Navigating Wikipedia for Scientific Answers: Tips and Tricks for Laymen

[Whitefire]

I have a question: can the mechanism behind Pauli's exclusion principle be considered a fundamental force, like gravitational, electromagnetic, nuclear weak or strong? Why?

Thx.

 

[Bill_K]

I have a question: can the mechanism behind Pauli's exclusion principle be considered a fundamental force, like gravitational, electromagnetic, nuclear weak or strong?

No, it's a fundamental restriction on the states that are available. Calling it a force would be like saying there's some force that pushes a particle into states with integer angular momentum. Same thing - states with integer angular momentum simply happen to be the only ones available!

 

[HeavyMetal]

I've seen that question on here before a few times. It has come up relatively frequent. Interesting how everyone keeps coming to that conclusion! A good mathematical explanation is given at http://en.wikipedia.org/wiki/Identical_particles#Symmetrical_and_antisymmetrical_states.

In short, as Bill_K says, there is simply a restriction on the states that are available. The formula describing an antisymmetric combination of quantum states cannot be normalized, because if the quantum states are the same, the solution would be 0. Zero is not normalizable, and therefore, this is not a valid solution.

In math terms...

Given two identical particles, 1 and 2, and the complete set of each of their quantum numbers, n, then the quantum state of the system must be given by the equation:

\mid\!n_{1}n_{2}\rangle\,\,\pm\mid\!n_{2}n_{1}\rangle

However, with electrons being fermions, we employ the antisymmetric combination:

\mid\!n_{1}n_{2}\rangle\,\, - \mid\!n_{2}n_{1}\rangle

State vectors need to be able to be normalized, so we multiply this entire expression by a constant. If n_{1}=n_{2}, then the solution is 0 and cannot be normalized. Therefore, the quantum state for "electron one" and "electron 2" must be different.

 

[WannabeNewton]

I have a question: can the mechanism behind Pauli's exclusion principle be considered a fundamental force, like gravitational, electromagnetic, nuclear weak or strong? Why?

No. Are you thinking of degeneracy pressure?

 

[bhobba]

No. Are you thinking of degeneracy pressure?

I suspect it may be, as I think Dyson first showed, that solidity is in fact a result of the Pauli Exclusion principle. Since many of the forces that occur in the everyday world are reaction forces due to this solidity in a sense it could be considered a fundamental force.

I don't agree with that view, but understand some may look at it that way.

Thanks
Bill

 

[Zarqon]

I suspect it may be, as I think Dyson first showed, that solidity is in fact a result of the Pauli Exclusion principle. Since many of the forces that occur in the everyday world are reaction forces due to this solidity in a sense it could be considered a fundamental force.

I don't agree with that view, but understand some may look at it that way.

Thanks
Bill

I think a tricky thing about it is that from a bottom up approach they certainly look different as they arise from different things, as was already mentioned in this thread. However, from a top down approach, looking at the effects of them they look very much the same.

Consider for example the situation of two electrons (of the same spin) in some potential well. If we just write down all forces acting on them to see how they will move, we will get to a point where the sum of the fundamental forces are balanced, at least in part, due to the repulsion of the Pauli exclusion principle. In such a situation the effect from the PEP has the unit of a force and thus appear like a force.

I'm not sure myself how to get from one picture to the other but I guess a good explanation of why the PEP is not a force should include bridging the gap between the bottom up and the top down approaches. Anyone know of a good explanation for this?

 

[Whitefire]

No, it's a fundamental restriction on the states that are available.

States are states. How the restriction happens is the mechanics I was asking about.

I've seen that question on here before a few times. It has come up relatively frequent. Interesting how everyone keeps coming to that conclusion!

That's probably because fermions' inability to occupy the same quantum state leads to inability to occupy the same space, which sets balance against fundamental forces. And the forces can intuitivelly be understood as reactions to

fundamental restriction on the states that are available

It seems that Pauli's principle is about the relation between states (percieved simultaneously); while forces are relations percieved/measured with the passing time. Seems simple enough, except that the word 'simultaneously' is no longer simple when contrasted with relativity theories.

I want a Theory of Everything in my lifetime :)

 

[HeavyMetal]

Was WannabeNewton correct in his assumption that your idea was born while reading about degeneracy pressure?

 

[Bill_K]

No, it's a fundamental restriction on the states that are available.

States are states. How the restriction happens is the mechanics I was asking about.

And forces are forces. You asked if the Pauli exclusion principle can be considered a force, and the answer is no, it cannot.

 

[tom.stoer]

States are states. How the restriction happens is the mechanics I was asking about.

It's a build-in mechanism.

Introducing fermionic creation operators an electron with quantum numbers denoted by 'r' is described by a fock space state

|r\rangle = a_r^\dagger|0>

where $|0>$ is the vacuum state. All states can be constructed that way. And all interactions can be expressed in terms of these creation (and annihilation) operators.

