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Lagrangian density for a complex scalar field (classical)

by Trave11er
Tags: classical, complex, density, field, lagrangian, scalar
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Dec7-12, 10:44 AM
P: 71
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.
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Dec7-12, 12:27 PM
Sci Advisor
P: 2,470
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.
Dec7-12, 01:02 PM
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P: 11,950
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.

Dec7-12, 01:46 PM
Sci Advisor
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.
Dec7-12, 03:56 PM
P: 71
Thank you for the answers - they are very insightful.

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