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Suppose $\phi(x)$ is a real Klein-Gordon field, then the single-particle wave function $\psi(x)$ corresponding to a momentum $p$ is given by (QFT, Ryder) $$\psi_p(x)=\langle0|\phi(x)|p\rangle.$$ The calculated result is just as expected ($e^{-ipx}$). But is this by definition or does it come from somewhere else?

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  • $\begingroup$ Related: physics.stackexchange.com/questions/127796/… $\endgroup$ – innisfree Aug 26 '14 at 13:42
  • $\begingroup$ Could you make your question a little more precise? I don't quite understand what you're confused about. You can plug in the mode expansion for $\phi(x)$ and derive the result. Do you want to know why we should expect this result? $\endgroup$ – Edward Hughes Aug 26 '14 at 14:47
  • $\begingroup$ @Ed, I guess (idk) that whilst $e^{ipx} = \langle0|\phi(x)|p\rangle$ is straightforward, it's unclear why it should be that $\langle0|\phi(x)|p\rangle = \psi_p(x)$. $\endgroup$ – innisfree Aug 26 '14 at 15:20
  • $\begingroup$ @innisfree Yes, this is exactly what I meant $\endgroup$ – M. Zeng Aug 26 '14 at 16:07

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