Now because of the Pauli principle (or the more fundamental spin-statistics theorem) we construct the creation operators such that the following holds

a_r^\dagger\,a_s^\dagger = -a_s^\dagger\,a_r^\dagger

For $r \neq s$ this means that a state with two different fermions is anti-symmetric w.r.t. to interchange of to particles.

But for $r = s$ this means

(a_s^\dagger)^2 = 0

So we construct the mathematical formalism such that no two identical fermions can be described by the formalism. It's like asking "what shall I do when I see a blue traffic light?" Blue traffic lights do not exist by construction, so there's no rule what to do when you see one ;-)

 

[Ken G]

What's interesting about asking if the Pauli exclusion principle is a fundamental force in regard to degeneracy pressure is that the degeneracy, and the PEP, it not actually playing any fundamental role in that question. The real underlying question there is, is pressure itself a fundamental force? When you ask this question, you see that degeneracy is not adding anything, the question is actually, what is pressure?

Pressure is a lot of things to a lot of people, but there is some sense to which it is a fundamental force, it is the fundamental force you get when you average over the behavior of individual particles and speak about a fluid or solid picture of the combined system. So it is not a "fundamental force" on particles, which is what most mean by fundamental, but it certainly is a "fundamental force" on solids and liquids, which is what others might mean by fundamental (connecting to the point bhobba made about what we call normal forces). So what the whole question about the PEP being a fundamental force really gets at is whether apparent forces that appear when you average over the behavior of large systems should count as fundamental, and that really depends on whether you regard such systems as themselves fundamental!

For most people, fundamental is taken in the reductionist sense, so you have to be talking about forces on particles, and then the PEP is not a force on particles. But to a solid-state physicist or a hydrodynamicist, averaging over the particles is "fundamental" to their craft, and they might not buy off on the particle physicist perspective that only particles are fundamental!

Also, note that if you hold that pressure forces, and the "normal force", can only be there because of interparticle forces, the latter of which are fundamental, realize that this is simply not true. Basic hydrodynamics normally assumes there are no forces between the particles at all, and not even any need for collisions, once one has already bought off on the fluid averaging that is fundamental to hydrodynamics. Also, when you walk on the ground and receive a normal force, it is likely that the PEP is all you need to have that, you don't need any actual interparticle forces there. Indeed, the interparticle forces tend to attractive, not repulsive! The repulsion that is the normal forces comes from dE/dz, the way the energy changes when you compress a solid, and that comes from the PEP, but it's no different than situations where dE/dz comes from more mundane forms of pressure, say in an ideal gas.

So is dE/dz a "fundamental force"? Not in the reductionist particle-physics sense, where only the particles are "fundamental." But we never actually solve solid-state or gas problems by solving for the particles, so what is "fundamental" to those kinds of solutions is not taking the particle picture! Instead, issues like dE/dz become what is fundamental to those applications. So I'm not sure there is a lot to be gained by defining "the 4 fundamental forces" when it comes to applications where it just doesn't matter at all the nature of the interparticle forces (and it's hugely important that this doesn't matter at all), what matters is what physics controls dE/dz.

 

[Jabbu]

Pauli exclusion principle is directly related to spin. Spin is just another name for magnetic dipole moment. Is magnetic force between two dipoles not part of derivation for Pauli exclusion principle?

 

[Nugatory]

Spin is just another name for magnetic dipole moment

No, and therefore

Is magnetic force between two dipoles not part of derivation for Pauli exclusion principle?

No

 

[Jabbu]

No, and therefore

No

http://en.wikipedia.org/wiki/Electron_magnetic_dipole_moment

Wikipedia redirects "electron spin" to "electron magnetic dipole moment" article. How are they not the same thing, what are you talking about?

 

[Ken G]

Fermions have half-integer spin, but they don't have to have charge so they don't have to have a magnetic dipole moment. Moreover, since electrons have charge, they experience electrostatic forces, but that also has nothing to do with the PEP. The PEP would be true even for chargeless Fermions, as in neutron stars, so the PEP has nothing to do with electrostatic or magnetic forces, even though they are in principle effects that are present for electrons.

 

[HeavyMetal]

From what I understand, Pauli's exclusion principle is merely a mathematical statement of fact explaining that no more than two fermions can occupy the same quantum state. The force of repulsion between two electrons during pairing IS a force, but I'm not sure about neutrons.

Anyone? Do paired neutrons repel each other, and if so, is this the basis of degeneracy pressure in neutron stars?

