Spin about an arbitrary axis This is based off question 4.30 from Griffith's Introduction to Quanum Mechanics. It asks for the matrix $\textbf{S}_r$ representing the component of spin angular momentum about an axis defined by: $$r = \sin{\theta}\cos{\phi}\hat{i}+\sin\theta\sin\phi\hat{j}+\cos\theta\hat{k}$$
for a spin = $1/2$ particle.The problem is, I can't visualize how the spin vectors relate to spatial coordinates. $$\chi_+ = \begin{pmatrix} 1 \\ 0\end{pmatrix}\ \text{ and } \chi_- = \begin{pmatrix} 0 \\ 1 \end{pmatrix}. $$
So these form a basis of some kind of space, but this is a space I don't understand, for example, why does: 
$$\chi^{(x)}_+ = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\1 \end{pmatrix} ?$$
I was going to try to use a rotation matrix with angles corresponding to my axis of rotation to try and "rotate" the spin vector onto it, but then I realized that not only do the dimensions not match up, but the angle is π/4, which doesn't make much sense to me. So I guess my question is, how does the 'geometry' of spin work and how can I transform the spins corresponding to a transform I do in space?
Thanks.
 A: Great and important question; I hope this response is illuminating and encourages you and others to explore Lie Groups, Lie Algebras, and their representations.
When you want to rotate a vector $\mathbf v$ in three dimensions, then you act on that vector with a rotation matrix $R$ to obtain a rotated vector $\mathbf v'$ related to the original vector by
$$
  \mathbf v' = R\mathbf v
$$
It is a mathematical fact that every rotation (special orthogonal transformation) $R$ is a rotation by some angle $\theta$ about some axis defined by a unit vector $\mathbf n$, and that each such rotation can be written as the matrix exponential of a particular linear combination of certain 3-by-3 matrices $J_i$ called rotation generators.  Explicitly
$$
  R(\theta, \mathbf n) = e^{-i\theta n_i J_i}, \qquad (J_i)_{jk} = i\epsilon_{ijk}
$$
So we can write the rotation of a vector as
$$
  \mathbf v' = e^{-i\theta n_i J_i} \mathbf v
$$
It turns out, that we can define the rotation of spinors in an analogous way.  Instead of the rotation generators $J_i$ which are 3-by-3 matrices, we choose the pauli matrices which are 2-by-2, and for a given spinor $\xi$, we define a rotated spinor by
$$
  \chi' = e^{-i\frac{\theta}{2} n_i\sigma_i}\chi
$$
Notice how this is basically the same as rotating a vector in 3 dimensions, it's just that we have represented rotations acting on the vector space of spinors in a different way.
The math behind all of this is called representation theory.  In particular, when we are talking about spin, the representation theory is related to that of the Lie groups $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$ and their so-called Lie algebras and their representations.
A: In the $z$ basis, any spinor of the form
$$ s=\frac{1}{\sqrt 2} \begin{bmatrix} e^{i\alpha} \\ e^{i\beta}\end{bmatrix} $$
can be an $x$ basis vactor. If we measure $s$ along the $z$ axis, we have 50%-50% chance to get up or down spins. Hence, $s$ is in the $xy$ plane. It is our choice how we want to define the basis vector.
A: 
The problem is,
I can't visualize how the spin vectors relate to spatial coordinates.

A spinor is a "vector" with two complex components.
$$\chi = \begin{pmatrix}a \\ b \end{pmatrix}
= a \chi_+ + b \chi_-$$
From such a spinor you can always calculate the following 3D vector (i.e. with three real components):
$$\begin{align}
\mathbf{n}
 &= \langle\chi|\boldsymbol{\sigma}|\chi\rangle \\
 &= \langle\chi|(\sigma_x\hat{\mathbf{i}}+\sigma_y\hat{\mathbf{j}}+\sigma_z\hat{\mathbf{k}})|\chi\rangle \\
 &= \langle\chi|\sigma_x|\chi\rangle \hat{\mathbf{i}}
  + \langle\chi|\sigma_y|\chi\rangle \hat{\mathbf{j}}
  + \langle\chi|\sigma_z|\chi\rangle \hat{\mathbf{k}} \\
 &= \begin{pmatrix}a^*&b^*\end{pmatrix} \begin{pmatrix}0& 1\\1&0\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} \hat{\mathbf{i}}
  + \begin{pmatrix}a^*&b^*\end{pmatrix} \begin{pmatrix}0&-i\\i&0\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} \hat{\mathbf{j}}
  + \begin{pmatrix}a^*&b^*\end{pmatrix} \begin{pmatrix}1& 0\\0&-1\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} \hat{\mathbf{k}} \\
 &= (a^*b+b^*a) \hat{\mathbf{i}} + (-ia^*b+ib^*a) \hat{\mathbf{j}} + (a^*a-b^*b) \hat{\mathbf{k}}
\end{align}$$
When the spinor $\chi$ is normalized, then the vector $\mathbf{n}$ is a unit vector.
You can visualize it as being the axis of rotation of the spinning particle.
Examples:

*

*for $\chi=\begin{pmatrix}1\\0\end{pmatrix}$
you get $\mathbf{n}=\hat{\mathbf{k}}$,

*for $\chi=\begin{pmatrix}0\\1\end{pmatrix}$
you get $\mathbf{n}=-\hat{\mathbf{k}}$,

*for $\chi=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$
you get $\mathbf{n}=\hat{\mathbf{i}}$.

