Complex scalar field - Feynman integral

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SUMMARY

The propagator for a complex scalar field, defined as \(\phi = \frac{1}{\sqrt{2}} (\phi_1 + i \phi_2)\), is derived from the individual propagators of the real fields \(\phi_1\) and \(\phi_2\) with masses \(m_{\phi_1}\) and \(m_{\phi_2}\). The correct expression for the propagator is \(\frac{1}{2} \left[ \frac{i}{(k^2-m_{\phi_1}^2)} + \frac{i}{(k^2-m_{\phi_2}^2)} \right]\). The presence of the imaginary unit 'i' does not alter the fundamental structure of the propagator but indicates the complex nature of the field. The Euler-Lagrange equations confirm that the propagator retains the characteristics of the Klein-Gordon equation.

PREREQUISITES
  • Understanding of complex scalar fields in quantum field theory
  • Familiarity with the Klein-Gordon equation
  • Knowledge of propagators in quantum mechanics
  • Basic grasp of the Euler-Lagrange equations
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  • Study the derivation of the Klein-Gordon equation for complex fields
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  • Investigate the relationship between the Euler-Lagrange equations and propagators
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ryanwilk
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Homework Statement



For a real scalar field \phi, the propagator is \frac{i}{(k^2-m_\phi^2)}.

If we instead assume a complex scalar field, \phi = \frac{1}{\sqrt{2}} (\phi_1 + i \phi_2), where \phi_1,\phi_2 are real fields with masses m_{\phi 1},m_{\phi 2}, what is the propagator?

Homework Equations



N/A

The Attempt at a Solution



Is this true?: There's a 1/2 probability that the propagator has mass \phi_1, \phi_2 so the propagator should just be \frac{1}{2} [ \frac{i}{(k^2-m_{\phi_1}^2)} + \frac{i}{(k^2-m_{\phi_2}^2)} ]. But how does the 'i' change things?

Any help would be appreciated,

Thanks.
 
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The Euler-Lagrange eqns (which typically determine the Feynman propagator) are completeley separated and identical with the original KG equation. So indeed the propagator will be written as a sum b/w a term with i and the other with i2.
 

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