## Archive for May, 2010

### Variational transition state theory

Monday, May 17th, 2010

This is my note from Prof. Ron Elber’s lecture on “varational transition state theory”.

In the system, there are more than one minimum. If we can find a surface on which the probability of state going to one minimum is equal to going to another. Finding such a surface can be formulate as$min[I=\int_{S}e^{-\beta U(x)}d\sigma]$

The surface S is usually in high dimension and complicated. A simplification is that the surface is a plane, that is, $S: \vec{n} \cdot \vec{x} - b=0$.

Therefore the problem for finding the S is equivalent to $min[I(\vec{n},b)]$.

We reformulate the integral,

$I_{approximate}=\int_{V}\delta(\vec{n} \cdot \vec{x} - b)dx\int_{S}e^{-\beta U(x)}|\nabla S|d\sigma$,

because it is a plane, no Jacob transformation, so $\nabla S = 1$.

So the minimization is to make sure, $\frac{\partial I}{\partial n_{i}}=0, \frac{\partial I}{\partial b}=0$.

The index i here is due to the fact that in general one plane is not a good separator, so multiple planes (piece-wise) are needed, i is just the i-th plane. So reorganize the definition of I, we have

$I_{approximate}= \sum_{i=1}^{m}\int_{V}\delta(\vec{n_{i}} \cdot \vec{x} - b_{i})dx\int_{S}e^{-\beta U(x)}d\sigma$ For simplicity, I consider $I_approximate$ as I.

The $\delta$ function can be rewritten as a limitation, so we have $I = \lim_{k \to \infty}\int \sqrt{\frac{k}{\pi}}e^{-k(\vec{n_{i}} \cdot \vec{x}-b_{i})^2}e^{-\beta U(x)}dx$.

Some tricks here:

$\frac{d log(I)}{db}=\frac{1}{I}\frac{\partial I}{\partial b} = \frac{1}{I}\lim_{k \to \infty} \int \sqrt{\frac{k}{\pi}} [2k(\vec{n_{i}} \cdot \vec{x}-b_{i})]e^{-k(\vec{n_{i}} \cdot \vec{x}-b_{i})^2}e^{-\beta U(x)}dx$

define some new symbols

$x=(q,y), q=\vec{n} \cdot \vec{x}, \delta q = \vec{n} \cdot \vec{x}-b$

rewrite the above with the new symbols:

$\frac{d log(I)}{db}=\frac{1}{I}\lim_{k \to \infty} \int \sqrt{\frac{k}{\pi}}(-2k\delta q)e^{-k(\delta q)^{2}}e^{-\beta U(q,y)}d\delta q dy$

Apply Talyor expansion to U, we have

$e^{-\beta U(q,y)}=e^{-\beta U - \beta \frac{\partial U}{\partial \delta q}|_{\delta q =0}\delta q}=e^{-\beta U} (1-\beta (\frac{\partial U}{\partial \delta q})_{\delta q=0}\delta q)$

Now we have

$\frac{d log(I)}{db}=\frac{1}{I}\lim_{k \to \infty} \int \sqrt{\frac{k}{\pi}}(-2k\delta q)e^{-k(\delta q)^{2}}e^{-\beta U} (1-\beta (\frac{\partial U}{\partial \delta q})_{\delta q=0}\delta q)d\delta q dy$

There exists such a simplification:

$\frac{1}{I}\lim_{k \to \infty}\int \sqrt{\frac{k}{\pi}}(-2k\delta q)e^{-k(\delta q)^{2}}e^{-\beta U}d\delta q dy=0$

So we have

$\frac{d log(I)}{db}=-\frac{\beta}{I}\lim_{k \to \infty}\int \sqrt{\frac{k}{\pi}}(-2k\delta q)e^{-k(\delta q)^{2}}e^{-\beta U} (\frac{\partial U}{\partial \delta q})_{\delta q=0}\delta q d\delta q dy$

$\frac{d log(I)}{db}=-\frac{\beta}{I}\lim_{k \to \infty}\int \sqrt{\frac{k}{\pi}}(-2k(\delta q)^{2})e^{-k(\delta q)^{2}} d\delta q \int e^{-\beta U} (\frac{\partial U}{\partial \delta q})_{\delta q=0} dy\\=-\frac{\beta}{I}\int e^{-\beta U} (\frac{\partial U}{\partial \delta q})_{\delta q=0} dy=\ll \vec{F} \gg = \bar F$

The above is simplified because

$\lim_{k \to \infty}\int \sqrt{\frac{k}{\pi}}(-2k(\delta q)^{2})e^{-k(\delta q)^{2}} d\delta q = \lim_{k \to \infty}\sqrt{\frac{k}{\pi}} \frac{\partial }{\partial k}\int e^{-(k \delta q)^{2}} d\delta q=\lim_{k \to \infty}\sqrt{\frac{k}{\pi}} \frac{\partial }{\partial k} \frac{\sqrt{\pi}}{k}=-\lim_{k \to \infty}\sqrt{\frac{k}{\pi}}\frac{\sqrt{\pi}}{k^2}=must=-1$The must here means, in previous introduction of limit function for the $\delta$function, the coefficient is not right. Anyway, the trick is that in the end,all k terms are reduced. I am too lazy to revisit each term and change them.

Because we expect $\frac{\partial I}{\partial b}=0\Rightarrow \bar{F} =0$. So just make sure the total force along the surface, here under the approximation the plane the total force is equal to zero.

I don’t know what it will turn out after we do $\frac{\partial I}{\partial n_{i}}=0 \forall i \in [1,m]$ though.