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.


marked as duplicate by Aaron Stevens, Kyle Kanos, Jon Custer, AccidentalFourierTransform, ZeroTheHero Dec 4 '18 at 0:30

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  • $\begingroup$ Yep, that does answer my question, thank you! So the completeness relation as written is correct! I searched and didn't see that post. It's a different question overall but the answer is included. $\endgroup$ – doublefelix Dec 2 '18 at 21:59
  • $\begingroup$ Revisiting this question, I no longer understand how (or even if) the expansion works in the interacting theory. For the free theory I mostly have no questions. But the momentum eigenstates are defined using $a, a^{\dagger}$, which come from the free theory. What guarantee do we have that these states span the whole space in the interacting theory, or even that they lie in the same hilbert space as the interacting theory does? $\endgroup$ – doublefelix Mar 14 at 18:11