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My QFT lecturer said:

  1. Particle number is fixed in QM (I understand this.)
  2. Particle number can vary in relativistic QFT, but not in non-relativistic QFT. (and he said '$E=mc^2$ is at the root of this)

Why can't particle number vary in non-relativistic QFT? Why can't you use a non-relativistic Lagrangian with coupling terms that cause particle number to not be conserved?

I got confused by this question: When particle number can change in quantum physics?.

Is the reason just that the energy scales needed to create a new particle are typically too large unless you account for $E=mc^2$ (what I think the above question suggestions), or is there a more rigerous reason, or is it not the case at all?

Obviously quasi-particle numbers can vary in non-relativistic QFT, but what about actual particles? My assumption is that the QFT framework facilitates varying particle number, regardless of the Lagrangian.

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Usually when we say that e.g. an electron is assumed non-relativistic, we mean that it has a velocity $v \ll c$.

If "non-relativistic QFT" means QFT for energies much smaller than $m c^2$, then there are no physical mechanisms allowing the creating or annihilation of physical fundamental particles.

Why can't you use a non-relativistic Lagrangian with coupling terms that cause particle number to not be conserved?

I suppose you can, it would just not be a physical Lagrangian.

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    $\begingroup$ Okay so its not something intrinsic (somehow) in the framework of a non-relativistic QFT. Its just that when you use the correct (physical) Lagrangian and take the non-relativistic limit, the energies are (by definition of non-relativistic) too small to create new particles? $\endgroup$
    – Alex Gower
    Commented Dec 16, 2021 at 14:59
  • $\begingroup$ I would say so. As you say, you can always write up terms which are non-conserving, but they just don't describe anything physical. I might suggest asking your lecturer for clarification on what exactly they meant, if that is possible. $\endgroup$
    – Codename 47
    Commented Dec 16, 2021 at 15:01
  • $\begingroup$ Yeah you're answer makes perfect sense. I think I assumed he meant that if you don't include an $mc^2$ term in the Lagrangian, somehow particle nunber must be conserved in the framework. But I think he really meant that, for physical Lagrangians, you obviously need sufficient energy to create new particles $\endgroup$
    – Alex Gower
    Commented Dec 16, 2021 at 15:02
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    $\begingroup$ What about the "creation" of phonons? $\endgroup$
    – Quantumwhisp
    Commented Dec 16, 2021 at 15:10
  • $\begingroup$ @Quantumwhisp Phonons are quasiparticles, not fundamental particles. $\endgroup$
    – Codename 47
    Commented Dec 16, 2021 at 23:18

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