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.

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    $\begingroup$ $\uparrow$ Which book? $\endgroup$ – Qmechanic Mar 24 '14 at 17:50
  • $\begingroup$ @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. $\endgroup$ – Andrew McAddams Mar 24 '14 at 17:57
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    $\begingroup$ Is the author of the book at your university? $\endgroup$ – rob May 6 '14 at 22:02
  • $\begingroup$ @rob : no, he already doesn't working. But I have invented an answer. $\endgroup$ – Andrew McAddams May 7 '14 at 11:20
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    $\begingroup$ It'd be great if you could write up your answer here, in case anyone else is searching for a similar problem later. $\endgroup$ – rob May 7 '14 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 $\gamma_{5}G_{ab}(-p)\gamma_{5} = -G_{ab}(p)$.


Not sure. Completely lost about the equations. However, have you seen the experiment in which electrons that pass through one slit are rotated by 2 pi. The interference pattern shifts. The min and maximum flip. I'm not exactly sure why an electron that is in different state would show any kind of interference pattern at all. In any case the experiment might have something to do with why the pauli exclusion principle takes place. I'm not sure that the exclusion principle has anything to do with relativity. The wave function of a two electron system is anti symmetric. That's a statement of the principle. I don't know if you a have a wave equation made up of two spinors, must that wave equation be anti symmetric given the way individual spinors transform? I don't care to work it out. Its an interesting question. I suspect that a proof could be that easy.


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