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Lagrangian density for a complex scalar field (classical)

  1. Dec 7, 2012 #1
    Let's say we have a complex scalar field [itex]\varphi[/itex] and we separate it into the real and the imaginary parts:
    [itex]\varphi[/itex] = ([itex]\varphi1[/itex] + i[itex]\varphi2[/itex])
    It's Lagrangian density L is given by:
    L = L([itex]\varphi1[/itex]) + L([itex]\varphi1[/itex])
    Can you tell the argument behind the idea that in summing the densities of cpts. we treat the imaginary part on equal basis with the real.
  2. jcsd
  3. Dec 7, 2012 #2


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    Do you mean to say L([itex]\varphi[/itex]) = L([itex]\varphi1[/itex]) + L([itex]\varphi2[/itex])? That's because L([itex]\varphi[/itex]) = L([itex]\varphi1[/itex]) + L([itex]i\varphi2[/itex]) due to superposition principle, and L([itex]i\varphi2[/itex])=L([itex]\varphi2[/itex]) due to U(1) symmetry. Neither are absolutely generally true. Former requires a linear Lagrangian, later requires it to be symmetric under U(1) transformations. Both of these are true in Quantum Mechanics, but not necessarily in general field theory.
  4. Dec 7, 2012 #3


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    U(1) symmetry follows from the general requirements for a Lagrangian field theory. The action must be real under complex conjugation, hence the lagrangian density must contain matched products of phi and phi star and subsequent spacetime derivatives.
  5. Dec 7, 2012 #4


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    You are right, it does follow from L = L*. I never really thought of it that way.
  6. Dec 7, 2012 #5
    Thank you for the answers - they are very insightful.
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