As per Rob's suggestion, I decided to make this an answer.
(Addendum: I've been meditating on this very topic for some time, and have been directed to some interesting literature referenced on Streater's webpage. As per Rococo's comment, I've updated my answer, but kept the old version for posterity.)
The Answer
To talk about "spin-1/2 particles", we need the spin-statistics theorem.
The spin-statistics theorem doesn't hold for non-relativistic QM. Or more precisely, the "naive" spin-statistics theorem doesn't hold, and if we try to contrive one...it's nothing like what we'd expect.
In brief, the reason is: spin-statistics theorem depends critically on microlocality (i.e., the commutator of spacelike separated conjugate operators vanishes identically). This property holds relativistically, but not non-relativistically. (The other proofs similarly do not hold, since Lorentz invariance fails.)
But there is a trick around this. Just take a relativistic spin-1/2 particle, then take the nonrelativistic limit.
But does the Schrodinger Equation Hold?
Now, the question originally asked is: does the Schrodinger equation hold for spin-1/2 particles? To talk about spin-1/2 particles, we really are working with representations of the Lorentz group. A nonrelativistic spin-1/2 particle is obtained by taking the nonrelativistic limit (i.e., the $c\to\infty$ limit) of the Dirac equation, which is the Pauli equation:
$$\left[ \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi \right] |\psi\rangle = i \hbar \frac{\partial}{\partial t} |\psi\rangle$$
where $\boldsymbol{\sigma}$ are the Pauli matrices, $\mathbf{A}$ an external vector potential, $\phi$ an external electric potential, and $q$ the electric charge of the particle. But observe when we "turn off" electromagnetism (setting $\mathbf{A}=\phi=q=0$) we recover
$$\left[ \frac{1}{2m}(\boldsymbol{\sigma}\cdot\mathbf{p})^2 \right] |\psi\rangle = i \hbar \frac{\partial}{\partial t} |\psi\rangle$$
Uh, I leave it as an exercise for the reader to show the left hand side of this equation is $(1/2m) \mathbf{p}^{2}\otimes\boldsymbol{1}_{2}$ where $\boldsymbol{1}_{2}$ is the 2-by-2 identity matrix.
References
For more thorough reviews on this matter, I can heartily refer the reader to:
- A.S. Wightman, "The spin-statistics connection: Some pedagogical remarks in response to Neuenschwander's question" Eprint, 7 pages
- R. E. Allen, A. R. Mondragon, "No spin-statistics connection in nonrelativistic quantum mechanics". Eprint arXiv:quant-ph/0304088, 2 pages
The (Old) Answer
The "vanilla" Schrodinger's equation (from non-relativistic QM) does not describe a spin-1/2 particle.
The plain, old Schrodinger's equation describes a non-relativistic spin-0 field.
Case Studies
If we pretend the wave function is a classical field (which happens all the time during the "second quantization" procedure), then it turns out to describe a spin-0 field. See Brian Hatfield's Quantum Field Theory of Point Particles and Strings, specifically chapter 2 --- on "Second Quantization".
But wait, there's more! If we consider other non-relativistic fields and attempt quantizing, e.g. the Newton Cartan theory of gravity, we also get spin-0 boson! For this result (specific to quantizing Newtonian gravity), see:
The Reason
Well, this should not surprise us, since the Schrodinger equation is the nonrelativistic limit to the Klein-Gordon equation. And we should recall the Klein-Gordon equation describes spin-0 bosons!
The nonrelativistic limit $c\to\infty$ should not affect the spin of the particles involved. (That's why the Pauli model is the nonrelativistic limit of the Dirac equation!)
Best Answer
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.