Not understanding cosmological constant in field theory

[geoduck]

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[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 canceled by the bare cosmological constant in the Lagrangian?

Thanks.

 

[fzero]

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.

You are not alone. The above picture that you give is not a description that is supported by much evidence. It is simply one possible explanation of how the observed cosmological constant/vacuum energy emerges from the microscopic physics. In models with supersymmetry, what you are calling the bare cosmological constant, $\Lambda_0$, ( $\Omega_0$ is usually reserved for the critical density in cosmological models, so I don't want to use it here) is actually the vacuum energy of the superpartners, which comes with the opposite sign to ordinary matter.

The truth is that we don't know what is going on in the real world here. It is plausible that the vacuum energy of field theory is a source in the Einstein equations, but we don't know why the observed value of the cosmological constant is so small. Even if we accept that the observed $\Lambda$ is "renormalized", we have an enormous fine-tuning problem.

First, if someone talks about the energy of the vacuum, which of those 3 quantities are they talking about?

This depends on the context. If the discussion is about field theory, it is what you're calling the zero-point energy. If it is about cosmology, then it might be the observed value of the CC. If the discussion is about the interplay between the two, then the author should be careful to distinguish what terms they're using to refer to which concept.

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?

As I mentioned above, we are really not sure. In the specific cosmological model, the observed value of $\Lambda$ that is reported corresponds to what you're calling the renormalized value above. But we don't know whether or how the field theory vacuum energy contributes to this for certain.

Third, when they say the cosmological constant is small, they mean Ω right?

That's the only value that we can measure.

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[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 canceled by the bare cosmological constant in the Lagrangian?

In field theory, the reference point of the energy can be freely shifted to set the vacuum energy to zero. The whole problem in the cosmological setting is that general relativity does not allow you to do this. So if we're trying to compute the contributions to the stress-energy tensor for the Einstein equation, we have to be more careful than if we were ignoring gravity. If there is a large bare cosmological constant, then we should really be doing QFT in curved spacetime, include all the vacuum diagrams, and tune $\Lambda_0$. Another complication is that perturbative physics alone is not enough to properly discuss this. I think this sort of perturbative field theory argument gives a cartoon picture of what might be happening, but is inevitably leaving out important details.

 

[geoduck]

I'm not familiar with supersymmetry, but my understanding is that the superpartners are heavier, and the vacuum energy does depend on mass (in zeroth order, the sum sqrt[k^2+m^2] over all 3-momentum modes). So that seems like it doesn't require fine tuning.

But I thought bare quantities could be interpreted as renormalized quantities except at a high scale. If that's the case, $$\Lambda_0$$ should be interpreted as coming from both particles and their superpartners at high energies. And at high energies, the masses shouldn't matter. So their zero-point energies should cancel exactly, giving $$\Lambda_0=0$$?

What troubles me is that renormalized parameters aren't physical. The charge of the electron depends on scale. I don't see how a renormalized $$\Lambda$$ can have a cosmological effect. Maybe one can define a physical $$\Lambda_p$$, i.e. like when you define a physical mass that's different from bare and renormalized mass?

 

[fzero]

I'm not familiar with supersymmetry, but my understanding is that the superpartners are heavier, and the vacuum energy does depend on mass (in zeroth order, the sum sqrt[k^2+m^2] over all 3-momentum modes). So that seems like it doesn't require fine tuning.

But I thought bare quantities could be interpreted as renormalized quantities except at a high scale. If that's the case, $$\Lambda_0$$ should be interpreted as coming from both particles and their superpartners at high energies. And at high energies, the masses shouldn't matter. So their zero-point energies should cancel exactly, giving $$\Lambda_0=0$$?

In a theory with unbroken SUSY, the vacuum energies do cancel. This is a huge improvement over an infinite vacuum energy, but still doesn't explain a small non-zero cosmological constant. For broken SUSY the vacuum energy should behave like $M_\mathrm{SUSY}^4$, which is still too large when compared with observation.

What troubles me is that renormalized parameters aren't physical. The charge of the electron depends on scale. I don't see how a renormalized $$\Lambda$$ can have a cosmological effect. Maybe one can define a physical $$\Lambda_p$$, i.e. like when you define a physical mass that's different from bare and renormalized mass?

It is precisely the renormalized parameters that are physical in a QFT. The scale-dependent electric charge is usually considered in the form of the fine-structure constant. At very low energies $\alpha \sim 1/137$, but it has been measured that agreement with scattering events at high energies requires that $\alpha \sim /128$ at energies comparable with the W-boson mass.

As a technical exercise, a renormalization scheme involves a bare parameter $\lambda_b$ and a counterterm $\lambda_c$. There are divergent diagrams at any given order in $\lambda_b$. The counterterm parameter $\lambda_c# is chosen to be divergent in such a way to cancel the divergence coming from the bare theory. When the bare terms and counterterms are taken together, we have the renormalized couplings$\lambda_r## which are scale dependent. This scale dependence is precisely what it required by renormalization in order that these parameters are physcial.

I would suggest learning more about the renormalization group before worrying too much about the cosmological constant. Since we don't have a fundamental quantum theory of gravity, it is not even clear that the cosmological constant should be renormalized in the same fashion as a parameter in a QFT. As I tried to explain earlier, the kind of "renormalization" that you're talking about is an educated guess at best; a plausible starting point from which we can try to derive reasonable consequences.