The vacuum is filled with quantum bubbles. In a Feynman diagram this would correspond to a closed one particle propagator line, a circle, or a bubble. I'm curious how this is described mathematically. So without the context of vertices and external legs. Is the vacuum energy derived from these? Are negative energy solutions involved (as energy and momentum are not on mass shell)?
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You already said that these are closed propagators: this is the mathematical description. – Jeanbaptiste Roux Feb 07 '22 at 08:56
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@JeanbaptisteRoux Is it a superposition of all energies and momenta? – MatterGauge Feb 07 '22 at 09:10
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It is the integral over the whole four-momentum of the $p$ representation of the Green function of the considered field. For the vacuum energy density, you might be interested in the first few pages of this thesis – Jeanbaptiste Roux Feb 07 '22 at 09:34
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"The vacuum is filled with quantum bubbles."
The precise technical meaning of this is that the vacuum energy (density) of a free and an interacting quantum field theory are different, and that the way to compute the interacting vacuum energy in a scheme where the free vacuum energy has been renormalized to be zero is to evaluate all "bubble" Feynman diagrams, i.e. diagrams with zero external legs. This is the only significance of these bubble diagrams in actual quantum field theory. Saying "the vacuum is filled with quantum bubbles" sounds impressive, but we must refrain from interpreting any additional significance into this phrase, as unfortunately non-technical presentations of QFT often do.
Technically, this is because the sum of all bubble diagrams represents the expression $\langle 0\vert T\mathrm{e}^{-\mathrm{i}\int H_I\mathrm{d}t}\vert 0\rangle$ (any insertion of a field in front of the exponential would correspond to an external leg), which by a standard computation in the interaction picture (see e.g section 2.8.1 of Weigand's QFT notes [pdf link]) is proportional to $\mathrm{e}^{-E_\Omega 2T}$, where $E_\Omega$ is the (infinite) energy of the interacting vacuum. What really matters in an infinite universe is the energy density, not the energy, and it turns out that the energy density $\rho_\Omega$ is given by $$\rho_\Omega = \mathrm{i}\frac{\sum_j V_j}{\delta^{(4)}(0)},$$ where $V_j$ are the values of all bubble diagrams with a single connected component, freed of an infrared divergence of the form $\delta(0)$. This value is still UV divergent and needs to be renormalized at all orders in perturbation theory just like the free energy density.
Note that your description of bubble diagrams as "In a Feynman diagram this would correspond to a closed one particle propagator line, a circle, or a bubble" is incorrect. A single propagator is not a bubble diagram, and in fact the free theory, which only has propagators, doesn't have any bubbles at all. The bubble diagrams are given by drawing the vertices of the interacting theory, with all their lines connected to other vertices, and none to an external line. For $\phi^4$ theory with a single 4-vertex, the sum of all bubbles looks like this:
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