Energy optimization

Question 1

Why does one need to do energy optimization?

Answer 1

When starting with atoms randomly placed in space they may be very far from the lowest energy state and to start with such a configuration in the simulation will make it very hard to do any calculations.

Question 2

Explain the simplex method.

Answer 2

If have an energy landscape in N-dim, take N+1 points.
Eg. in 2dim. take 3 points on the energy landscape and calc. the energy in these points. Then reflect the point with highest energy through the average of the others. This gives a new point and the procedure is repeated until the energy is below som threshold.

Question 3

Explain the gradient method: steepest descent method.

Answer 3

Start by calc. the energy and the gradient g_k=1 in the starting point. The step as taken in the negative gradient direction of length lambda:
R_k+1 = R_k + lambda*g_K / abs(g_k)

Then repeat this until in region of low enough energy.

Question 4

Give examples of how to optimize the steepest descent method.

Answer 4

• optimize the step length lambda, if the new point has lower energy this is in the right direction and lambda is increased for the next time, and vice versa.
• calc. a few points along the gradient direction or fit quadratic function to find the min along this direction, the min is then the new point.

Question 5

Explain the conjugated gradients method and what the advantage is compared to the steepest descent method.

Answer 5

In this method is also the previous gradient included in the algorithm. The new search direction for the new point is:
V_k = -g_k + y_k*V_k-1
where the weight y_k depends on the magnitude of the new gradient g_k compared to the previous g_k-1.

The advantage compared to steepest descent method is that when one has a very elongated landscape the steepest descent only jumps back and forth and takes very long time to reach the low energy region, but this method finds it much faster.

Question 6

Explain the second derivative method: Newton-Raphson. And how this can be made less expensive in the quasi-Newton method.

Answer 6

Taylor expand the energy change in the vicinity of a point. For a quadratic function one can use that when the first derivative is =0 this is the min and this gives the min in one step:
x_min = x_k - dV/dx(x=x_k) / (d^2V/dx^2(x=x_k))

In higher dim. the second derivative is the Hessian matrix H_xk and the gradient g_xk the first derivative.
The Newton-Raphson formula is then:
R_min = R_k - g_xk*(H_xk)^-1

But this is costly since one has to invert the Hessian matrix.
So instead one can do quasi-Newton, first only use the gradient and then gradually approx. the Hessian by finite difference.
In the beginning the method is more steepest descent and in the end more like Newton-Raphson.