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

  1. May 23, 2014 #1
    Mega quick question: must the Lagrangian (density) be hermitian?

    In other words, must [itex]\mathcal{L}=\mathcal{L}^\dagger[/itex] always be valid?
    Last edited: May 23, 2014
  2. jcsd
  3. May 23, 2014 #2


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    This is discussed at the opening chapter of "Symmetry and Its Breaking in Quantum Field Theory" By T. Fujita
  4. May 23, 2014 #3


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    :uhh: Can you tell us what it says there?
  5. May 23, 2014 #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?
    Last edited: May 23, 2014
  6. May 23, 2014 #5


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    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.
  7. May 23, 2014 #6


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    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.
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