This is a followup to a question I read recently: What does the identity operator look like in Quantum Field Theory?
Out of curiosity, I was writing down what I figured to be the momentum basis vectors in QFT, ignoring for now internal degrees of freedom (spin etc). If each momentum label signifies an additional particle, and $|\Omega\rangle$ is the vacuum state, I wrote:
$$|\Omega\rangle, |p_1\rangle, |p_1 p_2\rangle, |p_1 p_2 p_3\rangle...$$
where actually each of the items in this list contains "infinitely more" than the last, because there is one basis vector for each value of momentum in $\mathbb{R}^3$. I also assumed all states had been normalized.
My question is: How does one write the identity down, in this basis? My guess is:
$$1 = |\Omega\rangle\langle\Omega| +\int_{-\infty}^{\infty}|p_1\rangle\langle p_1|d^3p + \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|p_1 p_2\rangle\langle p_1 p_2 |d^3p_1 d^3p_2 \,+\, ...$$
however the lack of integration on the vacuum, and varying numbers of integrals seem a bit unusual to me. Maybe I have misunderstood something fundamental.
