# 1-particle momentum eigenfunction in terms of field operator for real Klein-Gordon field

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?

• – innisfree Aug 26 '14 at 13:42
• 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? – Edward Hughes Aug 26 '14 at 14:47
• @Ed, I guess (idk) that whilst $e^{ipx} = \langle0|\phi(x)|p\rangle$ is straightforward, it's unclear why it should be that $\langle0|\phi(x)|p\rangle = \psi_p(x)$. – innisfree Aug 26 '14 at 15:20
• @innisfree Yes, this is exactly what I meant – M. Zeng Aug 26 '14 at 16:07