# Is Pauli-repulsion a "force" that is completely separate from the 4 fundamental forces?

You can have two electrons that experience each other's force by the exchange of photons (i.e. the electromagnetic force). Yet if you compress them really strongly, the electromagnetic interaction will no longer be the main force pushing them apart to balance the force that pushes them towards each other. Instead, you get a a repulsive force as a consequence of the Pauli exclusion principle. As I have read so far, this seems like a "force" that is completely separate from the other well known forces like the strong, electroweak and gravitational interaction (even though the graviton hasn't been observed so far).

So my question is: is Pauli-repulsion a phenomenon that has also not yet been explained in terms of any of the three other forces that we know of?

Note: does this apply to degenerate pressure too (which was explained to me as $\Delta p$ increasing because $\Delta x$ became smaller because the particles are confined to a smaller space (Heisenberg u.p.), as is what happens when stars collapse)?

• If you like this question you may also enjoy reading this Phys.SE post. Nov 20, 2012 at 19:42
• No matter what anyone says about this, I don't think we can truly know the answer to this until we have a theory of quantum gravity and understand what is actually at the center of a black hole. If you look at a neutron star or hypothetical quark star, there's nothing AFAIK stopping further collapse other than Pauli. Yet at black hole densities Pauli seems to be overcome, which sure makes it seem like a force, yet there's no field or force carrier boson associated with it. Dec 14, 2022 at 23:31

The QM notion of a "force" is highly technical jargon that doesn't match up with how the word force is used in the world at large. Basically, the notion of a "force" in QM is defined to be an interaction mediated by force carrier particles and therefore the exchange-interaction is arbitrarily defined not to be a force. Likewise, gravity is not a "force" because it doesn't work via exchange particles but instead works through curving space-time.

So yes the exchange interaction is a force. No the exchange interaction is not a QM force. Physicists who argue that the exchange interaction is not a force have confused the meaning of their jargon with the meaning of the word force in the minds of the common person and are making a prescriptavist mistake.

• Indeed, this is IMO a big part of the confusion. Oct 26, 2016 at 16:12

I'd like to add a different take on essentially the same as Anna's answer. I am writing to clarify the especial things that gave rise to my own personal misconceptions, so likely not everybody is going to benefit from this answer.

What is the fundamental force involved when you try to squash fermions together? It's whatever is confining fermions! You need to think of an ideal gas. When the molecules in the ideal gas are in flight between the walls, there is no force on them. Whatever force holds the walls and piston in place is the force that imparts the impulse to them to rebound them back into container. Likewise, the particles in the Fermi gas in the infinite square well are in the quantum mechanical equivalent of motion. It is the force shaping the infinite potential well that imparts the impulse needed to rebound them back into the well. Sometimes like you I feel the need for a Pauli "force". I think where I trip up here is that my everyday experience of solids is, well, they are solid!

So I kind of forget that the particles in them are in a dynamic state. They are in motion and by Newton II they need force to keep them confined. I and civil engineers tend to think of squashing a block of iron as a problem in statics, so we tend to think that if we are crushing something seemingly "static", there must be a balancing force thrusting back. Actually, when the crushing force gets really big, this everyday viewpoint breaks down: the problem is more one of dynamics. There is no backthrusting force. The particles are in motion and the crushing force from the outside is continually changing their state of motion, by Newton II, and thus keeping them confined. Or, if you like, in a D'Alembertian mindset, there is a backthrusting force but it is an inertial force arising from thinking of the problem from an accelerated frame.

Actually the ideas work equally well for bosons as for fermions. If you confine light in a perfect resonant cavity, there is a backthrusting force and it varies inversely with the resonant cavity's volume.

The difference is that, owing to the Pauli exclusion, the force needed to keep particles in their dynamic confined state in a given volume is MUCH bigger for fermions than for bosons.

The Fermi gas in a container model applies when all other forces (electromagnetic, for example, that help arrange solids into crystals and so forth) are very small relative to outside forces crushing the solid. I have heard (I don't have direct experience of this) that in certain kinds of explosive analysis, especially the worthwhile analysis of the evolution of a nuclear explosion in a weapon, you can just assume that everything is a gas, with negligible error. You can simply think of the crushing force and the particle dynamics in the presence of this force and ignore everything else: interactions between the particles fade into the background.

Lastly let's look at the core of a body made up of charged fermions. I am NOT an astronomer so I've no idea of the exact numbers in the following diagram, which is to be taken with a grain of salt given that I heavily mix classical (GR) ideas with quantum interactions. Let's assume our fermions are at first so energetic that a swarm of them doesn't collapse, and at first let's assume a low density. So we look at the top graph in my diagram, where I have drawn some paths of classical particles near the centre of a uniform swarm of them. You may know that in Newtonian gravity, near the centre of a uniform density body, the gravitational force makes for a simple harmonic oscillator potential, so the particles are taking the sinusoidal paths with time.

