# What are the calculations for Vacuum Energy?

In wiki the Vacuum Energy in a cubic meter of free space ranges from $10^{-9}$ from the cosmological constant to $10^{113}$ due to calculations in Quantum Electrodynamics (QED) and Stochastic Electrodynamics (SED).

I've looked at Baez and references given on the wiki page but none of them give a clear working for how these values are derived.

Can someone point me in the right direction, as to how values like $10^{-9}$ are derived from the cosmological constant; OR $10^{113}$ due to calculations in Quantum Electrodynamics?

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The cosmological constant is measured from WMAP data, baryonic density and standard candle distances/blueshifts; see en.wikipedia.org/wiki/Lambda-CDM_model . The so-called "vacuum energy" that one "calculates" from the Standard Model is nonsense; see the lead up to section 2.3.1 in damtp.cam.ac.uk/user/tong/qft/qft.pdf . For more what the cosmological constant with less brain-deadness, see arxiv.org/abs/1002.3966 – genneth Mar 18 '12 at 0:49
I think section IV of this link may contain the sort of "calculation" you're asking about, although I don't have a working knowledge of the effective action formalism. – twistor59 Mar 18 '12 at 10:21

You can understand the origin of these numbers from a simple consideration of dimensional analysis, and the cosmological data available. This keeps the answer intuitive, and any more complicated derivation will not change the answer substantially.

The first of your numbers, $10^{-9}$ Joules per cubic meter, is simply an empirical measurement in the framework of the Lamda-CDM model. Measurements of the CMB (WMAP), combined with type Ia supernovae, tell us that this is about the energy density of the universe, and that most of the energy density is in the form of dark energy. We assume that the dark energy comes from a cosmological constant $\Lambda$.

In natural units where $\hbar$ and $c$ are set equal to 1, a length is essentially an inverse energy. So in these units $\Lambda$ is about $10^{-46}$ GeV$^{4}$.

Here comes the essential point: If we consider the Planck mass to be the natural energy scale for the vacuum energy, then the ratio of the observed energy density in the cosmological constant is too small by 122 orders of magnitude (and this is the origin of the second number - it just comes from taking the planck mass to the fourth power in natural units).

So, the fundamental puzzle is, why is $\Lambda$ so small compared to the 'natural' scale we would expect? One way out is to argue that a different energy scale other than the planck mass is what we should be comparing $\Lambda$ to.

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You If is important: the Planck mass is neither large nor very small: it is roughly the amount of Vitamin D you should consume a day, and many living things are smaller, so it is not obviously fundamental, except as a indicator of where general relativistic effects and quantum mechanical effects start to confuse each other. – Henry Mar 20 '12 at 22:51
Again, the question I'm asking about isn't about the cosmological constant, it's the Quantum Field Theory derivation. I have tried to be clear about this, see the question in the bounty. All the same, thanks for your replies. – metzgeer Mar 21 '12 at 2:28
@metzgeer, there simply is no consensus for how to derive the vacuum energy from first principles. It is one of the biggest outstanding puzzles in physics, and a first principles derivation would revolutionize the field. Prior to the measurements of the CMB and type Ia supernovae, there was no clear indication that it would turn out to be $10^{-9}$ Joules per cubic meter. That's just what we measure. Its exceedingly small value (compared to the Planck scale) causes many physicists to invoke anthropic explanations. – kleingordon Mar 21 '12 at 6:17
@metzgeer And then there is also the perspective in the paper linked from the comments below your question: our current understanding of QFT can't predict the measured value of $10^{-9}$ Joules / m$^3$, but there's no "grand mystery" because the existence of $\Lambda$ in Einstein's equations can be considered to be a fundamental law in its own right. The key quote, however, is "there is no known natural way to derive the tiny cosmological constant that plays a role in cosmology from particle physics. And there is no known understanding of why this constant is not renormalized to a high value." – kleingordon Mar 21 '12 at 8:44

The first number is not vacuum energy actually. It is the energy of the cosmological repulsive field. This field is one of the forces of matter, possibly a repulsive component of gravitation or a separate force.

