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The zero point energy is usually supposed to contribute to the cosmological constant. And the mismatch between the small cosmological constant compared with the huge zero point energy is deemed as one of the most serious problems in physics.

However, the zero point energy and the cosmological constant are totally different animals.

The energy-momentum tensor $T^{\mu\nu}_\Lambda$ of the cosmological constant is of the form: $$ T^{00}_\Lambda = \rho_\Lambda, $$ and $$ T^{11}_\Lambda = T^{22}_\Lambda = T^{33}_\Lambda= p_\Lambda $$ with $$ p_\Lambda = -\rho_\Lambda. $$

And what does the zero point energy look like? Let's take a massless fermion for example, the vacuum energy-momentum tensor $T^{\mu\nu}_F$ can be calculated as (see details on page 55 of Philip Mannheim's paper here): $$ T^{00}_F = \rho_F = -\frac{2\hbar}{(2\pi)^3}\int k d^3k, $$ and $$ T^{11}_F = T^{22}_F = T^{33}_F= p_F = -\frac{2\hbar}{3(2\pi)^3}\int k d^3k $$ Therefore, with a proper regularization/cutoff, one has $$ p_F = \frac{\rho_F}{3} $$ which is categorically different from the case of cosmological constant $p_\Lambda = -\rho_\Lambda$.

Hence there is no similarity between the zero point energy and the cosmological constant at all!

In cosmological nomenclature $$ p = w\rho $$ where $w$ is called equation of state parameter, which is $-1$ for the cosmological constant and $1/3$ (radiation-like) for the above massless fermion example.

Changing to massive fermion/boson will not help the case either. You would get a $w$ somewhere between mater-like ($w=0$) and radiation-like ($w=1/3$) depending on the energy-momentum level, but you would not get a cosmological-constant-like ($w=-1$) equation of state parameter. Interested readers are encouraged to verify independently.

Some readers may challenge the above calculation, specifically about the Lorentz invariance of the regularization. Well I would like to challenge anyone to provide a specifically Lorentz invariant regularization and prove that equation of state parameter $w$ is the same between cosmological constant and zero point energy of massless mater.

To corroborate the notion let's quote Jerome Martin's paper (page 12) Everything You Always Wanted To Know About The Cosmological Constant Problem (But Were Afraid To Ask);

It is clear from the previous expressions that $p/\rho \neq -1$ which indicates that the stress energy tensor is not of the form ∝ $-\rho g_{\mu\nu}$. In the limit m → 0, as can be easily shown from Eqs. (75) and (78), the equation of state is in fact $p/\rho = 1/3$. This would mean that the zero point fluctuations do not behave like a cosmological constant but rather like radiation.

The paper goes on to discuss how to fix this with dimensional regularization. But the dimensional regularization usually kills off non-logarithmic divergences and the divergent integral in hand is quartical, so I am not particularly convinced.


Added note.

Note that the Higgs potential : $$ V_{H} \sim (-m_H^2 |\phi|^2 + \lambda |\phi|^4) $$ may indeed contribute to cosmological constant since it has a equation of state parameter $w=-1$, should the Higgs field $\phi$ develop a non-zero VEV upon spontaneous symmetry breaking. Mind you that the Higgs potential contribution is a separate story from the zero point energy ppl usually talk about.

MadMax
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1 Answers1

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I think you and your sources are right, in the sense calculations of $T_{\mu\nu}$ for zero point fluctuations of a field have trouble to recover anything resembling the cosmological term.

Introducing frequency cutoff means we have a preferred frame, so energy density isn't Lorentz invariant. Such cut fluctuations can't have the form of the cosmological term.

If we allow the integration to run to infinity, then we get something that formally looks Lorentz invariant. For example, zero point fluctuations of EM field have spectral function $k\omega^3$ and its value at any frequency is Lorentz invariant.

For such infinite fluctuations all diagonal entries of $T$ have the same magnitude... Perhaps this gives some motivation to the idea that cosmological term is due to vacuum fluctuations.

But they are infinite! Clearly, explaining the cosmological term in terms of vacuum fluctuations is not very functional mathematically, and is completely inconsistent with observations (which imply zero or small cosmological term).

I think it better to lay this idea that zero point fluctuations have real big positive energy to rest. There is no known phenomenon which requires existence of big positive energy of zero point fluctuations. Even the Casimir effect turned out to be better explained in terms of interaction forces between material bodies (retarded EM forces, sometimes called London forces). Besides the classical Jaffe paper, I like how Grundler explains this. [1],[2] and regarding the stress-energy tensor not having the form of the cosmological term, see [3].

[1] Grundler G., The Casimir-Effect: No Manifestation of Zero-Point Energy, https://arxiv.org/abs/1303.3790

[2] Grundler G., What the Casimir-Effect really is telling about Zero-Point Energy, https://www.astrophys-neunhof.de/mtlg/se93327.pdf

[3] Grundler G., Canonical quantization of elementary fields, https://arxiv.org/abs/1506.08647

One possibility which I sometimes wonder about is that maybe zero point fluctuations of EM field do exist, like postulated in Stochastic Electrodynamics, in the sense $\langle E^2\rangle$ is non-zero even at $T=0K$ and has the above mentioned spectrum. But perhaps Poynting formula does not correctly give EM energy on the microscale, so the field does not need to have substantial positive energy to fluctuate and there is no UV catastrophe. If that was so, we could have zero-point fluctuations with the Lorentz invariant spectrum, keep the SED results and there would be no associated energy/cosmological term problem.

Urb
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