# Kris's Research Notes

## May 19, 2011

### More on the new energy barriers

Filed under: GaAs Simulations — Kris Reyes @ 6:51 am

In the previous post, we described how detailed balance can be violated for certain events. For such an event, we had assigned an incorrect energy barrier, and hence detailed balance was not satisfied. To address this, we now assign the correct energy barrier: $E(X) - E(X\wedge Y)$, the energy difference between a state $X$ and the resulting state when the diffusing atom is removed: $X\wedge Y$. In this note, we explore how this change affects certain events.

### Example: Breaking from step-edge

Here is a concrete example of how the energy barrier is changed when we make this correction — breaking from a step-edge.

Here is an As atom attempting to break from a step-edge to become an adatom. The left figure shows the state $X$ while the right figure shows the state with that atom removed, $X\wedge Y$. Prior to the above change, the energy barrier associated to this was $\gamma(G4,A2) + \gamma(G3,A2)$, the sum of the bonds about this atom. However, this is not the energy $E(X)-E(X\wedge Y)$ that is required in order to satisfy detailed balance. To compute this, we tabulate the relevant bonds in both states:

$\displaystyle E(X) = 2\gamma(G4,A2) + 2\gamma(G4,A4) + 2\gamma(G3,A4) + \gamma(G3,A2) + \text{common bonds}$

$\displaystyle E(X\wedge Y) = \gamma(G3,A2) + 2\gamma(G3,A4) + 2\gamma(G2,A4) + \text{common bonds}$

If we use the following energy definition:

$\displaystyle \gamma(G(i), A(j)) = \begin{pmatrix} \gamma_u & \gamma_u & \gamma_u & \gamma_u \\ \gamma_u & \gamma_u & \gamma_u & \gamma_p \\ \gamma_u & \gamma_u & \gamma_p & \gamma_p \\ \gamma_u & \gamma_p & \gamma_p & \gamma_f\end{pmatrix}$,

then

$\displaystyle E(X) - E(X\wedge Y) = 2\gamma_p + 2\gamma_f + 2\gamma_p + \gamma_u - \gamma_u - 2\gamma_p - 2\gamma_p = 2 \gamma_f$.

Compare this to the previous barrier of $\gamma(G4,A2) + \gamma(G3,A2) = \gamma_p + \gamma_u$. If we set $\gamma_f = 1.0$ eV, $\gamma_p = 0.6$ eV and $\gamma_u = 0.3 eV$, we see that the correct barrier is 2 eV, while the incorrect one is 0.9 eV.

### Example: Exchanges at droplet/substrate interface

The effect of the new energies is quite apparent in exchange events — those events in which two non-vacuum atoms exchange positions. e.g. rates for an As diffusing through a droplet and exchanges at the Ga droplet/GaAs substrate interface. Here we consider the former case.

We briefly recall how rates for such events are calculated. This, however, turns out to be incorrect and does not satisfy detailed balance. To exchange two atoms, $A, B$ we used the rates

$\displaystyle r_{A \leftrightarrow B} = R_0 \exp\left[-\beta \epsilon (E(A) + E(B))\right]$

with the incorrect assumption that the $E(X) - E(X\wedge Y) = E(A) + E(B)$, where $X$ is the state prior to the exchange and $X\wedge Y$ is the intermediate state with the two atoms removed:

N.B. Setting the intermediate state to be one with the two exchanging atoms removed is probably incorrect. A later post will address work done in introducing “hybrid” species in which the two exchanging atoms are replaced with an intermediate third species.

With the old method, the barrier $\Delta E$ was the sum of the bond strengths about the two exchanging atoms, namely:

$\displaystyle \Delta E = 2\gamma(G4, A4) + 3\gamma(G2, A4) + 2\gamma_{G^\prime},$

which equals 4.3 eV if we substitute in the energies described above.

The correct barrier $\Delta E = E(X) - E(X\wedge Y)$ adds the difference in the bonds about the neighboring atoms:

$\displaystyle E(X) - E(X\wedge Y) = 2\gamma(G4, A4) + 3\gamma(G2, A4) + 2\gamma_{G^\prime} + 4(\gamma(G4,A4) - \gamma(G3,A4)) + 2(\gamma(G4, A4) - \gamma(G3, A3)) + (\gamma(G4, A4) - \gamma(G4, A3)) + (\gamma(G2,A4) - \gamma(G2, A3)) + (\gamma(G2, A4) - \gamma(G1, A3))$

Using the above energy definitions, this becomes 7.7 eV, an increase of 3.4 eV from the old method.

