Pauli's exclusion principle bosons

[misogynisticfeminist]

I've heard quite abit of things about bosons and am quite confused. The biggest thing which distinguishes fermions from bosons, would be Pauli's exclusion principle. But I've also heard things about bosons having half- integer, while fermions have interger spin, among many others. I've also heard also that liquid helium can be considered a boson, but I once thought that bosons can be applied only to fundamental particles.

What are the properties which distinguishes bosons and fermions?

 

[Mk]

Bosons, named after Satyendra Nath Bose, are particles which form totally-symmetric composite quantum states, they have integer quantum spin.

All elementary particles are either bosons or fermions, so everything not a fermion is a boson

Gauge bosons are elementary particles which act as the carriers of the fundamental forces/messenger particles

http://newsimg.bbc.co.uk/media/images/39882000/gif/_39882466_standard_model2_416.gif

Particles composed of a number of other particles (such as protons or nuclei), like atoms, can be either fermions or bosons, depending on their total spin. Hence, many nuclei are bosons. While fermions obey the Pauli exclusion principle(no more than one fermion can occupy a single quantum state) there is no exclusion property for bosons, which are free to crowd into the same quantum state. This explains the spectrum of black-body radiation and the operation of lasers, the properties of liquid Helium-4 and superconductors and the possibility of bosons to form Bose-Einstein condensates, a particular state of matter.

Because bosons do not obey the Pauli exclusion principle, it is much harder to form stable structures with only bosons than with fermions. This difference accounts for the difference between what we think of as matter and things that confuse people between if it's matter and not sometimes, such as light.


Fermions, named after Enrico Fermi, are particles that obey the Pauli exclusion principle, and Fermi-Dirac statistics. The spin-statistics theorem states that fermions have half-integer spin. One possible way of visualizing spin is that particles with a 1/2 spin, i.e. fermions have to be rotated by two full rotations to return them to their initial state.

The elementary particles which make up matter are fermions, belonging to either the quarks (which form protons and neutrons) or the leptons (such as electrons). The Pauli exclusion of fermions is responsible for the stability of the electron shells of atoms, making complex chemistry possible. It also allows the stability of degenerate matter under extreme pressures.

Fermions:
* electrons
* quarks
* protons
* neutrons
* neutrinos

 

[mikeyork]

All identical particles, of whatever spin, obey the same general exclusion rule:

"When all other quantum numbers are the same, only even eigenstates of composite spin are allowed."

It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle.

So the notion that there are two types of particles, fermions and bosons, is misleading. It just happens that only those particles which obey the Pauli principle can give any properties to matter characterized by a specific spatial location. Other particles just form an undifferentiated soup.

The generalized rule is a fundamental property of QM and can be proved assuming only the obvious:

1. Unique state vectors require a uniquely described physical state.
2. Particle permutation is an artifice of state descriptions and not physically observable.

The rule then follows as an interference effect.

 

[Norman]

It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle.

Hmmm... Clebsch-Gordon might say different?

So the notion that there are two types of particles, fermions and bosons, is misleading. It just happens that only those particles which obey the Pauli principle can give any properties to matter characterized by a specific spatial location. Other particles just form an undifferentiated soup.

Maybe I don't know what you mean by an "undifferentiated soup", but bosons are real and they exist in our every day lives- Read Mk's post for examples. I also have no idea what this undifferentiated soup has to do with the OP question about bosons and what they are. Maybe your ideas are better kept in Theory Development.

misogynisticfeminist- I would stick with Mk's information- it is sound.

 

[ZapperZ]

It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle.

Then you will have a heck of a time explaining the existence of spin-triplet superconductors like the ruthenates and superfluidity of He3.

Zz.

 

[mikeyork]

Hmmm... Clebsch-Gordon might say different?

Not for identical particles when all other quantum numbers are the same and "permutation" is not a physically observable operation. Please see the context of the remark you are questioning (only even eigenstates allowed) and you will understand it better. The generalized exclusion rule I quote is verifiable throughout physics. Identical particle scattering (p-p, pi-pi, d-d, alpha-alpha, etc) is the obvious example. Even when symmetry breaking is present, particles which belong to the same multiplets obey this rule approximately, whether bosons or fermions.

