34
$\begingroup$

Why is the Pauli exclusion principle not considered a sixth force of nature, given it produces such things as repelling of atoms and molecules in solids?

$\endgroup$
8

11 Answers 11

29
$\begingroup$

Since Pauli exclusion is an inherently quantum phenomenon, let's talk about what the most general effect of a force of nature is, or how we would most generally categorize what a force of nature is, in that context. Specifically, let's talk about scattering theory in quantum mechanics.

In scattering theory we start with some particles that are in wave packets that are narrowly peaked around some momenta $\mathbf k_1,\dots$ at some early time $t=-T\to - \infty$. This is an "in state." We evolve this state for a long time $2T$ using the usual evolution operator constructed from the Hamiltonian. Then at the far future time $t=+T\to +\infty$, we look for the overlap of the evolved in state with an "out state" that has particles in wave function narrowly peaked around some (possibly different) momentum $\mathbf k'_1,\dots$ constructed at that late time.

What happens if we work in a free theory, that is, one with no forces? In that case the only overlap of the in state with the out state will occur when the $\mathbf k_i$ and $\mathbf k'_i$ are the same. If we find that they are not always the same, then some force of nature is present. It contributes some piece that we might call an "interaction term" to the Hamiltonian. The effect of that interaction term is what gives us the non-trivial overlap of the in and out states.

Pauli exclusion, on the other hand, is not something that affects how the Hamiltonian acts on a state. It is simply a statement about what states are allowed.

That is (obviously) not to say that it has no effect on the outcome of some process. For a bosonic system, I could find that some system has a nonzero overlap with an out state $_\mathrm{out}\langle \mathbf k_1,\mathbf k_2,\dots|$ where $\mathbf k_1=\mathbf k_2$ (and any other quantum numbers are the same). This state is of course not an option for fermions, so a similar experiment performed with fermions would have a different outcome. But that would simply be a statement about what states are available, and not a statement about what interactions (or forces of nature) were present in the Hamiltonian, the thing that defines the dynamics of the system.

$\endgroup$
16
  • $\begingroup$ Pauli exclusion is a consequence of the Poincare group. It is a not a consequence of quantum mechanics. $\endgroup$
    – my2cts
    Dec 29, 2020 at 19:26
  • 8
    $\begingroup$ @my2cts does Pauli exclusion occur in a classical system that is Poincaré invariant? $\endgroup$
    – kaylimekay
    Dec 29, 2020 at 19:32
  • $\begingroup$ Classical systems are not Poincaré invariant. However (non-relativistic) quantum mechanical systems do display Pauli exclusion effects which in their origin are spin related and thus Poincaré related. $\endgroup$
    – ohneVal
    Dec 31, 2020 at 10:44
  • 2
    $\begingroup$ @ohneVal Classical systems can be invariant under boosts, rotations, and translations. $\endgroup$
    – kaylimekay
    Jan 2, 2021 at 8:09
  • 2
    $\begingroup$ @ohneVal classical EM. Or take any Poincaré invariant quantum system that you like and then take $\hbar\to 0$. :) $\endgroup$
    – kaylimekay
    Jan 2, 2021 at 11:14
9
$\begingroup$

The Pauli exclusion principle as understood nowadays, is a consequence of the spin-statistics behavior of fermions. We know that Spin representations are related to wave-functions with specific commutation relations which enforce anti-symmetry (all of this encoded in the Dirac-equation, which already accounts for spin) of the wave-function, thus not allowing for two fermionic particles to have the same quantum states in a system. As you can see, it is a statement related to statistics and spin. Both are very well understood and are a consequence of the commutation relations, not the fundamental interactions between single particles (I would prefer fields) which is what we call a force.

Forces on the other hand, are of a different nature. They are not a statistical effect but concern the point-like interactions among fields. Except for gravity, we describe all of them by a gauge (mediator) boson. We don't need to include such a bosonic field to account for the Pauli exclusion principle since the fermionic commutation relations already encode this effect.

