The first version is the usual one we're taught. But there's this other version too : A quantum non-relativistic field theory.

Take a non-relativistic classical field, like the non-relativistic limit of the Klein Gordon field. And then quantise this field.

You end up with a theory that, unlike the usual NR quantum mechanics, allows for a probabilistic particle number. We have a Fock space in this theory.

If our universe were Galilean, would it follow the usual Non Relativistic QM, or this Quantum Non Relativistic Field theory?

OR MAYBE Both these theories are identical in practice. The second theory does have a Fock space. But I think most of the vectors in that Fock space are useless in practice.

Because to actually create or annihilate a particle, you'd need infinite energy in this theory by $E=mc^2$, $c=\infty$. So this theory does have a Fock space, but the interactions don't actually allow for particle creation and annihilation. So we're left with the usual Hilbert space of a fixed number of particles, identical to the ordinary NR Quantum Mechanics. Is this reasoning correct?

  • 1
    $\begingroup$ AFAIK quantum field theories of all types have to be based on the plane wave solutions of the corresponding equations of the particle fields in order to be able to calcualte numbers to compare with data. $\endgroup$
    – anna v
    May 29, 2022 at 6:22
  • $\begingroup$ @annav Are you saying the second theory would predict the wrong data? I'm reading David Tong's notes, and there they derive that the second theory predicts the same data as the usual Schrodinger equation (in the free particle case). But the notes don't say anything about this theory serving as a replacement for the usual Non relativistic QM. $\endgroup$
    – Ryder Rude
    May 29, 2022 at 7:49
  • $\begingroup$ as they must be discussing validated theories, I mean that any theory must be based on the same basic quantum mechanics and the "new and different" will be only the form of mathematics. The way you can have a simple function also described by an expansion of a series, except in more complicated ways. $\endgroup$
    – anna v
    May 29, 2022 at 8:15
  • 5
    $\begingroup$ These two are equivalent ways of writing the same thing. In this context they're generally called the "first quantized" and "second quantized" versions. $\endgroup$
    – knzhou
    Jun 14, 2022 at 3:19
  • 5
    $\begingroup$ There's no problem with indefinite particle number in a nonrelativistic theory... in condensed matter physics they create and destroy quasiparticles all the time. There is also no problem with indefinite particle number in a wavefunction formalism -- your wavefunction is just a direct sum of zero-particle, one-particle, two-particle, etc. wavefunctions. $\endgroup$
    – knzhou
    Jun 14, 2022 at 5:33

2 Answers 2


Short answer: There is only one theory, but there are different ways to analyze the theory, some of which may be more convenient in practice.

  1. Consider a Galilean invariant quantum field theory. As usual, you can analyze the theory using a Hamiltonian or Lagrangian formalism.

  2. The theory has a conserved particle number (this is a statement about representations of the Galilean group), and in both the Lagrangian and Hamiltonian formulation we can couple a chemical potential and consider the partition function $Z(\mu)$ or $Z(\mu,T)$.

  3. We can also consider the theory in the $N$-particle sector. In the Hamiltonian formulation, $H$ is block-diagonal in $N$, and diagonalizing $H$ in one of these sub-spaces is just ordinary $N$-particle QM. In the Lagrangian formulation we can compute the $N$-particle Green function, and because there is no particle creation the $N$-particle Green function decouples from the $N+1,N-1$ particle Green fcts etc. Summing all $N$-particle diagrams (by solving a Bethe-Salpeter equation, for example) is equivalent to solving the $N$-particle Schroedinger equation.

  4. The Hilbert space of the Hamiltonian theory can be constructed as a Fock space, that is as a tensor product of single particle states (with some symmetrization requirements). If the Hamiltonian only contains local fields with conserved charges interacting via potentials that are local in time (but may be non-local in space) then the Hamiltonian in the $N$ particle sector can be written as $H(x_1,\ldots,x_N)$. However, I can also consider a local field $\psi(x,t)$ with a conserved particle number interacting via non-relativistic fields $\phi(x,t)$ (phonons, pions, etc), and then the representations of the $N$ particle Hamiltonian may be more complicated.

  • $\begingroup$ "Short" as it may be, this is a densely information-packed answer, making very clear the difference between the relativistic and non-relativistic approach. Is a non-conservation of particles a statement about the non-Galilean group (Poincaré or Lorentz)? $\endgroup$ Jun 16, 2022 at 15:41

I think you're getting tripped up by a common use of sloppy language by physicists, which is implying that the phrase "quantum field theory" is just shorthand for "relativistic QFT". But you can certainly have non-relativistic QFTs (which can have indefinite particle number) that work fine. People use these all the time in condensed matter theory. Search for "dynamical exponent"; loosely speaking, a dynamical exponent of $z=1$ for the gapless excitations means that a theory is (Lorentz-)relativistic, while any other value of $z$ means that it's non-relativistic.

Sometimes, a non-Lorentz-relativistic theory displays emergent Lorentz symmetry in its gapless excitations, so the effective QFT that describes those excitations is (Lorentz-)relativistic even though the underlying theory isn't. But this certainly isn't always the case.

(A related sloppy claim that physicists often make is that "all massless field theories are conformal". This is incorrect: massless field theories are generically non-conformal, unless you also assume Lorentz invariance.)

If our universe were Galilean-invariant, then it could still be described by non-(Lorentz-)relativistic QFT, although it's also possible that the emergent effective field theory would turn out to be Lorentz-invariant. That would depend on the details of the microscopic physics.

Indeed, some people have even proposed that at the ultra-microscopic level, far below the length scales that our particle accelerators can probe - e.g. the Planck scale - our universe actually is non-relativistic, and its apparent Lorentz invariance is just an approximate symmetry that emerges at the (relatively) "large" length scales that we can access. So your hypothetical may not be counterfactual after all.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.