Please inform me when you find a pair of indistinguishable electrons forming a composite spin triplet.

Maybe I don't know what you mean by an "undifferentiated soup", but bosons are real and they exist in our every day lives- Read Mk's post for examples.

Of course. They just don't give any structure to matter. That is why they are called "force" particles and fermions are called "matter" particles.

I also have no idea what this undifferentiated soup has to do with the OP question about bosons and what they are. Maybe your ideas are better kept in Theory Development.

misogynisticfeminist- I would stick with Mk's information- it is sound.

There is nothing contentious or "unsound" about this difference. The differentiation of matter is critically dependent on the fact that "matter" particles have spin 1/2 and obey the Pauli rule. For example, without the Pauli rule we would have no chemistry and no life.

 

[mikeyork]

Then you will have a heck of a time explaining the existence of spin-triplet superconductors like the ruthenates and superfluidity of He3.

Zz.

Please explain why. Are you claiming a violation of the Pauli principle or, like Norman, ignoring the context of the remark you quote?

 

[ZapperZ]

Please explain why. Are you claiming a violation of the Pauli principle or, like Norman, ignoring the context of the remark you quote?

I'm sorry, but it is you who have to do the explanation. Nothing is being violated here, least of all the pauli exclusion principle (you seem to be forgetting that the total asymmetric wavefunction can be satisfied with either asymmetric spin OR asymmetric spatial part). All you need to do is look up the superfludity of He3 and the superconducticity of Sr2RuO4 as examples. They are well-verified spin-triplet pairings, or else they won't have given Tony Leggett his Nobel Prize a couple of years ago.

Zz.

 

[James Jackson]

mikeyork, you are clearly confusing the term 'boson' with the force carrying 'vector bosons'.

 

[dextercioby]

Not for identical particles when all other quantum numbers are the same and "permutation" is not a physically observable operation. Please see the context of the remark you are questioning (only even eigenstates allowed) and you will understand it better. The generalized exclusion rule I quote is verifiable throughout physics. Identical particle scattering (p-p, pi-pi, d-d, alpha-alpha, etc) is the obvious example. Even when symmetry breaking is present, particles which belong to the same multiplets obey this rule approximately, whether bosons or fermions. Please inform me when you find a pair of indistinguishable electrons forming a composite spin triplet.Of course. They just don't give any structure to matter. That is why they are called "force" particles and fermions are called "matter" particles.There is nothing contentious or "unsound" about this difference. The differentiation of matter is critically dependent on the fact that "matter" particles have spin 1/2 and obey the Pauli rule. For example, without the Pauli rule we would have no chemistry and no life.

Good.
1.First of all,composing 2 spins 1/2 will give 2 irreducible spacesne of weight 0 which is unidimensional and one irreducible space of weight 1 which is 3 dimensional.
2.Bosons can be "matter particles".

Daniel.

 

[DB]

One possible way of visualizing spin is that particles with a 1/2 spin, i.e. fermions have to be rotated by two full rotations to return them to their initial state.

I like that, good statement it helps me visualize the concept of intrinsic spin.

 

[Astronuc]

What are the properties which distinguishes bosons and fermions?

Try -

Fermions

Bosons

And while at it try -

Spin - one the distinguishing characteristics that differentiates fermions and bosons.

 

[mikeyork]

I'm sorry, but it is you who have to do the explanation. Nothing is being violated here, least of all the pauli exclusion principle (you seem to be forgetting that the total asymmetric wavefunction can be satisfied with either asymmetric spin OR asymmetric spatial part).
Zz.

Since the statement you questioned was none other than the Pauli principle, you seem to be contradicting yourself now. It's you that seems to have forgotten the import of the Pauli rule (no two electrons can be in the same state). If the spatial part is asymmetric, then the spatial quantum numbers cannot be the same and the Pauli rule does not apply.

I cited a more general rule which I'll repeat here:

"For all identical particles, of whatever spin,... when all other quantum numbers are the same, only even eigenstates of composite spin are allowed."

