# Lagrangian density for a complex scalar field (classical)

1. ### Trave11er

71
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.

2. ### K^2

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.

3. ### dextercioby

12,314
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.

4. ### K^2

2,470
You are right, it does follow from L = L*. I never really thought of it that way.

5. ### Trave11er

71
Thank you for the answers - they are very insightful.

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