Does the Schrödinger equation apply to spinors? I was reading about Larmor precession of the electron in a magnetic field in Griffiths QM when I came across the equation
$$
i\hbar \frac{\partial \mathbf \chi}{\partial t} = \mathbf H \mathbf \chi,
$$
where $\mathbf\chi(t)$ is a 2D vector that represents only the spin state and does not include information of the wave function. The Hamiltonian is
$$
\mathbf H = - \gamma \mathbf B \cdot \mathbf S = - \frac{\gamma B_0 \hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}
$$
for a uniform magnetic field $\mathbf B = B_0 \hat k$. Why should these spinors also obey the Schrödinger equation? The book does not provide any further information as to why this should hold.
 A: The correct wave equation for an electron is the famous Dirac equation (in fact it is correct for all spin $\frac{1}{2}$ particles).
The non-relativistic limit of the Dirac equation is called Pauli equation. The derivation can be found here. In this equation $|\psi \rangle$ is a two-component spinor wavefunction.
Pauli equation (general)
$${\displaystyle \left[{\frac {1}{2m}}({\boldsymbol {\sigma }}\cdot (\mathbf {\hat {p}} -q\mathbf {A} ))^{2}+q\phi \right]|\psi \rangle =i\hbar {\frac {\partial }{\partial t}}|\psi \rangle }$$
If you substitute $\mathbf {A}=0$ above you get back the Schrödinger equation. In fact, a relativistic wave equation for any spin should under some limit give back the Schrödinger equation be it Klein–Gordon equation or Proca action or Rarita–Schwinger equation.
A: The Schrödinger equation applies to any quantum system or quantum field theory, as long as you have a continuous time dimension. The equation
$$\hat H\psi = i\hbar\frac{\partial \psi}{\partial t} $$
is just the statement that the Hamiltonian $\hat H$ is the infinitesimal translation operator in time. By Noether's theorem, it corresponds to the total energy of your system.
In order to describe a quantum system, you need to first decide what the degrees of freedom are. This decides which Hilbert space this equation is defined on. For example, for a single particle in $\mathbb R^d$, the Hilbert space is $\mathcal H = L^2(\mathbb R^d)$. And thus the wavefunction is $\psi(\mathbf r,t)$. Since the Hamiltonian is the total energy, we have (here for a non-relativistic particle)
$$\left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r)\right]\psi(\mathbf r,t) =  i\hbar \frac{\partial}{\partial t}\psi(\mathbf r, t).$$
Now imagine that the particle has some internal degree of freedom, such as spin. For a spin-$\tfrac 12$ particle, the Hilbert space is two dimensional $\mathbb C^2$. Quantum mechanics tells us that the Hilbert space of the full system is a tensor product
$$ \mathcal H = L^2(\mathbb R^d)\otimes\mathbb C^2$$
and wavefunctions are thus
$$ \psi(\mathbf r,t) = \begin{pmatrix} \psi_\uparrow(\mathbf r,t) \\ \psi_\downarrow(\mathbf r,t)\end{pmatrix}.$$
The Hamiltonian will now also contain terms coupling spins together, but it depends on which forces are present. For example in $d=3$ you could have
$$\left[-\frac{\hbar^2}{2m}\nabla^2\,\mathbf 1 + V(\mathbf r)\,\mathbf 1 + \alpha \,\mathbf B\cdot\mathbf\sigma\right]\psi(\mathbf r,t) =  i\hbar \frac{\partial}{\partial t}\psi(\mathbf r, t).$$
Here $\mathbf 1$ is a $2\times 2$ identity matrix and $\mathbf\sigma = (\sigma_x,\sigma_y,\sigma_z)$ are the Pauli matrices.
Sometimes people write down a Schrödinger equation for only position or only spin degrees of freedom, if they don't couple to other degrees of freedom in the problem at hand.
For example, if the Hamiltonian only has $\mathbf 1$ for the spin then the Hamiltonian will be block diagonal and $\psi_\uparrow$ and $\psi_\downarrow$ will be completely decoupled. You could therefore forget spin, and only keep position in your description. Or if the particle is trapped somewhere and cannot move, then only the spin degrees of freedom are relevant for the description.
But the Schrödinger equation is valid for any quantum system, it just describes how the system evolves in time.
A: If you want to describe non-relativistic, interacting spins you need to extend the Schrödinger equation with the Pauli interaction, $H_{spin} = -\gamma {\bf B} \cdot {\bf S}$. This is the simplest of a class of hamiltonians known as spin hamiltonians. For non-translating spins, as in your case, you only need the Pauli term.