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You can see it this way, but it is also correct to view it as a positive energy density in the vacuum. The two points of view only differ in which side of Einstein's equation you put the cosmological term, as stress-energy, or as a modification of the gravitational force law. The reason people nowadays put it in the stress-energy column is because a constant scalar field with a nonzero potential energy at its value will produce a cosmological constant, and in modern understanding, any such constant can be thought of as the result of at least an effective scalar field. – Ron Maimon Mar 21 '12 at 0:43
Okay and thanks Anixx and Ron, but I'm really looking for a derivation from Quantum Field Theory, not using the cosmological constant. – metzgeer Mar 21 '12 at 2:26

The vacuum energy for a free field is the ground state energy of each field oscillator, ${1\over 2} \omega$, summed over all the modes. For a cubic periodic box of side-length L, you get

$$\sum_k {1\over 2} \sqrt{k^2+m^2}$$

Where the sum is all k's in an infinite size 3d cubic lattice where each k component is an integer multiple of $2\pi\over L$. When you make L big, this makes the k-lattice continuous, and the sum turns into an integral:

$$({L\over 2\pi})^3 \int \sqrt{k^2+m^2} d^3k$$

If you put a cutoff $\Lambda$, the result diverges as

$$E\propto V \Lambda^4$$

so that the energy density is proportional to the fourth power of the momentum cutoff. This integral reproduces the dimensional expectation when $\Lambda$ is the Planck length.

For interacting field theories, the vacuum energy is the sum of all vacuum loop Feynman diagrams. In the free case, the loop is just a single propagator joined to itself (this is a very degenerate Feynman diagram). The sign of Fermion and Boson loops are opposite, and Fermionic oscillators with the most natural definition of energy give an opposite sign vacuum energy in each oscillator. In a supersymmetric theory, when the Hamiltonian is presented in the form that preserves supersymmetry, the vacuum energy is zero. This is the only principle that we know today that can control the cosmological constant.

the problem is that SUSY is broken in our world at a cutoff scale of about a Tev, so the cancellations in SUSY are not exact. This means that the residual non-SUSY vacuum energy has to cancel from the scale of the Higgs (at least) to the scale of the observed cosmological constant, which is many orders of magnitude smaller.

You can't just get rid of vacuum energy by a natural statement that the vacuum has zero energy, because the vacuum in QCD (and in the Higgs mechanism) is full of crud. There is a pion condensate, a gluon condensate, and a Higgs condesate at the very least, and if you make everything cancel for our exact values of the masses of the quarks and leptons, if you change the mass of the quark, the condensate energy density changes in extremely complicated ways, so that the subtraction constant must be tuned to a magical value with no dynamical explanation.

Weinberg suggested that this is an anthropic accident--- that we need to have a low cosmological constant to evolve intelligent life. This predicts that the cosmological constant should be of the exact same order as the density of matter today, after life has evolved, but no lower, since it doesn't need to be any lower. This is what is observed, so Weinberg might be right, and there might be no explanation for the cosmological constant.

If Weinberg is right, the string vacuum that describes our universe will be very special--- it will be a non-supersymmetric vacuum with an accidentally small cosmological constant. If this is an accident with no rhyme or reason, then it will be very useful in picking out the right vacuum. We'll know we have it when it produces the right cosmological constant.

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This is very close to what I'm looking for, it's just the actual calculations that lead to such high values as 10^113 Jm^-3 would be interesting, but if no one else can improve on this, then this is all I want. None of the references on the wiki page offer much in the way of an extensive answer. Thanks Ron. – metzgeer Mar 22 '12 at 3:40
@metzgeer: The "actual calculations" are exactly what I did--- if you plug in $\Lambda=10^{19} GeV$, you get the order of magnitude. Since it is so absurd, nobody bothers to do any more elaborate calculations. The value of $\Lambda$ is also sometimes taken to be $1 TeV$, this assumes that SUSY cancels vacuum energy contributions past this point. – Ron Maimon Mar 22 '12 at 3:43
Ah... in that case all kudos to the bunny of love, question answered. Thanks. – metzgeer Mar 22 '12 at 5:46
maybe some unaccounted fields have negative contributions that are not accounted in our positive energy theories – diffeomorphism Dec 18 '13 at 3:05