For identical spin 1/2 particles, when all other quantum numbers are the same, as I said before, this implies a spin singlet. This, in turn, implies the Pauli rule

When you refer to spatial asymmetry this means that not "all other quantum numbers are the same" and so the spin singlet rule does not apply.

If this condition does not hold, the rule needs to be re-expressed. For example, when it comes to pairs of particles in their CM frame, since their momenta are opposite, the rule needs to be transformed to the CM frame. In fact it becomes:

"For all pairs of identical particles, of whatever spin, in their CM frame, when all other quantum numbers are the same, only even sums of composite orbital and spin angular momentum are allowed."

i.e. L+S must be even. This rule is the same for bosons and fermions and has been verified in identical particle scattering experiments.

Even for non-identical particles belonging to the same isospin multiplet (e.g. p.n) it is possible to derive that L+S+I must be even as an approximation and this rule is used to help understand the structure of nuclei.

 

[mikeyork]

Good.
1.First of all,composing 2 spins 1/2 will give 2 irreducible spacesne of weight 0 which is unidimensional and one irreducible space of weight 1 which is 3 dimensional.

In general, yes of course, but not when all other quantum numbers are the same and permutation is not observable.

2.Bosons can be "matter particles".

Some forms of matter are indeed bosons. But even they are fermion composites. Bosons cannot give the structure which differentiates matter. That is the point I am trying to make.

 

[dextercioby]

It doesn't matter whether the bosons are composed from 2 fermions.They're still described by a massive scalar field,electrically charged or not.

Daniel.

 

[mikeyork]

It doesn't matter whether the bosons are composed from 2 fermions.They're still described by a massive scalar field,electrically charged or not.

Even though this is a theory-dependent statement, not one that is directly observable, let's assume it is true. I still have to say "so what?".

Let me remind you what my original post was about:

The notion that bosons and fermions are two different types of particles is misleading. They both obey the same generalised exclusion rule (which follows from basic QM when you require unique state vectors for uniquely described states and unobservability of particle permutation). The essential difference between fermions and bosons is that, as a result of the exclusion rule, only fermions can give the differentiated structure to matter that we see in the periodic table, atomic nuclei and the tables of sub-nuclear hadrons. Without fermions there would be no chemistry or biology or nuclear energy.

 

[ZapperZ]

Since the statement you questioned was none other than the Pauli principle, you seem to be contradicting yourself now. It's you that seems to have forgotten the import of the Pauli rule (no two electrons can be in the same state). If the spatial part is asymmetric, then the spatial quantum numbers cannot be the same and the Pauli rule does not apply.

I cited a more general rule which I'll repeat here:

"For all identical particles, of whatever spin,... when all other quantum numbers are the same, only even eigenstates of composite spin are allowed."

For identical spin 1/2 particles, when all other quantum numbers are the same, as I said before, this implies a spin singlet. This, in turn, implies the Pauli rule

When you refer to spatial asymmetry this means that not "all other quantum numbers are the same" and so the spin singlet rule does not apply.

If this condition does not hold, the rule needs to be re-expressed. For example, when it comes to pairs of particles in their CM frame, since their momenta are opposite, the rule needs to be transformed to the CM frame. In fact it becomes:

"For all pairs of identical particles, of whatever spin, in their CM frame, when all other quantum numbers are the same, only even sums of composite orbital and spin angular momentum are allowed."

i.e. L+S must be even. This rule is the same for bosons and fermions and has been verified in identical particle scattering experiments.

Even for non-identical particles belonging to the same isospin multiplet (e.g. p.n) it is possible to derive that L+S+I must be even as an approximation and this rule is used to help understand the structure of nuclei.

And you seem to be IGNORING an important fact. If YOUR interpretation of what the Pauli Exclusion principle is is TRUE, then it has been severely and SPECTACULARLY violated by SEVERAL experimental observations! Go open Leggett's paper on superfluidity of He3, the paring state of the ruthenates, how the d-orbital is filled, and a whole zoo of other observations.

