Mega quick question: must the Lagrangian (density) be real valued?

In summary, the discussion in "Symmetry and Its Breaking in Quantum Field Theory" by T. Fujita states that the Lagrangian does not necessarily have to be hermitian, as it is not an observable quantity. However, for theories that have CPT symmetry, the Lagrangian must be hermitian. This can be seen in the fact that the Hamiltonian, which is an observable, must be hermitian. The use of a non-hermitian Lagrangian may simplify calculations, but the overall theory must still be consistent with the requirement that the Hamiltonian is hermitian.
  • #1
guest1234
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1
Mega quick question: must the Lagrangian (density) be hermitian?

In other words, must [itex]\mathcal{L}=\mathcal{L}^\dagger[/itex] always be valid?
 
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  • #2
This is discussed at the opening chapter of "Symmetry and Its Breaking in Quantum Field Theory" By T. Fujita
 
  • #3
UltrafastPED said:
This is discussed at the opening chapter of "Symmetry and Its Breaking in Quantum Field Theory" By T. Fujita
:uhh: Can you tell us what it says there?
 
  • #4
It basically says that the Lagrangian must not be hermitian since it's not observable.
But I still have mixed feelings about it. After googling "hermicity of the Lagrangian" this popped up stating that the Lagrangian of the theory invariant under CPT symmetry must be hermitian (don't know whether the statement is invertible).
The reason why I'm asking this is that the term [itex]F^{\mu\nu}V^{\mu}V_{\mu}^{\dagger}[/itex] is not hermitian and constructing such Lagrangian with the term is impossible.
So.. what I am asking now is: what is gained (lost) in the theory if the Lagrangian is (non-)hermitian?
 
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  • #5
Often we work with a non-Hermitian Lagrangian just because it's simpler, but it's understood that tacked on at the end is the admonition "+ h.c.".

For sure, the Hamiltonian must be Hermitian, since it's an observable. Proofs of the CPT Theorem utilize this fact explicitly.
 
  • #6
The Lagrangian of a field theory must be real with respect to the involution and have Grassmann parity 0, if the fields containing it form a Grassmann algebra with involution. For complex scalar fields, the involution on the associative algebra of the fields is the complex conjugation.
 

1. Is it necessary for the Lagrangian to be real valued?

Yes, the Lagrangian must be real valued in order for the equations of motion to be well-defined and physically meaningful. A complex-valued Lagrangian can lead to non-physical solutions.

2. What does it mean for the Lagrangian to be real valued?

A real-valued Lagrangian means that all of its components (kinetic and potential energy terms) are real numbers. This ensures that the Lagrangian is invariant under real transformations, which is necessary for physical laws to remain consistent.

3. What happens if the Lagrangian is not real valued?

If the Lagrangian is not real valued, the equations of motion will not be well-defined and may lead to non-physical solutions. This can result in inconsistencies and contradictions within the physical laws being described.

4. Are there any exceptions to the requirement of a real-valued Lagrangian?

In certain cases, such as in quantum mechanics, a complex-valued Lagrangian may be used. However, this requires a more sophisticated mathematical approach and should be used with caution.

5. How can one ensure that the Lagrangian is real valued?

The Lagrangian can be checked for real-valuedness by inspecting each component individually and making sure they are all real numbers. Additionally, using established mathematical techniques and principles can help ensure that the resulting Lagrangian is real valued.

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