# Spin-statistics theorem proof details

Recently I have read one book where there was some incomprehensible proof of the Pauli's spin-statistics theorem. I want to ask about a few details of the proof.

First, the author derives commutation (anticommutation) relations like $[\hat {\psi} (x), \hat {\psi} (x')]_{\pm}$ for arbitrary moments of time for scalar, E.M. and Dirac theory cases. He notices that all of them depend on the function $$D_{0} = \int e^{i(\mathbf p \cdot \mathbf (\mathbf x - \mathbf x'))}\frac{\sin(\epsilon_{\mathbf p}(t - t'))}{\epsilon_{\mathbf p}}\frac{d^{3}\mathbf p}{(2 \pi)^{3}}, \quad \epsilon^{2}_{\mathbf p} = \mathbf p^{2} + m^{2},$$ which (as it can be showed) is Lorentz-invariant. For example, it is not hard to show that for fermionic field $$[\Psi (x), \Psi^{\dagger } (x')]_{+} = \left( i\gamma^{\mu}\partial_{\mu} + m\right)D_{0}(x - x').$$

Second, he assumes that for the case of arbitrary integer spin $2n$ there exists a function $\Psi(x)$, for which $$[\hat {\Psi}_{a} (x), \hat {\Psi}^{\dagger}_{b}(x')]_{\pm} = F^{\ 2n}_{ab}\left(\frac{\partial}{\partial x}\right)D_{0}(x - x'),$$ and for the case $s = 2n + 1$ there exists a function $\Psi(x)$, for which $$[\hat {\Psi}_{a} (x), \hat {\Psi}^{\dagger}_{b}(x')]_{\pm} = F^{\ 2n + 1}_{ab}\left(\frac{\partial}{\partial x}\right)D_{0}(x - x'),$$ where $F_{ab}^{\\ k}$ refers to the $\frac{\partial}{\partial x}$ polynomial of rank $k$ and the author (at this stage of the proof) doesn't clarify the sign of commutator.

How can one argue such a generalization from spin $0, \frac{1}{2}$ and $1$ cases on the arbitrary cases of spin value? It is a very strong assumption, because formally it almost proves Pauli's theorem.

• $\uparrow$ Which book? Mar 24, 2014 at 17:50
• @Qmechanic : this is a book published by my university. Unfortunately, this book is not in the public domain, and also it isn't written in English. Mar 24, 2014 at 17:57
• Is the author of the book at your university?
– rob
May 6, 2014 at 22:02
• @rob : no, he already doesn't working. But I have invented an answer. May 7, 2014 at 11:20
• It'd be great if you could write up your answer here, in case anyone else is searching for a similar problem later.
– rob
May 7, 2014 at 11:50

1. 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): $$\psi^{\mu_{1}..\mu_{l}} = \begin{pmatrix}\psi_{a}^{\mu_{1}...\mu_{l}} \\ \kappa^{\dot {b}, \mu_{1}...\mu_{l}}\end{pmatrix},$$ $$\tag 4 (i\gamma^{\mu}\partial_{\mu} - m) \psi = 0, \quad \gamma_{\mu_{i}}\psi^{\mu_{1}...\mu_{i}...\mu_{l}} = 0, \quad g_{\mu_{i}\mu_{j}}\psi^{\mu_{1}...\mu_{i}...\mu_{j}..\mu_{l}} = 0.$$
2. 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 $[\psi_{a}(x), \psi_{b}^{\dagger}(y)]_{\pm}$.
3. 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.
4. 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 $\gamma_{5}G_{ab}(-p)\gamma_{5} = -G_{ab}(p)$.