I have read that fermions cannot exist in the same state simultaneously. I understand why indistinguishable particles with an antisymmetric superposition of states can't exist in the same state simultaneously, but why must fermions have an antisymmetric superposition of states?

The only characterising property I know of fermions having besides antisymmetry is spin, for which they have half-integer. I understand that this is the case simply because particles with half-integer spin and particles with zero or integer spin were defined as fermions and bosons respectively.

My perusal of the Wikipedia page on the spin statistics theorem leaves me under the impression that spin has nothing to do with the wavefunction's symmetry properties:

Naively, neither has anything to do with the spin, which determines the rotation properties of the particles, not the exchange properties.

Are antisymmetric wavefunctions simply classified as fermions, in the way half-integer spin particles were? I don't see how this could be the case, as, if spin and symmetry were independent, half-integer spin particles with symmetric wavefunctions (and antisymmetric integer spin particles) would be possible.


closed as unclear what you're asking by ACuriousMind, Martin, Sebastian Riese, Bill N, Kyle Kanos Sep 18 '15 at 11:35

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  • $\begingroup$ The beginning of this nlab page may provide you with some additional detail, and clarify your ideas. $\endgroup$ – yuggib Sep 17 '15 at 6:00
  • 2
    $\begingroup$ The spin statistics theorem DOES say that a fermion (=spin 1/2, 3/2, 5/2,...) particle has an antisymmetric wavefunction under identical particle exchange. It is not a result one can prove in QM, one has to use relativity. In fact, the wikipedia entry has a proof... $\endgroup$ – Omry Sep 17 '15 at 6:22
  • $\begingroup$ I don't understand this question. You show that you have read the page on the spin-statistics theorem. It proves the very thing you are asking about! $\endgroup$ – ACuriousMind Sep 17 '15 at 13:17
  • $\begingroup$ Unless I'm misaken, your question is basically, "Why is this definition defined this way?" which doesn't really make sense. Also, seems to be essentially a duplicate of this question $\endgroup$ – Kyle Kanos Sep 18 '15 at 11:35

You can find the right statistics using QFT.

In QFT you can write down the fields for spin-$0$ and spin-$\frac{1}{2}$, then you can show that fermions have to be antisymmetric otherwise there will be an infinity number of negative energy states (if they are finite you can always shift your definition of ground state with the lower state). In this sense you can find a proof in Peskin and Schroder - QFT page 52-58.

Another way is to require the propagators to be Lorentz invariant. You can find more details in Schwartz - QFT and SM chapter 12.4.

You can even use the C symmetry that for spin-$\frac{1}{2}$ reads $\psi_C=\gamma_C\psi^{\dagger*}$. You can find the properties of $\gamma_C$ by imposing that it must satisfy the same Dirac equation as $\psi$. If you then try to compute the energy of this particle you will find that it is negative unless $\phi$ anticommute with itself. You can have a taste of it in Peskin and Schroder - QFT page 70, although they don't exactly do this you can easily understand the argument.

This are three ways that came to mind, I'm pretty sure that there are other since it's really a fundamental property of our equation. You can see that in all this proofs we needed the relativistic description: Klein-Gordon's equation for scalars, Dirac's equation for fermions (only spin-$\frac{1}{2}$). While in the classical limit (not relativistic but still QM) you treat all the particles in the same with the Schrödinger's equation, in the relativistic limit you have to use two different set of equations for fermions and scalars. This leads to the two different statistics.


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