$\endgroup$
5
  • $\begingroup$ This does not explain where the field commutation relations come from. $\endgroup$
    – my2cts
    Dec 29, 2020 at 19:52
  • $\begingroup$ The question is not asking about the origin of the commutation relations, is just asking for why it is not considered a force. I could add some comments on it if you want. Or you can post a new question about them, I just thought it would be a bit far from the question. $\endgroup$
    – ohneVal
    Dec 31, 2020 at 10:41
  • $\begingroup$ The question is about the physical origin of the PEP. The commutation relations are equivalent to the PEP. Hence stating that the PEP is "a consequence of the commutation relations" does not bring the question any closer to an answer. $\endgroup$
    – my2cts
    Jan 4, 2021 at 11:25
  • $\begingroup$ It does I believe. At the end of the day the only reason we have for the commutation relations or namely QM is observation. Depends at which level do you want to stop, if you think the above answer is not good quality or is wrong in a more specific way please let me know. $\endgroup$
    – ohneVal
    Jan 4, 2021 at 12:56
  • $\begingroup$ The OP asks to go one level deeper than a general statement principle or a mathematical notation. You have not given this. Hence my comment here. I also do not have it, hence my non-answer, which review has deleted. $\endgroup$
    – my2cts
    Jan 4, 2021 at 16:47
2
$\begingroup$

The Pauli exclusion principle is a rule, similar to the Heisenberg uncertainty "principle", an observational encapsulation of quantum mechanical theory.An easy overview in organizing interactions. The Pauli exclusion follows the spin statistics theorem.

The spin–statistics theorem implies that half-integer–spin particles are subject to the Pauli exclusion principle, while integer-spin particles are not. Only one fermion can occupy a given quantum state at any time, while the number of bosons that can occupy a quantum state is not restricted.

A force in quantum mechanical terms is a $dp/dt$. One could describe the exchange of a virtual electron in Compton scattering as a force, but not as a fundamental force, because it is not a gauge boson, and the couplings are the electromagnetic couplings. But the Pauli exclusion does not involve a dp/dt in any way.

Exclusion exists classically too, though we do not call them principles. Take the seats in a theater. Only one person can sit in a seat , no force there except you try to sit on someone :). An electron scattering off a filled slot will go away with a dp/dt scatter on the atom, but the force will be the electromagnetic virtual photon exchange.

$\endgroup$
3
  • $\begingroup$ I am surprised by the phrase "the exchange of a virtual electron in Compton scattering". Can you elaborate this? $\endgroup$
    – my2cts
    Dec 29, 2020 at 19:37
  • $\begingroup$ @my2cts Look at my answer here physics.stackexchange.com/q/445774, it has the lowest order feynman diagrams. It is the vrtual electron that carries the dp/dt $\endgroup$
    – anna v
    Dec 29, 2020 at 19:52
  • 2
    $\begingroup$ The fact that I can't sit where another person is already sitting, is already the quantum mechanical Pauli exclusion: My electrons simply don't want to share space with the electrons that make up the other person. Not without providing enough energy to lift them to the corresponding energy states. Since I'm scared of supplying that kind of energy, I'd rather sit somewhere else... $\endgroup$ Dec 30, 2020 at 17:06
2
$\begingroup$

You could say that the list of phenomena that we currently call fundamental forces is just a matter of convention. In reality it is more complicated.

All the currently accepted fundamental forces in the Standard Model, have a mediator, that mediates the interaction. Photons for EM, gluons for the strong force, hypothetical gravitons for gravity, and W and Z bozons for the weak force.

On the other hand, you have a list(please note that this is a very diverse list, these might have nothing to do with the PEP) of other phenomena, that in our current understanding do not need a mediator, or we do not have any idea what what the mediator could be, this includes the HUP, the PEP, van der waals (please note this might be based on EM), and dark energy (completely not understood), and the Higgs mechanism, and entanglement. It is very important to understand that the Higgs boson is not a mediator of any kind regarding the Higgs mechanism, but it is just the excitation of the Higgs field.

All gauge fields can be interpreted this way - and all 4 "fundamental forces" are in fact gauge fields.

Why do we still need to think of gravity as a force?

So to have a phenomenon be called in our current understanding a fundamental force, you need a fundamental field (needs to be a gauge field) associated with it, and a mediator particle.

This says no two electrons are allowed in the same state, and this is essentially an entanglement phenomenon.

