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In flat-space QFT, consider a spinor operator $\phi_i$, taken to lie at the origin. Given a Lorentz transformation $g$, we have $$\tag{1} U(g)^\dagger \phi_i U(g) = D_{ij}(g)\phi_j$$ where $D_{ij}$ is some spinor projective representation of $SO^+(3,1)$, and $U$ is an infinite-dimensional unitary, perhaps projective, representation of $SO^+(3,1)$, satisfying $$\tag{2} U(gh)= e^{i\alpha(g,h)}U(g)U(h)$$ for some real phases $\alpha(g,h)$. Now, using (1) and (2) we have: \begin{align} D_{ij}(gh)\phi_j &\stackrel{(1)}{=} \tag{3}U(gh)^\dagger\phi_iU(gh) \\ \tag{4} &\stackrel{(2)}{=}e^{-i\alpha(g,h)}U(h)^\dagger U(g)^\dagger\phi_iU(g)U(h)e^{i\alpha(g,h)}\\ \tag{5} &\stackrel{(1)}{=} U(h)^\dagger \big((D_{ik}(g)\phi_k\big)U(h)\\ \tag{6} &\stackrel{(1)}{=} D_{ik}(g)D_{kj}(h)\phi_j. \end{align}

Hence $$\tag{7} D_{ij}(gh) = D_{ik}(g)D_{kj}(h).$$ That is, $D$ is a true, non-projective, representation. Clearly this is false for spinors, so where did I go wrong?

Qmechanic
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nodumbquestions
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1 Answers1

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Your observation is correct, which is why careful statements of the Wightman axiom eq. (1) (see e.g. Wiki, nLab) stipulate that $D$ is a representation of $\mathrm{SL}(2,\mathbb{C})$, the universal cover of the Lorentz group, not a representation of the Lorentz group itself.

Note that classical fields - the things on which $D$ acts - are not necessarily complex-valued, so the notion of a projective representation on them does not even make sense in general.

ACuriousMind
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