When particle number can change in quantum physics? 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.

 A: 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.
