout << endl;
}

The routine for the steepest descent method

Vector SteepestDescent(Matrix A, Vector b, Vector x0){
  int IterMax, i;
  int dim = x0.Dimension();
  const double tolerance = 1.0e-14;
  Vector x(dim),f(dim),z(dim);
  double c,alpha,d;
  IterMax = 30;
  x = x0;
  r = A*x-b;
  i = 0;
  while (i <= IterMax){
    z = A*r;
    c = dot(r,r);
    alpha = c/dot(r,z);
    x = x - alpha*r;
    r =  A*x-b;
    if(sqrt(dot(r,r)) < tolerance) break;
    i++;
  }
  return x;
}

Conjugate gradient method

In the CG method we define so-called conjugate directions and two vectors $\hat{s}$ and $\hat{t}$ are said to be conjugate if

$\begin{equation*} \hat{s}^T\hat{A}\hat{t}= 0. \end{equation*}$

The philosophy of the CG method is to perform searches in various conjugate directions of our vectors $\hat{x}_i$ obeying the above criterion, namely

$\begin{equation*} \hat{x}_i^T\hat{A}\hat{x}_j= 0. \end{equation*}$

Two vectors are conjugate if they are orthogonal with respect to this inner product. Being conjugate is a symmetric relation: if $\hat{s}$ is conjugate to $\hat{t}$ , then $\hat{t}$ is conjugate to $\hat{s}$ .

Conjugate gradient method

An example is given by the eigenvectors of the matrix

$\begin{equation*} \hat{v}_i^T\hat{A}\hat{v}_j= \lambda\hat{v}_i^T\hat{v}_j, \end{equation*}$

which is zero unless $i=j$ .

Conjugate gradient method

Assume now that we have a symmetric positive-definite matrix $\hat{A}$ of size $n\times n$ . At each iteration $i+1$ we obtain the conjugate direction of a vector

$\begin{equation*} \hat{x}_{i+1}=\hat{x}_{i}+\alpha_i\hat{p}_{i}. \end{equation*}$

We assume that $\hat{p}_{i}$ is a sequence of $n$ mutually conjugate directions. Then the $\hat{p}_{i}$ form a basis of $R^n$ and we can expand the solution $ \hat{A}\hat{x} = \hat{b}$ in this basis, namely

$\begin{equation*} \hat{x} = \sum^{n}_{i=1} \alpha_i \hat{p}_i. \end{equation*}$

Conjugate gradient method

The coefficients are given by

$\begin{equation*} \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}. \end{equation*}$

Multiplying with $\hat{p}_k^T$ from the left gives

$\begin{equation*} \hat{p}_k^T \hat{A}\hat{x} = \sum^{n}_{i=1} \alpha_i\hat{p}_k^T \hat{A}\hat{p}_i= \hat{p}_k^T \hat{b}, \end{equation*}$

and we can define the coefficients $\alpha_k$ as

$\begin{equation*} \alpha_k = \frac{\hat{p}_k^T \hat{b}}{\hat{p}_k^T \hat{A} \hat{p}_k} \end{equation*}$

Conjugate gradient method and iterations

If we choose the conjugate vectors $\hat{p}_k$ carefully, then we may not need all of them to obtain a good approximation to the solution $\hat{x}$ . We want to regard the conjugate gradient method as an iterative method. This will us to solve systems where $n$ is so large that the direct method would take too much time.

We denote the initial guess for $\hat{x}$ as $\hat{x}_0$ . We can assume without loss of generality that

$\begin{equation*} \hat{x}_0=0, \end{equation*}$

or consider the system

$\begin{equation*} \hat{A}\hat{z} = \hat{b}-\hat{A}\hat{x}_0, \end{equation*}$

instead.

Conjugate gradient method

One can show that the solution $\hat{x}$ is also the unique minimizer of the quadratic form

$\begin{equation*} f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n. \end{equation*}$

This suggests taking the first basis vector $\hat{p}_1$ to be the gradient of $f$ at $\hat{x}=\hat{x}_0$ , which equals

$\begin{equation*} \hat{A}\hat{x}_0-\hat{b}, \end{equation*}$

and $\hat{x}_0=0$ it is equal $-\hat{b}$ . The other vectors in the basis will be conjugate to the gradient, hence the name conjugate gradient method.

