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!)
Why are fondamental particles either bosons or fermions?
A particle specie is either a boson or a fermion depending on how the wave function changes when permuting two particles of the same specie. A general two particles wave function $\Psi(x_1,x_2)$ can be acted on by the operator $P$ which permutes the two particles. Of course, $P^2\Psi(x_1,x_2) = \Psi(x_1,x_2)$ so $1$ is an eigenvalue of $P^2$ and so the only two possible eigenvalues for $P$ are $\pm1$, $+$ for bosons and $-$ for fermions.
Why are composites also either bosons or fermions?
To answer this, I will cite the excellent answer to the question "How to combine two particles of spin $\frac{1}{2}$?". The main equation to remember is following with it's interpretation:
$$
(2j_a+1)\otimes(2j_b+1) = \bigoplus_{i=1}^n(2j_i+1),
$$
On the left-handed side, the object describes the hilbert space of two particles, one with spin $j_a$ the other with spin $j_b$. The particles are decoupled so they have a fix total spin and can have any corresponding projection.
On the right-handed side, the object describes the hilbert space of a single composite particle which have can change total spin $j_i \in \{|j_a-j_b|, |j_a-j_b+1|, ..., j_a+j_b\}$.
Both spaces are related by a change of basis. What you need to see is that the total spin, even if it changes, is either an integer or an half-integer. If both particles are fermions or bosons, the composite particle is a boson, but else, the composite particle is a fermion.
Best Answer
The Dirac equation does describe composite spin-1/2 fermions - namely, baryons like the proton and the neutron. Conversely, future experiments might reveal the electron to be composite even though it's described by the Dirac equation (plus perturbative corrections).
The vertex term you describe does appear in the scattering cross-section for proton-photon scattering, but it's corrected by loop-level renormalization terms that stem from interactions, which are tiny (but measurable) for the electron but large for the proton.