Does the Schrodinger Equation care about spin?

I have taken the non-relativistic limit of the Klein-Gordon and Dirac equation, and both have brought me to the Schrodinger equation.

The Klein-Gordon equation describes spin 0 particles, and the Dirac equation describes spin $$\frac12$$ particles, and yet in the non-relativistic case they are equal.

I'm starting to wonder if spin is a relativistic thing, but I know that I studied it in Griffiths' Introduction to Quantum Mechanics, which is all non-relativistic.

Why is it that the Schrodinger equation can describe particles of seemingly any spin, and in the relativistic case we have to be so careful about it?

• What do you mean by "the Schrödinger equation" here? In its most abstract form, it's just $-\mathrm{i}H\lvert\psi(t)\rangle = \partial_t\lvert \psi(t)\rangle$. – ACuriousMind Jun 29 at 11:47
• I mean $$i\partial_t\phi=-\frac{1}{2m}\partial_i\partial^i\phi$$ where $\phi$ is a field, not a wavefunction. – user45757 Jun 29 at 11:50
• – AccidentalFourierTransform Jun 29 at 12:12
• Well if you go from Klien Gordan or Dirac Equation to Schrodinger Equation with some EM interaction then you will see that the Spin only arises as a first order perturbation contribution to Schrodinger equation. The second term in $\sigma \cdot (p - \frac{eA}{c})$ contributes to the spin term. The Schrodinger equation really doesn't address spin unless you artificially rewrite the momentum term as I just mentioned above. – Kolandiolaka Jun 29 at 12:22
• What distinguishes a field from a wave function? – my2cts Jun 29 at 15:38

The Schrödinger equation is the nonrelativistic limit of the Klein-Gordon - not of the Dirac - equation, hence it does not contain spin. You can add spin by interpretating the Schrödinger wave function as a sponsor and adding the Pauli spin interaction. This equation is the nonrelativistic limit of the squared Dirac equation. In Itzykson&Zuber the Dirac equation is solved for hydrogenic atoms via the squared Dirac equation. Highly recommended.

• I found this after I took the limit of the Dirac equation, where $\chi$ is a spinor field. $$i\partial_0\chi=-\frac{1}{2m}\partial_i\partial^i\chi$$ Does this mean that this equation (which looks identical to the Schrodinger), is the Pauli equation (with 0 external magnetic field)? Does this mean that it is the Schrodinger equation for the spinor field component-wise? – user45757 Jul 1 at 15:08

There are some misconceptions in what you wrote.

There are four (sets of) equations, all discovered in 1926-1928. The Schrödinger equation (1926, spin $$0$$, non-specially relativistic), the Pauli equation(s) (1927, spin $$1/2$$, non-specially relativistic), the Klein-Gordon equation (1926, spin $$0$$, specially relativistic), the Dirac equation(s) (1928, spin $$1/2$$, specially relativistic). So you can now understand how they are related, which simplification/generalization you need to do, in order to move from one to another.

Spin is a relativistic thing, either Galilean relativistic, or specially-relativistic.

If you say "Why is it that the Schrödinger equation can describe particles of seemingly any spin?", then this is true, only if the Schrödinger equation is cast into the form written in ACuriousMind's comment, that is the Hamiltonian is not defined to be the one taken by Schrödinger in 1926, but can be any of them, including the Dirac Hamiltonian.

• Any equation with a Dirac hamiltonian cannot be a Schrödinger equation. There are limits. – my2cts Jul 1 at 18:39