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

enter image description here

  • 2
    $\begingroup$ 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. $\endgroup$
    – Sam Palmer
    Commented May 29, 2021 at 21:51
  • $\begingroup$ 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. $\endgroup$
    – Mauricio
    Commented Jun 1, 2021 at 17:08

2 Answers 2


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.