In the bottom two diagrams, I zoom in on a few "periods" of our classical particles. In a GR description, there is no force on them: we assume the swarm is like a GR "dust" so their aggregate effect is to curve spacetime as though they were a fluid continuum. Their energies will "thermalise" (i.e. follow roughly a Boltzmann distribution) and if the density is low, there are very few interactions between them. As I said, I am not an astronomer so I don't know whether we one can have a gas of fermions that is both (i) dense enough to curve spacetime so that the gas stays confined (bottom right diagram) and (ii) yet sparse enough that there are few interactions between particles. But what's important here is what happens as we "turn up the density" of our swarm by adding more matter. Now the fermions follow their spacetime geodesics only in little hops (bottom right diagram): very often they interact by swapping $$\gamma$$s. If you zoom in on these interactions, you get Anna's diagram: the "blue" and "green" fermions swap "red" (colours added only to tell the classical particles apart in my diagram) $$\gamma$$s. The fermions thus kick each other from one spacetime geodesic to another, and thus make jagged zigzag, highly non-geodesic paths through spacetime. So the shape and distribution of the swarm will change from being having its distribution being scaled down inversely with density as in the classical, noninteracting swarm case to something that is limited in density.

The Pauli exclusion principle governs how often these interactions (in this case electromagnetic) happen and can thus be thought of a constraint on a given problem - akin to boundary conditions and other information needed to fully define a situation - in this picture.

• Ok, you show charged particles can bounce off each other using electric repulsion. But that has nothing to do with Paili Exclusion, or explain why neutral atoms seem to have a hard-contact size like little balls. Dec 26, 2016 at 6:02
• @JDługosz Actually, that's kind of the point, in a way. As I discuss, the principles of the answer hold for both bosons and fermions: the "force" that needs confining by an applied force (or a gravitational well) is not a fundamental interaction. Rather, the applied force is simply changing the dynamics of the system, and the exclusion principle (as well as much else) sets the equation of state that determines the amount of force you will need to disturb the dynamics. It's the same principle for an ideal gas, a gas of bosons or a gas of fermions - the PEP just makes the last case .... Dec 27, 2016 at 0:05
• .... particularly demanding of confining force. Dec 27, 2016 at 0:07
• As written, it seems to describe ordinary pressure in a non-ideal gas, and furthermore implies that the hard-contact bouncing is due to electromagnetism but that’s circular. Dec 27, 2016 at 0:15
• @JDługosz well yes, but the question is whether the PEP is a separate fundamental force. An I'm saying that no it isn't - the apparent "force" is a very familiar effect that you can understand in thinking about very simple things like an ideal gas. The PEP is part of the details that define the size of that force - like a boundary condition. That's the essential insight that I certainly found myself forgetting when pondering - for example - the degeneracy pressure of a star. I guess the name "degeneracy pressure" itself skews ones thinking because the name implies the PEP is "making" .... Dec 27, 2016 at 0:20

So my question is: is Pauli-repulsion a phenomenon that has also not yet been explained in terms of any of the three other forces that we know of?

$$\def\ket#1{|#1\rangle} \let\up=\uparrow \let\dn=\downarrow \def\PD#1#2{{\partial#1\over\partial#2}}$$ There is no repulsion and no unexplained force. I would also add that PEP is an outdated way of describing the matter. In QM you should rather speak of antisymmetry of fermion states. It's only when we build up a many particle state as a tensor product of one particle states that antisymmetry forces us to keep only different states for each single particle. A simple example with two particles will explain this (I hope).

Two identical fermions in an infinite well

Consider particles in one dimension, constrained in a segment $$0\le x\le L$$ (what is usally called an "infinite potential well"). Energy eigenfunctions (standing waves) are sinusoidal waves vanishing at boundaries: $$\psi_n = \sin {n\,\pi\,x \over L} \qquad (n = 1,2,\ldots)$$ (these aren't normalized, but it's of no consequence for my present purposes.) The corresponding energy eigenvalues are $$E_n = {n^2 h^2 \over 8\,m\,L^2}.\tag1$$ A short derivation follows, which you may skip with no harm.

$$\psi_n$$ has wavelength $$2L/n$$, then momentum $$p = {h \over \lambda} = {n\,h \over 2\,L}.$$ Then energy (only kinetic) is $$E_n = {p^2 \over 2\,m} = {n^2 h^2 \over 8\,m\,L^2}.$$

Assume your particles are non-interacting spin 1/2 fermions. Then above expression for energy eigenfunction is to be supplemented by specifying the spin state. Then Dirac's ket notation is preferable: $$\ket{n\up} \quad \hbox{or} \quad \ket{n\dn}$$ both belonging to $$E_n$$ eigenvalue.

