Locating saddle points of any index on potential energy surfaces by the generalized gentlest ascent dynamics

Abstract

The system of ordinary differential equations for the method of the gentlest ascent dynamics (GAD) is tested to determine the saddle points of the potential energy surface of some molecules. The method has been proposed earlier [E and Zhou in Nonlinearity 24:1831 (2011)]. We additionally use the metric of curvilinear internal coordinates. By a number of examples, we explain the possibilities of a GAD curve; it can find the transition state of interest by a gentlest ascent, directly or indirectly, or not. A GAD curve can be a model of a reaction path, if it does not contain a turning point for the energy. We further discuss generalized GAD formulas for the search of saddle points of a higher index. We calculate diverse examples.

Appendix: Hessian matrix of the Lennard-Jones cluster

The potential energy surface (PES) of an n-atomic Lennard-Jones (LJ) cluster is as follows:

$$\begin{aligned} V(x_1,y_1,z_1,\ldots,x_n,y_n,z_n) = 4 \epsilon \sum_{i=1}^{n} \sum _{j=i+1}^{n}\, \left[ \sigma ^{12}\left(\frac{1}{r_{ij}}\right)^{12}- \sigma^{6}\left(\frac{1}{r_{ij}}\right)^{6} \right] \end{aligned}$$

(32)

where the double sum can be shortened to $\sum _{i<j}$, and

$$\begin{aligned} r_{ij}^2 = (x_i-x_j)^2 + (y_i-y_j)^2 +(z_i-z_j)^2 \end{aligned}$$

(33)

is the square of the distance between atoms i and j. $\epsilon , \sigma$ are parameters of the current cluster.

$V(\mathbf{q})$ is a $3n$-dimensional surface over $I\!\!R^{3n}$. Let $\mathbf{grad}(\mathbf{q})$ be its gradient vector, $\nabla _\mathbf{q}V(\mathbf{q})$, and let $\mathbf{H}(\mathbf{q})$ be the matrix of its second derivatives, the Hessian $\nabla _\mathbf{q} \mathbf{grad}^T$. We could not find the calculation of the Hessian in the older literature; thus, we give it here in the appendix. We get for the distance between two atoms

$$\begin{aligned} \frac{\partial r_{ij}}{\partial x_i} = \frac{x_i-x_j}{r_{ij}}, \quad \frac{\partial r_{ij}}{\partial x_j} = \frac{x_j-x_i}{r_{ij}}. \end{aligned}$$

(34)

A single component of the gradient becomes

$$\begin{aligned} \frac{\partial V}{\partial x_i} = 4 \epsilon \sum _{i\ne j}^n \, (x_i-x_j)\ \left[ -12 \sigma ^{12}\left(\frac{1}{r_{ij}}\right)^{14} +6 \sigma ^{6}\left(\frac{1}{r_{ij}}\right)^{8} \right].\end{aligned}$$

(35)

If we put the abbreviation

$$\begin{aligned} T1(ij) =: \left[ -12 \sigma ^{12}\left(\frac{1}{r_{ij}}\right)^{14} +6 \sigma ^{6}\left(\frac{1}{r_{ij}}\right)^{8} \right] \end{aligned}$$

(36)

and if we set for simplification of the summation $T1(ii)=0$, then we can write

$$\begin{aligned} \frac{\partial V}{\partial x_i} = 4 \epsilon \sum _{j=1}^n \, (x_i-x_j)\ T1(ij), \end{aligned}$$

(37)

and analogously we have for $i=1,\ldots ,n$

$$\begin{aligned} \frac{\partial V}{\partial y_i}=4 \epsilon \sum _{j=1}^n \, (y_i-y_j)\ T1(ij) \ \quad \hbox {and}\ \quad \frac{\partial V}{\partial z_i} =4 \epsilon \sum _{j=1}^n \, (z_i-z_j)\ T1(ij). \end{aligned}$$

(38)

We further define the abbreviation

$$\begin{aligned} T2(ij) =: \left[ 12*14 \sigma ^{12}\left(\frac{1}{r_{ij}}\right)^{16} -6*8 \sigma ^{6}\left(\frac{1}{r_{ij}}\right)^{10} \right] \ . \end{aligned}$$

(39)

Again, we set $T2(ii)=0$. Then, single components of the Hessian are for the same atom

$$\begin{aligned} H(x_i,x_i)=\frac{\partial ^2 V}{\partial x_i^2} = 4 \epsilon \sum _{j=1}^n \,\left( T1(ij) + (x_i-x_j)^2\ T2(ij) \right), \end{aligned}$$

(40)

and

$$\begin{aligned} H(x_i,y_i)=\frac{\partial ^2 V}{\partial x_i\,\partial y_i} = 4 \epsilon \sum _{j=1}^n \, (x_i-x_j)* (y_i-y_j)\ T2(ij), \end{aligned}$$

(41)

and analogously one gets $H(x_i,z_i)$ and $H(y_i,z_i)$ and their symmetric counterparts. If different atoms with numbers i and j are involved, then we get

$$\begin{aligned} H(x_i,x_j)=\frac{\partial ^2 V}{\partial x_i\,\partial x_j} = -4 \epsilon \,\left( T1(ij) + (x_i-x_j)^2\ T2(ij) \right), \end{aligned}$$

(42)

and

$$\begin{aligned} H(x_i,y_j)=\frac{\partial ^2 V}{\partial x_i\,\partial y_j} = -4 \epsilon \, (x_i-x_j)* (y_i-y_j)\ T2(ij), \end{aligned}$$

(43)

and analogously one gets $H(x_i,z_j)$ and $H(y_i,z_j)$ and their symmetric counterparts.