According to the Wikipedia article on the Bloch sphere, a pure state of a qubit can always be represented as $$| \psi \rangle = \cos \left( \frac{\theta}{2} \right)| 0 \rangle + e^{i \phi} \sin\left(\frac{\theta}{2}\right)|1 \rangle$$
The parameters $\theta$ and $\phi$ specify a point $$\vec{a} = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$ on the unit sphere in $\mathbb{R}^3$, as depicted below
Question:
If $\hat{z} = |0 \rangle$ and $-\hat{z} = |1 \rangle$ then wouldn't any linear combination of $|0 \rangle$ and $|1 \rangle$ be along the $z$ axis again with complex coefficients? Also, is $| \psi \rangle = \vec{a}$?
Best Answer
The Bloch sphere is not a vector space. In particular, the "vectors"/arrows you draw on it cannot be added like usual vectors.
The Bloch sphere is what you get when you take the two-dimensional complex vector space $\mathbb{C}^2$ and ask what you get when you make no distinction between vectors that only differ by multiplication with a complex scalar. This is called a projective Hilbert space.
Note that $\hat{z}$ and $-\hat{z}$ differ only by multiplication by the scalar $-1$, so they would be identified if the points on the Bloch sphere were still vectors in the original Hilbert space. They are not and the only points corresponding to pure states are really on the sphere, neither in its interior nor its exterior.
Where a linear combination of $\lvert 0\rangle$ and $\lvert 1 \rangle$ sits on the Bloch sphere is exactly what your formula for $\lvert \psi\rangle$ tells you - the linear combination $\lvert \psi\rangle$ sits at the coordinates $(\phi,\theta)$ on the sphere, where those are the usual polar coordinates on a sphere.
Writing $\lvert 0\rangle = \hat{z}$ and $\lvert 1\rangle = -\hat{z}$ is very confusing - the equality is not an equality as vectors, it is just an equality as points on the sphere. You get to writing this by looking at your general form for $\lvert \psi\rangle$ and defining the map $\mathbb{C}P^2\to S^2, \lvert \psi\rangle\mapsto (\phi,\theta)$ using polar coordinates for the $S^2$, then by embedding the $S^2$ into an $\mathbb{R}^3$ and then observing that the two poles get mapped to the $\hat{z}$ and $-\hat{z}$ in the $\mathbb{R}^3$.