Does $\langle\Omega|\mathrm{e}^{\mathrm{i}Px}=\langle\Omega|\mathrm{e}^{\mathrm{i}0x}$? $(\langle\Omega| =$ ground state of the interacting theory)

Question

Let $\langle\Omega|$ be the ground state of an interacting theory, just as Peskin & Schroeder(PS) describes on page 82 and page 213. On page 213 PS do the following

$$\tag{1}\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle = \langle\Omega|\exp(+\mathrm{i}Px)\phi(0)\exp(-\mathrm{i}Px)|\lambda_{\vec{p}}\rangle \\ =\langle\Omega|\phi(0)|\lambda_{0}\rangle\exp(-\mathrm{i}px)\bigg|_{p^0=E_{\vec{p}}} $$

In the above $|\lambda_0\rangle$ is an eigenstate of the full interacting hamiltonian $H$. The ket $|\lambda_{\vec{p}}\rangle$ is the boost of $|\lambda_0\rangle$ with momentum $\vec{p}.$ Furthermore $E_{\vec{p}} = \sqrt{p^2+m_\lambda^2}$ and $\vec{P}|\lambda_0\rangle = 0.$

I assume PS used $$\tag{2}\langle\Omega|\mathrm{e}^{\mathrm{i}Px}=\langle\Omega|\mathrm{e}^{\mathrm{i}0x}$$ in $(1)$.

Is $(2)$ true?

I thought $\langle0|\mathrm{e}^{\mathrm{i}Px}=\langle0|\mathrm{e}^{\mathrm{i}0x}$ where $\langle0|$ is the ground state of the free theory.

I hope I have provided enough info for the question to make sense.

Answer 1

Both the free and interacting vacuum are invariant under translations, assuming that translation invariance isn't spontaneously broken. Usually we expand around spatially homogeneous and time-independent field configurations, so that you don't have to worry about spontaneously breaking translations.

There are some cases where translations are broken in the ground state, and this leads to Goldstone bosons that are the transverse fluctuations. For example one could take a ground state which has a long QCD string, and there are massless bosons on the string which correspond to the transverse fluctuations.

Another example is single-field inflation, where the vacuum expectation value of the inflaton $\phi$ is spatially homogeneous but depends on time. In this case we get a Goldstone boson for the breaking of time diffeomorphisms.

Edit: Here is a formula:

\begin{align*} P|\Omega\rangle=P|0\rangle=0,\text{ usually. } \end{align*}