8

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
  • 786
  • 1
  • 5
  • 20

3 Answers3

8

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
  • 16,527
4

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
  • 58,971
  • 4
  • 53
  • 158
0

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
  • 5,451
  • 13
  • 42