### Example: As diffusion through Liquid Ga

Not all the barriers increase when the correct energy differences are calculated. Consider an As atom diffusing through liquid Ga:

As before, we can calculate the difference between the new and old barriers by summing the differences in the bonds about the neighboring atoms between states $X$ and $X\wedge Y$. That difference is:

$\displaystyle \Delta E_{new} - \Delta E_{old} = 9(\gamma_{G^\prime} - \gamma_{G_0}) = -0.45 eV$.

So in this case, the new barrier is lower than the old one.

### Disparity of Rates: Exchange events

Consider again the exchange at the droplet/substrate interface:

Here we depict the states $X, Y$ before and after an exchange on the interface, respectively. Regardless of how we define $X\wedge Y$ (which we have previously stated to be incorrect), we may ask what the difference is in the barriers for the event $X\rightarrow Y$ and the reverse event $Y \rightarrow X$. This is independent of the intermediate state $X\wedge Y$, since this difference is:

$(E(X) - E(X\wedge Y)) - (E(Y) - E(X\wedge Y)) = E(X) - E(Y)$

This quantity measures the disparity of rates and if it is too large, the system equilibrates so fast in this respect that we do not see etching. That is, if the barrier $E(X) - E(X\wedge Y)$ is large while the barrier $(E(Y)-E(X\wedge Y)$ is small, then an instability of the substrate underneath a Ga droplet becomes a rare event that gets reversed on a much shorter time scale.

Tabulating the relevant bonds in state $X$, we obtain:

$\displaystyle E(X) = 9\gamma(G4,A4) + 5\gamma(G2,A4) + 5\gamma(G,G) + 4\gamma(G0,G0) + \text{common bonds}$.

For state $X\wedge Y$, we obtain:

$\displaystyle E(Y) = \gamma(G4,A3) + 5\gamma(G3,A4) + \gamma(G3,A3) + \gamma(G2,A3) + \gamma(G1,A4) + 3\gamma(G1,A3) + 10\gamma(G,G) + \gamma(A,A) + \text{common bonds}$.

With the above energy definitions, the difference between these two energies becomes:

$\displaystyle E(X) - E(Y) = 9\gamma_f - 2\gamma_p - 5\gamma_u + 4\gamma(G0,G0) - 5\gamma(G,G) - \gamma(A,A)$.

Using the above values for $\gamma_f, \gamma_p$ and $\gamma_u$ and setting $\gamma(G0,G0) = 0.30 eV$, $\gamma(G,G) = 0.25 eV$ and $\gamma(A,A) = 0.1 eV$, we obtain the difference

$E(X) - E(Y) = 6.15 eV$.

Thus the two events $X\rightarrow Y$ and $Y\rightarrow X$ have large difference in energy barrier, which makes sense as $X$ should be much more stable than $Y$.

To put this difference into perspective, if the transition $Y \rightarrow X$ had no barrier so that it occurs with rate 1, then the event $X\rightarrow Y$ would have a barrier of 6.15 eV — making this transition essentially impossible. Specifically, the expected time for the event $X\rightarrow Y$ is equivalent to $6.5\times 10^{48}$ adatom hops at a temperature of $573 K$.

This is clearly not the case. Experiments show that there is a significant amount of etching at this temperature within a reasonable time-scale. There are a couple of ways to address this. First, we may forfeit detailed balance for these two events. We essentially do this when we use exchange coefficients, as I will describe in a later post.

Alternatively, we can change the energies so that this difference is not so large. Indeed, the solution to this linear program suggests we make $\gamma_f, \gamma_p, \gamma_u$ as close to each other and as small as possible and $\gamma(G,G), \gamma(G0,G0)$ as close to each other and as large as possible. For example, if we use

$\gamma_f = 0.6, \gamma_p = 0.55, \gamma_u = 0.5, \gamma(G0, G0) = 0.40, \gamma(G,G) = 0.3, \gamma(A,A) = 0.2$,

then the difference in energy barriers is $E(X) - E(Y) = 1.45 eV$.