Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Highly related to A reading list to build up to the spin statistics theorem

I see 2 parts to the spin statistics theorem:

  • (spin $n$ or $n+\frac{1}{2}$)
  • step 1 given that a spin is integral or non-integral conclude symmetry or antisymmetry of wave function
  • ([anti]symmetry)
  • step 2 using symmetry or antisymmetry of wavefunction conclude that it must obey BE or FD statistics.
  • (BE/FD statistics)

I have never seen a clear explanation of 1 nor 2...

The referenced question author seems to be satisfied with just 1... and seems to believe 2 by naming association: symmetric or assymetric wavefunction corresponding to boson or fermion, and BE statistics corresponds to boson, FD statistics to fermion. i.e. enough to be able to decide what to use without necessarily understanding how statistics is associated with wave function

share|improve this question
The definition of BE and FD statistics is property 2--- BE is symmetric wavefunctions and FD is antisymmetric. There is othing to prove. –  Ron Maimon Jan 23 '12 at 4:38
note that neither of the 3 pieces of information is a name: ([half]integerness of spin),([anti]symmetry),(statistical distribution for bosons [fermions]) none of them is a name, so I dont care about proving the name, which is convention, but I'd like to see a proof circle that [half]integer spin => [anti]symmetric wavefunction => statistical distribution for bosons [fermions] => [half] integer spin, such that knowing one of the 3 properties implies the others –  propaganda Jan 23 '12 at 4:57
are you claiming you cant prove the statistic from the [anti]symmetry of the wavefunction? if so how do you think they are associated, through measured observation alone? –  propaganda Jan 23 '12 at 4:59
I guess I just dont see how to calculate the statistic for half integer and integer spin... (without remembering that [half]integral represents [fermions]bosons and then selecting the statistic for [fermions]bosons) –  propaganda Jan 23 '12 at 6:35
im not sure if youre really finding the calculation of statistical state distribution from a bunch of [anti]symmetric wavefunctions a trivial exercise, or if you up till now like me were taught to associate halfinteger spin to one statistical distribution and integral spin to another. if its really trivial this part would already help me a lot –  propaganda Jan 23 '12 at 6:38
show 5 more comments

2 Answers

up vote 3 down vote accepted

Antisymmetric wave functions instantly imply the Pauli exclusion principle – essentially because $\psi(x,x)=-\psi(x,x)=0$, to write the concept schematically – which implies that the occupation numbers are $N=0,1$ and statistical physics is therefore inevitably governed by the Fermi-Dirac statistics which may be derived from Boltzmann/statistical physics for these occupation numbers.

Similarly, symmetric wave functions imply that particles are indistinguishable but the occupation numbers may be $N=0,1,2,3,\dots$. That implies the Bose-Einstein distribution by applying the Boltzmann steps to the multiparticle states with these occupation numbers. For the proofs of the first two paragraphs of my answer, see

How to derive Fermi-Dirac and Bose-Einstein distribution using canonical ensemble?

The bulk of the spin-statistics theorem is to link the antisymmetric functions with the half-integer spin and symmetric functions with integer spin. It was proved by Pauli and all the evidence available to me suggests that you haven't seen a clear proof because you haven't tried. I won't reproduce the full proof here because I don't believe it would be a good investment of time but I will give a sketch. The Lagrangian for a spin-0 real field $\phi$ has to contain the kinetic term $$\frac{1}{2} \partial_\mu \phi \partial^\mu \phi $$ which is dictated by the Lorentz symmetry etc. If $\phi$ were anticommuting, the object above would identically vanish and there would be no dynamics. For spin-1/2 fields, it's the other way around (the Majorana kinetic term would vanish if the fields were not anticommuting), and so on. For the wrong combinations of spin and statistics, Pauli actually showed one can't have positive norms and/or Hamiltonians bounded from below.

share|improve this answer
what does phi(x,x) represent? a wavefunction of a system of 2 particles? what if there are three particles:phi(x,y,z) in general; now what is the symmetry of this system when all positions of particles 1,2,3 are equal? if just 2 are equal? for large N this analysis seems hard, or am I missing something (equivalence, approximation,...) –  propaganda Jan 23 '12 at 7:08
Dear @Propaganda, yes, if there are more particles, this gets extended to $\psi(x_1,x_2,\dots , x_n)$. It is completely antisymmetric, so it's enough for two of the arguments to coincide for the function to vanish (all the other arguments that don't participate in the permutations just sit at their place and don't do anything). There's nothing new happening about these simple rules for large $N$, and surely nothing hard. –  Luboš Motl Jan 23 '12 at 11:22
add comment

For spin 1/2, there is a simple answer through animations. Look up http://vimeo.com/tag:belttrick and you will see how a spin 1/2 object behaves, and also that it automatically behaves as a fermion.

share|improve this answer
Although this is a nice demonstration, one needs to have a link to Dirac's belt trick: en.wikipedia.org/wiki/Plate_trick –  gns-ank Mar 23 '13 at 18:37
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.