In summary: If the SE doesn't work with SR, what specifically, in terms of the math, is the reason for this? I'm specifically interested in how this relates to the creation and annihilation operators of QFT, as I previously was under the impression that those operators didn't work with the Schrödinger equation, but my professor recently told me they're the same as the ladder operators used in non-relativistic quantum mechanics.
I posted a few questions a couple months ago, about the relationship between QM and QFT, such as What is the mathematical relationship between the wave functions of QM and the fields in QFT? and I got some very helpful explanations about how QFT requires the creation and annihilation operators, and I've also read (can't remember where) that the problem with the Schrödinger equation formulation of QM (as opposed to, e.g., the Dirac formulation) is that it assumes a single static reference frame, which obviously doesn't work with special relativity, and hence it's replaced by the Dirac equation in QFT. I also saw Schrödinger Equation and Special Relativity where the OP refers to a relativistic version of the Schrödinger equation and is told there is no such version, but no explanation is given for why.
However, there are two things I'm not understanding:
The Schrödinger equation is perfectly compatible with changes of basis, such as converting between position and momentum space via the Fourier transform. The way I understand it from linear algebra and SR, a change of basis is mathematically equivalent to changing reference frames/coordinate systems. And, I mean, obviously changing between position and momentum space isn't the same thing as applying the Lorentz transform to change coordinates in spacetime, both the Lorentz transform and the quantum mechanical operators are linear operators over vector spaces, and the wavefunctions that are solutions to the Schrödinger equation are vectors in the vector space the operators act on. So if we can apply, e.g., the momentum operator to them, shouldn't we also be able to apply the Lorentz transform to them? If not, why?
A couple months ago, when I was reading those responses (such as the second answer to the above linked question) about how what makes QFT fundamentally different from QM is the creation and annihilation operators allowing us to switch reference frames (because, since energy and hence mass is relative, that means particle number is also relative), I got the impression that these operators were something specific to QFT that wouldn't work with the Schrödinger equation. However, in my QM course, we just finished the unit on the quantum harmonic oscillator, including using the ladder operators to find a recursive formula for the wavefunction at an arbitrarily high energy level. Griffiths' covers this in detail in chapter 2 of his QM text, and, at that point in the text at least, he's still starting every derivation with some expression he's previously derived from the Schrödinger equation. So clearly the ladder operators work fine with the Schrödinger equation and I thought nothing of it, until I read the Wikipedia article on those operators, which calls them creation and annihilation operators. I asked my QM professor in office hours and he said that, yes, they're the same creation and annihilation operators used in QFT, and then he started talking about how QED in particular is basically built-up from the quantum harmonic oscillator. I then asked a friend who previously did research in QFT if that was true what my professor said about QED being built up from the harmonic oscillator and he said yes, absolutely. So my question then is this: if QED, at least, is built up from the quantum harmonic oscillator, and QFT more generally uses the creation annihilation operators to change reference frames (which, if I understand correctly, is the main thing the Schrödinger formulation of QM lacks), but Griffiths' shows both that we can model the quantum harmonic oscillator using a wavefunction derived from the Schrödinger equation and that this wavefunction works perfectly well with the aforementioned operators, then how can the Schrödinger equation be incompatible with SR?
On a side note, I did see 'Why' is the Schrödinger equation non-relativistic?, however, while that sounds like the same question as mine, reading the question details, (and a later comment by the OP) what the OP is actually asking about is why the De Broglie equations apparently are compatible with SR while the SE seems not to be. They also appear to already be familiar with the technical reasons the Schrödinger equation is apparently non-relativistic (I say "apparently" and still have my question worded as "is it" because one of the answers to that question says the SE itself IS compatible with relativity but that the Hamiltonian is the problem, and the OP comments saying they already knew that, which has me even more confused because of their initially saying they knew SE wasn't relativistic), whereas those reasons are what I'd like to know in the first place. Hence, please don't merge these questions together, as they're only superficially similar and definitely not duplicates, for the reasons stated above.
Edit: As explained above, this question is not a duplicate of the linked question and is only superficially similar to it. Please actually read the entire question before voting to close, rather than merely going by the title.