Identical particles, eigenstates, empirical or theoretical

[jostpuur]

My question is that is the fact, that all particles are either boson or fermions, only an empirical fact, or can it be argued theoretically too.

The reason why I'm asking this is that I have not encountered anyone stating honestly, that it is an empirical fact only. But on the other hand, I have not encountered a theoretical proof either. An most importantly, I have encountered one incorrect theoretical proof repeatedly.

The incorrect proof goes like this. If the Hamiltonian commutes with the particle swapping operator $\psi(x_1,x_2)\mapsto \psi(x_2,x_1)$, then the eigenstates of the Hamiltonian must be eigenstates of the swapping operator. But possible eigenvalues of the swap are only $\pm 1$.

The truth is that the eigenstates of the Hamiltonian don't need to be eigenstates of the swapping operator. The Hamiltonian will be degenerate, and an eigenstate of the Hamiltonian can be written as a linear combination of other eigenstates, which are common eigenstates of the Hamiltonian and the swapping operator.

 

[DrDu]

I have never seen any proof that only fermions and bosons are possible in principle. In 3 dimensions, also para-fermions and para-bosons are possible which correspond to non-trivial representations of the permutation groups.

The swapping argument (in the case of non-degenerate Hamiltonians) also does not exclude other possibilities like parastatistics as these only differ from bosonic and fermionic behaviour when more than two particles are interchanged.

 

[xepma]

The question why particles of the same type behave as indistinguishable particles is emperical indeed. It's one of the pillars of modern day physics, and does not come from quantum theory itself.

The question: "what happens when particles are identical?" is of a different nature. And the very partial answer is given by the proof you have given. In three dimensions and higher you get that only bosonic and fermionic statistics are truly fundamental -- parastatistics turns out to be equivalent to these forms of statistics. Swapping in 3 dimensions comes down to the action of the permutation group, so the Hilbert space should be a representation of this group (emperical assumption).

In two dimensions the story is not finished. In that case the swapping of particles is history dependent as it depends on the order in which the particles are interchanged and in what manner this interchange happens. For instance, swapping two particles clockwise or anti-clockwise is not equivalent in two dimensions (it is in three).

In two dimensions the permutation group is therefore replaced by what is called the Braid group. The representation theory of this group is massively complicated, and gives rise to new types of particles called anyons. Their statistics are very interesting. It is believed these particles (effectively) emerge in low-dimensional condensed matter systems such as the quantum Hall effect.

But back to your original question: the fact that all particles are bosons or fermions comes from the fact that they are indistinguishable. The fact that they are indistinguishable cannot be proven from first principles of quantum mechanics.

 

[element4]

Xepma gave a very nice answer, I can give you a few more technical details.

As Xepma said, the indistinguishably of particles is indeed an experimental fact and cannot be proven theoretically using other axioms. But how is this implemented? I will sketch the argument.

The configuration space for $$N$$ distinguishable particles in $$d$$-dimensions is $$\mathbb R^{dN}-\Delta$$, while for indistinguishable particles it must be $$(\mathbb R^{dN}-\Delta)/S_N$$, where $$S_N$$ is the permutation group (all points where particles are swapped are identified). The singular points where two particles occupy the same position, $$\Delta$$, are subtracted. (Thanks to weejee for the correction, this point is crucial!). When particles are exchanged, they must be "dragged" along continuous curves. One must then classify all the topologically different curves, which is done by the first homotopy group. For two and three dimensions the result is

$$\pi_1((\mathbb R^{3N}-\Delta)/S_N) = S_N$$, for d = 3,

$$\pi_1((\mathbb R^{2N}-\Delta)/S_N) = B_N$$, for d = 2,
where $$B_N$$ is the braid group. I don't remember why, but the wave function has to carry a representation under these groups.

In three dimensions, there are two one-dimensional representations of $$S_N$$, which gives bosons and fermions. Higher dimensional representations are called para-statistics, as mentioned by DrDu. But as far as I know they can always be decomposed into fermions or bosons, with internal degrees of freedom (like color) and thus give no new type of particles.

I two dimensions , the story is too rich to discuss here. Particles in this class can potentially become the bones in the so called topological quantum computers.

(I must mention that recently it was discovered that more exotic statistics is possible in three dimensions, under the name "Projective Ribbon Permutation Statistics". But this is a whole new story. See http://arxiv.org/abs/0909.4741" .)

 

[inempty]

Things about parastatics have been demonstrated clearly by xepma and element4. Only what I want to tell is an answer, which is simpler and straighter but not as complete as above, to the question and what's written below can also be regarded just as a sidenote. We needn't hamilitonian be non-degenerate and it is naturally assumed that possiblities, the square of the wave function's magnitude, are invariant when swapping two of indistinguishable particles, so permutation should only change the wave function with a global-invariant phase factor like exp(ia). In 3 or more dimensions, swapping two identical particles twice turn out the same wave function as before, then it's obvious that exp(ia)=1 or -1. But in 2-dimensions it isn't like that, so exp(ia) can be anything and there are anyons in 2-dimensions.

 

[Dickfore]

Look for Spin-Statistics Theorem.

 

[DrDu]

It is true that parastatistics can be reduced to ordinary boson or fermion statistics upon introduction of an additional unobservable degree of freedom. In fact, when people first thought about quarks, they thought that they behave like para fermions, the internal degree of freedom is now called colour. Later, it was proposed that this degree may have observable consequences which turned out to be true.

There is another possibility of which I never have seen any proof how to rule it out. In principle, it would be sufficient to have a projective representation of the permutation group and they are well known under the name "double group" in molecular and solid state physics although there they only appear for parts of the wavefunction, the total wavefunction of the electrons being obviously fermionic.

 

[element4]

There is another possibility of which I never have seen any proof how to rule it out. In principle, it would be sufficient to have a projective representation of the permutation group and they are well known under the name "double group" in molecular and solid state physics although there they only appear for parts of the wavefunction, the total wavefunction of the electrons being obviously fermionic.

I think this idea was first put forward by Frank Wilczek in a http://arxiv.org/abs/hep-th/9806228" . Read says in the abstract:

We point out that projective permutation statistics is not possible in a local quantum field theory as it violates locality,...

I haven't read the papers, so I don't know much about this.

 

[DrDu]

Thank's for the very interesting links!

 

[weejee]

The configuration space for $$N$$ distinguishable particles in $$d$$-dimensions is $$\mathbb R^{dN}$$, while for indistinguishable particles it must be $$\mathbb R^{dN}/S_N$$, where $$S_N$$ is the permutation group (all points where particles are swapped are identified).

Actually we need to remove all the points in $$\mathbb R^{dN}$$ such that positions of any two of the N particles coincide. (Physically, this may be realized by hard-core repulsion between particles). If this is not the case, any closed path in the configuration space can be deformed to a point by moving through those points of coincidence. That is, only bosons can exist.

 

[element4]

Actually we need to remove all the points in $$\mathbb R^{dN}$$ such that positions of any two of the N particles coincide. (Physically, this may be realized by hard-core repulsion between particles). If this is not the case, any closed path in the configuration space can be deformed to a point by moving through those points of coincidence. That is, only bosons can exist.

You are correct, this point is very important. I have corrected my post. Thank you.

 

[vanhees71]

The "classical" paper about all these issue is

Laidlaw, M. G. G., and DeWitt, Cécile Morette: Feynman Functional Integrals for Systems of Indistinguishable Particles , Phys. Rev. D 3, volume 3, 1375, 1970

There the discussed issues are derived within the path-integral formalism of quantum mechanics.