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It is assumed that a valid QFT, must have states/observables in a Hilbert space, transform according to the projective unitary representations of the Poincare group.

I can understand why the observables must transform according to unitary representations of the Poincare group, but don't see any compelling reason for that symmetries must be projective as well.

Some have explained it, as it is required for dealing with the phase factor redundancy in the wavefunction,but I guess this redundancy is dealt by the unitary representation itself, as the phase factors cancels out. I don't see why a phase factor redundancy demands the symmetry be projective.

Eden Zane
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    What do you mean "the phase factors cancels out"? The prime example of a projective but not unitary representation is the spin-1/2 representation of $\mathrm{SO}(3)$. Are you trying to argue there should be no half-spin objects in QM? See also this answer of mine for the relation between projective representations and unitary representations of larger groups – ACuriousMind Sep 16 '23 at 16:45
  • when we have a state, $\psi$ and another $e^{i\theta}\psi$, they are related by a phase factor $e^{i\theta}$, and when we calculate the probabilities, these phase factors gets cancels out.

    this is what I mean.

    Regarding the spin -1/2, I am not. I just want to make my conceptions clear.

     – Eden Zane Sep 16 '23 at 17:14
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    Then you'll have to be more specific what bothers you here - the usual arguments for why we need projective and not just unitary representation are precisely because that phase factor cancels out of the Born rule so the elements of the representation can fulfill the more "relaxed" group composition rule $\Sigma(g)\Sigma(h) = C(g,h)\Sigma(gh)$ (notation see my linked answer) instead of the usual $U(g)U(h) = U(gh)$ for a strictly unitary representation. – ACuriousMind Sep 16 '23 at 17:18
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    A projective representation of an algebra (e.g., group) reproduces the algebra up to a complex phase, as you noted. Demanding a projective representation is less stringent of a requirement than demanding an ordinary one. As mentioned above, this can be done to allow the use of universal covering groups, like SU(2) for SO(3), where the half-odd-integer irreps of SU(2) are projective irreps of SO(3), while the integer irreps of SU(2) are ordinary irreps of SO(3). I don't have an physical explanation for why this should be so. I imagine it's an observation, and not a priori justified. – just a phase Sep 16 '23 at 20:14
  • I was thinking of the term "projective" from the projective geomtry point of view, in which case, the main feature is of the inversion of the orientation as illustrated here. While I think it is used just as a projection map in the sense of a fibre. May be I am illegally mixing ideas from manifolds and groups – Eden Zane Sep 17 '23 at 01:03

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