How Quantum-entangled particles communicate with each other

The PEP is fundamentally different from all other phenomena, in that it shows some similarities with one of the building blocks of QM, that is entanglement.

So the answer to your question is, that the PEP does not have its own fundamental gauge field, nor a mediator particle, and thus in our current understanding we do not rank it as a fundamental force (rather a phenomenon that shows some similarities with entanglement).

$\endgroup$
5
  • $\begingroup$ How is the exclusion principle supposed to be "based on entanglement"? $\endgroup$
    – ACuriousMind
    Dec 29, 2020 at 17:17
  • 1
    $\begingroup$ I think this is as courageous list: "the HUP, [...], van der waals, and dark energy, and the Higgs mechanism, and entanglement." These very diverse phenomena have nothing in common with the PEP. The van der Waals interaction is fully explained by standard quantum mechanics as an electronic effect. HUP and entanglement are direct consequences of standard quantum mechanics. Dark energy is completely not understood. $\endgroup$
    – my2cts
    Dec 29, 2020 at 19:43
  • $\begingroup$ @my2cts correct, I am adding a note to clarify that. I hope if this is better this way, can you please retract the downvote (I don't know if you did or not)? $\endgroup$ Dec 29, 2020 at 20:05
  • $\begingroup$ @ÁrpádSzendrei I did not downvote. $\endgroup$
    – my2cts
    Dec 29, 2020 at 20:07
  • $\begingroup$ @my2cts thank you I edited anyway. $\endgroup$ Dec 29, 2020 at 20:08
1
$\begingroup$

Let me try to put my answer in a more general perspective than the already many others.

In our experience, we are so used to considering every correlation as induced by interactions that it is sometimes difficult to consider the presence of effects (correlations) without introducing a direct interaction as their cause. Still, this is the case of the correlations induced by the fermionic or bosonic nature of the particles.

Pauli's exclusion principle is a consequence, at the level of a one-particle description of the quantum states of many-body systems, of the antisymmetric nature of their states. It has an obvious counterpart in the case of bosons. Looking at this property as due to some interaction is not consistent with the definition we have of an interaction. In quantum physics, interactions are always introduced by coupling terms added to the sum of the non-interacting Hamiltonians for free particles.

It is a fact that we can see the effects of the statistics even in the case of no interaction in the Hamiltonian. The pair distribution functions of two fermions or two bosons in a perfect gas of the same kind of particles differ from the uniform, uncorrelated result at short distances. The exact result is (see section 5.5 of Pathria's textbook on Statistical Mechanics): $$ \langle {\bf r_1 r_2}|e^{-\beta \hat H}| {\bf r_1 r_2}\rangle = \frac{1}{2 \lambda^6} \left( 1 \pm \exp(-2 \pi r^2_{12}/\lambda^2) \right). $$ Sing plus corresponds to bosons, while sign minus to fermions. It is clear that such an effect on the pair distribution function can always formally reinterpreted as due to some inter-particle interaction. But its derivation from the states of a non-interacting Hamiltonian should make clear the artificial role of such interpretation.

Notice that the appeal for the introduction of new interactions to explain the observed correlations is not limited to the quantum regime. Osmotic forces or depletion forces in liquid solutions are a more complex example of the same attitude in classical systems.

$\endgroup$
1
$\begingroup$

There are many good answers, but I would like to add a short one: the fundamental forces are the "gauge fields", nothing more than that. So no, the Pauli principle is not a force at all (it is not even a field). See e.g. https://physicstoday.scitation.org/doi/10.1063/1.2911184

One may discuss if the Higgs field is a fifth fundamental force: this is a matter of language. However, the Higgs is introduced "by hand" and not not by "symmetry requirements", like gauge fields. So, historically, the "fundamental forces" are the ones associated with the symmetries of your (classical or quantum) field model (colloquially, 99% of the time this model is the standard model plus the still debated gravity).

