1. An energy minimization can stop either because a stationary point is found or the program ran out of steps. It’s important that you know why it stopped. If it ran out of steps you have to do another energy minimization starting with the geometry that has the lowest energy. This is not always the last geometry!
2. Larger molecules will require more steps to reach a minimum than smaller molecules. The relationship is roughly linear, so if you double the size of the molecules you will probably take at least twice as many steps, and maybe more. The reason is that each step explores one degree of freedom the molecules, all degrees of freedom must be explored to find the minimum (unless you get lucky), and the number of degrees of freedom increases linearly with the number of atoms (3N-6). This is not quite true for Newton-Raphson energy minimizations if the Hessian is computed, but this is rarely done in practice (see point 5).
3. Floppy molecules with relative flat PESs require more steps than rigid molecules, e.g. a linear hydrocarbon takes longer to optimize than the corresponding cyclic structure. The reason is that flat surfaces mean small gradients and therefore small steps. Also, flat surfaces tend to be non-quadratic so the Newton-Raphson steps have to be scaled back.
4. It is sometimes more efficient to use internal coordinates (bond lengths and angles) for the optimization than Cartesian coordinates, since internal coordinates are more quadratic than Cartesian coordinates. However, if the molecule has many fused rings it might be better to use Cartesians coordinates because of linear-dependence issues.
5. For Newton-Raphson minimizations the Hessian is usually estimated rather than computed exactly because it is expensive to calculate. Attempts are usually made to improve or update the Hessian after each step. So, if you run 50 Newton-Raphson steps from an initial geometry, take the last geometry as the new guess, and run another 50 Newton-Raphson steps you will take a different path towards the minimum than if you run 100 Newton-Raphson steps from the initial geometry. This is because you start with a new guess at the Hessian in step “51” in the first approach. This usually works better in my experience.
6. If you want to find a transition state you must use the Newton-Raphson method, since only the Hessian can tell you in which direction you have to go uphill (the direction is defined by the normal mode corresponding to the imaginary frequency). For the same reason you must compute the Hessian for the initial guess geometry of your transition state.
7. For molecules with more than ~1000 atoms the algorithm needed to invert the Hessian (i.e. produce the inverse of H) becomes very expensive, so steepest descent or conjugate gradient methods are usually used to energy minimize such big molecules.
8. If your starting geometry for ammonium is planar (e.g. in the xy plane), it will stay planar. The reason for this is that the gradient perpendicular to the plane is zero because of symmetry. In essence there is two opposing ways of making ammonia non-planar and the net effect is a cancellation of the force out-of-the plane. Notice that the gradient components in the x and y directions are small (below the optimization threshold) but non-zero, while the gradient in the z-direction is exactly zero. Of course, a zero gradient means no displacement in the out-of-plane direction and ammonia stays planar during the energy minimization.
More generally, the molecule will retain the symmetry of the starting guess structure during the minimization, whether that symmetry corresponds to an energy minimum or not.
2. Larger molecules will require more steps to reach a minimum than smaller molecules. The relationship is roughly linear, so if you double the size of the molecules you will probably take at least twice as many steps, and maybe more. The reason is that each step explores one degree of freedom the molecules, all degrees of freedom must be explored to find the minimum (unless you get lucky), and the number of degrees of freedom increases linearly with the number of atoms (3N-6). This is not quite true for Newton-Raphson energy minimizations if the Hessian is computed, but this is rarely done in practice (see point 5).
3. Floppy molecules with relative flat PESs require more steps than rigid molecules, e.g. a linear hydrocarbon takes longer to optimize than the corresponding cyclic structure. The reason is that flat surfaces mean small gradients and therefore small steps. Also, flat surfaces tend to be non-quadratic so the Newton-Raphson steps have to be scaled back.
4. It is sometimes more efficient to use internal coordinates (bond lengths and angles) for the optimization than Cartesian coordinates, since internal coordinates are more quadratic than Cartesian coordinates. However, if the molecule has many fused rings it might be better to use Cartesians coordinates because of linear-dependence issues.
5. For Newton-Raphson minimizations the Hessian is usually estimated rather than computed exactly because it is expensive to calculate. Attempts are usually made to improve or update the Hessian after each step. So, if you run 50 Newton-Raphson steps from an initial geometry, take the last geometry as the new guess, and run another 50 Newton-Raphson steps you will take a different path towards the minimum than if you run 100 Newton-Raphson steps from the initial geometry. This is because you start with a new guess at the Hessian in step “51” in the first approach. This usually works better in my experience.
6. If you want to find a transition state you must use the Newton-Raphson method, since only the Hessian can tell you in which direction you have to go uphill (the direction is defined by the normal mode corresponding to the imaginary frequency). For the same reason you must compute the Hessian for the initial guess geometry of your transition state.
7. For molecules with more than ~1000 atoms the algorithm needed to invert the Hessian (i.e. produce the inverse of H) becomes very expensive, so steepest descent or conjugate gradient methods are usually used to energy minimize such big molecules.
8. If your starting geometry for ammonium is planar (e.g. in the xy plane), it will stay planar. The reason for this is that the gradient perpendicular to the plane is zero because of symmetry. In essence there is two opposing ways of making ammonia non-planar and the net effect is a cancellation of the force out-of-the plane. Notice that the gradient components in the x and y directions are small (below the optimization threshold) but non-zero, while the gradient in the z-direction is exactly zero. Of course, a zero gradient means no displacement in the out-of-plane direction and ammonia stays planar during the energy minimization.
More generally, the molecule will retain the symmetry of the starting guess structure during the minimization, whether that symmetry corresponds to an energy minimum or not.
Posts
