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In a general QFT we say that the vacuum state $| \Omega \rangle$ is a state that is invariant under the poincare action, that is $U(\Lambda , a) | \Omega \rangle = |\Omega \rangle$ (wightmann axioms).

In particular consider just an infinitesimal spacetime translation (no boost) then $U( \mathbb{I} , a )| \Omega \rangle = \big(\mathbb{I} - ia.P + \mathcal{O}(a^2) \space\big)| \Omega \rangle = | \Omega \rangle $ $\Rightarrow$ $P^{\mu} |\Omega \rangle = 0$.

However, often the idea of a zero point energy is discussed, we say quantum fields fluctuate even in a vacuum because the heisenberg uncertainity principle says the 'fields can not stay still' and this yields a non-zero vacuum energy density.

Also by considering the zero point energy of our fields we predict the casimir effect which has been expeirmentally verified.

My question is, how do these two ideas relate? On the face of it, it seems contradictory. We must have the vacuum with zero energy else the theory would violate the principle of special relativity, yet it is often said an interacting QFT holds a zero point energy.

ColourConfined
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3 Answers3

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There is no non-zero vacuum energy density (at least in Wightman QFT), it is a myth. All attempts to extract physics from this myth have failed:

  • Measured Casimir force very likely has nothing to do with vacuum fluctuations, the explanation that involves vacuum fluctuations is one that works with an effective QFT model that isn't really the fundamental QFT under consideration.
  • Cosmological constant is predicted by some 120 orders of magnitude off.

Take free QFT for example. It is only well-defined mathematically on the Fock space when energy and momentum operators are normal-ordered:

$$ E = \sum_i \omega_i a_i^{\dagger} a_i. $$

If you add the metaplectic correction $1/2 \sum_i$ of ground state energies of the oscillator, you will get an infinite value which obviously doesn't make sense.

Actually, you can add a finite constant term to $E$ that corresponds to some ordering ambiguity, but it is always possible to redefine the states such that its value is zero (and hence the vacuum is Poincare-invariant) without changing any of the real predictions of the theory. The same wouldn't be possible if the ordering term was infinite (and perhaps cut off at high energy, like the vacuum energy fairy tales make you believe), because such an operator wouldn't act on the Fock space.