Conjugate gradient method

Let $\hat{r}_k$ be the residual at the $k$ -th step:

$\begin{equation*} \hat{r}_k=\hat{b}-\hat{A}\hat{x}_k. \end{equation*}$

Note that $\hat{r}_k$ is the negative gradient of $f$ at $\hat{x}=\hat{x}_k$ , so the gradient descent method would be to move in the direction $\hat{r}_k$ . Here, we insist that the directions $\hat{p}_k$ are conjugate to each other, so we take the direction closest to the gradient $\hat{r}_k$ under the conjugacy constraint. This gives the following expression

$\begin{equation*} \hat{p}_{k+1}=\hat{r}_k-\frac{\hat{p}_k^T \hat{A}\hat{r}_k}{\hat{p}_k^T\hat{A}\hat{p}_k} \hat{p}_k. \end{equation*}$

Conjugate gradient method

We can also compute the residual iteratively as

$\begin{equation*} \hat{r}_{k+1}=\hat{b}-\hat{A}\hat{x}_{k+1}, \end{equation*}$

which equals

$\begin{equation*} \hat{b}-\hat{A}(\hat{x}_k+\alpha_k\hat{p}_k), \end{equation*}$

or

$\begin{equation*} (\hat{b}-\hat{A}\hat{x}_k)-\alpha_k\hat{A}\hat{p}_k, \end{equation*}$

which gives

$\begin{equation*} \hat{r}_{k+1}=\hat{r}_k-\hat{A}\hat{p}_{k}, \end{equation*}$

Simple implementation of the Conjugate gradient algorithm

# Implementing the conjugate gradient descent for the example function

# start simple gradient descent with random guess again
x = np.random.randn(2,1)

# The residual
r = b.T-A @ x
v = r
c = r.T @ r
i = 0
while (i <= IterMax):
    z = A @ v
    t = c/(v.T @ z)
    x += t*v
    r -= t*z;
    d = r.T @ r
    if(np.sqrt(d) < tolerance):
      break
    v = r + (d/c)*v
    c = d
    i+=1

print("Absolute error for the Conjugate gradient solution",np.abs(x-exact))

The result is much better than those from the steepest descent or the plain gradient descent.

Broyden–Fletcher–Goldfarb–Shanno algorithm

The optimization problem is to minimize $f(\mathbf {x} )$ where $\mathbf {x}$ is a vector in $R^{n}$ , and $f$ is a differentiable scalar function. There are no constraints on the values that $\mathbf {x}$ can take.

The algorithm begins at an initial estimate for the optimal value $\mathbf {x}_{0}$ and proceeds iteratively to get a better estimate at each stage.

The search direction $p_k$ at stage $k$ is given by the solution of the analogue of the Newton equation

$B_{k}\mathbf {p} _{k}=-\nabla f(\mathbf {x}_{k}),$

where $B_{k}$ is an approximation to the Hessian matrix, which is updated iteratively at each stage, and $\nabla f(\mathbf {x} _{k})$ is the gradient of the function evaluated at $x_k$ . A line search in the direction $p_k$ is then used to find the next point $x_{k+1}$ by minimising

$f(\mathbf {x}_{k}+\alpha \mathbf {p}_{k}),$

over the scalar $\alpha > 0$ .

See notes from whiteboard

Using gradient descent methods, limitations

Gradient descent (GD) finds local minima of our function. Since the GD algorithm is deterministic, if it converges, it will converge to a local minimum of our energy function. If we deal with extremely rugged landscapes with many local minima, this can lead to poor performance.

GD is sensitive to initial conditions. One consequence of the local nature of GD is that initial conditions matter. Depending on where one starts, one will end up at a different local minima. Therefore, it is very important to think about how one initializes the training process. This is true for GD as well as more complicated variants of GD.

Gradients are computationally expensive to calculate. In fields like statistics and Machine learning, the function to optimize is a sum of terms, with one term for each data point. For example, in linear regression, $E \propto \sum_{i=1}^n (y_i - \mathbf{w}^T\cdot\mathbf{x}_i)^2$ ; for logistic regression, the square error is replaced by the cross entropy. To calculate the gradient we have to sum over all $n$ data points. Doing this at every GD step becomes extremely computationally expensive. An ingenious solution to this, is to calculate the gradients using small subsets of the data called "mini batches". This has the added benefit of introducing stochasticity into our algorithm.

GD is extremely sensitive to the choice of learning rates. If the learning rate is very small, the training process take an extremely long time. For larger learning rates, GD can diverge and give poor results. Furthermore, depending on what the local landscape looks like, we have to modify the learning rates to ensure convergence. Ideally, we would adaptively choose the learning rates to match the landscape.

GD treats all directions in parameter space uniformly. Another major drawback of GD is that unlike Newton's method, the learning rate for GD is the same in all directions in parameter space. For this reason, the maximum learning rate is set by the behavior of the steepest direction and this can significantly slow down training. Ideally, we would like to take large steps in flat directions and small steps in steep directions. Since we are exploring rugged landscapes where curvatures change, this requires us to keep track of not only the gradient but second derivatives. The ideal scenario would be to calculate the Hessian but this proves to be too computationally expensive.