 

[Ken G]

Neither electrons in a white dwarf, nor neutrons in a neutron star, "repel each other" in the sense of a force between the particles (the electron-electron electric forces are highly shielded by protons). The only sense in which there is a repulsion is the same way that ping pong balls bouncing around inside a box will "push on" the walls of the box-- they have kinetic energy, and that energy increases if you contract the box, so there is a dE/dV where E is the kinetic energy and V is the volume of the box. -dE/dV is what we call pressure, and it is there because the particles carry momentum, not because they repel each other. Exactly the same is true of the PEP for either electrons or neutrons. So that's why I say there is only a "force" in a globally averaged sense, accounting for the way the particles carry momentum around, not in the reductionist sense of how individual particles are interacting with each other.

Indeed, the PEP is playing no role at all in the pressure, once you specify the energy situation. All the PEP does is make it impossible to remove heat, which has a significant impact on the energetics.

 

[HeavyMetal]

Excellently explained. So, is this the same reason that when we deal with quantum mechanics problems such as particle in a box, as the box decreases in size, allowed energy levels for the system spread out and get higher?

 

[Ken G]

Yes, it stems from the uncertainty principle, if you like. The states always behave like that, but in a classical gas, the states are sparsely populated, so the particle behaviors are not constrained by what the states are doing, if you move around heat. In a degenerate gas, they are, so you have less freedom to move heat around. Yet either way, if you move no heat around, then both the classical and degenerate gases will have their particles follow the changes in their respective states, they follow the behavior of the state they are in-- the state that the particles in are "adiabatic invariants", so adiabatic compression of a degenerate gas is just like adiabatic compression of an ideal gas. Differences between "degeneracy pressure" and "ideal gas pressure" are often overstated.

 

[HeavyMetal]

Yes, it stems from the uncertainty principle, if you like. The states always behave like that, but in a classical gas, the states are sparsely populated, so the particle behaviors are not constrained by what the states are doing, if you move around heat. In a degenerate gas, they are, so you have less freedom to move heat around. Yet either way, if you move no heat around, then both the classical and degenerate gases will have their particles follow the changes in their respective states, they follow the behavior of the state they are in-- the state that the particles in are "adiabatic invariants", so adiabatic compression of a degenerate gas is just like adiabatic compression of an ideal gas. Differences between "degeneracy pressure" and "ideal gas pressure" are often overstated.

When you say "move heat around," do you just mean heat transfer throughout the medium? Also, do states mandate particle behaviors, vice versa, or are they directly related?

 

[Jabbu]

Fermions have half-integer spin, but they don't have to have charge so they don't have to have a magnetic dipole moment.

Neutron has no charge and yet it has magnetic moment. Can you name a particle with 1/2 spin which has no magnetic moment? What other physical property there is to spin beside magnetic moment?

Moreover, since electrons have charge, they experience electrostatic forces, but that also has nothing to do with the PEP.

Wouldn't magnetic dipole force between two electrons overcome their electric repulsion if they came close enough together?

 

[HeavyMetal]

What other physical property there is to spin beside magnetic moment?

Spin is a purely quantum mechanical property and there is no classical analogue.

 

[Jabbu]

Spin is a purely quantum mechanical property and there is no classical analogue.

http://en.wikipedia.org/wiki/Electron_magnetic_dipole_moment
http://en.wikipedia.org/wiki/Angular_momentum_coupling
http://en.wikipedia.org/wiki/Magnetism

- "In principle all kinds of magnetism originate (similar to Superconductivity) from specific quantum-mechanical phenomena (e.g. Mathematical formulation of quantum mechanics, in particular the chapters on spin and on the Pauli principle)."


It's not any analogue, it's a part of quantum mechanics.

 

[WannabeNewton]

Anyone? Do paired neutrons repel each other, and if so, is this the basis of degeneracy pressure in neutron stars?

They don't repel one another per say. The effect is quite easy to understand. Consider a degenerate Fermi gas e.g. say we have a gas of Fermions and we lower the temperature to $T = 0$.

Some fermions will settle into the ground state of the system but unlike a Bose gas this state has a maximum occupancy due to the exclusion principle so the other particles of gas will start filling in states which are higher in energy than the ground state, always in accordance with the Fermi statistics.

Eventually, after all the fermions have filled the lowest possible energy states, they will form what is called a Fermi sea with the fermions of the highest energy at $T = 0$ living on the so-called Fermi surface. The Fermi surface of this ground state configuration therefore corresponds to a non-vanishing energy, the Fermi energy $\mathcal{E}_F (V)$, where $V$ is the volume of the Fermi gas, so that there is a non-zero ground state pressure $P = -\partial_V \mathcal{E}_F |_{T,N}$ (under changes in volume or temperature it is usually the states on and directly below the Fermi surface that are affected since the lower energy states are completely filled).

So the pressure is really just a result of the exchange interaction between the fermions that forces the fermions to higher and higher energies above the ground state energy even at $T = 0$, resulting in force exerted on the "walls" of the "container" of gas.

