The following is take directly from this website: http://www.astro.ucla.edu/~wright/cosmo_constant.html

Vacuum Energy Density, or How Can Nothing Weigh Something?

Recently two different groups have measured the apparent brightness of supernovae with redshifts near z = 1. Based on this data the old idea of a cosmological constant is making a comeback.

Einstein Static Cosmology

Einstein's original cosmological model was a static, homogeneous model with spherical geometry. The gravitational effect of matter caused an acceleration in this model which Einstein did not want, since at the time the Universe was not known to be expanding. Thus Einstein introduced a cosmological constant into his equations for General Relativity. This term acts to counteract the gravitational pull of matter, and so it has been described as an anti-gravity effect.

Why does the cosmological constant behave this way?

This term acts like a vacuum energy density, an idea which has become quite fashionable in high energy particle physics models since a vacuum energy density of a specific kind is used in the Higgs mechanism for spontaneous symmetry breaking. Indeed, the inflationary scenario for the first picosecond after the Big Bang proposes that a fairly large vacuum energy density existed during the inflationary epoch. The vacuum energy density must be associated with a negative pressure because:

The vacuum energy density must be constant because there is nothing for it to depend on.
If a piston capping a cylinder of vacuum is pulled out, producing more vacuum, the vacuum within the cylinder then has more energy which must have been supplied by a force pulling on the piston.
If the vacuum is trying to pull the piston back into the cylinder, it must have a negative pressure, since a positive pressure would tend to push the piston out.

The magnitude of the negative pressure needed for energy conservation is easily found to be P = -u = -rho*c² where P is the pressure, u is the vacuum energy density, and rho is the equivalent mass density using E = m*c².

But in General Relativity, pressure has weight, which means that the gravitational acceleration at the edge of a uniform density sphere is not given by

g = GM/R² = (4*pi/3)*G*rho*R

but is rather given by

g = (4*pi/3)*G*(rho+3P/c²)*R

Now Einstein wanted a static model, which means that g = 0, but he also wanted to have some matter, so rho > 0, and thus he needed P < 0. In fact, by setting

rho(vacuum) = 0.5*rho(matter)

he had a total density of 1.5*rho(matter) and a total pressure of -0.5*rho(matter)*c² since the pressure from ordinary matter is essentially zero (compared to rho*c²). Thus rho+3P/c² = 0 and the gravitational acceleration was zero,

g = (4*pi/3)*G*(rho(matter)-2*rho(vacuum))*R = 0

allowing a static Universe.

Einstein's Greatest Blunder

However, there is a basic flaw in this Einstein static model: it is unstable - like a pencil balanced on its point. For imagine that the Universe grew slightly: say by 1 part per million in size. Then the vacuum energy density stays the same, but the matter energy density goes down by 3 parts per million. This gives a net negative gravitational acceleration, which makes the Universe grow even more! If instead the Universe shrank slightly, one gets a net positive gravitational acceleration, which makes it shrink more! Any small deviation gets magnified, and the model is fundamentally flawed.

In addition to this flaw of instability, the static model's premise of a static Universe was shown by Hubble to be incorrect. This led Einstein to refer to the cosmological constant as his greatest blunder, and to drop it from his equations. But it still exists as a possibility -- a coefficient that should be determined from observations or fundamental theory.

The Quantum Expectation

The equations of quantum field theory describing interacting particles and anti-particles of mass M are very hard to solve exactly. With a large amount of mathematical work it is possible to prove that the ground state of this system has an energy that is less than infinity. But there is no obvious reason why the energy of this ground state should be zero. One expects roughly one particle in every volume equal to the Compton wavelength of the particle cubed, which gives a vacuum density of

rho(vacuum) = M⁴c³/h³ = 10¹³ [M/proton mass]⁴ gm/cc

For the highest reasonable elementary particle mass, the Planck mass of 20 micrograms, this density is more than 10⁹¹ gm/cc. So there must be a suppression mechanism at work now that reduces the vacuum energy density by at least 120 orders of magnitude.

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