Why are all particles bosons or fermions?

[deblimp]

Why is it not possible for a particle to be neither symmetric nor antisymetric on exchange? If a particle can have 1/2 integer spin why can't it have 1/3 , 1/4 etc. I know it's a weird question to ask but I've been wondering about it for a while.

 

[DaveC426913]

Science does not attempt to answer questions of why, it only attempts to observe and model those observations as best as possible.

It's not that they can't have fractional spins; it's that they simply don't.

 

[ZapperZ]

Why is it not possible for a particle to be neither symmetric nor antisymetric on exchange? If a particle can have 1/2 integer spin why can't it have 1/3 , 1/4 etc. I know it's a weird question to ask but I've been wondering about it for a while.

Take note that this is not really true for quasiparticles in condensed matter physics. We have anyons, which are strongly tied to the fractional quantum Hall effect phenomenon.

Zz.

 

[HomogenousCow]

Why is it not possible for a particle to be neither symmetric nor antisymetric on exchange? If a particle can have 1/2 integer spin why can't it have 1/3 , 1/4 etc. I know it's a weird question to ask but I've been wondering about it for a while.

Any state containing identical particles will be either symmetric or antisymmetric on exchange, there is simply no third possibility. Have you read a treatment on the matter?

 

[The_Duck]

Why is it not possible for a particle to be neither symmetric nor antisymetric on exchange? If a particle can have 1/2 integer spin why can't it have 1/3 , 1/4 etc. I know it's a weird question to ask but I've been wondering about it for a while.

The fact that the only possible spins are 0, 1/2, 1, 3/2, ... comes from the group theory of 3D rotational symmetry. This is worked out in any undergraduate quantum mechanics textbook. [Actually when you account for special relativity there is another possibility, so-called "continuous spin" particles, which don't seem to be realized in nature.]

The fact that all integer-spin particles are bosons and all half-integer-spin particles are fermions is deeper and harder to show. It can be proved in quantum field theory, where it is called the spin-statistics theorem.

 

[DrDu]

Any state containing identical particles will be either symmetric or antisymmetric on exchange, there is simply no third possibility. Have you read a treatment on the matter?

That's not true in general. Once you consider the exchange of more than two particles, there exist more possibilities: In two dimensions there are more complex statistics possible, so called "braiding statistics". Also in 3 dimensions there are more possibilities, besides true representations of the permutation group - namely parabosons and parafermions - there are also projective representations of the symmetric group.

 

[vanhees71]

This is a very deep question. If the particles move in a space with 3 or more dimensions, there are only bosons and fermions, i.e., all multiparticle states are even or odd under exchange of two-particles. The reason is the topology of the multi-particle configuration space, using the assumption that two particles must not be at the same position. This assumption is due to the fact that many fundamental interactions are singular for such a representation (e.g., the Coulomb interaction of two electrically charged particles has a singularity, if the two particles are at the same place, because the interaction potential is $\propto 1/|\vec{x}_1-\vec{x}_2|$. The proof, using path integrals, can be found in

Laidlaw, M. G. G., DeWitt, Cécile Morette: Feynman Functional Integrals for Systems of Indistinguishable Particles, Phys. Rev. D 3, 1375, 1970
http://link.aps.org/abstract/PRD/v3/i6/p1375

 

[DrDu]

This is a very deep question. If the particles move in a space with 3 or more dimensions, there are only bosons and fermions, i.e., all multiparticle states are even or odd under exchange of two-particles. The reason is the topology of the multi-particle configuration space, using the assumption that two particles must not be at the same position. This assumption is due to the fact that many fundamental interactions are singular for such a representation (e.g., the Coulomb interaction of two electrically charged particles has a singularity, if the two particles are at the same place, because the interaction potential is $\propto 1/|\vec{x}_1-\vec{x}_2|$. The proof, using path integrals, can be found in

Laidlaw, M. G. G., DeWitt, Cécile Morette: Feynman Functional Integrals for Systems of Indistinguishable Particles, Phys. Rev. D 3, 1375, 1970
http://link.aps.org/abstract/PRD/v3/i6/p1375

This does not seem to be what Laidlaw and DeWitt wanted to say:
arxiv.org/pdf/math-ph/0406011

 

[vanhees71]

Very interesting! But now again Einstein has a good advice, concerning (theoretical physicists): "Don't listen to their words but look at their deeds." Laidlav and deWitt have written in the abstract that the path-integral quantization excludes parastatistics. It doesn't make sense to me, why I should write something in a paper (even in the abstract), if I don't want to. Where they forced by the referees at the time? So that sounds very strange.

The paper by Greenberg and Mishra, I've to study in detail to understand their point. At the moment I can only speculate that the reason for the apparent contradiction to the findings of Laidlav and deWitt and their demonstration of the possibility to describe parastatistics in terms of path integrals is that in the former paper, they considered Feynman path integrals of indistinguishable many-body systems in "first quantization" (i.e., path integrals in (mn)-dimensional configuration space for a system of n particles in m-dimensional space), while the latter paper addresses the question in terms of quantum-field theory ("second quantization").

 

[ZapperZ]

Very interesting! But now again Einstein has a good advice, concerning (theoretical physicists): "Don't listen to their words but look at their deeds."

But for theorists, since they don't do anything else, aren't their words equivalent to their deeds?

