how can a unit vector n with 3 dimensions be translated to $|+n\rangle$ This question comes a from problem in a quantum physics book. In Townsend's modern quantum book (A modern approach to quantum mechanics) in chapter 1 problem 1.3 there is a unit given that is $\boldsymbol n = \sin(\Delta)\cos(\phi)\,\boldsymbol i  + \sin(\Delta)\sin(\phi)\, \boldsymbol j + \cos(\Delta)\, \boldsymbol k$
then it specifies that by looking at figure 1.11 you are given 
$|+n\rangle = \cos(\Delta/2)|+z\rangle + e^{i\phi}\sin(Δ/2)|-z\rangle.$ 
This brings me to my question, how is that so? how can you get $|+n\rangle$ from $n$? is there a way to do it or is it just by definition? excuse my ignorance, i'm just not seeing the connection. thanks in advance for anyone that can help me   
 A: The way to prove this is to use the fact that angular momentum (including spin) is the generator of rotations. So, a rotation operator is given by
$$\hat{R}(\theta,\mathbf{a}) = \exp\left(i \theta \frac{\hat{\mathbf{L}}\cdot\mathbf{a}}{\hbar}\right),$$
where $\mathbf{a}$ is the unit vector axis of rotation and $\theta$ is the angle.
For spin-$^1/_2$, we have the components of $\hat{\mathbf{L}}$ in the $\hat{S}_z$-basis are
\begin{align}
  \hat{L}_x =\hat{S}_x &= \frac{\hbar}{2} \left[\begin{array}{cc}
      0 & 1 \\
      1 & 0 \end{array}\right], \\
  \hat{L}_y =\hat{S}_x &= \frac{\hbar}{2} \left[\begin{array}{cc}
      0 & -i \\
      i & 0 \end{array}\right],\ \text{and} \\
  \hat{L}_z =\hat{S}_x &= \frac{\hbar}{2} \left[\begin{array}{cc}
      1 & 0 \\
      0 & -1 \end{array}\right],
\end{align}
which is just $\hbar/2$ times the Pauli matrices.
The rest is figuring out what combination of $\mathbf{a}$ and $\theta$ rotates the positive $z$-direction unit vector $\mathbf{k}$ into $\mathbf{n}$, constructing the matrix exponential, and applying it to $|z+\rangle$ and $|z-\rangle$ to get your new basis. You can also do it in two steps/rotations, requiring you to compute two matrix exponentials but simplifying the algebra of each considerably.
That is left as an exercise to the reader, since I know for a fact that this is a homework problem from Townsend.
The relationship to $3$-d vectors comes from using a different set of $\hat{\mathbf{L}}$ opertors. See, $3$-d vectors are spin-$1$, so you need $3\times3$ matrices to rotate them. They are described in a funny basis, though. For spin-$^1/_2$ standard practice is to work in an eigenbasis of exactly one component of the spin, $S_z$. If the spin-1 $\hat{L}_z$ eigenbasis is given by $|z_+\rangle$, $|z_0\rangle$, and $|z_-\rangle$, then the standard basis we use for ordinary vectors is: $|x_0\rangle$, $|y_0\rangle$, and $|z_0\rangle$. In that basis we have:
\begin{align}
  \hat{L}_x &= \frac{\hbar}{i} \left[\begin{array}{ccc}
      0 & 0 & 0 \\
      0 & 0 & -1 \\
      0 & 1 & 0\end{array}\right], \\
  \hat{L}_y &= \frac{\hbar}{i} \left[\begin{array}{ccc}
      0 & 0 & 1 \\
      0 & 0 & 0 \\
      -1 & 0 & 0\end{array}\right],\ \text{and} \\
  \hat{L}_z &= \frac{\hbar}{i} \left[\begin{array}{ccc}
      0 & -1 & 0 \\
      1 & 0 & 0 \\
      0 & 0 & 0 \end{array}\right],
\end{align}
as ca be seen by taking the derivative of the three basic rotation matrices with respect to $\theta$ in $3$-d, doing the same for $\hat{R}(\theta,\mathbf{a})$ above, setting them equal to each other, and then setting $\theta=0$. For practice, I would recommend reconstructing at least one of those three matrices using the technique above. Hint: start by calculating $\hat{\mathbf{L}}_i^1$, $\hat{\mathbf{L}}_i^2$, etc, until you find the pattern, then use that pattern to group like terms in $\exp(x)=1+x+x^2/2 +\ldots$.
A: I don't know the book, but I guess that the context is the Bloch sphere, which is a model of a state space of a 1-qubit system. 
A given point n on the sphere is described by a pair of coordinates $(\Delta, \Phi)$, and the state it encodes is denoted $|\mathbf{n}\rangle$. The only thing that misses is what state this corresponds to in the ordinary (vector space) representation of the state space: if $|+z\rangle, |-z\rangle$ form an orthonormal basis of your state space, then this state, corresponding to the point n on the sphere, corresponds to the linear combination of the basis states that you wrote down. 
In the end this is just a convention, but a nice one when $|+z\rangle$ and $|-z\rangle$ correspond to spin up and spin down in the $z$-direction. In that case any pair of antipodal points form the orthonormal basis of eigenstates of spin along the line that connects them. 
