Lagrangian, scalar or pseudo-scalar?

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A Lagrangian density is typically required to be a scalar to ensure proper transformation under spacetime symmetries. The discussion raises the question of whether a Lagrangian could be a pseudo-scalar, noting that this would still allow it to be an eigenfunction of the parity operator. However, the consensus emphasizes that for quantum field theory in Minkowski spacetime, the Lagrangian must adhere to the requirements of the Poincare group, CPT, and gauge invariance. Each of these symmetries necessitates that the Lagrangian density is a scalar. Therefore, while the idea of a pseudo-scalar Lagrangian is intriguing, it does not align with established principles in quantum field theory.
salparadise
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Hi,

My question is. Can in principle, a Lagrangian density for some theory be a pseudo-scalar. Normally people say that the Lagrangian needs to be a scalar, but it case it is a pseudo-scalar it would also be a eigaen function of the parity operator.

This topic could well be on the classical physics section, but as this is more relevant in quantum theory I decided to place it here.

Thanks
 
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The normal requirement is that the fields transform irreducibly under the spacetime symmetry group and that the lagrangian density is a scalar under the spacetime symmetry group. The symmetry requirement for QFT in Minkowski spacetime is restricted Poincare group plus CPT plus gauge invariance. The Lagrangian density should be a scalar wrt to the 3 altogether or each taken separately.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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