GD can take exponential time to escape saddle points, even with random initialization. As we mentioned, GD is extremely sensitive to initial condition since it determines the particular local minimum GD would eventually reach. However, even with a good initialization scheme, through the introduction of randomness, GD can still take exponential time to escape saddle points.

Codes from numerical recipes

You can however use codes we have adapted from the text Numerical Recipes in C++, see chapter 10.7. Here we present a program, which you also can find at the webpage of the course we use the functions dfpmin and lnsrch. This is a variant of the Broyden et al algorithm discussed in the previous slide.

The program uses the harmonic oscillator in one dimensions as example.

The program does not use armadillo to handle vectors and matrices, but employs rather my own vector-matrix class. These auxiliary functions, and the main program model.cpp can all be found under the program link here.

Below we show only excerpts from the main program. For the full program, see the above link.

Finding the minimum of the harmonic oscillator model in one dimension

//   Main function begins here
int main()
{
     int n, iter;
     double gtol, fret;
     double alpha;
     n = 1;
//   reserve space in memory for vectors containing the variational
//   parameters
     Vector g(n), p(n);
     cout << "Read in guess for alpha" << endl;
     cin >> alpha;
     gtol = 1.0e-5;
//   now call dfmin and compute the minimum
     p(0) = alpha;
     dfpmin(p, n, gtol, &iter, &fret, Efunction, dEfunction);
     cout << "Value of energy minimum = " << fret << endl;
     cout << "Number of iterations = " << iter << endl;
     cout << "Value of alpha at minimum = " << p(0) << endl;
      return 0;
}  // end of main program

Functions to observe

The functions Efunction and dEfunction compute the expectation value of the energy and its derivative. They use the the quasi-Newton method of Broyden, Fletcher, Goldfarb, and Shanno (BFGS) It uses the first derivatives only. The BFGS algorithm has proven good performance even for non-smooth optimizations. These functions need to be changed when you want to your own derivatives.

//  this function defines the expectation value of the local energy
double Efunction(Vector  &x)
{
  double value = x(0)*x(0)*0.5+1.0/(8*x(0)*x(0));
  return value;
} // end of function to evaluate

//  this function defines the derivative of the energy 
void dEfunction(Vector &x, Vector &g)
{
  g(0) = x(0)-1.0/(4*x(0)*x(0)*x(0));
} // end of function to evaluate

You need to change these functions in order to compute the local energy for your system. I used 1000 cycles per call to get a new value of $\langle E_L[\alpha]\rangle$ . When I compute the local energy I also compute its derivative. After roughly 10-20 iterations I got a converged result in terms of $\alpha$ .

Bringing back the full code from last week

General expression for the derivative of the energy

We have

$\bar{E}_{\alpha} = 2\left( \langle \frac{\bar{\psi}_{\alpha}}{\psi[\alpha]}E_L[\alpha]\rangle -\langle \frac{\bar{\psi}_{\alpha}}{\psi[\alpha]}\rangle\langle E_L[\alpha] \rangle\right).$

Python program for 2-electrons in 2 dimensions

# 2-electron VMC code for 2dim quantum dot with importance sampling
# Using gaussian rng for new positions and Metropolis- Hastings 
# Added energy minimization with gradient descent using fixed step size
# To do: replace with optimization codes from scipy and/or use stochastic gradient descent
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,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 the 2-electron quantum dot 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)

# Derivate of wave function ansatz as function of variational parameters
def DerivativeWFansatz(r,alpha,beta):
    
    WfDer  = np.zeros((2), np.double)
    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
    WfDer[0] = -0.5*(r1+r2)
    WfDer[1] = -r12*r12*deno2
    return  WfDer

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

    qforce = np.zeros((NumberParticles,Dimension), np.double)
    r12 = sqrt((r[0,0]-r[1,0])**2 + (r[0,1]-r[1,1])**2)
    deno = 1.0/(1+beta*r12)
    qforce[0,:] = -2*r[0,:]*alpha*(r[0,:]-r[1,:])*deno*deno/r12
    qforce[1,:] = -2*r[1,:]*alpha*(r[1,:]-r[0,:])*deno*deno/r12
    return qforce
    

# Computing the derivative of the energy and the energy 
def EnergyMinimization(alpha, beta):

    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.zeros((2), np.double)
    DeltaPsi = np.zeros((2), np.double)
    DerivativePsiE = np.zeros((2), np.double)
    #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,beta)
    QuantumForceOld = QuantumForce(PositionOld,alpha, beta)