Pauli exclusion principle requires an antisymmetric TOTAL wavefunction. This means that you can have an asymmetry spatial*symmetric spin OR symmetric spatial*asymmetric spin! A spin-triplet state requires that the spatial part is asymmetric, meaning the two spins cannot be very close to each other! It means that any pairing of the two will be considerably over a larger extent. This is why He3 becomes a superfluid at a significantly lower temperature than He4! It requires a much lower temperature to reduce the thermal fluctuation to be able to maintain coherence between two spins being paired over a LARGER distance!

Hey, don't take my word for it, read it yourself! If you still maintain your postion, then the next posting you give on here better be to wiggle your way out of these whole bunch of experimental results that clearly have demolished your version of the exclusion principle.

Zz.

 

[mikeyork]

mikeyork, you are clearly confusing the term 'boson' with the force carrying 'vector bosons'.

Not me -- although I understand that a selective reading of my posts in this thread, focussed only on where I used the very loose terms "force particle" and "matter particle" could lead you to that confusion.

I'd also suggest that pions (which are scalar bosons) are also considered "force particles". This useage pre-dates the standard model and unified QFT. Back in the days of Yukawa, pions were considered to be the unique carriers of the strong force.

 

[ZapperZ]

Not me -- although I understand that a selective reading of my posts in this thread, focussed only on where I used the very loose terms "force particle" and "matter particle" could lead you to that confusion.

I'd also suggest that pions (which are scalar bosons) are also considered "force particles". This useage pre-dates the standard model and unified QFT. Back in the days of Yukawa, pions were considered to be the unique carriers of the strong force.

OK, so where is the "selective reading" here that will cause a confusion? You clearly are claiming that pions, just because it was MISTAKENLY taught to be the strong force carrier, is STILL a force particle? Whoa!

And what force is it carrier of? And while you're at it, what force is the kaon and the J/psi particle a carrier of?

Zz.

 

[mikeyork]

This means that you can have an asymmetry spatial*symmetric spin OR symmetric spatial*asymmetric spin! A spin-triplet state requires that the spatial part is asymmetric, meaning the two spins cannot be very close to each other!

I am well aware of the fact that the whole wavefunction needs to be taken into account. This is precisely why I talked about the necessity of equality of "all other quantum numbers" in what I wrote. Spatial asymmetry implies that there are quantum numbers which are not the same. So the Pauli rule does not apply. Please read it again.

Although there is a slight technical difference between the observable rule I give and the conventional expression via the Symmetrization Postulate (which actually needs a technical qualification before it is true, see below) this difference has no observable effects. The difference you are trying to create is non-existent.

Hey, don't take my word for it, read it yourself! If you still maintain your postion, then the next posting you give on here better be to wiggle your way out of these whole bunch of experimental results that clearly have demolished your version of the exclusion principle.
Zz.

My version of the Pauli exclusion principle is identical to Pauli's (no two identical spin 1/2 particles can be in the same state). No wiggling is necessary. You have just misunderstood what I wrote.

BTW the technical problem with the conventional Symmetrization Postulate is that it requires an additional qualification regarding the construction of the wavefunctions. It is trivially easy to construct wavefunctions for fermions that are permutation symmetric using a different method, but giving the same observable results. They are related to the conventional wavefunctions by an order-dependent phase so that exchange results in a 2pi relative rotation of one particle's state with respect to the other in one case but not in the other. This has been known for more than 30 years. My first paper on this (unpublished pre-print from Istituto di Fisica, Rome) was written in 1975. The next year Broyles published his attempted proof of the spin-statistic theorem (Am. J. Phys. 44 (4), 340-343, (1976)) using essentially the same method. Berry and Robbins published a variation using configuration space (Proc. R. Soc. London Ser. A 453, 1771-1790 (1997)). However those proofs neglected to provide a definitive theoretical basis for deciding which wave functions are necessarily symmetric and which anti-symmetric. The definitive theoretical basis was spelled out in my own paper presented at the Spin2000 conference in May 2000 and published in the proceedings (AIP Proceedings 545, pp 104-110). AFAIK this is the only published paper which gives a complete proof of the spin-statistics theorem. You can view it here:

http://xxx.lanl.gov/abs/quant-ph/0006101

All the standard field theory proofs and variations thereof, ignore the additional qualification needed regarding the way the wavefunctions (or creation operators) are constructed and are therefore incomplete.

