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Let me write a paragraph from D.Tong lecture notes on QFT-chapter2 when he is talking about non-relativistic limit of scalar quantum field theory :

A related fact is that the conserved charge $Q=\int{d^3x : \psi^\dagger \psi:}$ is the particle number. This remains conserved even if we include interactions in the Lagrangian of the form:

$L = V (\psi^\star \psi)$

So in non-relativistic theories, particle number is conserved. It is only with relativity, and the appearance of anti-particles, that particle number can change.

My question is what forbids us from adding a term like $\psi^4$ or $\psi^3$ to the Lagrangian of this non-relativistic scalar field theory? Because it seems that such term can change the particle number.

In the old version of question I had guessed that the thing that forbids us is Galilean invariance (or classical nature of theory. Is it right?

Update: the question is reformulated (butchered!) completely.

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  • $\begingroup$ For interacting particles, states with definite particle number aren't energy eigenstates. This means that there exist macroscopic systems, such as the photon gas in a laser or the atoms in a Bose-Einstein condensate, where "particle number" isn't a good quantum number for the system. $\endgroup$ – rob Jul 3 '16 at 11:59
  • $\begingroup$ So non-relativistic quantum mechanics can handle variable particle number? Seems strange. I don't think it is possible. $\endgroup$ – user121238 Jul 3 '16 at 12:08
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    $\begingroup$ Second quantization in non-relativistic quantum mechanics is just a device for doing calculations in a more convenient manner as mentioned in many body texts. But the second quantization in quantum field theory is really a new physical idea. $\endgroup$ – user121238 Jul 3 '16 at 12:25
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    $\begingroup$ For real particles to be "born", one needs the sufficient energy and this implies a relativistic consideration. But often we deal with quasi-particles whose occupation numbers change and the total number of quasi-particles is not conserved. Phonons, for example. $\endgroup$ – Vladimir Kalitvianski Jul 3 '16 at 14:30
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    $\begingroup$ That charge number is conserved in the non relativistic regime is an observational fact. Any lagrangian that aims to describe the data has to have this property by choice of construction. $\endgroup$ – anna v Jul 11 '16 at 7:14
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Whether or not you're allowed to include non-particle-number-conserving terms in your Hamiltonian has nothing to do (at the mathematical level) with whether your system is relativistic or not - it has to do with what Hilbert space you're using. If your Hilbert space takes the form $\mathcal{H} = \otimes_{i=1}^n \mathcal{H}_1$, where $H_1$ is a single-body Hilbert space (e.g. the space of $L^2$-norm complex functions on $\mathbb{R}^d$), then it can only describe states with exactly $n$ particles. Therefore a term like $a^4$ isn't even a linear operator on this Hilbert space at all (because it takes you out of the Hilbert space) so it wouldn't make any sense to include it in your Hamiltonian, which must be a linear operator on the Hilbert space.

On the other hand, if your Hilbert space, say, takes the form $\mathcal{H} = \oplus_{n=0}^\infty (\otimes_{i=1}^n \mathcal{H}_1)$, then it contains different "sectors," each of which consists of a definite number of particles. On this Hilbert space, a term like $aaaa$ makes perfect sense - it takes a state in the $n$-particle sector to the $n-4$-particle sector. All this is equally true whether or not your system is relativistic.

That's all math, now here's the physics. Fundamental particles with mass $m$ can only be created or destroyed in processes involving energies greater that $m c^2$ - scales that are common in relativistic situations. So empirically, elementary particle number changes all the time in high-energy situations. So a fixed-particle-number Hilbert space just isn't powerful enough to accurately describe high-energy physics.

On the other hand, if you're working with an $n$-body nonrelativistic system where $n$ is reasonably small (like, say, five), then you can describe the system using the usual many-body Hilbert space $(\mathbb{R}^{d})^n$. Since the system is nonrelativistic, the particle number won't change, so you can get away with just using a fixed-particle-number Hilbert space.

If you want to tackle a truly many-body nonrelativistic condensed-matter system, where $n \sim 10^{23}$, then particle number will be conserved but the Hilbert space will be completely intractably gigantic. So in practice you restrict yourself to the lowest few excited states which only have a few quasiparticles, and you work in the many-quasiparticle Hilbert space (where now "many" means "more than one, but not a huge number"). However, empirically, quasiparticle number can change in many condensed-matter systems (most notably superconductors), so you again need to work in the indefinite-particle-number Hilbert space.

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  • $\begingroup$ TLDR: relativistic system = fundamental particle number not conserved; nonrelativistic system = fundamental particle number conserved, quasiparticle number not (necessarily) conserved. $\endgroup$ – tparker Jul 11 '16 at 7:31
  • $\begingroup$ It's kind of a shame that many people don't encounter indefinite-particle-number Hilbert spaces until they study relativistic QFT. They get the impression that there's something inherently relativistic about the notion. But in fact it's just a practical matter: you don't always need such fancy Hilbert spaces in nonrelativistic contexts, but there's nothing stopping you from using them. $\endgroup$ – tparker Jul 11 '16 at 7:44
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    $\begingroup$ In fact, in my experience, long after you first study QFT, you realize that a lot of the concepts are just advanced concepts from nonrelativistic QM, that ideally you should have mastered before you started studying QFT ... okay, sorry, now I'm just rambling. $\endgroup$ – tparker Jul 11 '16 at 7:47
  • $\begingroup$ Interesting. If nothing new shows up I will mark it as the answer. $\endgroup$ – user121238 Jul 11 '16 at 20:26
  • $\begingroup$ The mass, as a central charge of the Galileu álgebra, would not prevent interactions that not conserved particles? $\endgroup$ – Nogueira Sep 18 '16 at 2:02

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