# Tag Info

31

Neutrons (and protons) being spin 1/2 fermions, must fit antisymmetric wavefunctions. This "wavefunction" doesn't always involve waves, though. For nucleons - the generic term for neutron or proton - this wavefunction for the pair is a produce of (1) a spatial part, (2) a spin part, and (3) an isospin part. The isospin part is a clever way to describe ...

19

Neutrons have spin 1/2 and therefore obey the pauli exclusion principle, meaning two neutrons cannot occupy the same space at the same time. When two neutrons' wavefunctions overlap, they feel a strong repulsive force. See http://en.wikipedia.org/wiki/Exchange_interaction .

16

OK, I'll make a complete revision of my original reply since it was quite sloppy. First, I originally confused two issues that are nevertheless related, I confused stability of matter and the impenetrability of matter. But, it must be clear that the two questions are related. If I have two chunks of matter of the same type on top of one another, one can't ...

15

Any answer based on analogies rather than mathematics is going to be misleading, so please bear this in mind when you read this. Most of us will have discovered that if you tie one end of a rope to a wall and wave the other you can get standing waves on it like this: Depending on how fast you wave the end of the rope you can get half a wave (A), one wave ...

14

Ok, I'll bite the bullet and get the downvotes, but my answer is EM. Why? Well, you can't stand on water and you can't stand on air. The Pauli principle applies to those cases but doesn't make matter solid. It's the (fundamentally) cristalline structure that makes an object solid enought to stand on. This is yes, related to QM (what is not) but certainly EM ...

12

Er ... nothing prevents this. That's what a Bose-Einstein condensate is: lots of bosons in the same place and quantum state. You are observing that the sate is not perfectly localized, but that is a consequence of the state not being exactly zero momentum. Ultimately the Heisenberg principle puts a lower limit on how localized they could be. If the bosons ...

12

I don't have a very satisfactory description of the microscopic picture, but let me share my thoughts. The Pauli exclusion doesn't quite say that fermions can't be squeezed together in space. It says that two fermions can't share the same quantum state (spin included). A black hole has an enormous amount of entropy (proportional to its area, from the ...

11

Holy moly, there are a lot of confused parts of answers out there. Here is a way to begin to sort out the different related principles, physically. The question currently asks "EM or/and Pauli?". Short answer: Neither, although it is true that electromagnetism is the force involved (rather than the strong or weak nuclear forces, the only other ...

10

How does the Pauli Exclusion Principle actually create a force? The Pauli exclusion principle doesn't really say that two fermions can't be in the same place. It's both stronger and weaker than that. It says that they can't be in the same state, i.e., if they're standing waves, two of them can't have the same standing wave pattern. But for bulk matter, ...

9

Well, we could say, yes, that is simply how quantum mechanics works. But these are not the axioms of quantum mechanics, and the exclusion principle in particular is really only understood in the context of quantum field theory. The electron does not "spiral in" because it doesn't move in the classical sense at all. At the scale of the size of an atom, ...

9

Indistinguishableness of particles is formulated in QM in terms of the total wave function symmetry. Then the wave functions can be symmetric or antisymmetric on their arguments. In case of antisymmetry ($\psi(x_1,x_2) = -\psi(x_2,x_1))$ the particles are called fermions and $\psi(x_1,x_2=x_1) = 0$. They say "the particles cannot occupy the same state". It's ...

8

The exchange interaction is, on the contrary, one of the approximate low-energy consequences of the Pauli exclusion principle and the identical character of the particles. The fundamental justification of these facts is offered by quantum field theory. Relativity requires observables to be linked to regions of spacetime - so that they don't communicate over ...

8

The Pauli exclusion principle can be stated as "two electrons cannot occupy the same energy state", but this is really only a rough way of stating it. It's more precise to say that the wavefunction of a system is anti-symmetric with respect to exchange of two electrons. The trouble is that now I have to explain to a non-physicist what "anti-symmetric" means ...

8

This is a nice puzzle--- but the answer is simple: the composite bosons can occupy the same state when the state is spatially delocalized on a scale larger than the scale of the wavefunction of the fermions inside, but they feel a repulsive force which prevents them from being at the same spatial point, so that they cannot sit at the same point at the same ...

8

Suppose you take two helium atoms and try to push them together. There will be a short distance repulsion and we'd normally describe this as the Pauli exclusion force, so it does indeed sound as if the exclusion principle generates a fundamental force. However suppose one of the He atoms has both electrons excited to the $2s$ orbital (not physically likely, ...

