Eigenvector of spin half particle in applied magnetic field at angle I am very new to this field of physics so sorry if this is basic. I was recently trying understand how you go about calculating energy splits of electrons in applied fields. I understand that given a magnetic field is along the z axis the wave the eigenvectors [0.5;0] and [0;0.5] still hold for spin up, spin down, and from that I can easily calculate the Eigenvalues and hence the energy levels of the two states. 
However where I am stuck is when I introduce another component of the magnetic field so that my Hamiltonian is then proportional to BzIz+BxIx. Previously when the field was only applied along one axis it was simply proportional to BzIz. The reason this confuses me is that now the states [0.5;0] and [0;0.5] are no longer eigenstates. Instead the Eigenstates are in the form A[0.5;0]+B[0,;0.5], where A and B are constants depending on Bz and Bx.
I believed a wave function like this to be a superposition of two states? This must be wrong as a wave function shouldn't collapse into a superposition... But then again I believed it to be the case that wave functions always collapse into an eigenstate of the operator being used... Safe to say I am confused, apologies if this is basic...
 A: If you have an external magnetic field of strength $B$ in the direction $(n_x,n_y,n_z)$ then the Hamiltonian has a term proportional to $$B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z.$$
And I put the hats on the operators so you know they are operators on a Hilbert Space rather than matrices.  Indeed you could choose a basis for your Hilbert Space that happens to be eigenvectors of $\hat\sigma_z$ and then all vectors in the Hilbert Space are just column vectors of complex numbers.
And then you can use the matrix representation $$B_x\sigma_x+B_y\sigma_y+B_z\sigma_z.$$
And the energy eigenstates can now have parts that are eigen to these operators/matrices. But the basis you choose to write something is just as arbitrary as the coordinate axis.
So in general, the natural basis is in terms of the eigenvectors to the actual Hamiltonian, but you can use any basis. And if you choose the wrong basis, your states can look more complicated, but they aren't different.
It's like if someone moved in a line at constant speed. You could pick your x axis to point in that direction, and the motion would look simple, e.g. $\vec r(t)=vt\hat i+0\hat j$, but you could also pick a coordinate system where it looks like $\vec r(t)=\frac{\sqrt 2}{2}vt\hat i'\frac{\sqrt 2}{2}vt\hat j'.$ And I don't mean like as in an analogy. I mean literally you are choosing a basis.
What can be confusing is that there is a basis for physical space and a basis for your Hilbert Space and they are related but different. When you write something as a bunch of scalars in a particular order then you picked a basis. The basis might make your math looks cluttered or simple, but it doesn't change the physics.
So you really have a magnetic field, there really is an operator $$B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z.$$ And it is a Hermitian operator and it has eigenvectors and writing it in terms of that basis (the basis of eigenvectors) might make your equations look the nicest just like $ \vec r(t)=vt\hat i+0\hat j$ looked nicer than $\vec r(t)=\frac{\sqrt 2}{2}vt\hat i'\frac{\sqrt 2}{2}vt\hat j'.$  But no physics is different.

I believed a wave function like this to be a superposition of two states? 

The word superposition is just the word linear combination. Any vector is a superposition of other vectors. It isn't physically meaningful. It just means your basis didn't include the vector in question.
When someone makes a big deal about a superposition, they are actually trying to make a big deal about it being a superposition of a particular set of basis vectors.