If your system consists of just two particles, a set of base kets would be obtained by taking tensor products, which in Dirac's notation are written just putting two kets one after another. E.g. $$\ket{m\up} \ket{n\up} \quad \ket{m\up} \ket{n\dn} \quad \ket{m\dn} \ket{n\up} \quad \ket{m\dn} \ket{n\dn}$$ for all positive integers $$m$$, $$n$$. A shorthand may be used: $$\ket{m\up\,;\,n\up} \ \ket{m\up\,;\,n\dn} \ \ket{m\dn\,;\,n\up} \ \ket{m\dn\,;\,n\dn} \tag2$$ where labels preceding ";" refer to first particle, those following to the second.

But states in (2) are wrong for identical fermion particles, as they aren't antisymmetrized. The right ones are \eqalign{ &\ket{m\up\,;\,n\up} - \ket{n\up\,;\,m\up} \qquad \ket{m\up\,;\,n\dn} - \ket{n\dn\,;\,m\up} \cr &\ket{m\dn\,;\,n\up} - \ket{n\up\,;\,m\dn} \qquad \ket{m\dn\,;\,n\dn} - \ket{n\dn\,;\,m\dn} \cr} (once again I'm neglecting normalization).

Observe however that if $$m=n$$ first and fourth expressions are identically zero, whereas second and third are the same apart for sign, thus representing the same state. This is the mathematical form PEP assumes in QM: for $$m=n$$ just one state exists for two particles, for $$m\ne n$$ there are four.

For more particles we would proceed analogously, with a somewhat higher complication.

Let's compute pressure

First of all let me remark that not fermions alone exert a pressure when confined in a finite volume. Bosons do as well. Radiation pressure is an example, and photons are bosons. So let's compute the pressure exerted by a gas of non-interacting bosons at $$0\,$$K, when all particles are in the ground state (this isn't forbidden for bosons).

If we have $$N$$ particles, overall energy is given by (1) taken for $$n=1$$ and multiplied by $$N$$; $$E = {N h^2 \over 8\,m\,L^2}.$$ As we are in one dimension we'll speak of force, not of pressure. It's most easily computed by $$F = -\PD EL = {N h^2 \over 4\,m\,L^3}.\tag3$$

For those who find too abstract the above derivation I'll add a semiclassical one. In our box we have free particles bouncing back and forth between boundaries. Their momentum is $$p=h/(2L)$$. A particle hits one boundary (e.g. the left one) once in a time $${2L \over v} = {2mL \over p} = {4 m L^2 \over h}$$ and every time it exchanges with the boundary a momentum $$2p$$. Then the momentum exchanged per unit of time, i.e. the force, is $$f = 2p\, {h \over 4 m L^2} = {h^2 \over 4 m L^3}.$$ This holds for one particle. It's only left to multiply by $$N$$ to get (3).

Now for fermions

What's the difference? Simply that even at $$0\,$$K a fermion gas doesn't have all particles in ground state. We've seen why it's forbidden by antisymmetry. So we have the task to arrange an antisymmetrical ket for $$N$$ particles, which sounds prohibitive. Actually it's not so much so, but we'll follow a roundabout way, in principle an approximated one but absolutely adequate to our purposes.

For each $$n$$ there are two states allowed, spin up and spin down. We already saw that for $$m=n=1$$ and two particles only one state is possible, wheres none is possibile for three. If we accept values 1 and 2 for $$m$$, $$n$$ we can accomodate up to four particles $$\ket{1\up;1\dn;2\up;2\dn}$$ (to be antisymmetrized). So we see that for $$N$$ particles all states from 1 to $$N/2$$ will be occupied, each by two particles with opposite spins.

And now we are able to compute the energy: $$E = 2\,\sum_{n=1}^{N/2} E_n = 2\,\sum_{n=1}^{N/2} {n^2 h^2 \over 8\,m\,L^2} = {h^2 \over 4\,m\,L^2} \sum_{n=1}^{N/2} n^2$$

(the sum has to be multiplied by 2 since for every $$n$$ there are two spin states). If $$N$$ is large we may approximate the sum to $${1 \over 24}\,N^3$$ and get $$E = {N^3 h^2 \over 96\,m\,L^2}.$$ As before $$F = -\PD EL = {N^3 h^2 \over 48\,m\,L^3}.\tag4$$

You can see the difference between (3) and (4). Whereas for bosons force is $$\propto N$$, for fermions it's $$\propto N^3$$, then much larger if $$N$$ is large. Actually extremely larger for a white dwarf: try to estimate how much is $$N$$ (number of electrons) for a star having Sun's mass.