8

Great question that exposes some really confusing terminology. This is a rather long answer, and the punchline is basically in the second-to-last paragraph, but I think (hope) it's worthwhile to read the whole answer because I tried to give a somewhat systematic description of fermionic states using a specific, simple example along the way. Firstly, let's ...

8

Take for example a 2 particle fermion system that are not interacting. Because they're not interacting we can assume the two-particle wavefunction can be written as a product of the single particle wavefunctions. Let's label the two single particles with $a_1$ and $a_2$, we have: $$... 7 First of all, strictly speaking, electron shells (as well as atomic orbitals) do not exist in atoms with more than one electron. Such physical model of an atom is simplified (and often oversimplified), it arises from a mathematical approximation, which physically corresponds to the situation when electrons do not instantaneously interact with each other, but ... 7 Here's a long answer leading up to a mathematical measure for the relevance of the Pauli principle. The quantum state of an electron is not just determined by its energy, angular momenta (quantities that lead to good or approximate quantum numbers). Position is also part of the equation. Now, position in QM is not just a simple vector like it is in classical ... 7 As suggested by M.Sameer I convert my comment into an answer: Dear M.Sameer: It seems that you are missing that the n+\ell Madelung rule is not an exact result derived from first principles, but rather a rule of thumb, that holds for, say, approximately 95 percent of all the elements, with important exceptions, cf. this wikipedia page. Nevertheless, ... 6 The use of the term "force" in quantum mechanics can be misleading, since the macroscopical classical force does not directly translate at the quantum level. That's why I prefer to speak about the 4 fundamental interactions than about the force. And the answer to your question is : the degeneracy pressure is not linked to any of the four fundamental ... 6 They are distinguishable so of course the number of states doubles if you double the number of species, i.e. if you use both electrons and muons. So with 2 muons and 2 electrons, 2 muons would first sit to the 1s orbitals very close to the nucleus, resembling a "small helium". The electrons are 200+ times lighter so their orbitals are 200+ times larger. (In ... 6 The atomic orbital description of an atom is only an approximation for any atom with more than one electron. Atomic orbitals only exist when the potential is centrally symmetric. With a centrally symmetric potential the wavefunction of the atom factors into separate atomic orbitals so we can talk about 1s, 2s, 2p etc orbitals. However, when there is ... 6 This argument just replaces one axiom by another. It assumes that if a quantum system consists of identical particles, then the state of the system should not change (it get's multiplied by a phase) under exchange of quantum numbers. Although this is (perhaps) a more intuitive way of thinking about states of identical particles, it's still a strong ... 6 Keeping it simple, let's asume that \psi(a) creates a particle in the state a (i.e., characterized by some collection of quantum numbers that we call a),$$ \psi(a)|0\rangle=|a\rangle .$$and \psi(b) does the same for b. We can create a state with two particles:$$ \psi(b)\psi(a)|0\rangle = \psi(b)|a\rangle = |a;b\rangle  ...

6

If what you mean by conservation of "number of fermions" is that the number of fermions in the initial state must be equal to the number of fermions in the final state, well, it's wrong. Consider for example this nuclear reaction (which exists): $$p + p \to \pi^+ + d$$ where $p\equiv$ proton (spin 1/2), $\pi^+$ pion (spin 0) and $d\equiv$ deuterium (spin ...

5

Yes, neutrinos should obey Fermi-Dirac statistics and yes, the Pauli Exclusion Principle should operate for neutrinos. But let's examine how dense the neutrino population has to be for this to be important. The Fermi momentum is given by $$p_F = \left( \frac{3}{8\pi}\right) h n_{\nu}^{1/3}$$ where $n_{\nu}$ is the neutrino number density. In order to be ...

5

You are correct to point out that there's no symmetry that forbids a state with isospin 3/2 and spin 1/2; in the nomenclature, this is also called a $\Delta$ resonance. The Particle Data Group lists two such particles, with mass 1620 MeV and 1910 MeV. They exist, but they are heavier than the spin-3/2 $\Delta$ at 1232 MeV. The reason why is isospin, ...

5

I think the answer is it depends on distance (relative to the size of your system). Another well known example of a boson which is comprised of fermionic components is the helium-4 atom, which has integer spin (both the nucleus and the neutral atom itself). Fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared ...

5

Well, if you run over a sufficiently deep hole, you WILL fall to center. So why don't you fall to the center anyway? Because there already is something thats blocks the space - namely "the earth". As most of it is either heavier than you are (so switching places would need energy to lift the it up) or at least is solid and sticks together, it's on you to ...

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