Part of the reason that relativistic QFT is so hard to learn is that there are piles of 'no-go theorems' that rule out simple physical examples and physical intuition. A very common answer to the question "why can't we do X simpler, or think about it this way" is "because of this no-go theorem".

To give a few examples, we have:

- the [Reeh-Schlieder theorem](https://en.wikipedia.org/wiki/Reeh%E2%80%93Schlieder_theorem), which I'm told forbids position operators in relativistic QFT
- the [Coleman-Mandula theorem](https://en.wikipedia.org/wiki/Coleman%E2%80%93Mandula_theorem), which forbids mixing internal and spacetime symmetries
- [Haag's theorem](https://en.wikipedia.org/wiki/Haag%27s_theorem), which states that naive interaction picture perturbation theory cannot work
- the [Weinberg-Witten theorem](https://en.wikipedia.org/wiki/Weinberg%E2%80%93Witten_theorem), which among other things rules out a conserved current for Yang-Mills
- the [spin-statistics theorem](https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem), which among other things rules out fermionic scalars
- the [CPT theorem](https://en.wikipedia.org/wiki/CPT_symmetry), which rules out CPT violation
- the [Coleman-Gross theorem](http://cds.cern.ch/record/363169/files/9808154.pdf), which states the only asymptotically free theory is Yang-Mills

Of course all these theorems have additional assumptions I'm leaving out for brevity, but the point is that Lorentz invariance is a crucial assumption for every one.

On the other hand, nonrelativistic QFT, as practiced in condensed matter physics, doesn't have nearly as many restrictions, resulting in much nicer examples. But the only difference appears to be that they work with a rotational symmetry group of $SO(d)$ while particle physicists use the Lorentz group $SO(d-1, 1)$, hardly a big change. Is there a fundamental, intuitive reason that relativistic QFT is so much more restricted?