Prof. Legolasov
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    Great answer! I will just add that it is not possible to add a constant $E$ in supersymmetric or gravitational theories. In SUSY theories, the Hamiltonian is the square of the SUSY generator so its "constant value" is fixed by the algebra (it's zero, as you mentioned). In gravitational adding such a term corresponds to adding a cosmological constant so it definitely has a measurable effect. – Prahar Feb 07 '22 at 08:46
  • Is it currently a definite consensus in the physics community that the Casimir effect is not due to vacuum fluctuations? (Of course, it is not to say that there any fluctuations when there is nothing around, which would be consistent with $P^\mu | \Omega \rangle = 0$.) – Mahir Lokvancic Feb 08 '22 at 23:06
  • In the 20th century, we learned that the kinematic symmetry group for non-relativistic theory is not the Galilei group, but the Bargmann group, which has an 11th generator. This requires upgrading Relativity, too, to continue maintaining the Correspondence Principle; thus leading to the conclusion that the symmetry group for relativity is not the Poincaré, but the extended Poincaré group. The 11th generator, gives you extra room for a "vacuum energy". Redo Wightman with the extended Poincaré group, and it will be there. Note the diagrams 8.3.1 and 8.3.2 in https://arxiv.org/abs/2104.02627 – NinjaDarth Oct 06 '23 at 04:19
  • @NinjaDarth “ Redo Wightman with the extended Poincaré group, and it will be there. “ no, I don’t think you could do that consistently. Bargmann is a trivial extension of Poincaré, which means a Bargmann QFT necessarily is also Poincaré-invariant, so a regular Wightman QFT, with extra structure – Prof. Legolasov Oct 06 '23 at 06:03
  • First of all, the Bargmann group is not an extension of Poincaré at all, but of the Galilei group. You have your terms mixed up. Second, the extension of Poincaré gives you the very "extra structure" we're talking about - an extra generator for vacuum energy! So, when you re-do the Wightmann formalism, but with the extended Poincaré, instead of Poincaré, then you have extra place for vacuum energy. – NinjaDarth Oct 06 '23 at 16:10
  • @NinjaDarth I won’t argue about terminology, but I don’t see how adding an extra $u(1)$ central charge gives you vacuum energy. Energy is a spectral property of the time translation generator, which commutes with your $u(1)$ charge. – Prof. Legolasov Oct 07 '23 at 01:23
  • It bears closer examination. It's not just a matter of throwing in the extra generator and saying "Scalar. Done." The real issue is to do this in a way such that the interpretations respects the correspondence limit. We can take closer look and run through this in some detail. Here are the axioms. https://ncatlab.org/nlab/show/Wightman+axioms At the very least Axioms 2.1, 2.2, 2.3 and 2.5 will be affected; particularly 2.2. What corresponds to "energy operator" in 2.2. becomes the "relativistic mass operator", while time translation goes with "kinetic energy", which would be distinct. – NinjaDarth Oct 09 '23 at 00:05
  • Following up: call the "energy" of Wightman $E$. It now becomes the relativistic mass $M = E/c^2$, while the time translation goes with kinetic energy, denoted $H$. The central charge appear as $μ = M - H/c^2$. The extra generator crops up as a "internal" value $U$ for $H$, with the rest mass $m = μ + U/c^2$. In the non-relativistic limit $H = ½mv^2 + U$. The restriction from extended Poincaré to ordinary Poincaré corresponds to setting $U = 0$. The Wigner classification is affected too. For the vacuum class, $U$ appears as an extra energy term. Coordinate reps change, too, which affects 2.5. – NinjaDarth Oct 09 '23 at 00:12
  • @NinjaDarth wouldn’t (2.5) change by simply replacing Poincaré with the extended group/algebra; if that’s the case, it still contains Poincaré as a subalgebra, correct? If I understand you correctly, the time translation generator doesn’t correspond to energy in the non relativistic limit anymore due to the central charge appearing there, so basically you’ve redefined “energy”. – Prof. Legolasov Oct 09 '23 at 11:33
  • @NinjaDarth even then, I have some doubts that this program can be consistently done. Non-zero evergy density means infinite Hamiltonian (due to a trivial infrared divergence), I admit I haven’t heard of this extended Poincaré approach to vacuum energy before, but my experience with constructive QFT leaves me very sceptical. Do you have references to where this approach was applied to QFT? (I don’t just mean modifying the axioms, I mean an actual physical model that satisfies the modified axioms) – Prof. Legolasov Oct 09 '23 at 11:37
  • Ok, well this is starting to get a little serious. We might need to take this to chat; and give me a little time to wrap up a project (a publication) to get a clear mind. Are there references pertaining to things related to Bargmann? Yes. We can talk about it there. Has anyone actually tried doing the surgery on Wightman? Not yet. That's still unexplored territory. In the meanwhile, you might want to carefully go through section 4 of this link with a fine-toothed comb, keeping in mind the issues raised by our conversation. https://philsci-archive.pitt.edu/2673/1/earmanfraserfinalrevd.pdf – NinjaDarth Oct 12 '23 at 23:22
  • Also, the other thing I (indirectly) alluded to we can also discuss: the lifting of Galilei to Bargmann added a central charge, which when shifting over to extended Poincaré and then dropping down to Poincaré remained intact! Meanwhile, the 11th item dropped out from extended Poincaré was something present in unextended Galilei. So, there's "lift-shift-drop" and it's at a slant. An underpinning to Foldy-Wouthuysen. I demonstrate that vividly (in both directions) in my reply here. https://physics.stackexchange.com/questions/694598/issue-in-deriving-non-relativistic-dirac-equation/782473#782473 – NinjaDarth Oct 12 '23 at 23:28
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If the vacuum energy is non-zero then then, to maintain the Lorentz invariance of the vacuum state, the vacuum energy-momentum tensor will be of the form $T_{\mu\nu}=\lambda g_{\mu\nu}$. This means that in addition to positive energy density $E=\lambda g_{00}$there will be negative pressure $P= \lambda g_{11}=-\lambda$. As the effective source of gravity is $E+3P$, the negative $P$ wins out and leads to the expansion ("inflation") of space-time.

