The bare cosmological constant in field theory is needed to cancel the infinite vacuum zero-point energy. Then you get a renormalized cosmological constant. There are three quantites at play, Ω=E+Ω0, where E is the infinite vacuum zero-point energy, and Ω is the renormalized cosmological constant, and Ω0 is the bare cosmological constant. E diverges, Ω0 cancels some of the E such that the sum Ω is finite. I have some conceptual difficulties with this. First, if someone talks about the energy of the vacuum, which of those 3 quantities are they talking about? Second, which of those quantities enters into the field equations of Einstein? If Ω, then does that even make sense, since Ω is not a physical quantity but runs with coupling μ? Third, when they say the cosmological constant is small, they mean Ω right? Fourth, in field theory, the choice is usually made such that Ω=E+Ω0=0. This ensures that the vacuum to vacuum amplitude <0(∞)|0(-∞)> is equal to 1 rather than a phase. Does this mean field theory has to be changed if Ω is really big such that <0(∞)|0(-∞)> is a phase factor not equal to 1? What's usually done is we calculate the generating function Z[J] and by hand enforce Z=1 by ignoring the vacuum connected diagrams. So instead of Z=1 we would have to have Z=phase due to cosmological constant. How would this change the equations of field theory? Also, would that mean that the physical interpretation of setting Z=1 is that all the vacuum graphs get cancelled by the bare cosmological constant in the Lagrangian? Thanks.