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.
Best Answer
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 $\gamma_{5}G_{ab}(-p)\gamma_{5} = -G_{ab}(p)$.