 

[mikeyork]

OK, so where is the "selective reading" here that will cause a confusion? You clearly are claiming that pions, just because it was MISTAKENLY taught to be the strong force carrier, is STILL a force particle? Whoa!

Please don't shout at me or treat me like a horse. I deliberately said they were "loose terms" for obvious reasons. But they were appropriate in the wider context of expressing the essential difference between bosons and fermions in building structure into matter.

Please show a little more respect for your colleagues.

 

[ZapperZ]

I am well aware of the fact that the whole wavefunction needs to be taken into account. This is precisely why I talked about the necessity of equality of "all other quantum numbers" in what I wrote. Spatial asymmetry implies that there are quantum numbers which are not the same. So the Pauli rule does not apply. Please read it again.

Although there is a slight technical difference between the observable rule I give and the conventional expression via the Symmetrization Postulate (which actually needs a technical qualification before it is true, see below) this difference has no observable effects. The difference you are trying to create is non-existent.

My version of the Pauli exclusion principle is identical to Pauli's (no two identical spin 1/2 particles can be in the same state). No wiggling is necessary. You have just misunderstood what I wrote.

This is rather "amazing". It appears that you have somehow missed the generalized pauli exclusion principle completely. It explains why you said something like this:

All identical particles, of whatever spin, obey the same general exclusion rule:

"When all other quantum numbers are the same, only even eigenstates of composite spin are allowed."

It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle.

So the notion that there are two types of particles, fermions and bosons, is misleading. It just happens that only those particles which obey the Pauli principle can give any properties to matter characterized by a specific spatial location. Other particles just form an undifferentiated soup.

The generalized rule is a fundamental property of QM and can be proved assuming only the obvious:

1. Unique state vectors require a uniquely described physical state.
2. Particle permutation is an artifice of state descriptions and not physically observable.

The rule then follows as an interference effect.

Now tell me exactly where I misunderstood you when you said "It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle."

Did you not claim that a composite boson MUST (your word) have a spin of 0? You even padded that claim by saying the spins MUST be opposite?

So now, will you please tell me, before we proceed any further, how you would reconcile the existence of spin-triplet superconductors[1,2,3], where the composite boson are made up of a pair of electron with spins PARALLEL to each other? You continue to act as if these things do not exist! Did we just spectacularly destroyed the validity of your version of Pauli Exclusion Principle?

Please don't shout at me or treat me like a horse. I deliberately said they were "loose terms" for obvious reasons. But they were appropriate in the wider context of expressing the essential difference between bosons and fermions in building structure into matter.

But you contined by saying "I'd also suggest that pions (which are scalar bosons) are also considered "force particles". This useage pre-dates the standard model and unified QFT. Back in the days of Yukawa, pions were considered to be the unique carriers of the strong force."

You'd suggest? Which is why I asked about other bosons which are not "force carriers"! Why is this not a valid question?

Zz.

[1] K. Ishida et al., PRL v.84, p.5387 (2000).
[2] http://www.ss.scphys.kyoto-u.ac.jp/res-sub/contents/sr2ruo4/index-e.html
[3] F. Laube et al, PRL v.84, p.1595 (2000).

 

[dextercioby]

Hmmm,interesting,i remember being taught the difference between orthohelium & parahelium a year before the course on QM ever dealt with Clebsch-Gordan's theorem and symmetrization/VI-th postulate...Come on,Mike,you must remember how to add spins.

Daniel.

 

[mikeyork]

This is rather "amazing". It appears that you have somehow missed the generalized pauli exclusion principle completely. It explains why you said something like this:

Now tell me exactly where I misunderstood you when you said "It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle."

Because you have neglected that this sentence that you quote was written in the context of the rule in the previous sentences, which you persist in ignoring, despite my repeated attempts to draw your attention to it:

***
All identical particles, of whatever spin, obey the same general exclusion rule:

"When all other quantum numbers are the same, only even eigenstates of composite spin are allowed."
***

It clearly results in the Pauli rule, as I have been trying several times to explain, only "when all other quantum numbers are the same", because it is only in this context that odd composite spin is forbidden.

