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

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

• You are projecting onto two planes, you can basically think of this as spinning around vertically and spinning around horizontally, therefore you only need two angles in polar coordinates to span the space. So even though there are 3 axis you only need two rotations. May 29 at 21:51
• I would naively have said that there is just an angle missing for y instead, as the polar angle theta already is defined with respect to z axis. But another angle from y axis to the dotted line would be equivalent to the azimuthal angle phi. Jun 1 at 17:08

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.
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.