# How is the interacting vacuum defined in QFT?

I have seen this in a couple of textbooks (Schwartz and Zee), where the author would use the interacting vacuum $$|\Omega \rangle$$ in a calculation, but would never mention how the state is defined.

What does

$$|\Omega \rangle = ?$$

Equal to?

• It is probably just the vacuum in the $\textit{interaction}$ picture Commented May 9, 2023 at 0:29
• Typically via Gell-mann-Low theorem Commented May 9, 2023 at 1:02

Interacting vacuum is defined by $$H|\Omega\rangle=0,$$ where $$H$$ is total Hamiltonian. Normally, we don’t expand $$|\Omega\rangle$$ because we don’t need it. This definition is enough. If you want to expand $$|\Omega\rangle$$ in some basis, just take the inner product of the basis and $$|\Omega\rangle$$.

This is distinguished by free vacuum $$|0\rangle$$ which is defined by free Hamiltonian $$H_0|0\rangle=0.$$ Note that $$|\Omega\rangle\neq|0\rangle$$ unless we don’t impose this.

1. In the interaction picture, we use both of $$|\Omega\rangle$$ and $$|0\rangle$$, but its definition does not change from above. Also, the interacting vacuum is a general terminology used for other than interaction picture.
2. Gell-man low’s theorem is a little bit different context. Roughly speaking, this theorem asserts that the interacting vacuum $$|\Omega\rangle$$ can be identified to the free vacuum $$|0\rangle$$ in some limit, i.e., the interacting vacuum can be projected to the free vacuum under certain conditions. Also, this theorem plays an important role in operator formalism based on the interaction picture when writing n-point functions as expectation values for the free vacuum. A similar projection to the vacuum also appears in path integral formalism (See ch.6 of Srednicki). Note that this theorem relates two vacuums, but it never state they are same: i.e. $$|\Omega\rangle\neq |0\rangle$$. It is simply stating that only $$|0\rangle$$ can be taken out of $$|\Omega\rangle$$ (i.e. projection into the free vacuum) in a situation where we consider a certain limit with the appropriate operators acting to $$|\Omega\rangle$$.