Did you not claim that a composite boson MUST (your word) have a spin of 0? You even padded that claim by saying the spins MUST be opposite?

I don't believe I made any such statements. They are clearly nonsense. I have no idea why you imagine I said this.

So now, will you please tell me, before we proceed any further, how you would reconcile the existence of spin-triplet superconductors[1,2,3], where the composite boson are made up of a pair of electron with spins PARALLEL to each other? You continue to act as if these things do not exist! Did we just spectacularly destroyed the validity of your version of Pauli Exclusion Principle?

No. Because the spatial quantum numbers of the constituent fermions are not the same -- as I have repeatedly pointed out. In fact, I think you will find that such composite spin-triplet superconductors have odd L (e.g. L=1) in their CM frame, because in the CM frame the rule becomes "L+S must be even". Odd L is equivalent to an antisymmetric spatial part of the wave function.

 

[mikeyork]

Hmmm,interesting,i remember being taught the difference between orthohelium & parahelium a year before the course on QM ever dealt with Clebsch-Gordan's theorem and symmetrization/VI-th postulate...Come on,Mike,you must remember how to add spins.

I've no idea what you are trying to say that is of relevance to this thread. Although I can think of various possibilities, I prefer not to speculate. Please explain.

 

[ZapperZ]

This is getting severely ridiculous.

I don't believe I made any such statements. They are clearly nonsense. I have no idea why you imagine I said this.

So you're saying that I "imagined" that you actually said this:

It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle.

And not to mention, you also challenged Norman with this:

Please inform me when you find a pair of indistinguishable electrons forming a composite spin triplet

Even AFTER I first mentioned about the triplet paring superconductors, you even asked if I'm claiming a "violation" of the exclusion principle.

But oh, I'm sure this is all fine since we're restricting ourselves to "when all other quantum numbers are the same...", etc. Tell me honestly, if you're trying to explain this to the OP, do you think the person will have the IMPRESSION that ALL a pair of electrons CAN do is pair up in a singlet state? Really now! Reread your first few replies, especially when you break off a paragraph and put a statement by itself to say:

"It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle."

This is extremely misleading at best especially if one doesn't know that this is only for a highly special case and NOT the general case. I have no clue what made you do such a thing.

Zz.

 

[mikeyork]

This is getting severely ridiculous.

That is the most appropriate thing you have written so far.

So you're saying that I "imagined" that you actually said this:

No. What you imagined I said was quite different, as you well know.

And not to mention, you also challenged Norman with this:
"Please inform me when you find a pair of indistinguishable electrons forming a composite spin triplet"

I am still waiting. Particles which have different quantum numbers, such as the superconducting triplets you keep referring to, are distinguishable (by their quantum numbers).

Really now! Reread your first few replies, especially when you break off a paragraph and put a statement by itself to say:

"It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle."

This is extremely misleading at best especially if one doesn't know that this is only for a highly special case and NOT the general case. I have no clue what made you do such a thing.

I suggest you re-read all those replies, because you will find that in each case I took great pains to very explicitly point out the context you were missing (see below) yet you kept ignoring this. I have no clue what made you do such a thing.

Examples:
Reply #1 (to Norman):
**
Not for identical particles when all other quantum numbers are the same and "permutation" is not a physically observable operation. Please see the context of the remark you are questioning (only even eigenstates allowed) and you will understand it better."
**

Reply #2 (to ZapperZ):
**
Are you claiming a violation of the Pauli principle or, like Norman, ignoring the context of the remark you quote?
**

Reply #3 (to ZapperZ -- and which you quoted in your next post):
**
I cited a more general rule which I'll repeat here:

"For all identical particles, of whatever spin,... when all other quantum numbers are the same, only even eigenstates of composite spin are allowed."

