The bare cosmological constant in field theory is needed to cancel the infinite vacuum zero-point energy. Then you get a renormalized cosmological constant.(adsbygoogle = window.adsbygoogle || []).push({});

There are three quantites at play, Ω=E+Ω_{0}, where E is the infinite vacuum zero-point energy, and Ω is the renormalized cosmological constant, and Ω_{0}is the bare cosmological constant. E diverges, Ω_{0}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+Ω_{0}=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[0]=1 by ignoring the vacuum connected diagrams. So instead of Z[0]=1 we would have to have Z[0]=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[0]=1 is that all the vacuum graphs get cancelled by the bare cosmological constant in the Lagrangian?

Thanks.

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# Not understanding cosmological constant in field theory

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