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.
For context:
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:
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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?
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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.
Best Answer
You ask many questions here and they conflate many things. The answers you were given are partially correct, but not very precise. For example, things being similar does not mean that they are the same. So, I can totally understand why you are confused. I will try to answer at least part of your questions.
The Schrodinger equation and relativistic invariance.
SE describes how the wave function $\psi(x,t)$ evolves with time. Already this means that it attributes a special role to time. Time and coordinates are not treated on the same footing by SE. Thus, one says that SE and the whole approach of quantum mechanics is not manifestly relativistically invariant.
Still, SE is a completely universal equation for evolution in quantum systems. It can be applied -- an is frequently used -- to deal with relativistic quantum systems, including QFT (the Schrodinger picture in canonical quantization). Thus, despite the approach of SE is inherently not relativistic, it can still be used to describe relativistic systems. Due to this drawback of the SE manifest relativistic invariance gets lost and one should check if it is there manually.
For example, one can consider a free non-relativistic particle with the Hamiltonian $\frac{p^2}{2m}$. Using the standard quantization $p\to \hat p = -i\partial_x $ one gets a quantum Hamiltonian, which can be used to write SE in a given case. In this example, neither the system nor its description are relativistically invariant.
In contrast, one can consider a relativistic particle with the Hamiltonian $\sqrt{p^2+m^2}$. Replacing $p\to -i\partial_x $ one gets the respective quantum Hamiltonian. The associated SE reads \begin{equation} i\frac{\partial}{\partial t}\psi = \sqrt{-\partial^i \partial_i +m^2} \;\psi. \qquad (1) \end{equation} This equation is relativistically invariant. This means that when we act with translations \begin{equation} P_\mu \psi \equiv \partial_\mu \psi \end{equation} and Lorentz transformations \begin{equation} J_{\mu\nu} \psi \equiv (x_\mu\partial_\nu-x_\nu \partial_\mu )\psi \end{equation} on the wave function, then solutions of (1) map into solutions of (1). Still, obviously, time derivatives and spatial derivatives enter (1) differently, so this equation is not manifestly relativistically invariant -- the consequence of using SE.
(Let me note that up to factors, $P_i$ that I defined above are just the operators of momenta. Then $P_0$ is the Hamiltonian, once (1) is imposed. Analogously, $J_{\mu\nu}$ are just quantum operators of rotations and boosts)
The system (1) describes a free relativistic particle of mass $m$ and spin 0. It is a completely legitimate quantum-mechanical theory as is. Let me also note, that this is not the Dirac equation. Historically, Dirac derived his equation trying to make the Schrodinger equation Lorentz-covariant. What he got in the end was SE for a free spin-$\frac{1}{2}$ particle (the Dirac equation is often regarded as a classical field equation, which is not the same as a single-particle SE, but let's not get into this), not for a spin-$0$ particle. There are some reasons why he did not get (1), but let us not go into this. This is more on a history side than about physics. The current state of affairs is that you can have consistent relativistic quantum mechanics of free particles of any spin and mass, see e.g. section 2 "Relativistic quantum mechanics" of the book by Weinberg.
Creation and annihilation operators in QM and QFT
Again, there are many things conflated here. In quantum mechanics, creation and annihilation operators are primarily used is only one context: it is a simple trick to solve one particular problem, namely, finding the eigenstates and the eigenvalues of the Hamiltonian for the harmonic oscillator. This is it. In this context, the creation operator applied to an eigenstate maps it to the next eigenstate, while the annihilation operator acts in the opposite direction. In QFT one also uses creation annihilation operators, but their meaning is very different.
Before explaining what they do in QFT, let me first recall that there are two ways to get to QFT: by quantizing a classical field theory or by adding interaction in relativistic quantum mechanics, the latter inevitably requires to create and annihilate particles.
The first approach is more standard. The classical field theory is first of all a classical theory and to get to QFT it needs to be quantized. The classical field theory may be regarded as a limit of a classical mechanics, in which one has infinitely many point particles. With this amendment, classical field theory can be quantizing along the same lines as a system of classical particles. For example, one uses the canonical quantization: finds coordinates and conjugate momenta and then postulates that these no longer commute, but satisfy the canonical commutation relations. In doing so, it is convenient to make the Fourier transform of the field with respect to spatial directions. Then, one finds that different Fourier modes decouple from each other, while each one has the Hamiltonian of a harmonic oscillator with the frequency related to the momentum as $\omega_p = \sqrt{p^2+m^2}$. This is how the harmonic oscillator from quantum mechanics becomes relevant. You also end up dealing with creation and annihilation operators. But their meaning in free QFT is that the creation operator $a^\dagger_p$ creates a particle with momentum $p$, while $a_p$ annihilates it. Note, that there was no notion of a particle in field theory before it was quantized. In fact, this is the whole reason why quntization of field theory was needed: we do see that, say, energy from E-M waves is absorbed in chunks, as if it consists of particles, while classically E-M waves do not have them.
The second approach is followed, e.g. by Weinberg. One starts from a relativistic quantum mechanics, which describes a free single particle. What is missing here is that the theory describes only a single particle, while we want to describe also multi-particle states. We also know that when energies are high enough, processes of creation and of annihilation of particles become unavoidable, so one needs a language to deal with this. What one does is that instead of a single-particle Hilbert space (of quantum mechanics), considers the multi-particle Hilbert space. Creation and annihilation operators are introduces as operators that create or annihilate a particle of a given momentum. One can show that any Hamiltonian can be written in terms of creation and annihilation operators and thus they are used in the following.
To summarize, one does have creation and annihilation operators in QFT. However, unlike in quantum mechanics, in which they raise/lower energy of a single particle in one particular model of the quantum oscillator, in QFT $a$ and $a^\dagger$ annihilate and create particles. So, $a$ and $a^\dagger$ from QFT are completely out of the physical realm of quantum mechanics, as the latter deals only with a single particle, while in QFT $a$ and $a^\dagger$ create and annihilate particles. Still, as I explained above, the Hamiltonian of the free field theory is that of an infinite family of decoupled oscillators. This is how QFT creation and annihilation operators are relevant to the QM ones for the harmonic oscillator.
PS. I believe, it should follow from what I wrote above, $a$ and $a^\dagger$ are introduced in QFT not to be able to change reference frames. Similarly, the usage of $a$ and $a^\dagger$ for the harmonic oscillator does not say anything about its relativistic invariance. Harmonic oscillator in QM is not relativistically invariant.