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I am a newbie in DFT calculations. I did potential energy scans with GAUSSIAN 09 for different dimers, i.e.
# b3lyp/6-31G(d) scan test
title
0 1
Ni 0.0 0.0 0.0
Ni 0.0 0.0 r
r 20.0 19 -1.0
Unexpectedly, the energy at large distances does not converges to zero, but some finite negative value. Why does this happen?
The phenomenon seems to be independent of chemical element, functionals, basis set. I am obviously missing something here.
Thanks!
I assume you are asking about the total energy which is printed as the final result of your DFT calculation.
This energy represents the kinetic energy of all your particles (atoms and electrons) and their coulomb interaction. It does not have much physical meaning by itself. Usually, the more particles and the larger your basis set, the smaller (more negative) the total energy. This is true for all methods (DFT, CC, CI ...), although the actual numbers will be different.
To get something meaningful out of it, you need to compare with something on the same (or similar) level of theory. For example you can calculate a single nickel atom and subtract twice its total energy to get the dimers binding energy. This is what you expect going to 0 for large distances.
You always need a reference to compare your value with. Just as with voltage, pressure or the height of a mountain.
I think that the problem encountered in this question is not specificly for DFT. But more in general for QM methods. You say that $E(r\rightarrow \infty) \neq0$. Now let us consider the molecular Hamiltonian:
$$\hat{H}=\hat{T}_N+\hat{T}_e+\hat{U}_{NN}+\hat{U}_{Ne}+\hat{U}_{ee}$$
Here:
The kinetic energy of the nucleis: $$\hat{T}_N=-\sum_i\frac{\hbar^2}{2M_i}\nabla^2_{R_i}$$
The kinetic energy of the electrons: $$\hat{T}_e=-\sum_i\frac{\hbar^2}{2m_e}\nabla^2_{r_i}$$
The potential energy between the nucleis: $$\hat{U}_{NN}=\sum_i\sum_{i>j}\frac{Z_iZ_je^2}{4\pi\epsilon_0|R_i-R_j|}$$
The potential energy between the electrons: $$\hat{U}_{ee}=\sum_i\sum_{i>j}\frac{e^2}{4\pi\epsilon_0|r_i-r_j|}$$
The potential energy between the nucleis and electrons: $$\hat{U}_{Ne}=-\sum_i\sum_{j}\frac{Z_ie^2}{4\pi\epsilon_0|R_i-r_j|}$$
We would usually make a calculation under the clamped nuclei approximation, thus we will let the term for the kinetic energy of the nuclei fall out, $\hat{T}_N=0$. Now if we consider the Hamiltonian being an operator depending on the distance between two nucleis. We can write it as:
$$\hat{H}(|R_1-R_2|)=-\sum_i\frac{\hbar^2}{2m_e}\nabla^2_{r_i}+\frac{Z_1Z_2e^2}{4\pi\epsilon_0|R_1-R_2|}+\sum_i\sum_{i>j}\frac{e^2}{4\pi\epsilon_0|r_i-r_j|}+\left[ -\sum_{j}\frac{Z_1e^2}{4\pi\epsilon_0|R_1-r_j|} -\sum_{j}\frac{Z_2e^2}{4\pi\epsilon_0|R_2-r_j|} \right]$$
Now we can consider the limit of $|R_1-R_2|\rightarrow \infty$. It can be seen in the above equation that only one term depends directly on $|R_1-R_2|$, i.e.:
$$\lim_{|R_1-R_2|\rightarrow \infty}\frac{Z_1Z_2e^2}{4\pi\epsilon_0|R_1-R_2|}=0$$
Therfore:
$$\hat{H}(|R_1-R_2|\rightarrow \infty)=-\sum_i\frac{\hbar^2}{2m_e}\nabla^2_{r_i}+\sum_i\sum_{i>j}\frac{e^2}{4\pi\epsilon_0|r_i-r_j|}+\left[ -\sum_{j}\frac{Z_1e^2}{4\pi\epsilon_0|R_1-r_j|} -\sum_{j}\frac{Z_2e^2}{4\pi\epsilon_0|R_2-r_j|} \right]$$
It can be seen that $\hat{H}(|R_1-R_2|\rightarrow \infty) \neq 0$. It have to be noted in the above I have made no assumptions regarding the electrons placements to the nucleis. The terms that depends on the electrons will not be constant when the distance between the nucleis are increased, but will not go to zero. I did not look more explicit at these terms, to conserve simplicity of the argument.
The expectation value will thus not go to zero for infinitly seprated atoms. As a fruit for thought, what is the energy of a single atom? What should then the energy of two atoms at infinite distance be?
You can use the Grimme's suggested functional instead of B3LYP to correct the the interaction between molecules. In Gaussian program, have a look at empiricaldispersion keyword. Read more details here.