:)

Zz.

 

[atyy]

I too thought like vanhees71, that in 3D or greater there are only fermions and bosons.

But it turns out, without contradicting that idea, that anyons may also be possible in 3D: http://arxiv.org/abs/1102.5742.

 

[vanhees71]

I'd say that the formulae and numerical results produced by theorists are their deeds. The words is the text around them

 

[DrDu]

I too thought like vanhees71, that in 3D or greater there are only fermions and bosons.

But it turns out, without contradicting that idea, that anyons may also be possible in 3D: http://arxiv.org/abs/1102.5742.

Wilczek has considered the possibility of projective reps of the permutation group which would behave like 3d anyons. However he also showed that they don't qualify as elementary particles as they violate the cluster decomposition principle i.e. locality.

 

[DrDu]

I have a physical argument why parafermions should be possible: if the strong coupling were zero, quarks would behave like parafermions, as colour wouldn't be observable.

 

[haael]

Mathematically, there are only two possibilities in quantum theory. If we accept that energy is quantized, then a finite region of phase space can either have infinite amount of energy, or finite. (Forget things like Bekenstein bound for now. We are in the linear quantum mechanics.)

The infinite case corresponds to bosons. We can stuff as many bosons in a region of space as we wish.
The finite case are fermions. You can only have one fermion (or zero) in one quantum state, thus finite number of fermions in finite capacity of quantum states.

Standard definition of a fermion states, that there can be only zero or one quantum in one state. You can also define that there can be many as long as the number is finite i.e. from 0 to 7, but this definition can be reformulated so there may be only 0 or 1 and the remaining information is a normal quantum number.

 

[Mark Harder]

Mathematically, there are only two possibilities in quantum theory. If we accept that energy is quantized, then a finite region of phase space can either have infinite amount of energy, or finite. (Forget things like Bekenstein bound for now. We are in the linear quantum mechanics.)

The infinite case corresponds to bosons. We can stuff as many bosons in a region of space as we wish.
The finite case are fermions. You can only have one fermion (or zero) in one quantum state, thus finite number of fermions in finite capacity of quantum states.

Standard definition of a fermion states, that there can be only zero or one quantum in one state. You can also define that there can be many as long as the number is finite i.e. from 0 to 7, but this definition can be reformulated so there may be only 0 or 1 and the remaining information is a normal quantum number.

It's a new question here, not quite on-topic, but I think it's relevant: speaking of the infinite case, I have long wondered about matter in zero-volume singularities like black holes and the pre-inflation universe: As you say, the density of bosonic matter can approach infinity. But how can a mass density approach infinity in a volume approaching zero with all those fermions stuck inside it? I vaguely recall learning once that (cryogenic?) experiments have demonstrated the purely quantum mechanical effects of fermionic symmetry, so fermions have been observed, or at least inferred by experiment. Same goes for experiments with boson condensates. I can think of only 2 explanations: Either the quantum symmetries that necessitate the existence of fermions break down under singular conditions, or under extreme pressure, fermions convert into bosons. Any thoughts, anyone?

 

[Mark Harder]

Hael, I just noticed that you were speaking of phase space, where 'volumes' of n-point bodies are computed from 6 n dimensions, counting linear momenta but ignoring rotations. I don't know if that makes a difference in what we are saying, but one must take it into account, I think.

 

[haael]

Yes, I was talking about phase space. The question is: how many quanta can occupy the same point in phase space.

Let's look at it a bit more generally. These are intuitions that helped me to understand that topic once. We can have 5 possibilities:

The possibility number 0 is that quanta have identity. That means, we can give them "name" or "individuality" and we can track the history of each individual quantum and not mistake one for another.
This is somewhat the most intuitive variant as it corresponds to normal Newtonian-like set of particles we learn in school.
Mathematically, it would mean that when we get creation operators for two particles, there is no rule regarding their commutativity. That means, the structure of creation operators is a free structure with respect to commutator.

Now let's say that quanta are indistinguishable. We can get their count, but not identity. If two quanta occupy the same point of phase space, they "merge" and can not be treated as two individual quanta any more.

There are 4 more possibilities.

A. There can be infinitely many quanta in one point and their spectrum is continuous. That means, there are uncountably many possibilities.
This case describes Newtonian-like infinitely divisible fluids. You can have a droplet of any size. The state of the phase space can be described by a real number.
The commutator relation is:
[A, B] = 0

B. There can be infinitely many quanta in one point but their spectrum is discrete. There are countably many possibilities.
This case describes bosons in quantum mechanics.
The commutator relation is:
[A, B] = 1

C. There can be finitely many quanta in one point. There is a finite number of possibilities.
We can say, for instance, that there can be only 7 quanta occupying the same state. However we can always change the definition of "state" and say that there can be only two possibilities, corresponding to vacuum (zero quanta) and particle (one quantum). The rest of the information goes to the definition of the "state".
This situation describes fermions in quantum mechanics.
The commutator relation is:
{A, B} = 1

D. There can be at most zero quanta in one point. There is only one possibility.
This is a degenerate case, not very interesting except the fact that the fermion case is a deformation of this.
The commutator relation is:
{A, B} = 0These cases span all possibilities. The most important thing to understand is that the case C is valid for any integer greater or equal to 2.