# Spin-Statistics Theorem (SST)

Please can you help me understand the Spin-Statistics Theorem (SST)? How can I prove it from a QFT point of view? How rigorous one can get? Pauli's proof is in the case of non-interacting fields, how it will be in the presence of interacting fields? The origins of the minus sign, when swapping the wave-function, it implies the CPT theorem in play (spinors, anti-articles)?

## 2 Answers

The classic place to start would be the book "PCT, Spin & Statistics, and All That", by R.F.Streater and A.S.Wightman. The spin statistics theorem can be proved as rigorously as you likely can want in the context of the Wightman axioms. The difficulty with this statement relative to your question is that we cannot prove that interacting fields satisfy the Wightman axioms.

To me an interesting way of looking at the spin-statistics theorem from QFT has always been the commutation relations one has to define in order to have a Hamiltonian which is bounded by below. I don't know how familiar you are with QFT so I will give a very introductory way of looking at it.

For example when one is only starting to learn QFT and you're quantizing the Klein-Gordon Field (Which describes a spin 0 boson) you start with the lagrangian density $$\mathcal{L} = \frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi) - \frac{1}{2}m^2\phi^{2}$$

and calculate the conjugate momentum $$\pi = \frac{\partial \mathcal{L}}{\partial(\partial_{t}\phi)}$$ With these you can calculate the usual Hamiltoninan density by doing $$\mathcal{H} = \dot \phi \pi - \mathcal{L}$$

If now you want to quantize this system you promote these functions into operators and enforce that they must satisfy the following commutation relations: $$[\phi(x), \pi(y)] = i \delta^{3}(x-y)$$ $$[\phi(x), \phi(y)] = 0$$ $$[\pi(x), \pi(y)] = 0$$

Now if you were to write $$\phi$$ and $$\pi$$ in momentum space and in terms of some operators $$a^\dagger$$ and $$a$$ you can obtain analogous commutation relations for these operators. These are called creation and annihilation operators. The will create or annihilate a particle state.

So imagine you have your vacuum state $$|0\rangle$$ you want your theory to be such that if you apply and annihilation operator to your vacuum state you get nothing that is $$a |0\rangle$$ = 0 because if you were to get a lower energy state then $$|0\rangle$$ wasn't really the vacuum state.

The thing is if you where to consider a similar procedure for the Dirac equation starting from $$\mathcal{L} =\overline{\psi}(-i\gamma^\mu \partial_\mu +m)\psi$$ where $$\gamma^{\mu}$$ are the Dirac gamma matrices. As you might know the Dirac field describes a spin-1/2 fermion. If you follow the exact same procedure and enforce the commutation relations you find that your theories energies aren't bounded by below. That is you can keep applying annihilation operators to the vacuum state and get lower and lower energies. What happens is if you now force the field and momentum to have anti-commutation relations instead $$\{\psi(x), \pi(y)\} = i \delta^{3}(x-y)$$ $$\{\psi(x), \psi(y)\} = 0$$ $$\{\pi(x), \pi(y)\} = 0$$

the theory is now bounded from below and you can no longer apply the annihilation operator to the vacuum state to obtain a lower energy state.

So in a way this means that for bosons you needed fields such that $$[\phi(x),\phi(y)] = 0$$ but for fermions you need a field that does the opposite $$\{\psi(x),\psi(y)\} = 0$$ which is what one usually defines fermions and bosons by in undergraduate QM. This for me has always been a way of looking at spin-statistics from the QFT perspective.

NOTE: I'm not claiming that this is a rigorous or formal proof of the Spin Statistics Theorem in anyway but it was just a bit big to leave as a comment. I'm sorry if this wasn't answer worthy.