Can the Schrodinger Equation be Relativistic in it's General Case?

Question

So as we all know the Schrodinger equation is not relativistic when written (constantless) as$$i\partial_t\psi=-\partial_{xx}\psi+V\psi$$due to it not being Lorentz invariant. However, if we consider the absolute general case of the equation namely$$i\partial_t\psi=\hat{H}\psi$$one could postulate that if $\hat{H}$ is chosen appropriately and carefully it would in theory be satisfied and furthermore allow for relativistic eigenvalues $E_n$ when solving the time-independent version $\hat{H}\psi=E_n \psi$.

My question is, is this a legitimate way of thinking about the concept? Or is there something else that breaks down in this case? I tried to look for answers but I cannot seem to find any that explain anything clearly enough.

Answer 1

The general Schrödinger equation $$ i\partial_t \psi= H\psi $$ is simply the infinitesimal version of the statement that time-evolution is unitary (preserves inner products). This is just as true in relativistic QFT as it is in non-relativistic QM. For example, quantum chromodynamics can be written in this form. The symbol $\psi$ denotes an element of the Hilbert space (a state-vector), and $ H$ is the generator of time-translations. This is very generic.

That's in the Schrödinger picture, where the state-vector is parameterized by time and the observables are not. Even if the theory is actually Lorentz symmetric, that symmetry is obscured in the Schrödinger picture. In relativistic QFT, Lorentz symmetry can be made manifest by working in the Heisenberg picture instead, where the field operators (used to construct observables) are parameterized by both time and space, and the state-vector is not. The Hamiltonian $ H$ is the same in either case. In the Heisenberg picture, a field operator $\phi(t,x)$ obeys $$ \partial_t\phi(t,x)\propto [ H,\phi(t,x)] $$ and $$ \partial_x\phi(t,x)\propto [ P,\phi(t,x)] $$ where $ H$ is the total-energy operator (aka Hamiltonian, the generator of time-translations) and $ P$ is the total-momentum operator (the generator of translations in space). For simplicity, I've only indicated one spatial dimension. The operator $H^2-P^2$ is the total-mass-squared operator. "Total" means the whole system, not just one particle; in relativistic QFT, the number of particles is generally not even well-defined.

Answer 2

You are correct to note that Lorentz invariance is a property that some equations have and some equations don’t have, and that the equation $$i\partial_t\psi=\hat{H}\psi \qquad(*)$$ can end up as invariant for certain choices of the Hamiltonian $\hat{H}$. In that sense, this way of thinking about invariance is “legitimate”.

The problem is that while being legitimate, it is also a rather unhelpful way of thinking about the issue of invariance, because any equation can be written in the form $(*)$ by defining $\hat{H}$ appropriately. Moreover, thinking about things that way would only be helpful if there was some nice description for the class of Hamiltonians that cause $(*)$ to be Lorentz invariant. And there isn’t such a nice description, because the equation singles out the time variable for special treatment which spoils the relativistic Lorentz symmetry, so the only description for this class would be the ugly (and again, completely unhelpful) one of “all operators that become Lorentz invariant after you subtract $i\partial_t$”. In other words, as @AccidentalFourierTransform said in a comment, for certain choices of $\hat{H}$ the equation would end up relativistic, but in a nonobvious, unintuitive way that is difficult to verify; physicists say in such cases that the equation is not manifestly invariant. And manifestly invariant equations are much easier to work with and understand than non-manifestly invariant ones. See this discussion for some related background.

Answer 3

A Bosonic QFT in a (periodic) box with a momentum cutoff is a multidimensional harmonic oscillator. It is possible to generalise this to an infinite dimensional "Q-Space" (no cutoffs), see Lon Rosen's discussion, p.73 .

Answer 4

There is no way that you can make Schrödinger's equation covariant. Its left hand side transforms as the time component of a four vector. The relativistic generalisation is the Klein-Gordon equation.

Note that the Schrödinger equation is not covariant even under Galileo transformations.