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People often say that the particles in ground state of interacting fields are virtual (for instance, from youtube video "What is 'Nothing'?" by Sabine Hossenfelder). However, from my understanding, this vacuum $|\Omega>\ \propto lim_{t\rightarrow \infty \ }e^{itH} |0>$ is different from the non-interactive vacuum $|0>$. I tend to think that $|\Omega>$ has non-zero expectation value in particle number of a specific particle type. If so, then isn't it wrong to call those particle virtual?

Bohan Xu
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  • The "particles in" a vacuum are virtual. Even if you write $H$ in terms of creation and annihilation operators, it is going to have them integrated over. All the particles are in loops/bubbles. – hft Sep 24 '22 at 20:55
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    If there is a non-zero expectation value of the particle number operator then we would not call the state a "vacuum." – hft Sep 24 '22 at 20:56
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    People are talking about virtual particles in bubble diagrams, not about expectation values of particle numbers when they say that, see e.g. this answer of mine – ACuriousMind Sep 24 '22 at 20:57
  • Nobody has ever seen a virtual particle in an experiment. What we are seeing in experiments are energy shifts and decay times. Virtual particles are a computational trick in perturbation theory that allows us to calculate these actual physical effects. One can find similar mathematical terms already in the perturbation theory for the Schroedinger equation. For all I know nobody calls those terms "virtual particles", even though they serve the exact same function. – FlatterMann Sep 24 '22 at 20:59
  • @hft Is the expectation value of the number operator of any field, zero under the vacuum of interacting fields? – Bohan Xu Sep 24 '22 at 21:09
  • @FlatterMann Is the expectation value of the number operator of any field, zero under the vacuum of interacting fields? – Bohan Xu Sep 24 '22 at 21:10
  • @BohanXu Does the number operator represent a physical quantity by counting the quantum density in the field? If it does and the ground state expectation value is >0, then you should be able to build a machine that can extract energy from the vacuum. If you can not build such a machine, then a ground state number operator expectation value >0 is obviously not a physical quantity. I am also not sure what you mean by "interacting fields". I am not aware of any physical way to make the known fields "non-interacting". That case only exists on paper but not in nature. – FlatterMann Sep 24 '22 at 21:28
  • @FlatterMann I don't know if there is some technical difficulty in observing particles in the vacuum because particles are massive compared to the strength of the interaction, so fluactuation should be small. This is why I focus my question on the math and measurability in principle, instead of actual experiment. I mention non-interacting vacuum because non-interacting vacuum has zero as the expectation value of number operators. Vacuum of interacting fields is different from the non-interacting vacuum, and I think it might have non-zero expectation value for number of operators. – Bohan Xu Sep 24 '22 at 21:38
  • @BohanXu There are no observable fluctuations in the physical vacuum. That is just another unfortunate meme that has established itself and that is simply not true. You can look at the night sky and see light from stars thousands of light years away. The CMB even comes from the other end of the observable universe. If there were any detectable quantum fluctuations over these distances, then this light would have long been scattered and the universe would look foggy. The actual effects of virtual particles are energy shifts and decay time constants, not fluctuations. – FlatterMann Sep 25 '22 at 00:41
  • @FlatterMann I see. – Bohan Xu Sep 27 '22 at 17:50

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