$\endgroup$
2
  • 3
    $\begingroup$ It is unclear why an interaction should only be fundamental if it is a gauge field. $\endgroup$
    – my2cts
    Dec 29, 2020 at 19:49
  • $\begingroup$ It's a definition. There is an incredible amount of interactions (or forces, like the one mediated by pions, or inter-atomic forces, or macroscopic forces, and even entropic effective forces.. ), so you need to set boundaries to pick up the "most fundamental". Given the way in which we construct field theories, which is based on symmetries, it'customary to define "fundamental forces" the ones related to the gauge fields (because are related to the "fundamental symmetries" of the standard model). $\endgroup$
    – Quillo
    Dec 31, 2020 at 14:14
0
$\begingroup$

Pauli's exclusion is not so much a force as a geometric requirement for the wavefunctions of certain kinds of particles -- fermions -- that arises naturally out of quantum mechanics. It is not a force like gravity or electromagnetism.

$\endgroup$
5
  • $\begingroup$ Still, it is the main source of force, like this one: nasa.gov/sites/default/files/thumbnails/image/… $\endgroup$
    – Anixx
    Dec 28, 2020 at 13:28
  • 1
    $\begingroup$ @Anixx it's still electromagnetic force (+gravity) depicted on your picture. $\endgroup$
    – Ruslan
    Dec 28, 2020 at 23:32
  • $\begingroup$ @Ruslan no. It is Pauli exclusion between electrons. $\endgroup$
    – Anixx
    Dec 28, 2020 at 23:34
  • 1
    $\begingroup$ @Anixx no, it's not exclusion. Pauli exclusion makes EM interaction more prominent, but it's not what directly makes the atoms repel each other. See also: How can I stand on the ground? EM or/and Pauli? $\endgroup$
    – Ruslan
    Dec 28, 2020 at 23:55
  • $\begingroup$ Surely (according to general relativity) gravity is also “a geometric requirement”, and yet it counts as a force. $\endgroup$
    – Mike Scott
    Dec 29, 2020 at 17:30
0
$\begingroup$

Consider a couple of hypothetical particles, all of the known properties of which, like mass, charge, spin etc., are the same as those of the electron, but one additional property $\zeta$ is different for each particle. We thus have a system of electron-like particles, which are, unlike the electrons, distinguishable via the interaction $Z$ associated with $\zeta$.

By construction, the particles under consideration don't have Pauli exclusion. Now consider an initial state $\psi(\vec r_1,\vec r_2,\dots,\vec r_N)$, where $\vec r_i$ is the position and spin of $i$th particle. Let $\psi$ be antisymmetric in the exchange of any pair of particles $\vec r_i\leftrightarrow\vec r_j$: we are constructing an initial state for the Schrödinger's equation, so we are allowed to impose this constraint on $\psi$, despite the inapplicability of the Pauli principle.

Now, since all the properties of our hypothetical particles (except $\zeta$) are the same as those of electrons, the evolution of $\psi$ in time would leave its exchange symmetry invariant—if not the $Z$ term of the Hamiltonian.

Consider now what happens when the strength $\sigma$ of the interaction $Z$ approaches zero. Although $Z$ generally breaks the exchange antisymmetry, in the limit of $\sigma\to0$ this symmetry will become conserved. But now we've constructed a system that behaves exactly as if there were Pauli exclusion: none of these particles can share quantum state (because of the form of $\psi$), and this restriction is preserved in time. In fact, what we have constructed is exactly a system of $N$ electrons.

Notice how we haven't introduced any force at any point in the above construction. Instead, we've obtained Pauli exclusion as a consequence of the initial state of the system. More generally, it's a consequence of the initial state of the Universe (at some finite time point), and of the form of the creation operators for fermions. All fermions are created antisymmetrized, and all their interactions preserve this antisymmetry. That's all what Pauli exclusion is: it's not an additional interaction, much less so an additional "force of nature".

$\endgroup$
0
$\begingroup$

While it is true that quantum mechanical effects can be used to describe the "effective" force seen as the pauli exclusion principle (as said in most of the positively-recieved answers) - at the end of the day it seems that there is always an extra rule that's added to QM to explain this effect. This extra rule, whether it's that "ferminions are antisymmetric upon exchange" or some sort of spin statistics theorem, is something extra that is added.

Maybe it's not the best to name it a fundamental force, but it, in my opinion, is a separate extra rule that is added to QM or QFT.

$\endgroup$
-4
$\begingroup$

... Pauli exclusion principle ... produces such things as repelling of atoms and molecules in solids

Your question boils down to the understanding how is works that the state of electrons in atoms is observed pairwise.

