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innisfree
innisfree
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Suppose $\phi(x)$ is thea real Klein-Gordon field, then the single-particle wave function $\psi(x)$ corresponding to a momentum $p$ is given by (QFT,Ryder Ryder) $$\psi(x)=<0|\phi(x)|p>$$$$\psi_p(x)=\langle0|\phi(x)|p\rangle.$$ The calculated result is just as expected ($e^{-ipx}$). But is this by definition or does it comescome from somewhere else?

Suppose $\phi(x)$ is the real Klein-Gordon field, then the single-particle wave function $\psi(x)$ corresponding to a momentum $p$ is given by (QFT,Ryder) $$\psi(x)=<0|\phi(x)|p>$$ The calculated result is just as expected ($e^{-ipx}$). But is this by definition or it comes from somewhere else?

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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M. Zeng
M. Zeng
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1-particle momentum eigenfunction in terms of field operator for real Klein-Gordon field

Suppose $\phi(x)$ is the real Klein-Gordon field, then the single-particle wave function $\psi(x)$ corresponding to a momentum $p$ is given by (QFT,Ryder) $$\psi(x)=<0|\phi(x)|p>$$ The calculated result is just as expected ($e^{-ipx}$). But is this by definition or it comes from somewhere else?