# How can Pauli's exclusion principle originate forces?

Recently I've seem two situation in which the Pauli exclusion principle is said to originate forces.

The first was a derivation of the Lennard-Jones potential in the context of Solid State Physics aimed to explain the molecular binding. The second is the exchange interaction in ferromagnetism.

The idea seems to be in the general case that the Pauli exclusion principle is the reason why some repulsive interactions appear.

Now, what I know is that the principle states that a system of fermions can't have two fermions with the same quantum numbers.

How can a qualitative principle like that originate one interaction? And furthermore, how can one derive a potential for such interaction? Because in either cases I cited, a potential is derived (the Lennard-Jones potential and the exchange interaction potential energy).

• there is nothing "qualitative" about $\psi(1,2)=-\psi(2,1)$; this is a very powerful restriction. Sep 18 '16 at 0:44
• @hyportnex it sounds like the OP is not aware of the origins of the pauli-principle in QM. Sep 18 '16 at 2:30
• – udrv
Sep 18 '16 at 5:28

The Pauli exclusion principle states that the overall quantum state of an isolated system has to be antisymmetric under exchange of its constituent Fermions. A consequence of this is that no two Fermions can occupy the same state, leading to phenomena like the periodic table of elements.

To understand what is meant by the claim that the Pauli exclusion principle leads to forces, put $4$ particles into a one dimensional square well of length $a$ with infinite height, and let them fall to the system's ground state. If the $4$ particles are Fermions, you'll have two particles (one of each spin orientation) in the state $\psi_0\propto\sin\left(\frac{\pi x}{a}\right)$ (a one bump state) and two in the state $\psi_1\propto \sin\sin\left(\frac{\pi x}{a}\right)$ (a two bump state. If the particles were Bosons or non-interacting they'd all be in the one bump state. The total energy of the Fermionic and Bosonic states are: \begin{align}E_F &= 2\frac{\hbar^2 \pi^2}{2m a^2} + 2\frac{\hbar^2 (2\pi)^2}{2m a^2} = 10\frac{\hbar^2 \pi^2}{2m a^2},\ \mathrm{and} \\ E_B & = 4 \frac{\hbar^2 \pi^2}{2ma^2},\end{align} respectively.

How much energy to I need to add to compress the system by a small amount, $\Delta a$? \begin{align} \Delta E_F & = 10\frac{\hbar^2 \pi^2}{2m (a-\Delta a)^2} - 10\frac{\hbar^2 \pi^2}{2m a^2} \approx 20 \frac{\hbar^2 \pi^2}{2m a^3} \Delta a,\ \mathrm{and} \\ \Delta E_B & \approx 8 \frac{\hbar^2 \pi^2}{2ma^3} \Delta a.\end{align} Notice how in this case it takes more than twice as much force to compress the Fermions than the Bosons. That difference is "due to" the Pauli exclusion principle in the sense that it forces the Fermions to exist in a higher kinetic energy state, causing them to exert a greater force on their surroundings.

The Pauli exclusion principle is not really the origin of forces (although it is discussed used that way). It influences the strength of other forces because it influences what states of multi-particle systems are possible.

The pauli principle is usually said to be the idea that two fermions cannot be in the same state. As a rough example, imagine two charged particles. If they could be in the same state, then they could be more on-top of each other than otherwise, corresponding to a stronger coulomb interaction between the particles.

As a separate consequence, even for completely non-interacting particles, the exclusion principle changes the energy of the ground state of the system. Some multi-particle states that are accessible to bosons are inaccessible to fermions, alowing bosons to reach lower energies in general than fermions. This again isn't really a force (although you could probably pull something out that looked like one if you look at some limiting cases), it is a limitation on the available states of a multiparticle system.

It's probably not too surprising that if you explore the origin of the pauli-principle in quantum mechanics that the principle comes from general statements about restrictions on wavefunction overlap for bosons and fermions (multiparticle wavefunctions have to be symmetric vs. antisymmetric correspondingly). Restrictions on wavefunction overlap correspond to restrictions on how far interacting particles can be from another, which directly corresponds to how strongly they interact with each other (via the coulomb force, usually). So while the pauli-principle is not the origin of any unique forces, it influences the strength of existing forces between particles.

A force is, by definition, minus the gradient of the effective potential, where this effective potential arises from treating the interactions in an approximate way (using e.g. perturbation theory). A simple example can be two electrons where you treat the Coulomb interaction as a perturbation. Then in the singlet state the spin state is asymmetric, this means that the spatial part of the wavefunction must be symmetric, while in the triplet state the spatial part of the wavefunction must be asymmetric, because in that case the spin part of the state is symmetric.

This then means that the first order perturbation of the energy due to the Coulomb interaction will be suppressed in the triplet state. If you asymmetrize the two particle wavefunction, you'll make the wavefunction tend to zero for relative positions tending to zero.

Now suppose that the single particle wavefunctions are Gaussians of the same width but which have their peaks at different positions. Then we can take two of these wavefunctions to represent two electrons, symmetrize or anti-symmetrize them depending on whether the spin state is a triplet state or a singlet state. Since the kinetic energy if the same in all cases, the total energy can be taken as the effective potential, minus the gradient of the total energy w.r.t. position then yields the force which will then be suppressed in the triplet state.