In VASP, how is the chemical potential of elements calculated?

Question

I would like to calculate the chemical potential of elements having different environmental condition (rich or poor) using VASP. How is this accomplished?

Answer 1

The chemical potential in VASP or any other computational chemistry software is calculated based on the theory of chemical potential relation to Helmholtz energy functional as:

$$\mu_{i} = \Bigg(\frac{\partial F(V,T,N_{1},N_{2},...,N_{i}...,N_{n})}{\partial N_{i}}\Bigg)_{V,T,N_{j \neq i}}$$

You can approximate the above equation up to first order as:

$$\mu_{i} \simeq F(V,T,N_{i}+1,N_{j \neq i}) - F(V,T,N_{i},N_{j \neq i})$$

Here we omitted the terms of $\mathcal{O}\Big( \frac{\partial^{2} F}{\partial N_{i}^{2}} \Big)$.

I call $F(V,T,N_{i}+1,N_{j \neq i}) - F(V,T,N_{i},N_{j \neq i}) = \Delta F_{N_{i} \rightarrow N_{i}+1}$

So, you have:

$$\mu_{i} = \Delta F_{N_{i} \rightarrow N_{i}+1} = \mu_{\text{XC}}^{i}+\mu_{\text{ideal}}^{i}$$

Where $\mu^{i}_{\text{ideal}}$ is the chemical potential of the ideal gas and $\mu^{i}_{\text{XC}}$ is the chemical potential of exchange-correlation. The ideal gas chemical potential is trivial:

$$\mu^{i}_{\text{ideal}} = -k_{B}T\ln{\Bigg( \frac{V}{\Lambda^{3} (N_{i}+1)}\Bigg)}$$

Where $\Lambda$ is the Broglie wavelength: $\Lambda = \sqrt{\frac{2\pi\hbar^{2}}{mk_{B}T}}$.

The exchange-correlation chemical potential is calculated as:

$$\mu^{i}_{\text{XC}} = -k_{B}T \ln{\Bigg ( \frac{1}{V} \frac{\int \exp{\Big(-\frac{U(\mathbf{r}^{N_{i}+1})}{k_{B}T}\Big)} d^{3}\mathbf{r}^{N_{i}+1}}{\int \exp{\Big(-\frac{U(\mathbf{r}^{N_{i}})}{k_{B}T}\Big)} d^{3}\mathbf{r}^{N_{i}}} \Bigg )}$$

Or in ensemble form:

$$\mu^{i}_{\text{XC}} = -k_{B}T \ln{\Bigg ( \frac{1}{V} \Bigg \langle \int \exp{\Big( -\frac{\Delta U_{N_{i} \rightarrow N_{i}+1}}{k_{B}T} \Big)} d^{3}\mathbf{r}^{N_{i}+1} \Bigg \rangle \Bigg )}$$