The vacuum-vacuum expectation value in the absence of a source is in general not equal to 1, but exp[-iEt], where E is the energy of the vacuum. For some reason in QFT, we say E=0 (i.e., we normalize Z[0]=1, the generating functional), but we don't need to do this and one can in fact calculate E through some clever tricks. For example in free-field theory E is the zero-point energy of a collection of harmonic oscillators. This zero-point energy is infinity, but can be renormalized by adding a constant term Ω to the Lagrangian of dimension mass(adsbygoogle = window.adsbygoogle || []).push({}); ^{4}.

The value of the cosmological constant is almost zero, and some say that implies fine tuning. My question is how can you say this? Isn't the fine structure constant ∞, but after renormalization, it's 1/137? I thought bare values don't matter?

Then I read Zee's chapter on this topic, and he says that because of spontaneous symmetry breaking, you get constant terms in the Lagrangian like μ^{4}, where μ^{2}is the negative mass^{2}term in the Lagrangian. He claims that this renormalized value of μ is 10^{17}GeV, so raising that to the 4th power, he claims that should equal the cosmological constant, unless there is a bare parameter Ω_{0}that can cancel it, and this is fine tuning.

What I don't understand is whether he's using BPH renormalization or not. It seems like he is, so that the constant term from spontaneous symmetry breaking makes a contribution equal to (10^17)^4. Then don't you still need to add to the Lagrangian not only that, but Ω_{R}, and an interacting counter-term?

So the extra Lagrangian would be μ^{4}+Ω_{R}+ΔΩ.

Now that entire sum, plus the full propagator evaluated at the same space-time point, should equal the cosmological constant. ΔΩ cancels the divergences of the full-propagator term and leaves a finite piece, giving: μ^{4}+Ω_{R}+Ω_{F}.

I still don't see the fine tuning. Why can't the sum of the last two terms cancel the first term μ^{4}?

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# Cosmological constant in qft

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