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)?
[Physics] Spin-Statistics Theorem (SST)
cpt-symmetrymathematical physicsquantum-field-theoryspecial-relativityspin-statistics
Related Solutions
I wrote the Wikipedia page in question, so I feel bad. I thought it was clear.
There is a recent textbook by Banks which covers the spin/statistics theorem pretty good. I hope it is ok. The main difficulty is that there is no quantum field theory book that covers analytic continuation to Euclidean space, and this is the essential thing.
This is worked out by each person on their own, as far as I know. The problem is that it is very easy to say "plug in i times t everywhere you see t" and get 90% of everything right, without understanding anything. Streater and Whitman do it, that's most of their book, but they are too formal to be comprehensible. Schwinger is too long ago (and ideosyncratic). Perhaps the statistical section of Feynman and Hibbs (Path integrals), where they actually rederive the path integral in imaginary time, will allow you to extrapolate to the general bosonic fields.
The Fermionic case requires the Euclidean continuation of Majorana spinors, and this was in the literature more recently: http://arxiv.org/abs/hep-th/9608174. This stuff is covered in none of the textbooks, and unfortunately I can't recommend any of them with a good conscience.
Later Edit: If you don't want to go to Euclidean space, you should avoid anything past Feynman/Schwinger. The best path then is possibly to work through Pauli's argument:W. Pauli, The Connection Between Spin and Statistics, Phys. Rev. 58, 716- 722(1940).
- Let's look to the expression for field with mass $m$ and spin $s$ (for massless case following statements exist in similar form): $$ \tag 1 \hat {\psi}_{a}(x) = \sum_{\sigma = -s}^{s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3} 2E_{\mathbf p}}}\left( u^{\sigma}_{a}(\mathbf p )e^{-ipx}\hat{a}_{\sigma}(\mathbf p ) + v^{\sigma}_{a}(\mathbf p )e^{ipx}\hat{b}^{\dagger}_{\sigma}(\mathbf p )\right). $$ This field refers to the $\left( \frac{m}{2}, \frac{n}{2}\right), n + m = 2s$ spinor representation of the lorentz group (so indice a means $\psi_{a} = \psi_{a_{1}...a_{m}\dot {b}_{1}...\dot{b}_{n}}$; field is symmetric in all indices) and obeys the equations $$ \tag 2 (\partial^{2} + m^{2})\psi_{a}(x) = 0, \quad \hat {p}^{\mu}(\sigma_{\mu})^{\dot {b}_{j}a_{i}}\psi_{a_{1}...a_{i}...a_{m}\dot {b}_{1}...\dot{b}_{j}...\dot{b}_{n}} = 0. $$ 1.1. If we have the field with integer spin $l$, we can convert $(1)$ (here I have missed some calculations, which is not importance) to the symmetric tensor rank $l$ $A_{\mu_{1}...\mu_{l}}$ (which refer to the $\left(\frac{l}{2}, \frac{l}{2}\right)$ and also we can convert $(2)$ to the form (our tensor is traceless and transverce in all indices) $$ \tag 3 (\partial^{2} + m^{2})A = 0, \quad \partial_{\mu_{i}}A^{\mu_{1}...\mu_{i}...\mu_{l}} = 0, \quad A_{\mu_{i}}^{ \mu_{1}...\mu_{i}...