# Cosmological constant in qft

1. Jun 21, 2013

### geoduck

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 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 μ4R+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: μ4RF.

I still don't see the fine tuning. Why can't the sum of the last two terms cancel the first term μ4?

2. Jun 21, 2013

### Zmunkz

So is your underlying question if the cosmological constant implies fine tuning? Or are you looking for feedback on your mathematical reasoning?

If the first, I'll refer you to Victor Stenger's "the fallacy of fine tuning", or Leonard Susskand's "the cosmic landscape". Both conclude, as did you, that no fine tuning is implied.

3. Jun 21, 2013

### The_Duck

Fine tuning is all about bare parameters. Specifically, it's about how you have to "fine-tune" the bare parameters in a theory with a large but finite cutoff in order to achieve the observed low-energy parameters.

If you write down QED with a large but finite cutoff (say, 10^19 GeV), the bare coupling isn't that different from the low-energy coupling, since the coupling constant only depends logarithmically on the energy scale. You end up with a dimensionless bare coupling of order 1, which seems "natural" and not fine-tuned.

They obviously *can*, but why *should* they? Why would we expect them to, unless something has "fine-tuned" them to cancel?

To expand, the standard argument goes something like this. I hope this doesn't just repeat what Zee said.

Forget about infinities, counterterms, and renormalization schemes. The standard model is expected to be an effective field theory with a finite cutoff. If we have a finite cutoff $\Lambda$, we can just write down the theory in terms of its bare parameters and calculate with these bare parameters to get unambiguous finite predictions. Let's imagine we know the bare couplings of the standard model.

When we calculate the vacuum energy density, we get three contributions:

* a contribution from the bare cosmological constant, $\Omega_0$.
* a contribution from spontaneous symmetry breaking. If a scalar field $\phi$ with potential $V(\phi)$ has a vacuum expectation value $\langle \phi \rangle$, it contributes an energy density $V(\langle \phi \rangle)$. In your notation $V(\langle \phi \rangle) = \mu^4$.
* a contribution from interactions, namely the sum of all vacuum bubble diagrams. The biggest contribution comes from quartically divergent diagrams. Except we have a finite cutoff $\Lambda$, so these diagrams are finite and proportional to $\Lambda^4$.

So the vacuum energy density looks like

$\Omega = \Omega_0 + \mu^4 + c \Lambda^4$

with $c$ some dimensionless coefficient that we can compute by summing the vacuum bubble diagrams.

The observed fact is that $\Omega$ is many orders of magnitude smaller than $\Lambda^4$. In fact, even if you ignore the $\Lambda^4$ term, $\Omega$ is still many orders of magnitude smaller than $\mu^4$ just counting the vacuum energy that should be there from the potential of the SM Higgs field.

So for $\Omega$ to come out small, it seems like $\Omega_0$ has to be "fine-tuned" to be almost (but not quite) exactly the negative of $\mu^4 + c \Lambda^4$.

The actual value of $\Omega_0$ is determined by the UV-completion of the theory, the true physics above the cutoff of the standard model. So one way you could resolve this issue is by giving a model for the UV completion that explains why $\Omega_0$ has to be fine-tuned in this way. But the usual argument is that all we should expect is that $\Omega_0$ is some number of order $\Lambda^4$ that generically will be unrelated to the other contributions to $\Omega$. In this case we should expect $\Omega$ to come out to be of order $\Lambda^4$, in contradiction with observation.

So, the standard argument concludes, one of the steps above is wrong. But no one is quite sure what.

Supersymmetry sets up some neat cancellations that force $c = 0$ and (I think) $\Omega_0 = 0$. But since the Higgs boson observably doesn't have an exact superpartner, $\mu^4 \neq 0$, so that $\Omega = \mu^4 \neq 0$, which is still in contradiction with observation by many orders of magnitude.

Last edited: Jun 21, 2013
4. Jun 21, 2013

### geoduck

Does this mean if a coupling is ever nonlogarithmic in scale (say its quadratic), then at a large finite cutoff, you'll get a bare coupling that is very large? So essentially everything has to be logarithmic.

I'm a little confused about whether it's vacuum graphs that contribute.
I've seen a calculation of this that goes like this for phi^4 theory:

$$<0+|0->=\int [d\phi] e^{-S[\phi]}$$

$$\frac{\partial}{\partial m^2}<0+|0->=-(1/2)\int [d\phi] e^{-S[\phi]} \int \phi(x)^2d^4x <0+|0->$$

$$\frac{\partial}{\partial m^2}ln(<0+|0->)=-(1/2)\int [d\phi] e^{-S[\phi]} \int \phi(x)^2 d^4x$$

Then the correlation function for phi^2 is calculated and substituted on the RHS, and then equation is integrated to get <0+|0->.

Since <0+|0->=exp[-ET], the E contributes to the cosmological constant.

So it seems to me it's not quite the vacuum graphs, but the correlation function of a composite operator phi(x)^2?

Last edited: Jun 21, 2013