To be sure we should reason about pressure, not about force. This requires leaving our naive 1D model for a more realistic 3D one. I'll content myself to give the result $$P = {(3\,\pi^2)}^{2/3} \left(\!{\hbar^2 \over m}\!\right)\,{N \over V}^{\!5/3}.$$

The most important difference is in the dependence on $$N$$: $$N^{5/3}$$ instead of $$N^3$$. I can't explain its origin (it has to do with the different accounting in 1D and in 3D for the one-particle states up to $$N/2$$). I'll only say that even with the smaller exponent resulting pressure is enough to counterbalance gravity for dwarfs of mass near Sun's and size about Earth's.

A final comment

It should be clear that no mysterious force could account for our results. Note that total energy of $$N$$ particles depends on a power of $$N$$ and it would be hard to explain that with some interaction between particles. Instead all depends on which and how many independent states are allowed when identical particles are concerned. In a different way for bosons against fermions and both different of the one that would be used for classical particles.

As Feynman liked to say, this is the way things are.

• As a summary, I would say that the need to occupy higher energy states for the fermions is what increases their kinetic energy and therefore its pressure causing the high pressures that balance the gravitational force inside the stars. Another way to see this is that when many fermions are confined to a smaller physical region, they must to occupy a great momentum space. Feb 8, 2022 at 22:30

The pauli exclusion principle is not a repulsive force. It applies to fermions. It says that two electrons cannot occupy an energy state in a potential well with exactly the same quantum numbers. They have to differ by at least one quantum number. It is the Pauli exclusion principle that organizes the electron shells filling them sequentially from low to higher energy levels in atoms, otherwise they would all pile up at the lowest energy level. Also the periodic table of elements filling the baryons in the strong potential well. It makes matter as we know it.

Yet if you compress them really strongly, the electromagnetic interaction will no longer be the main force pushing them apart to balance the force that pushes them towards each other. Instead, you get a a repulsive force as a consequence of the Pauli exclusion principle.

The above is a misunderstanding.

It is not a force, since at the particle level forces have carriers that are exchanged between particles so that momentum and energy change.

In your "compression" description there is a continuum and not a quantized state so the PEP does not apply. When one scatters an electron on an electron one can get very close until the exchange particle ( the photon in this case) transfers enough energy in the center of mass system to start creating other elementary particles. The process is accurately described by quantum electrodynamics.

• But how does this explain degenerate pressure? The name implies that a force is acting. Would it actually mean that adding electrons to a small space would force them (PEP) into higher energy levels because all the ground states are occupied and thus they actually have a higher kinetic energy (and thus also a higher momentum which results in a higher pressure) Nov 20, 2012 at 20:38
• I just have difficulty understanding the apparent link between energy (levels) of the contained particles and the way they exert a force on the "walls" of the container. The whole explanation tells something about particles being forced in higher energy levels, and somehow this results in a higher pressure, but I have trouble seeing that link. Nov 20, 2012 at 22:40
• @PatronBernard It is important to notice that in the case of, for instance, degenerate matter in a white dwarf there is an actual force (gravitation) trying to squeeze the electrons ever closer together. It is the tension between that force and the exclusion based limit on the number of low momentum electrons that can be in a particular volume that results in the electrons getting forced into high momentum states. The energy comes from gravity, not from the PEP. Nov 21, 2012 at 0:26
• @annav: It is funny, but Dirac represented it as an "exchange interaction", via effective potentials depending on spins. And "exchange" here means exchange with fermions ;-) Nov 21, 2012 at 18:20
• Chandrasekhar derived an expression for the radius of white dwarfs completely ignoring electron interactions.
– lcv
Dec 15, 2018 at 11:47

As with a lot of other things, the trick rests in the nuances of the definitions.

Force is a well-defined concept. What it means is an interaction between two or more objects which contributes to a change in their momentum. The quantity of force is defined as the vectored size of this change:

$$\mathbf{F} = \frac{d\mathbf{p}}{dt}$$

. Now this statement also carries over to quantum mechanics. The difference is momentum is now an operator on the Hilbert space, $$\hat{\mathbf{p}}$$, instead of a simple vector. But we can still define the quantity of force, $$\hat{\mathbf{F}}$$, in the same way, via the Heisenberg picture in which it is the operators that are changing and not the quantum state vector. Note that the momentum operator doesn't always change! For an isolated system where momentum is conserved, its non-doing so is exactly how conservation of momentum expresses itself in quantum theory.

