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Lagrangian density for a complex scalar field (classical) 
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#1
Dec712, 10:44 AM

P: 71

Hi.
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
Dec712, 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.



#3
Dec712, 01:02 PM

Sci Advisor
HW Helper
P: 11,895

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
Dec712, 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.



#5
Dec712, 03:56 PM

P: 71

Thank you for the answers  they are very insightful.



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