What is an electron

The electron was discovered as an electric charge. Because - in metals - electrons can easily be separated under the influence of an external electrical potential.
The electrons is a magnetic dipole with its own fundamental physical constant $g_{e^-}$. Electrons (their magnetic dipols) can easily be aligned under the influence of an external magnetic field. In permanent magnets the common field holds the magnetic dipoles aligned (as long as a rising temperature with its higher mobility of the atoms does not destroys the aligned dipols).

The order of the discoveries of the charge and the magnetic dipole of the electron have determined our view of the electron. However, it would be appropriate to treat the electron equally as a charge and as a magnetic dipole.

What is the spin of subatomic particles

A moving electron - under the influence of an external magnetic field - is deflected in the plane that is perpendicular to both the electron's direction of propagation and the direction of the external magnetic field. In the process, the electron emits photons and slows down until all its kinetic energy is exhausted.

This mechanism is responsible for both the Lorentz force and the Hall effect.

Instead of speaking of an inexplicable spin (it is usually said that nothing spins there, but this signals the analogy to the gyroscopic effect), it would be more purposeful to speak of the interaction between the external field and the magnetic dipole of the electron.
If one also takes into account that the emission of photons is accompanied by the exchange of an impulse to the electron, one could explain the deflection of the electron without a spin as follows.

The photon emission gives the electron a sideways momentum and at the same time disturbs the alignment of the magnetic dipole of the electron. The external magnetic field immediately realigns the magnetic moment and the process starts again - as long as the electron still has kinetic energy that can be converted into photons.

That I‘m talking about to show that the intrinsic (existing independent from external circumstances) magnetic dipole of the electron is a very important property of electrons.

The electrons magnetic dipole and the atomic structure

When you are just starting out in physics, it should be easier for you to replace the term spin with the term magnetic dipole. The longer you are a physicist, the less acceptable this seems.

But indeed, it is an easy model to imagine if two electrons in an atom, occupying the same first three quantum numbers, are aligned antiparallel by their spin (just as two permanent magnets might be paired with their poles in a rectangle (like the 69 resp face-to-foot).

Loss of energy as the electron approaches the nucleus

The electrons charge is a constant value in all measurement we have done. Indeed? what is about the electron as it appoaches the nuclues. Where the energy comes from which is emitted in form of photons? How the electrical neutrality of an atom is explained although the positive charges are concentrated in the nucleus?

The claimed charge constancy overshadows that for bounded electrons and protons in atomic interactions the electric charges are exhausted by photon emission and the magnetic dipoles are responsible for the atomic structures.

Conclusion There is a magnetic interaction and an electric interaction between electrons. For bound electrons, the magnetic interaction is the more important one and it explains the Pauli principle.

Although everything inferred here is based on existing observations and in particular the laws of conservation of energy and momentum remain completely valid, don't take it too seriously. Because, you know, old physicists are not amused and even annoyed by thinking in generally new ways.

$\endgroup$
1
  • 9
    $\begingroup$ I don't see what this answer has to do with the question. The exclusion principle is a general phenomenon for fermions, regardless of whether they have charges and magnetic moments or not. The notion that spin can be replaced with "magnetic dipole" is decidedly non-mainstream, since particles can have spin without interacting with the electromagnetic field at all. $\endgroup$
    – ACuriousMind
    Dec 29, 2020 at 17:21
-6
$\begingroup$

The fact that the Pauli exclusion principle is associated withn spin states, would seem to suggests to that it is a result of a magnetic interaction.

$\endgroup$
4
  • 2
    $\begingroup$ It is actually a consequence of the discrete spacetime symmetries CPT. $\endgroup$
    – kaylimekay
    Dec 28, 2020 at 14:10
  • $\begingroup$ R.W.Bird How acceptable or speculative is my answer? My answer corresponds with your answer? $\endgroup$ Dec 30, 2020 at 12:05
  • 2
    $\begingroup$ Absolutely not. $\endgroup$
    – my2cts
    Jan 1, 2021 at 12:13
  • 1
    $\begingroup$ Note to flaggers: please don’t use “not an answer” for posts which are answers but happen to be incorrect. $\endgroup$
    – rob
    Jan 4, 2021 at 2:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.