Vacuum energy of particle antiparticle pairs may be responsible for the value of cosmological constant?

In short, No. There is an enormous discrepancy. However, this did not strictly disprove the idea that vacuum energy contributes to the cosmological constant. It showed that the simplest quantum-field-theory estimate is catastrophically wrong. This is the cosmological constant problem.

1. Observed cosmological constant

Using the Planck ΛCDM values$$H_0=67.4\ \mathrm{km\,s^{-1}\,Mpc^{-1}}, \qquad \Omega_\Lambda \approx 0.685,$$

the cosmological constant is$$\Lambda=\frac{3\Omega_\Lambda H_0^2}{c^2}.$$

Converting the Hubble constant:$$H_0=2.184\times10^{-18}\ \mathrm{s^{-1}},$$

gives$$\boxed{\Lambda\approx1.09\times10^{-52}\ \mathrm{m^{-2}}}.$$

That is the geometrical value appearing in Einstein’s equation. The corresponding vacuum-energy density is$$\rho_\Lambda^{(E)} =\frac{\Lambda c^4}{8\pi G} =\Omega_\Lambda\frac{3H_0^2c^2}{8\pi G}.$$

Therefore,$$\boxed{\rho_\Lambda^{(E)} \approx5.25\times10^{-10}\ \mathrm{J\,m^{-3}}}.$$

As an equivalent mass density:$$\boxed{\rho_\Lambda^{(m)} =\frac{\rho_\Lambda^{(E)}}{c^2} \approx5.85\times10^{-27}\ \mathrm{kg\,m^{-3}}}.$$

That is only about 3.5 proton masses per cubic metre. Its characteristic particle-physics energy scale is$$\left[\rho_\Lambda(\hbar c)^3\right]^{1/4} \approx2.24\times10^{-3}\ \mathrm{eV}.$$

So dark energy corresponds to an extraordinarily small energy scale of approximately$$\boxed{2.2\ \mathrm{meV}}.$$

These numbers use the Planck Collaboration’s ΛCDM parameters and standard constants from NIST. Planck cosmological parameters, NIST physical constants

2. Quantum-field vacuum-energy estimate

In quantum field theory, every field mode behaves like a harmonic oscillator with zero-point energy$$E_0=\frac12\hbar\omega.$$

For one massless bosonic degree of freedom, summing all modes up to a maximum wave number $k_{\max}$ gives$$\rho_{\rm vac} = \int_0^{k_{\max}} \frac{d^3k}{(2\pi)^3} \frac12\hbar ck.$$

Evaluating the integral:$$\rho_{\rm vac} = \frac{\hbar c}{16\pi^2}k_{\max}^4.$$

The natural extreme cutoff is the Planck scale:$$k_{\max}=\frac{1}{\ell_P}, \qquad \ell_P=\sqrt{\frac{\hbar G}{c^3}} \approx1.616\times10^{-35}\ \mathrm m.$$

Therefore,$$\rho_{\rm vac}^{\rm QFT} = \frac{\hbar c}{16\pi^2\ell_P^4} = \frac{c^7}{16\pi^2\hbar G^2},$$

which gives$$\boxed{\rho_{\rm vac}^{\rm QFT} \approx2.93\times10^{111}\ \mathrm{J\,m^{-3}}}.$$

3. How far apart are they?

Compare this with the observed value:$$\frac{\rho_{\rm vac}^{\rm QFT}} {\rho_\Lambda^{(E)}} = \frac{2.93\times10^{111}} {5.25\times10^{-10}} \approx5.58\times10^{120}.$$

Thus,$$\boxed{\rho_{\rm vac}^{\rm QFT} \text{ is approximately }5.6\times10^{120} \text{ times too large}.}$$

That is a discrepancy of approximately$$\boxed{121\text{ orders of magnitude}.}$$

Using the dimensional “one Planck energy per Planck volume” estimate gives$$\rho_P=\frac{c^7}{\hbar G^2} \approx4.63\times10^{113}\ \mathrm{J\,m^{-3}},$$

and therefore$$\frac{\rho_P}{\rho_\Lambda} \approx8.8\times10^{122},$$

or approximately 123 orders of magnitude. This is why the problem is commonly described as a $10^{120}$-fold discrepancy; the exact exponent depends on the cutoff convention and numerical factors.

Quantity	Energy density
Observed dark-energy density	$5.25\times10^{-10}\ \mathrm{J\,m^{-3}}$
One-field Planck-cutoff estimate	$2.93\times10^{111}\ \mathrm{J\,m^{-3}}$
Planck energy per Planck volume	$4.63\times10^{113}\ \mathrm{J\,m^{-3}}$
Discrepancy	$10^{121}$–$10^{123}$

What the discrepancy actually means

The popular description of “particle–antiparticle pairs constantly appearing in empty space” is only a heuristic picture. The calculation is really a sum of the zero-point energies of quantum fields.

Bosonic and fermionic fields contribute with opposite signs, so some cancellation can occur. But known physics provides no mechanism that cancels these contributions to roughly 120 decimal places while leaving the tiny positive remainder we observe.

In Einstein’s equation, only the total effective value matters:$$\rho_{\Lambda,\rm observed} = \rho_{\Lambda,\rm bare} + \rho_{\rm vacuum}^{\rm quantum} + \rho_{\rm phase\ transitions} +\cdots$$

The mystery is why these individually enormous terms apparently cancel to produce$$5\times10^{-10}\ \mathrm{J\,m^{-3}}.$$

So the conclusion is not that vacuum energy has been ruled out. It is that our present understanding of how quantum vacuum energy gravitates is profoundly incomplete.