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I often hear that while QFT is a relativistic theory, standard quantum mechanics is not. But fields aren't inherently relativistic, you can easily construct non-relativistic QFTs, a relativistic QFT is one with a Lorentz invariant Lagrangian.

In the same way, it doesn't seem to me that there is anything inherently non-relativistic about the standard description of quantum mechanics, and that the actual non-relativistic part is the Schrödinger equation. As a matter of fact if we substitute the Schrödinger equation axiom with dynamics described purely by quantum channels, with the only requirement that an evolved quantum state is still a quantum state and that the theory is linear, we do recover some statements that resemble locality in the Lorentz sense, like the no communication theorem.

Are fields necessary to obtain a relativistic theory? To what extent is standard QM non-relativistic? If QM is inherently non-relativistic even without the Galilean dynamics given by the Schrödinger equation, why should one care about non-locality in Bell inequalities and "spooky action at a distance"? Those would be expected in a non-relativistic theory, just like the gravitation of the Sun on the Earth is spooky action at a distance in Newtonian gravity.

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  • $\begingroup$ What is spooky about action at a distance is not that is NOT that it needs to respect relativity. Of course a non relativistic theory doesn't. In the absence of no communication theorems, we are presented with a theory, that works in the non relativistic regime, and performing purely non relativistic experiments, we seem to violate relativity, through effects that can be measured in the non relativistic regime. This suggests that there is NO relativistic completion of the theory, which is an abhorrent idea. Luckily, we do have no communication theorem and we are saved. $\endgroup$ – Anonjohn Feb 6 '20 at 18:43
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    $\begingroup$ Not even the Schrödinger equation, which is still valid in QFT; what's non relativistic is the specific form of the Hamiltonian. $\endgroup$ – Javier Feb 6 '20 at 20:11
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the actual non-relativistic part is the Schrödinger equation

Indeed. Hence, people tried to come up with a Lorentz-invariant evolution equation for the wave function such as the Klein-Gordan and Dirac equations (as mentioned by anna v). However, interpretation of solutions of these equations as probability densities are problematic (in case of the Klein-Gordon equation, the predicted probabilities won't be positive definite - though the Dirac equation 'fixes' this particular problem).

We've come to realize that these equations are not to be understood as evolution equations for the quantum state, but in the sense of quantum field theory, where (in the Schrödinger picture), their (classical) solutions correspond to the configurations of the system (ie the equivalent of positions in single-particle QM), with states being wave-functionals on that solution space.

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The Klein Gordon equation , the Dirac equation were developed partially to give quantum mechanical equations that were also Lorenz invariant, to replace the Schrodinger one. Maxwell's equation when used for the photons as a quantized equation is of course Lorentz invariant.

QFT depends on the plane wave solutions of the above equations in its construction, that is why it is Lorentz invariant by construction.

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I am not an expert on this, and this only answers

Are fields necessary to obtain a relativistic theory?,

but I think there are (mathematically provable) inconsistencies (with physics that is) of relativistic single-particle QM.

The usual heuristic reasoning given is of course that relativity allows for particle creation and destruction, so the given formulation of relativistic QM must allow for nonfixed particle number.

From this point on one does not have to quantize fields. One can construct the single particle Hilbert space $\mathcal H_1$ with a representation of the Poincaré group on it, then one can construct $$ \mathcal H^S_k=\mathcal H_1\vee...\vee\mathcal H_1 $$ ($k$-fold product) for bosons ($\vee$ is the symmetric tensor product) or $$ \mathcal H^A_k=\mathcal H_1\wedge...\wedge\mathcal H_1 $$($k$-fold product) for fermions, then take $$ \mathcal H^S=\bigoplus_{k=0}^\infty\mathcal H^S_k,\quad \mathcal H^A=\bigoplus_{k=0}^\infty\mathcal H^A_k $$ to form the Fock space, which contains states with arbitrary particle number.

As it turns out the concepts of creation and annihilation operators as well as number states are fairly natural in these spaces, and when you express physical quantities as operators on this space, you get basically QFT.

This approach is what's followed in Weinberg's The Quantum Theory of Fields (vol 1). So basically, you don't come looking for field theory, field theory comes looking for you when you do relativistic QM.

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    $\begingroup$ It is common folklore but there is no mathematical inconsistency with relativistic QM. The point is that when interactions are included they are naturally described in a Fock space, but this is a physical fact. There is no problem in describing a non-interacting relativistic particle and its states are nothing but the states of the one-particle space in the Fock space if the corresponding qft. $\endgroup$ – Valter Moretti Feb 6 '20 at 18:25
  • $\begingroup$ @ValterMoretti Fair enough, there is a lot of folklore going on about QFT tbh. Most textbooks are ok to learn from but are horribly unsystematic in my opinion. But also this is kind of what I meant by "(with physics that is)". $\endgroup$ – Bence Racskó Feb 7 '20 at 14:42
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There are a lot of sub-questions in your question. First off, yes you can have a non relativistic quantum field theory (like many field theories describing solid state matter) and on the other hand you can have relativistic "standard quantum mechanics" like the Dirac equation (and Klein-Gordon equation although it has some problems as others have mentioned).

Then you say/ask:

As a matter of fact if we substitute the Schrödinger equation axiom with dynamics described purely by quantum channels, with the only requirement that an evolved quantum state is still a quantum state and that the theory is linear, we do recover some statements that resemble locality in the Lorentz sense, like the no communication theorem.

I don't think this is a good analogy for a relativistic behavior. But there others more precise. For example, many lattice models describing matter (for example the undoped version of some cuprate material can be described by a antiferromagnetic Heisenberg model) at low energy are described by relativistic field theory (in this case the non-linear sigma model with or without topological term). Another way in which relativity pops out in Galieleian models is provided by the Lieb-Robinson bound. Such result essentially asserts that in any Galileian quantum theory with local interaction there is a maximum propagation speed (the Lieb-Robinson speed) and so there are approximate light-cones.

Later you ask

Are fields necessary to obtain a relativistic theory?

No, Dirac's equation is a counter example

To what extent is standard QM non-relativistic?

Well, simply to the extent that it's not relativistic covariant.

But in your last question you make a good point. Why do we care about Bell's theorem if any Galieleian interaction also violates Lorentz causality. I think the point is precisely that even Galieleian theories are still local, and so their action decreases with the distance. Instead in Bell's experiment we are able, in principle, to put a particle on hearth and the other on Alpha Centauri and we would observe the same violation.

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