Does the magnetic anisotropy state only have two possible directions? Wikipedia says "The magnetic moment of magnetically anisotropic materials will tend to align with an easy axis". Does this mean that it is completely impossible to orient the magnetic moment with any direction no matter how strong the magnetic field is? Or it is only about spontaneous magnetization?
 A: Magnetocrystalline anisotropy is all about spontaneous magnetization. Considere a (single domain) ferromagnetic material. Depending on its crystal structure, its spontaneous magnetization will tend to be aligned with a direction that minimize the interaction energy between magnetic dipoles carried by the atoms contained in one unit cell of the lattice.
That's why, depending on the geometry of the lattice system, you will have different types of magnetic anisotropy : uniaxial (one single easy axe), cubic (2 axes, one easy and one hard), tetragonal, etc.
So far, the most common anisotropy one can meet is the uniaxial one, that is particulary useful for the Stoner-Wohlfarth model.
Now, how about when an external magnetic field is applied? Well, there are two kind of energies in competition : the anisotropy energy which is minimum when the magnetization is along the direction of the easy axis, and the Zeeman energy which is the interaction energy between the magnetization and the external field. This situation is well described by the Stoner-Wohlfarth model and consists basically in minimizing the energy of the ferromagnetic system :
$$
\mathrm{E}(\theta,\phi)=K\sin^2\theta-\mu_0M_sH\cos(\theta-\phi)
$$
The first term is the anisotropy energy ($\theta$ being the angle between the easy axis and the magnetization, and $K$ the anisotropy strength), the second correspond to the Zeeman energy ($H$ is the norm of the applied magnetic field and $\phi$ the angle that it forms with the easy axis).
For convinience, let be $\phi=\pi$ fixed as an example. Then the energy $\mathrm{E}$ has to be minimize according to the variable $\theta$. In dimensionless units, one have :
$$
\mathrm{e}(\theta)=\frac{1}{K}\,\mathrm{E}(\theta,\phi=\pi)=\sin^2\theta+2h\cos\theta\quad\text{with}\quad h=\frac{\mu_0M_sH}{2K}
$$
One can easily find the equilibrium states $\theta_{\mathrm{eq}}$ by computing $\partial_\theta\,\mathrm{e}(\theta=\theta_{\mathrm{eq}})=0$ :
$$
\theta_{\mathrm{eq}}\equiv\{0,\pi,\theta_h=\arccos(h)\}
$$
and then study their stability by looking at the sign of the quantity $\partial^2_\theta\,\mathrm{e}(\theta=\theta_{\mathrm{eq}})$, namely :
$$
\partial^2_\theta\,\mathrm{e}(\theta=0)=2(1-h)
$$
$$
\partial^2_\theta\,\mathrm{e}(\theta=\pi)=2(1+h)
$$
$$
\partial^2_\theta\,\mathrm{e}(\theta=\theta_h)=2(h^2-1)
$$
The take home message here is that it is possible to change the stability of the different equilibrium solutions by varying the magnitude $H$ of the magnetic field, i.e. by varying $h$. By changing $h$, one can actually flip the magnetization from one equilibrium orientation to another.