For identical spin 1/2 particles, when all other quantum numbers are the same, as I said before, this implies a spin singlet. This, in turn, implies the Pauli rule

When you refer to spatial asymmetry this means that not "all other quantum numbers are the same" and so the spin singlet rule does not apply.
**

Reply #4 (to ZapperZ):
**
I am well aware of the fact that the whole wavefunction needs to be taken into account. This is precisely why I talked about the necessity of equality of "all other quantum numbers" in what I wrote. Spatial asymmetry implies that there are quantum numbers which are not the same. So the Pauli rule does not apply. Please read it again.
**

Reply #5 (to ZapperZ):
**
My version of the Pauli exclusion principle is identical to Pauli's (no two identical spin 1/2 particles can be in the same state). No wiggling is necessary. You have just misunderstood what I wrote.
**

 

[ZapperZ]

No. What you imagined I said was quite different, as you well know.

Then tell me WHY a bunch of people came down on you and pointed out that you have incorrectly stated that only a spin-singlet state is possible? As I've said, AT BEST, you have stated something very misleading. We can't all "imagined" the same thing I have in my head. You somehow refused to acknowledge that what you have said is rather strange and vague. Saying

".. when all other quantum numbers are the same..."

says nothing, because this doesn't tell you WHAT quantum numbers are good and aren't, especially in a degenerate case. When I fill the d-shell with the first 2 electrons, they both have spins in the same directions. So what "quantum numbers" are different?

I am still waiting. Particles which have different quantum numbers, such as the superconducting triplets you keep referring to, are distinguishable (by their quantum numbers).

Oh, this is good. So what "quantum numbers" for a spin triplet pairing is different? And what makes them distinguishable? This is rather strange because if they are distinguishable, the BCS theory will never work since the fermionic ground state is the starting point! Are you saying that singlet paring cooper pairs are indistinguishable and are fine and dandy, but triplet pairing cooper pairs are distinguishable?

.. or maybe you're also arguing that one doesn't need indistinguishibility to make use of the fermionic statistics.

Zz.

 

[mikeyork]

Then tell me WHY a bunch of people came down on you and pointed out that you have incorrectly stated that only a spin-singlet state is possible? As I've said, AT BEST, you have stated something very misleading. We can't all "imagined" the same thing I have in my head.

I long ago gave up trying to understand why people jump to false conclusions and continue to do so even when it is pointed out.

In this particular context, I rather like the comment of Duck and Sudarshan in their book ("Pauli And The Spin-Statistics Theorem") where they say:

"Everyone knows the Spin-Statistics Theorem but no one understands it"

You somehow refused to acknowledge that what you have said is rather strange and vague. Saying

".. when all other quantum numbers are the same..."

says nothing, because this doesn't tell you WHAT quantum numbers are good and aren't, especially in a degenerate case.

The context from which this comes (the rule I first quoted) makes it quite plain that "all other quantum numbers" refers to all quantum numbers except the composite spin.

When I fill the d-shell with the first 2 electrons, they both have spins in the same directions. So what "quantum numbers" are different?

Probably the third component of orbital angular momentum.

Oh, this is good. So what "quantum numbers" for a spin triplet pairing is different? And what makes them distinguishable?

Depending on frame of reference, their position (or momentum) or third component of orbital angular momentum.

This is rather strange because if they are distinguishable, the BCS theory will never work since the fermionic ground state is the starting point! Are you saying that singlet paring cooper pairs are indistinguishable and are fine and dandy,

Singlet identical fermion pairs are distinguishable by their spin orientations.

but triplet pairing cooper pairs are distinguishable?

This time by their spatial quantum numbers.

Like I said, particle states are distinguishable by their quantum numbers. I would have thought this was a no-brainer.

.. or maybe you're also arguing that one doesn't need indistinguishibility to make use of the fermionic statistics.

Technically, the opposite: fermionic statistics require that all fermionic systems be distinguishable by their quantum numbers. That is just what the Pauli rule says.

 

[ZapperZ]

Can someone else explain to me the previous posting? Did Indistinguishable particle statistics just become distinguishable? Are two electrons in a spin-triplet state with l=1 angular momentum quantum number "distinguishable" by the "spatial quantum numbers"? Huh?

Zz.