The Hamiltonian of a spin 1/2 system in a magnetic field $\mathbf{B} = B \hat{\mathbf{n}}$ is \begin{equation}\hat{H} = – \frac{e}{mc} \hat{\boldsymbol{\sigma}} \cdot \mathbf{B} \end{equation}
where $\hat{\mathbf{n}}$ is an arbitrary vector and $\hat{\boldsymbol{\sigma}}$ the vector of Pauli matrices, i.e. $\hat{\boldsymbol{\sigma}} = (\sigma_1,\sigma_2,\sigma_3)$.
Now the problem is to find the eigenspinors of the Hamiltonian.
My first idea (which works fine) was to first consider the system with $\hat{\mathbf{n}} = (0,0,1)$:
\begin{equation}
\hat{\boldsymbol{\sigma}} \cdot \hat{\mathbf{n}} = \sigma_3
\end{equation}
In this case the eigenspinors are known and by rotating the system it is possible to find the eigenspinors for the system with arbitrary $\hat{\mathbf{n}}$.
More specifically, one eigenspinor (before rotation) is $\chi_+ = (1,0)$ and applying the rotation in SU(2):
\begin{equation}
e^{-\frac{i}{\hbar} \sigma_z \varphi} e^{-\frac{i}{\hbar} \sigma_y \theta}\chi_+ = \begin{pmatrix}e^{-i\varphi/2} \cos(\theta/2) \\ e^{i\varphi/2} \sin(\theta/2)\end{pmatrix}
\end{equation}
which is (up to a phase factor) the result from Wikipedia (https://en.wikipedia.org/wiki/Eigenspinor)
Another way is to start with an arbitrary magnetic field and then compute the eigenvectors of the Hamiltonian. That is, one rotates the vector $\hat{\mathbf{n}} = (0,0,1)$ such that afterwards $\hat{\mathbf{n}} = (\cos \varphi \sin \theta, \sin \varphi \sin \theta,\cos \theta)$. Doing this one finds the same eigenspinors.
So I see that the results are the same, however, I don't really understand how the 3-dimensional rotation (of the space) is related to the 2-dimensional one (of the spinor).
I know that SU(2) is a double cover of SO(3) but I don't see how one would formally relate the two in the above example.
I guess that the answer is somehow related to the homomorphism between the groups (or their Lie algebra) but I confused myself so much that I can't figure it out.
Best Answer
Let $$ \hat{\sigma}\cdot{\underline{n}}= n^1\hat{\sigma}_1 + n^2\hat{\sigma}_2+n^3\hat{\sigma}_3 = \hat{S}(\theta,\phi)\hat{\sigma}_3 $$ namely any combination of Pauli matrices can always be written as the product of one other Pauli matrix times another object in $SU(2)$. The matrix $\hat{S}(\theta,\phi)$ is, in the case at hand, the element $$ \hat{S}(\theta, \phi) = \begin{pmatrix} e^{-i\phi/2}\cos(\theta/2)\quad e^{-i\phi/2}\sin(\theta/2)\\ e^{i\phi/2}\sin(\theta/2)\quad e^{i\phi/2}\cos(\theta/2) \end{pmatrix}. $$ Now let us associate two vectors in $\mathbb{R}^3$ to the above equation, one to the left hand side and one to the right hand side, namely $$ (n^1, n^2, n^3)\mapsto(0,0,n'^3) $$ as the right hand side only possesses the $\sigma_3$. Let $\hat{R}(\theta, \phi)$ the transformation matrix such that $$ \hat{R}(\theta, \phi)\,(n^1, n^2, n^3) = (0,0,n'^3). $$ As such we have now two objects at our disposal: $$ \hat{S}(\theta, \phi)\in SU(2),\qquad \hat{R}(\theta,\phi)\in SO(3); $$ the map $\rho\colon S\mapsto R$ for each pair $(\theta, \phi)$ is the desired double covering map (one can show that both $S$ and $-S$ correspond to the same $R$).
Because of the above argument multiplying the $\sigma$ by $\hat{S}$ is equivalent to multiplying $\underline{n}$ by $\hat{R}$, that is why the eigenvalues are the same no matter whichever multiplication you perform first.