VASP calculation of beta-tin Si

Series of try VASP sample. “beta-tin Si” here.

Si has the diamond structure under air pressure. However, it has beta-tin structure under more than about 100 kbar. Can DFT support this phenomenon?

Especially, there is no additional explanation for the input files (all parameter are the same as those in the previous articles).

Results

The Energy was calculated for each lattice parameter. The results are shown in the following figure with diamond Si.

Fig. 1 E-V curve of beta-tin Si and diamond Si

Horizontal line shows the volume (par atom) and vertical line shows the energy. The volume and Energy are excerpt from “vasprun.xml” or “OUTCAR”.

According to the first law of thermodynamics for a closed system, any net change in the internal energy dE must be fully accounted for, in terms of heat \delta Q entering the system and work  \delta W done by the system:

(1)   \begin{equation*}  dE = \delta Q - \delta W  \end{equation*}

For a process in a closed system, \delta Q = 0 and \delta W = PdV. So,

(2)   \begin{equation*} dE = -PdV \end{equation*}

Therefore, pressure P is,

(3)   \begin{equation*} P = -\frac{dE}{dV} \end{equation*}

Equation (3) means the pressure is the gradient of E-V curve. When there are 2 E-V curves, the gradient of the common tangent is equal to the pressure of the phase transition. The unit of the gradient of the common tangent is eV/Å. To change the unit into “GPa”, the following equation is used:

(4)   \begin{equation*} 1 [GPa] = 0.00624150948 [eV/\AA] \end{equation*}

The each E-V curve in Fig. 1 was fitted by a quadratic curve and the gradient of the common tangent of about 10 GPa are found. The phase transition pressure between beta-tin Si and diamond Si is known as 100 kbar (10 GPa). The estimated value by DFT is accurate.

Birch-Murnaghan’s equation of state

Although I estimated the slope using a quadratic curve, a quadratic curve is available just for the limited range of equilibrium position, where the second-degree Taylor polynomial approximation is applicable. As shown in the following figure, a quadratic curve is not fitted.

Birch–Murnaghan’s equations of state has been well known for the relationship between energy and volume (lattice parameter):

(5)   \begin{equation*} E (V) = E_0 + \frac{9 V_0 B_0}{16} \left\{ \left[ \left( \frac{V_0}{V} \right)^\frac{2}{3} - 1 \right]^3 B'_0 + \left[ \left( \frac{V_0}{V} \right)^\frac{2}{3} - 1 \right]^2 \left[ 6 - 4 \left( \frac{V_0}{V} \right)^\frac{2}{3} \right] \right\} \end{equation*}

This equations fits to the results very well.

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