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=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 mass4. 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 μ4, where μ2 is the negative mass2 term in the Lagrangian. He claims that this renormalized value of μ is 1017 GeV, so raising that to the 4th power, he claims that should equal the cosmological constant, unless there is a bare parameter Ω0 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 ΩR, and an interacting counter-term? So the extra Lagrangian would be μ4+ΩR+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: μ4+ΩR+ΩF. I still don't see the fine tuning. Why can't the sum of the last two terms cancel the first term μ4?