Why is there no angle for the $z$ axis in the Bloch sphere?

Question

I see that in Bloch spheres, there is an angle for the $x$ and $y$ axes but not for the $z$ axis. Why?

Answer 1

This is because we are using a spherical coordinate system. You can think of this coordinate system as specifying a radius, then two angular coordinates $\theta$ and $\phi$. Pure single-qubit states have a Bloch vector with radius equal to unity, so they only need two more coordinates to fully specify the direction in which the vector points.

When you look at a map of the Earth, everyone moves around at approximately the same radius, but our two angular coordinates can change. You might be familiar with latitude and longitude - these are two coordinates describing where we are on the surface of the Earth, just like the two coordinates $\theta$ and $\phi$ describe where a vector is on the surface of the Bloch sphere.

Another way of approaching the problem: three-dimensional vectors have three free parameters. Once we specify a radius, we only need two more parameters to fully specify the vector.

Answer 2

Because for a general state $|\phi\rangle = \alpha |0\rangle + \beta |1\rangle$ with $\alpha,\beta \in \mathbb{C}$, we want to describe states with $\langle \phi|\phi\rangle = 1$. There are 4 degrees of freedom, two for the real and imaginary components of $\alpha$ and $\beta$ each.

The normalization condition $\langle \phi|\phi\rangle = 1$ reduces that to 3 DoF and restricts us to a sphere. Additionally, a global phase change (rotation of both $\alpha$ and $\beta$ in the complex plane) by $\theta$ leads to $|\psi\rangle = \exp(i \theta)|\phi\rangle$ and $$ \langle \psi | A | \psi \rangle = \exp(-i \theta)\exp(i \theta) \langle \phi | A | \phi \rangle = \langle \phi | A | \phi \rangle \text{,}$$ which means that this is an extra degree of freedom, that we also cannot measure in experiments. So we just don't use it and the description as a point on a sphere is valid.