## Lagrangian density for a complex scalar field (classical)

Hi.
Let's say we have a complex scalar field $\varphi$ and we separate it into the real and the imaginary parts:
$\varphi$ = ($\varphi1$ + i$\varphi2$)
It's Lagrangian density L is given by:
L = L($\varphi1$) + L($\varphi1$)
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.
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 Recognitions: Science Advisor Do you mean to say L($\varphi$) = L($\varphi1$) + L($\varphi2$)? That's because L($\varphi$) = L($\varphi1$) + L($i\varphi2$) due to superposition principle, and L($i\varphi2$)=L($\varphi2$) 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.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.

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