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  1. Is there any representation of the Lorentz group where $$U^{-1} f(x) U = f(\lambda^{-1}x)$$ other than the (0,0) representation?

  2. If not then is it possible for a field (with a well defined polynomial basis) to behave like a scalar field under the Lorentz group?

  3. Will such fields still be called the (0,0) representation of the Lorentz group?

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  • $\begingroup$ No finite dimensional representation of the Lorentz group superior to (0,0) transform as such, nor the continuous spin representation of the Lorentz group. $\endgroup$ – Slereah Oct 30 '15 at 8:28
  • $\begingroup$ You used the term finite dimensional can you please elaborate? $\endgroup$ – Abhishek Pal Oct 30 '15 at 8:37
  • $\begingroup$ The field exists in an infinite dimensional functional space spanned by a polynomial basis. Is it possible then ? $\endgroup$ – Abhishek Pal Oct 30 '15 at 10:35
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    $\begingroup$ I read your question 2. as: "Is it possible for a field to behave like a scalar field under the action of the Lorentz group?" The answer is obviously yes, if it is a scalar field. But I think you might be mixing the classification of fields based on their "point-group transformation behaviour" and the "global transformation behaviour". I.e. the non-relativistic wave-function is the scalar representation of the galiean group in the "local" point-group sense, but spherical harmonics and the corresponding wave-functions are non-scalar representations of rotations in the global trafo. sense. $\endgroup$ – Void Oct 30 '15 at 11:40
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It precisely one of the Wightman axioms that the infinite-dimensional unitary representation 1 $U : \mathrm{SO}(1,3)\to\mathrm{U}(\mathcal{H})$ on the space of states $\mathcal{H}$ of the theory upon which the field acts as operator is compatible with the field transformation law under the finite-dimensional representation $\rho_\text{fin}: \mathrm{SO}(1,3)\to\mathrm{GL}(V)$ where $V$ is the target space of the field. For a real scalar field, $V=\mathbb{R}$ and $\rho_\text{fin}$ is the trivial representation. Being "compatible" means that $$ U(\Lambda)^\dagger\phi_i(x)U(\Lambda) = \sum_j\rho_\text{fin}(\Lambda)_{ij}\phi_j(\Lambda^{-1}(x))$$ holds as an operator equation on the space of states.

Now, if $\phi$ is scalar, then $\rho_\text{fin}$ is trivial. However, this does not mean, in any way, that $U$ is trivial. The infinite-dimensional unitary representations of the Poincare group $\mathrm{SO}(1,3)\ltimes \mathbb{R}^4$ are given by Wigner's classification, and the scalar field creates particles with mass and momentum, so the unitary representation is not trivial - the trivial unitary representation is just the vacuum.


1No finite-dimensional representation can be unitary.

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