mike stone
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  • This looks like a cosmologist's answer, I think OP was looking more into QFT-ish explanations, but +1 :) – Prof. Legolasov Jan 31 '22 at 13:28
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    @Prof. Legoslov. Probably, but he seems worried about the apparant conflict between vacuum energy and Poincare invariance. That is not a big issue. The origin of "$E$" is a bg issue of course. – mike stone Jan 31 '22 at 13:30
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    OP is clearly interested in QFT on flat spacetime that isn't coupled to gravity, as a mathematical model. I like your answer because it is outside of the box, but I doubt it answers OP's question. – Prof. Legolasov Jan 31 '22 at 13:32
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Contrary to @mike stone's claim, the zero point energy is NOT Lorentz invariant. I would like to challenge anyone here at PSE to provide a specifically Lorentz invariant regularization scheme and calculate the Lorentz invariant zero point energy. See more details here.

MadMax
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  • @ MadMax. The vacuum bubble for both scalar and spin models is given by a manifestly Lorentz invariant expression $\propto \int d^Nk \ln (k^2+m^2)$ and can be regulated by Lorentz invariant Pauli-Villars . This gives a $\lambda g^{\mu\nu}$ for $T^{\mu\nu}$. – mike stone Feb 09 '22 at 13:05
  • @mikestone, instead of hand-waving, please give a concrete calculation of energy-momentum tensor $T^{\mu\nu}$ as the vacuum expectation of say a massless fermion: $\langle \Omega| i\hbar \bar{\psi}\gamma^\mu\partial^\nu\psi | \Omega \rangle$, as detailed as these papers: https://arxiv.org/abs/1610.08907 or https://arxiv.org/abs/1205.3365. – MadMax Feb 09 '22 at 14:32
  • Is this not exactly what is done in detail leading to eq 514 in arxiv.org/abs/1205.3365? They use zeta (proper time) regulator rather than Pauli-Villars, but there is still no problem with Lorentz invariance. I am confused.... – mike stone Feb 09 '22 at 15:38
  • See section IV A of arxiv.org/abs/1205.3365 where the issue with Lorentz invariance of the energy-momentum tensor is discussed. The point is that you have to calculate each component of the energy-momentum tensor explicitly, instead of looking at the effective action in general. – MadMax Feb 09 '22 at 16:54
  • They say that to get the Lorentz invariant $\langle \rho\rangle= -\langle p\rangle$ you must use a Lorentz invariant regulator. That I certainly agree with. Then $\langle T^{\mu\nu}\rangle \propto g^{\mu\nu}$, and is the derivative of $W$ with respect to the metric. – mike stone Feb 09 '22 at 17:08
  • The hard part is how to "use a Lorentz invariant regulator". Any body can begin the argument with "if I had a Lorentz invariant regulator". There are two very important points one has to keep in mind: 1) Zeta/Dimension regulation kills off non-logarithmic divergences, which is not applicable to calculating quantities such as the quartic divergent zero point energy. 1) Calculating the components of energy-momentum tensor and calculating the $W$ are different tasks, since the order of regulating divergences and taking derivative of $W$ DOES matter. This is at the center of the whole discussion. – MadMax Feb 09 '22 at 19:52
  • Let us continue this discussion in chat. – mike stone Feb 09 '22 at 20:42
  • I agree with MadMax. The ZPF are not Lorentz invariant. There is reason for that due contradiction with electro-weak interactions of charged particles via "virtual" photons: I am still not convinced that ZPF of vacuum can be regarded as Lorentz invariant. I mean we do notknow from where these came from? Could be at any energy state even superluminous before they enter our spacetime domain. Surely EM flux and interaction between particle charges is not done via "virtual" photons with "virtual" as you put it being formalism artifacts? Virtual particles are real. Math does not lie. – Markoul11 Feb 10 '23 at 09:22
  • IMO, virtual particles consisting electro-weak interaction are coherent distortions in a hidden from us in our known spacetime, form of unknown yet vacuum energy and ZPF is the incoherent noise of this untapped yet type of energy or field. – Markoul11 Feb 10 '23 at 09:30