    #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,beta)
            QuantumForceNew = QuantumForce(PositionNew,alpha, beta)
            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
        DeltaE = LocalEnergy(PositionOld,alpha,beta)
        DerPsi = DerivativeWFansatz(PositionOld,alpha,beta)
        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


#Here starts the main program with variable declarations
NumberParticles = 2
Dimension = 2
# guess for variational parameters
alpha = 0.9
beta = 0.2
# Set up iteration using gradient descent method
Energy = 0
EDerivative = np.zeros((2), np.double)
eta = 0.01
Niterations = 50
# 
for iter in range(Niterations):
    Energy, EDerivative = EnergyMinimization(alpha,beta)
    alphagradient = EDerivative[0]
    betagradient = EDerivative[1]
    alpha -= eta*alphagradient
    beta -= eta*betagradient 

print(alpha, beta)
print(Energy, EDerivative[0], EDerivative[1])

Using Broyden's algorithm in scipy

The following function uses the above described BFGS algorithm. Here we have defined a function which calculates the energy and a function which computes the first derivative.

# 2-electron VMC code for 2dim quantum dot with importance sampling
# Using gaussian rng for new positions and Metropolis- Hastings 
# Added energy minimization using the BFGS algorithm, see p. 136 of https://www.springer.com/it/book/9780387303031
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
from scipy.optimize import minimize
import sys


# Trial wave function for the 2-electron quantum dot 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 the 2-electron quantum dot 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)

# Derivate of wave function ansatz as function of variational parameters
def DerivativeWFansatz(r,alpha,beta):
    
    WfDer  = np.zeros((2), np.double)
    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
    WfDer[0] = -0.5*(r1+r2)
    WfDer[1] = -r12*r12*deno2
    return  WfDer

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

    qforce = np.zeros((NumberParticles,Dimension), np.double)
    r12 = sqrt((r[0,0]-r[1,0])**2 + (r[0,1]-r[1,1])**2)
    deno = 1.0/(1+beta*r12)
    qforce[0,:] = -2*r[0,:]*alpha*(r[0,:]-r[1,:])*deno*deno/r12
    qforce[1,:] = -2*r[1,:]*alpha*(r[1,:]-r[0,:])*deno*deno/r12
    return qforce
    

# Computing the derivative of the energy and the energy 
def EnergyDerivative(x0):

    
    # Parameters in the Fokker-Planck simulation of the quantum force
    D = 0.5
    TimeStep = 0.05
    NumberMCcycles= 10000
    # 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)

    energy = 0.0
    DeltaE = 0.0
    alpha = x0[0]
    beta = x0[1]
    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,beta)
    QuantumForceOld = QuantumForce(PositionOld,alpha, beta)

    #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,beta)
            QuantumForceNew = QuantumForce(PositionNew,alpha, beta)
            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
        DeltaE = LocalEnergy(PositionOld,alpha,beta)
        DerPsi = DerivativeWFansatz(PositionOld,alpha,beta)
        DeltaPsi += DerPsi
        energy += DeltaE
        DerivativePsiE += DerPsi*DeltaE
            
    # We calculate mean values
    energy /= NumberMCcycles
    DerivativePsiE /= NumberMCcycles
    DeltaPsi /= NumberMCcycles
    EnergyDer  = 2*(DerivativePsiE-DeltaPsi*energy)
    return EnergyDer


# Computing the expectation value of the local energy 
def Energy(x0):
    # 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)

    energy = 0.0
    DeltaE = 0.0
    alpha = x0[0]
    beta = x0[1]
    NumberMCcycles= 10000
    #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,beta)
    QuantumForceOld = QuantumForce(PositionOld,alpha, beta)

    #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,beta)
            QuantumForceNew = QuantumForce(PositionNew,alpha, beta)
            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
        DeltaE = LocalEnergy(PositionOld,alpha,beta)
        energy += DeltaE
            
    # We calculate mean values
    energy /= NumberMCcycles
    return energy




#Here starts the main program with variable declarations
NumberParticles = 2
Dimension = 2
# seed for rng generator 
seed()
# guess for variational parameters
x0 = np.array([0.9,0.2])
# Using Broydens method
res = minimize(Energy, x0, method='BFGS', jac=EnergyDerivative, options={'gtol': 1e-4,'disp': True})
print(res.x)

Note that the minimize function returns the finale values for the variable $\alpha=x0[0]$ and $\beta=x0[1]$ in the array $x$ .

Gradient Methods and energy optimization

February 28, 2025