======================
Solution to exercise 5
======================

--------------------------------
Nudged Elastic Band calculations
--------------------------------

.. contents::
.. section-numbering::
.. include:: ../charents.txt





Self-diffusion on the Al(110) surface
=====================================

The script `<neb1.py>`__ will find the minimum energy path (MEP) for
diffusion along the rows.  If you fill in 9 instead of ``??`` in the line::

  for i in range(??):

you will get 9 images between initial and final states, which is more
than enough.  The barrier is 0.11 eV:

|j1i| |j1t| |j1f|

In order to find the barrier for diffusion across the
rows, you must change this line defining the final state::

  final[-1].SetCartesianPosition((a / 2, 3 * b / 2, h))

to::

  final[-1].SetCartesianPosition((3 * a / 2, b / 2, h))

The barrier for this process is 0.56 eV:

|j2i| |j2t| |j2f|

The MEP for the exchange process is found by replacing the above
line by::

  final[24].SetCartesianPosition((3 * a / 2, 3 * b / 2, h))
  final[-1].SetCartesianPosition((a, b, 0))

The exchange process is more complex than the two simple hopping
events, and we need to relax the path using 15 x 10 steps - so change
5 to 15 in this line::

  for i in range(5):

The barrier is 0.15 eV:

|j3i| |j3t| |j3f|

Notice that the path for the exchange process goes through a meta
stable state.  Plot the energy along the path like this::

  plottrajectory jump3.traj

.. figure:: energy.gif
   :width: 500
   :align: center
   :alt: Energy barrier for exchange
   :figwidth: 500

   The energy barrier for exchange diffusion found using the NEB
   method with 22 images.


.. |j1i| image:: j1i.gif
.. |j1t| image:: j1t.gif
.. |j1f| image:: j1f.gif
.. |j2i| image:: j2i.gif
.. |j2t| image:: j2t.gif
.. |j2f| image:: j2f.gif
.. |j3i| image:: j3i.gif
.. |j3t| image:: j3t.gif
.. |j3f| image:: j3f.gif






Analytical problem on transition state theory (TST)
===================================================

The four basic assumptions of TST:

1) The dynamics of the atoms can be described by Newton's equation,
   i.e. classical dynamics.

2) The Born-Oppenheimer approximation is valid, i.e. the time scale of
   the motion of electrons and atoms is different enough that we can
   solve for the electronic degrees of freedom for fixed positions of
   the nuclei.  This gives a potential energy surface (PES) for the
   motion of the nuclei.

3) The transitions are slow enough that a Boltzmann distribution of
   energy is established and maintained for each degree of freedom of
   the reactant(s).

4) A region, the so called transition state (TS), of the PES in
   between reactants and products can be defined in such a way that
   any reactive trajectory must go through this region and if the
   system makes it there and is on its way towards products it will
   continue to go to the product region of the PES and stay there for
   an extended time.  The potential surface should have a simple and
   narrow barrier (without dips) in between the initial and final
   state.  Then TST can be expected to be a good approximation.

Here is how to plot the potential using ``gnuplot``::

  gnuplot> Vs=0.2
  gnuplot> b=3
  gnuplot> a=2
  gnuplot> V(x,y,z)=Vs*(exp(-cos(2*pi*x/b)-cos(2*pi*y/b)-2*a*z)-2*exp(-a*z))
  gnuplot> splot [-b:b] [-1:0] V(x,0,y)

.. figure:: gnuplot.gif
   :width: 400
   :align: center
   :alt: H/fcc(100) potential
   :figwidth: 400

   Potential energy in the *x, z* plane with *y* = 0.

The minima are at:

  (*x, y, z*) = (*ib*, *jb*, -2/|alpha|),

where *i* and *j* are integers.  The saddle points are located at:

  (*x, y, z*) = (*ib*/2, *jb*/2, 0),

where *i* + *j* is an odd integer.
 
The Taylor expansion close to the minima at (0, 0, -2/|alpha|) is:

  *V*\ (*x, y, z*) = *V*\ :sub:`s`\ *e*\ :sup:`2`\ [-1 + 2(|pi|/*b*)\ :sup:`2`\ (*x*\ :sup:`2` + *y*\ :sup:`2`) + |alpha|\ :sup:`2`\ (*z* + 2/|alpha|)\ :sup:`2`],

and close to the saddle point at (*b*/2, 0, 0) we have:

  *V*\ (*x, y, z*) = *V*\ :sub:`s`\ [-1 - 2(|pi|/*b*)\ :sup:`2`\ (*x* - *b*/2)\ :sup:`2` +  2(|pi|/*b*)\ :sup:`2`\ *y* :sup:`2` + |alpha|\ :sup:`2`\ *z*\ :sup:`2`]

The energy barrier is:

  *E*\ :sub:`a` = *V*\ :sub:`s`\ (*e*\ :sup:`2` - 1) = 1.28 eV

The frequencies can be found from the force constants like this:

  |nu| = (*k*/\ *m*)\ :sup:`1/2`\ /(2\ |pi|).

From the Taylor expansion close to the minimum, we get:

  *k*\ :sub:`x` = 4\ *V*\ :sub:`s`\ (|pi|\ *e*\ /*b*)\ :sup:`2`

  *k*\ :sub:`y` = 4\ *V*\ :sub:`s`\ (|pi|\ *e*\ /*b*)\ :sup:`2`

  *k*\ :sub:`z` = 2\ *V*\ :sub:`s`\ (|alpha|\ *e*)\ :sup:`2`

Plugging in the numbers, we get the frequencies: 
 
  3.98 x 10\ :sup:`13`/sec

  3.98 x 10\ :sup:`13`/sec

  5.38 x 10\ :sup:`13`/sec

At the saddle point (transition state) there are two positive force
constants:

  *k*\ :sub:`y` = 4\ *V*\ :sub:`s`\ (|pi|\ /*b*)\ :sup:`2`

  *k*\ :sub:`z` = 2\ *V*\ :sub:`s`\ |alpha|\ :sup:`2`

corresponding to the frequencies:

  1.46 x 10\ :sup:`13`/sec

  1.98 x 10\ :sup:`13`/sec

The third frequency is imaginary, because the force constant *k*\
:sub:`x` is negative.  This mode is along the *x*-axis which is the
reaction coordinate.

The prefactor in the Arrhenius expression is equal to the product of
the three frequencies at the minimum divided by the product of the two
frequencies at the transition state:

  (*V*\ :sub:`s`\ /*m*)\ :sup:`1/2`\ *e*\ :sup:`3`\ /*b* = 2.95 x 10\ :sup:`14`/sec

The correction to the activation energy from zero point motion is
-0.20 eV, so the final expression for the rate is

  k =  2.95 x 10\ :sup:`14`/sec exp[-1.07 eV/(*k*\ :sub:`B`\ *T*)]

The average length of time in between diffusion hops at room
temperature (298 K) is 4805 sec and 0.11 msec at 400 K.





