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So I'm aware that the correct relativistic approach to quantum mechanics is through quantum fields, but I'm still interested in the question that follows.

We know the Schrodinger equation in free space in $d=1$ is as follows: $$i\hbar \partial_t \psi=-\frac{\hbar^2}{2m}\partial_{xx}\psi.$$ Classically, the relativistic generalization of kinetic energy is $$E=\sqrt{m^2c^4+p^2c^2}-mc^2.$$ What happens if we expand up to order $4$ in $p/mc$, and substitute $p\to -i\hbar\partial_x$ to write the little bit more relativistic Schrodinger equation:

$$i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}\partial_{xx}\psi-\frac{\hbar^4}{8m^3c^2}\partial_{xxxx}\psi,$$ neglecting the higher order contributions. Is this new equation physically meaningful? We still have conservation of probability because the hamiltonian is still hermitian. I ask my question again: is this approach physically useful?

Qmechanic
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  • Possible duplicates: https://physics.stackexchange.com/q/194877/2451 , https://physics.stackexchange.com/q/346780/2451 and links therein. – Qmechanic May 17 '22 at 03:21
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    It's physically meaningful in the sense that it leads to verifiable corrections, but the theory is still not lorentz invariant or causal. – AfterShave May 17 '22 at 03:32

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