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I have a couple of questions about what kind of potentials can be used in Schrödinger's equation:

  1. How about the potential from a magnetic field? Isn't Dirac's equation more appropriate in that case, because it will take spin into account? If I choose to ignore spin and use Schrödinger's with a magnetic potential, do I still get useful results?

  2. How about gravitational potential? What happens if I use e.g., $mgh$ in Schrödinger's equation? Do I still get useful results for e.g, neutrons? Or do people normally don't do that, because gravitation is negligible compared to electric fields?

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In the Schrödinger equation you can introduce, in principle, whatever form of potential you like. All the question is whether it allows a physical solution.

About a particle in the magnetic field you can very well use the Schrödinger equation in which you introduce the interaction term $mB$, where $m$ is the magnetic dipole of the particle and $B$ the magnetic field.

If you ignore the spin, you ignore the magnetic dipole, s.t. what interacts with the magnetic field?

You can use $mgh$ in the Schrödinger equation if you describe free falling of particle in absence of other fields. There are also experiments of this type (see Esslinger in the arXiv quant-ph). As to movement in constant fields, the Schrödinger eq. becomes an Airy equation.

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To include magnetic field into non-relativistic Schrödinger's equation, you can consider Pauli's phenomenological equation, which is a non-relativistic approximation of Dirac equation. It includes both spin and vector potential. If you drop spin-dependent term from it, you'll get an equation for charged spinless particle in magnetic field. For a spinless particle you'll then only change kinetic energy term compared to usual Schrödinger's equation, so that

$$\hat T=\frac1{2m}(\hat{\vec p}-\vec A)^2,$$

where $\vec A$ is electromagnetic vector potential.

Gravitational potential is usually added just as $mgh$ term in usual potential energy operator, i.e. for a particle in electric field with gravitational field you'll get potential energy $$U(\vec r)=q\phi_e(\vec r)+m\phi_g(\vec r).$$

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