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When treating QFT we want our theory to be invariant under different symmetry groups, for example, the Standard Model is a non-abelian gauge theory with the symmetry group $U(1)×SU(2)×SU(3)$. Moreover, it is invariant under Lorentz transformations which form the Lorentz group $O(1,3)$ with the proper transformations subgroup $SO(1,3)$ In this example $S$ stands for Special, meaning that these transformations are represented by matrices with determinant $1$.

My questions are:

  1. why is this requirement so important?
  2. In particular, what would happen if we allow the determinant to be any real (or complex) number?
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    $\begingroup$ The determinant of an orthogonal matrix is already restricted to being either 1 or -1, so it couldn't be just any number! $\endgroup$
    – Danu
    Commented May 10, 2014 at 7:05
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    $\begingroup$ ok, but why not -1 then? $\endgroup$
    – Costantino
    Commented May 10, 2014 at 20:57

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To preserve the notion of probabilities in QFT, symmetries must be implemented as unitary or antiunitary operations. See also Wigner's theorem. For finite-dimensional representations, the determinant of such operations must be just a phase factor.

Example: $U(1)$ symmetry.

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