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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.

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    $\begingroup$ How can you hope to have Lorentz form-invariance when you have singled out $t$, just one of the four spacetime coordinates? $\endgroup$
    – G. Smith
    Jun 19, 2019 at 20:32
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    $\begingroup$ possible dup.? physics.stackexchange.com/q/257787/84967 $\endgroup$
    – AccidentalFourierTransform
    Jun 19, 2019 at 20:33
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    $\begingroup$ @G.Smith You can take a covariant equation, single out $t$, move the rest to the other side and call it $H$. There is nothing wrong with that: the equation is still covariant, albeit not manifestly so. E.g. the Dirac equation, see comment above. $\endgroup$
    – AccidentalFourierTransform
    Jun 19, 2019 at 20:35
  • $\begingroup$ @AccidentalFourierTransform Doh! Excellent point. I retract my dumb comment. $\endgroup$
    – G. Smith
    Jun 19, 2019 at 20:42
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    $\begingroup$ related: physics.stackexchange.com/q/345554 $\endgroup$
    – user4552
    Jun 19, 2019 at 22:28

4 Answers 4

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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.

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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.

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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 .

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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.

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    $\begingroup$ This doesn't seem quite right to me. The OP is making a correct distinction between two usages of the term "Schrödinger equation." Is the answer referring to only the first of these two? $\endgroup$
    – user4552
    Jun 19, 2019 at 22:29
  • $\begingroup$ The difference between the time dependent and independent form has no relevance for its behaviour under Lorentz transformation. $\endgroup$
    – my2cts
    Jun 19, 2019 at 22:44
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    $\begingroup$ @my2cts By "two usages", Ben means $i\dot\psi=-\psi''+V\psi$ vs $i|\dot\psi\rangle=H|\psi\rangle$, not the TDSE vs the TISE. $\endgroup$
    – AccidentalFourierTransform
    Jun 19, 2019 at 22:53
  • $\begingroup$ I have not read any remotely decent physical argument against my post, yet -2. Don't pretend my answer is wrong without arguments. If you have a better answer yourself, let's see it. $\endgroup$
    – my2cts
    Jun 20, 2019 at 21:55
  • $\begingroup$ @my2cts I didn't downvote, but it might relate to how the question is interpreted. Like you said, the general Schr. eq. (GSE) itself isn't Lorentz covariant, just like a single component of a four-vector isn't. But the GSE still applies in a theory that has Lorentz symmetry. The GSE is simply the statement that time-evolution is unitary. Expressing a Lorentz-symmetric QFT in the Schr. picture obscures that symmetry, but the theory itself still has that symmetry, just like special relativity still has Lorentz symmetry even when the metric is expressed in a symmetry-obscuring coordinate system. $\endgroup$
    – Chiral Anomaly
    Jun 21, 2019 at 1:36

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