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.


3 Answers 3


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Prahar
    Commented Feb 7, 2022 at 8:46
  • $\begingroup$ 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 $P^\mu | \Omega \rangle = 0$.) $\endgroup$ Commented Feb 8, 2022 at 23:06
  • $\begingroup$ 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 arxiv.org/abs/2104.02627 $\endgroup$
    – NinjaDarth
    Commented Oct 6, 2023 at 4:19
  • $\begingroup$ @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 $\endgroup$ Commented Oct 6, 2023 at 6:03
  • $\begingroup$ 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. $\endgroup$
    – NinjaDarth
    Commented Oct 6, 2023 at 16:10

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.

  • $\begingroup$ This looks like a cosmologist's answer, I think OP was looking more into QFT-ish explanations, but +1 :) $\endgroup$ Commented Jan 31, 2022 at 13:28
  • 1
    $\begingroup$ @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. $\endgroup$
    – mike stone
    Commented Jan 31, 2022 at 13:30
  • 3
    $\begingroup$ 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. $\endgroup$ Commented Jan 31, 2022 at 13:32

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.

  • $\begingroup$ @ 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 $T^{\mu\nu}$. $\endgroup$
    – mike stone
    Commented Feb 9, 2022 at 13:05
  • $\begingroup$ @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: arxiv.org/abs/1610.08907 or arxiv.org/abs/1205.3365. $\endgroup$
    – MadMax
    Commented Feb 9, 2022 at 14:32
  • $\begingroup$ 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.... $\endgroup$
    – mike stone
    Commented Feb 9, 2022 at 15:38
  • $\begingroup$ 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. $\endgroup$
    – MadMax
    Commented Feb 9, 2022 at 16:54
  • $\begingroup$ 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. $\endgroup$
    – mike stone
    Commented Feb 9, 2022 at 17:08

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