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- Evaluating a correlation function for the Klein-Gordon field I found a term like ##\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle## which should be equal to ##\langle\Omega|\phi(0)|\lambda_{\vec{p}}\rangle e^{-ipx}\left.\right|_{p^0=E_\vec{p}}##

First, let me introduce the notation; given a Hamiltonian ##H## and a momentum operator ##\vec{P}##, and writing ##P=(H,\vec{P})##. Let ##|\Omega\rangle## be the ground state of ##H##, ##|\lambda_\vec{0}\rangle## an eigenstate of ##H## with momentum 0, i.e. ##\vec{P}|\lambda_\vec{0}\rangle=0## and ##|\lambda_\vec{p}\rangle## a ##\vec{p}##-boost of ##|\lambda_\vec{0}\rangle##.

If ##\phi(x)## is the Klein-Gordon field, in the Heisenberg picture ##\phi(x)=e^{iPx}\phi(0)e^{-iPx}## and then $$\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle=\langle\Omega|e^{iPx}\phi(0)e^{-iPx}|\lambda_{\vec{p}}\rangle$$. Then I understand that ##e^{-iPx}|\lambda_{\vec{p}}\rangle=e^{-ipx}|\lambda_{\vec{p}}\rangle\left.\right|_{p^0=E_\vec{p}}## because ##|\lambda_{\vec{p}}\rangle## is a ##\vec{p}##-boost of a eigenstate of ##H##. But the term ##e^{iPx}## disapears and I don't understand why. The easy answer would be to suppose that ##H|\Omega\rangle=\vec{P}|\Omega\rangle=0##, the second one I think one can argue that must be satisfied by the ground state, but not the first, in general ##H|\Omega\rangle=E_0|\Omega\rangle\neq 0##.

Also, I for another step I need to use that ##|\Omega\rangle## is Lorentz invariant, what I think makes sense, but also I'm not sure how to prove it, so maybe both problems are related.

Thank you

If ##\phi(x)## is the Klein-Gordon field, in the Heisenberg picture ##\phi(x)=e^{iPx}\phi(0)e^{-iPx}## and then $$\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle=\langle\Omega|e^{iPx}\phi(0)e^{-iPx}|\lambda_{\vec{p}}\rangle$$. Then I understand that ##e^{-iPx}|\lambda_{\vec{p}}\rangle=e^{-ipx}|\lambda_{\vec{p}}\rangle\left.\right|_{p^0=E_\vec{p}}## because ##|\lambda_{\vec{p}}\rangle## is a ##\vec{p}##-boost of a eigenstate of ##H##. But the term ##e^{iPx}## disapears and I don't understand why. The easy answer would be to suppose that ##H|\Omega\rangle=\vec{P}|\Omega\rangle=0##, the second one I think one can argue that must be satisfied by the ground state, but not the first, in general ##H|\Omega\rangle=E_0|\Omega\rangle\neq 0##.

Also, I for another step I need to use that ##|\Omega\rangle## is Lorentz invariant, what I think makes sense, but also I'm not sure how to prove it, so maybe both problems are related.

Thank you