The Pauli exclusion principle is a principle regarding how a joint state vector of a composite system of a certain kind of identical particles is to be built up from those of the individuals. As it only is talking about state vectors, it does not imply any change in the operators, thus by implication no change in the momentum operator, and thus is not a force.

The normal force does indeed have something to do with the PEP, but it's more like a "team job" between both the PEP and electromagnetism. PEP sets a limit, and electromagnetism provides the actual force - the $$\frac{d\hat{\mathbf{p}}}{dt}$$ - to "enforce" (:D) it.

• Regarding the "team job": could we say that the exchange interaction is similar to a constraint force in classical mechanics, which is also a "team job"? Like gravity is the "real" force that drives a pendulum, while the centripetal force is just what is required to maintain the constraint (stay on the circle)? I remember that gauge fixing in Yang-Mills theory is a kind of constraint equation and that it leads to Fermion-like behavior (Grassmann numbers, Faddeev-Popov ghost fields). But true Fermions (Pauli principle) cannot be considered the result of a constraint, can they? Sep 7, 2021 at 21:42

## The Pauli exclusion principle is not a force

At first glance, it seems like Fermions repel and Bosons attract. But at no point does the exclusion principle apply any force!

Particles in a box push on the box (the L^2 term means that the energy of a given state is lower for bigger boxes). Higher energy quantum numbers push harder. This corresponds to the classical model of billiard balls hitting the walls.

If you compress the box adiabatically (not letting heat escape) the particles stay in their quantum states. However, the states have more and more energy as the box shrinks: the gas heats up. If you start with an (amost) ideal gas you could compress the gas all the way to white dwarf densities and it would stay ideal!

The Pauli exclusion principle kicks in when you cool the gas and particles drop to lower energy levels. Initially the pressure will drop. But eventually the electrons will fill up the lower energy levels. No amount of cooling can get them to drop any lower. The remaining pressure at zero temperature is the degeneracy pressure.

For Bosons there is no such limitation. Like Fermions, Bosons are identical particles: the microstate of the entire system can be described by how many particles are in each state. Doing so requires little information per particle at high occupancy; cold and dense Boson gases have very low entropy. At this point they can collapse all the way to the ground easily in a Bose Einstein condensate.

## So how does it hit you a home run?

Without the Pauli principle the atoms in the ball could find places inside the bat where they would be perfectly happy: the electric potential inside solid matter is a chaotic mess of hills and valleys, and there is somewhere where an atom's electrons and nuclei could find a decent compromise. The ball would stick to the bat and the runner would stick to the home plate. This would make hitting a home run hard (even setting aside the violent destruction of matter as we know it).

But the Pauli exclusion principle forces many electrons into anti-bonding orbitals. The wavefunction in such an orbital has opposite phase on each of the two atoms and a nodal plane in the middle. This anti-symmetry satisfies the requirement that the Fermion joint wave function must be anti-symmetric under exchange of any two electrons with the same spin. These anti-bonding orbitals have already been set up long before the ball hits the bat (when an electron is created, technically it must satisfy the Fermion exchange rule with all other electrons in the universe).

Anti-bonding orbitals, when the atoms get close, are high in energy. If the pair of nuclei is a crude "box" bonding orbitals are the ground state and antibonding orbitals are the first excited state. Also, anti-bonding electrons, forced away from the nodal plane, don't really "benefit" from being close to two nuclei; such a benefit is needed to offset the nucleus-nucleus repulsion.

Thus the Pauli exclusion principle sets up the joint wavefunction, of all the atoms in the ball and bat, in such a way that the bat will hit the ball normally. It's the electrostatic forces that do the actual "pushing".

Helium-4 atoms are Bosons so several can occupy a given state. However, their electrons are still fermions and subject to the exclusion principle. Bulk Helium-4 resists compression like any ordinary matter.

The Pauli Exclusion Principle isn't a fundamental force because it doesn't have the same origin as the 4 fundamental forces. It's like the pressure you feel from a normal gas in that it definitely exists, but it comes from the fact that you have many particles in the system and are averaging over their behavior. We usually call that an "emergent phenomenon," which is a property we detect at the macroscopic level but looks very different at smaller scales.

One of the other answers described the Pauli Exclusion Principle at the atomic scale for a particle in a well — you can see that it isn't treated like a force in the same way that electromagnetism is. An emergent example is the pressure that keeps a white dwarf from collapsing on itself.