 

[HeavyMetal]

So, because more than two bosons can occupy the same energy state, they demonstrate much lower pressure for the same number of particles as a fermion gas at any given temperature?

If $T = 0 K$, is the energy of a large number of bosons much lower than a group of fermions due to attraction?

 

[WannabeNewton]

So, because more than two bosons can occupy the same energy state, they demonstrate much lower pressure for the same number of particles as a fermion gas at any given temperature?

Not at any temperature. If the temperature is high enough for the thermal wavelength to be much smaller than the spacing between gas particles then we pass over into Maxwell-Boltzmann statistics which doesn't distinguish between fermions and bosons. However at extremely low temperatures, i.e. $T = 0$, we can say that for ideal Bose gas vs. ideal Fermi gas in the same volume the pressure of the latter is greater than that of the former. See below.

If $T = 0 K$, is the energy of a large number of bosons much lower than a group of fermions due to attraction?

Yes but again it's not an attraction. It's just an exchange interaction. The bosons all like to clump together in the ground state because of Bose statistics. There is no attractive potential between the bosons in an ideal gas.

For a Bose gas of conserved particle number, and for a Fermi gas, the equation of state $P = \frac{2}{3}\frac{E}{V}$ still holds. For a photon gas, where particle number isn't conserved, we instead have $P = \frac{1}{3}\frac{E}{V}$.

 

[HeavyMetal]

Not at any temperature. If the temperature is high enough for the thermal wavelength to be much smaller than the spacing between gas particles then we pass over into Maxwell-Boltzmann statistics which doesn't distinguish between fermions and bosons. However at extremely low temperatures, i.e. $T = 0$, we can say that for ideal Bose gas vs. ideal Fermi gas in the same volume the pressure of the latter is greater than that of the former. See below.
Yes but again it's not an attraction. It's just an exchange interaction. The bosons all like to clump together in the ground state because of Bose statistics. There is no attractive potential between the bosons in an ideal gas.

For a Bose gas of conserved particle number, and for a Fermi gas, the equation of state $P = \frac{2}{3}\frac{E}{V}$ still holds. For a photon gas, where particle number isn't conserved, we instead have $P = \frac{1}{3}\frac{E}{V}$.

I am not very familiar with the exchange interaction yet. It is something I've encountered but not used in my work. Is an exchange interaction similar to two potential wells that are next to each other, that gain just enough momentum to exchange? And as a result, the wavefunction becomes antisymmetric or stays symmetric depending on the nature of the particles?

 

[Ken G]

When you say "move heat around," do you just mean heat transfer throughout the medium? Also, do states mandate particle behaviors, vice versa, or are they directly related?

I mean adding heat, not just shifting the levels by slowly changing the volume. Adding heat shuffles the particles into new higher-energy states, whereas adiabatic compression simply moves all the levels up to higher energies but keeps the particles in the "same levels" as they shift up.

 

[Ken G]

Neutron has no charge and yet it has magnetic moment.

Good point, but that's because a neutron is not itself an elementary particle with zero charge, it is comprised of particles of positive and negative charges. In any event, the magnetic properties of the neutron have little or nothing to do with the pressure inside a neutron star.

Can you name a particle with 1/2 spin which has no magnetic moment?

Sure, any chargeless fermion that is itself an elementary particle. Say, a neutrino.

What other physical property there is to spin beside magnetic moment?

The quantum statistics that have to do with the Pauli exclusion principle and the way heat is partitioned among a set of fermions that are close to their ground state, like the electrons filling the levels in a molecule or atom. You don't think the electrons filling the shells of a noble gas atom have important magnetic effects do you? Yet, the level structure of that atom is completely ruled by the PEP.

Wouldn't magnetic dipole force between two electrons overcome their electric repulsion if they came close enough together?

Not generally. For example, the helium atom has two electrons, and their electric repulsion is more important than their magnetic interactions. The PEP appears in how differently those electrons act if they have the same or opposite spins, and that's way more important than any magnetic effects.

 

[Ken G]

So the pressure is really just a result of the exchange interaction between the fermions that forces the fermions to higher and higher energies above the ground state energy even at $T = 0$, resulting in force exerted on the "walls" of the "container" of gas.

Be careful though, there exist a lot of misconceptions about pressure-- the pressure is a result of whatever reasons the system has internal kinetic energy, as usual. It's true that if you put the system in a low-temperature bath, it will still draw a lot of heat from that bath because degeneracy does require a lot of internal energy to fill all those states, and it's because of the PEP that this is true, but that's not the exchange interaction. The exchange interaction is different, it requires that the particles exert a force on each other, i.e., it comes from the potential energy in the Hamiltonian. Degeneracy pressure would normally neglect any forces between the particles, it applies for the quantum mechanical equivalent of an ideal gas.