Graduate Correlation function of a Klein-Gordon field

[Gaussian97]

TL;DR

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

 

[vanhees71]

Note that $x$ are c-number valued four-vector comonents and $P$ is the operator of total momentum. Now you have
$$\exp(-\mathrm{i} \hat{P} \cdot x)|\lambda_{\vec{p}} \rangle=\exp(-\mathrm{i} p \cdot \vec{x}) |\lambda_{\vec{p}} \rangle$$
and
$$\exp(-\mathrm{i} \hat{P} \cdot x)|\Omega \rangle=|\Omega \rangle.$$
Now use this to evaluate the one-point function in question.

 

[Gaussian97]

$$\exp(-\mathrm{i} \hat{P} \cdot x)|\Omega \rangle=|\Omega \rangle.$$
Now use this to evaluate the one-point function in question.

I don't see why $e^{-\mathrm{i} \hat{P} \cdot x}|\Omega \rangle=|\Omega \rangle.$ It's because even if the energy of $|\Omega\rangle$ is not zero it's simply a constant phase and then its unobservable?

 

[Demystifier]

I don't see why $e^{-\mathrm{i} \hat{P} \cdot x}|\Omega \rangle=|\Omega \rangle.$ It's because even if the energy of $|\Omega\rangle$ is not zero it's simply a constant phase and then its unobservable?

Here $\hat{P}^{\mu}$ is assumed to be the normal-ordered operator, so $\hat{P}^{\mu}|\Omega \rangle=0$.

 

[vanhees71]

I don't see why $e^{-\mathrm{i} \hat{P} \cdot x}|\Omega \rangle=|\Omega \rangle.$ It's because even if the energy of $|\Omega\rangle$ is not zero it's simply a constant phase and then its unobservable?

Indeed. You don't get anything new when assuming ray representations of the proper orthochronous Poincare group. All are equivalent to the usual unitary representations. That's why it's just convenient to normal order energy, momentum and angular momentum, i.e., to make the vacuum the state where all these additive conserved quantum numbers are 0.