# Lagrangian density for a complex scalar field (classical)

by Trave11er
Tags: classical, complex, density, field, lagrangian, scalar
 P: 70 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.
 Sci Advisor P: 2,470 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.
 Sci Advisor HW Helper P: 11,863 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.
P: 2,470

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

You are right, it does follow from L = L*. I never really thought of it that way.
 P: 70 Thank you for the answers - they are very insightful.

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