Molecular Simulation/Repulsive Interactions

Pauli Exclusion Principle
For a system of two indistinguishable fermions (particles of half integer spin: protons, electrons,...etc.), the total wavefunction is a mixture of independent single-particle states $$\psi_a$$ and $$ \psi_b$$:

$$\psi(\mathbf{r_1},\mathbf{r_2})=A\left[\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})-\psi_b(\mathbf{r_1})\psi_a(\mathbf{r_2}) \right]$$, where $$\mathbf{r}$$ a spatial variable.

Notice that the wavefunction vanishes if $$\psi_a = \psi_b$$. In the case of an electron, the wavefunction also contains a spinor, $$\chi(s)$$, which distinguishes between spin up and spin down states. For simplicity, assume that both electrons have the same spin. This results in the Pauli Exclusion Principle:

Two electrons of the same spin cannot exist in the same state.

Since there are only two allowable spins (+1/2,-1/2), this guarantees that each orbital contains a maximum of two electrons of opposing spin.

Exchange Force
The apparent force that is observed when electron orbitals overlap originates from the requirement that the wavefunction be antisymmetric under exchange,

$$\psi(\mathbf{r_1},\mathbf{r_2})\chi(s_1,s_2) = - \psi(\mathbf{r_2},\mathbf{r_1})\chi(s_2,s_1)$$.

This is commonly known as the exchange force. When the spin is symmetric (ex. $$\uparrow \uparrow$$, $$\downarrow\downarrow$$) the spatial wavefunction must be antisymmetric (antibonding), resulting in a repulsive interaction. This is also the case for filled electron orbitals; helium with an electron configuration $$1s^2$$ does not bond with itself because such interacting electrons reside in the destabilized $\sigma^*$ antibonding orbital (verified with MO-diagram of He–He).

The Pauli repulsion energy is of the form

$$\mathcal{V}(r) = a e^{-b(r-c)}$$, where $$a$$, $$b$$, $$c$$ are constants.

Notice that this parallels with the radial distribution of electron density around a nucleus (Figure: Electron Density Repulsion). Intuitively, the phenomenon of Pauli repulsion should only exist in the environment of electrons.

It is of practical advantage to replace the exponential function with the simpler polynomial expression as in

$$\mathcal{V}(r) = \frac{C_{12}}{r^{12}}$$, where $$C_{12}$$ is a constant.

A comparison of the polynomial and exponential form of the potential is illustrated on the right (Figure: ''Comparison of Pauli Repulsion Formulae). While the functions appear to be very different for small $$\mathbf{r}$$'', the difference is immaterial considering that the probability of being in such a high energy regime is so low.

Comparison to Attractive Forces
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The potential for electrostatic interactions, such as charge-charge, charge-dipole, dipole-dipole, and quadrupole-quadrupole, always takes on some variation of the form above, where $$n$$ and $$m$$ are integers signifying the order of the pole in the multipole expansion (see Table: Range of Electrostatic Interactions).

For the sake of completeness also consider the charge-induced interactions

$$\mathcal{V}(r)=- \frac{\alpha}{2} \left(\frac{Q}{4 \pi \epsilon_0} \right)^2 \frac{1}{r^4} \propto \frac{1}{r^4}$$ ,

and and dipole-induced interactions

$$\mathcal{V}(r)=- \frac{C_6}{r^6}\propto \frac{1}{r^6}$$, $$C_6=\frac{3}{2}\alpha_1^'\alpha_2^' \frac{I_1 I_2}{I_1+I_2}$$, where $$\alpha$$ is the polarizability of the molecule, $$I$$ is the ionization potential, and $$Q$$ is the charge.

For electrostatic interactions following this general form, the potential energy for the interaction between oppositely-charged particles will approach negative infinity as the distance between approaches zero. This would suggest that if electrostatic interactions are the only forces in our model, oppositely-charged ions would collapse onto each other as $$r$$ approaches zero. Clearly this is an inaccurate representation of close range interactions, and some repulsive term must also be considered.

Pauli repulsion is a comparatively short range interaction, with $$V(r) \propto 1/r^{12}$$. The slope of this curve is very large for small internuclear separations, $$r$$. That is to say, the repulsive force rapidly diverges to infinity as the distance between nuclei approaches zero. For this reason it is said that the Pauli repulsive term is much "harder" than the other electrostatic forces.

Van der Waals Model of Fluid (Hard Sphere)
In liquids it is very improbable to have two particles at a highly repulsive distance (i.e., at distances where these functions diverge). Therefore, it is unnecessary to provide a perfect model at these distances, as long as it is repulsive. It is insightful to consider a fluid as collection of hard spheres, interacting with only Pauli repulsive forces. Neglecting the attractive forces between particles it justified by the following:


 * Attractive forces are weak and isotropic. For a uniform (constant density) fluid these forces will average to zero.
 * The cohesive force attributed to the change in entropy of gas to liquid phase is small for a dense gas; there is a small structural change between gas and liquid.
 * Repulsive forces are extremely strong to prevent any arrangement where atoms overlap.
 * Liquids arrange to minimize volume by maximizing packing of these hard spheres.