\mu_{l - 1}} = 0. $$ 1.2. In a case of half-integer spin $s = l + \frac{1}{2}$, if we want to get the theory which is invariant under time and spatial inversions we must introduce the direct sum $\left(\frac{l + 1}{2} , \frac{l}{2}\right) \oplus \left(\frac{l}{2} , \frac{l + 1}{2}\right)$ and then to construct the equation-projector which reduce the number of independent components. So we get from $(2)$ and requirement given above the following (the field is also symmetric, of course): $$ \psi^{\mu_{1}..\mu_{l}} = \begin{pmatrix}\psi_{a}^{\mu_{1}...\mu_{l}} \\ \kappa^{\dot {b}, \mu_{1}...\mu_{l}}\end{pmatrix}, $$ $$ \tag 4 (i\gamma^{\mu}\partial_{\mu} - m) \psi = 0, \quad \gamma_{\mu_{i}}\psi^{\mu_{1}...\mu_{i}...\mu_{l}} = 0, \quad g_{\mu_{i}\mu_{j}}\psi^{\mu_{1}...\mu_{i}...\mu_{j}..\mu_{l}} = 0. $$
- Here is one strong theorem: field $(1)$ is lorentz-covariant field if and only if $u_{a}^{\sigma}(\mathbf p)$ and $v_{a}^{\sigma}(\mathbf p)$ are connected through relation $$ v_{a}^{\sigma}(\mathbf p) = (-1)^{s + \sigma}u_{a}^{-\sigma}(\mathbf p). $$ This result is correct if $(1)$ transforms under the irreducible rep of the Lorentz group $T$ which contains of the irrep of the rotation group of spin $s$ only once. This is correct for 1.1, but it is incorrect in case of 1.2. For the last case the modification of the theorem gives $v_{a}^{\sigma}(\mathbf p ) = (-1)^{s + \sigma}\gamma_{5}u_{a}^{-\sigma}(\mathbf p ) $. Using 1 and 2 we can convert $[\psi_{a}(x), \psi_{b}^{\dagger}(y)]_{\pm}$.
Integer spin: $$[\psi_{a}(x), \psi_{b}^{\dagger}(y)]_{\pm} = \sum_{\sigma}\int \frac{d^{3}\mathbf p}{(2 \pi )^{3}2E_{\mathbf p}}G^{\sigma}_{ab}(p)\left(e^{-ip(x- y)} \pm e^{-ip(x - y)} \right), $$ where $G^{\sigma}_{ab}(p) = u^{\sigma}_{a}(p)(u^{\sigma}_{b}(p))^{\dagger}$. After that we may use following recipe: 1) we have symmetric tensor $G^{\sigma}_{ab}(p)$, so as lorentz covariant object it may be constructed only from $g_{\mu \nu}, p_{\nu}$ (the other object, the Levi-Civita symbol, is antisymetric). It means that it may be constructed only as polinome of rank $2s$ on $p_{\mu}$ which contains only the summands of even degree of $p$; so 2) $G^{\sigma}_{ab}(p)e^{-ipx} =G^{\sigma}_{ab}(\hat {p})e^{-ipx}$ and $G^{\sigma}_{ab}(p)e^{ipx} = G^{\sigma}_{ab}(-\hat {p})e^{ipx} = G^{\sigma}_{ab}(\hat {p})e^{ipx}$. If we don't need the invariance under spatial inversion and time inversion, we will get the result (by the same way) $G^{\sigma}_{ab}(-\hat {p}) = -G^{\sigma}_{ab}(\hat {p})$ for half-integer spin realisations.
Half-integer spin. For this case we have $$ [\psi_{a}(x), \psi_{b}^{\dagger}(y)]_{\pm} = \sum_{\sigma}\int \frac{d^{3}\mathbf p}{(2 \pi )^{3}2E_{\mathbf p}}\left(G^{\sigma}_{ab}(p)e^{-ip(x- y)} \pm \gamma_{5}G^{\sigma}_{ab}(p)\gamma_{5}e^{-ip(x - y)} \right). $$ By using eq. $(4)$ we can state that $G_{ab}(p) = (\gamma^{\mu}p_{\mu} + m)R_{ab}(p)$, where $R_{ab}(p)$ is constructed as the sum of products of even number of gamma-matrices and even number of momentums and products of odd number of gamma's and odd number of momentums. So by having the relation $[\gamma_{5}, \gamma_{\mu}]_{+} = 0$ and the statement above we may assume that $\gamma_{5}G_{ab}(-p)\gamma_{5} = -G_{ab}(p)$.
Best Answer
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.