Adding to what @pyramids conveyed in their answer:

A qubit's state is generally written as $\alpha|0\rangle + \beta|1\rangle$, where $\alpha, \beta \in \Bbb{C}$, and $|\alpha|^2+|\beta|^2=1$.

$\Bbb{C}^2(\Bbb{R})$ is a four-dimensional vector space, over the field of real numbers. Since any $n$-dimensional real vector space is isomorphic to $\Bbb{R}^n(\Bbb{R})$, you can represent any qubit's state as a point in a $4$-dimensional real space, too, whose basis vectors you can consider to be $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$. In such a case a qubit's state would be represented as $a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1)$.

Say, $\alpha = a + i b$ (where $a,b\in \Bbb{R}$) and $\beta = c + id$ (where $c,d\in \Bbb{R}$). You need the condition $|a+ib|^2+|c+id|^2=1\implies a^2+b^2+c^2+d^2=1$ to be satisfied, which implies the state of the qubit would be a point on a 3-sphere.

As you know, it is difficult to efficiently represent a $4$-dimensional space on a $2$-dimensional surface like a paper, or your screen. Hence, you don't see that representation used often. Bloch sphere is pretty much the **most efficient representation** out there (for a single qubit), since it reduces one degree of freedom (of the complex numbers $\alpha,\beta$ each of which have two degrees of freedom) due to the fact that a qubit's state is usually normalized to a magnitude of $1$ i.e. $|\alpha|^2+|\beta|^2=1$.

Now, using the Hopf
coordinates
let's say:

$$\alpha = e^{i\psi}\cos(\theta/2)$$

$$\beta = e^{i(\psi+\phi)}\sin(\theta/2)$$

Here, $\theta$ can run from $0$ to $\pi$ whereas, $\psi$ and
$\phi+\psi$ can take values between $0$ to $\pi$.

*In case you're wondering why $\theta/2$ is being used instead of $\theta$ have a look at the answers on this excellent thread on Physics Stack Exchange.*

Okay, even now you notice three degrees of freedom $\psi,\phi,\theta$, whereas in a unit radii sphere, you only have two angles which you can change to get the different states of a qubit.

Notice that $\phi$ is basically the "relative phase" between $\alpha$ and $\beta$. On the other hand $\psi$ does not contribute to the "relative phase" of $\alpha,\beta$. Also, neither $\phi$ nor $\psi$ contribute to the magnitude of $\alpha,\beta$ (since $|e^{i\varphi}|=1$ for any angle $\varphi$). Since $\psi$ contributes neither to "relative phase" nor to the "magnitudes" of $\alpha,\beta$ it is said to have **no physically observable consequences** and we can *arbitrarily choose* $\alpha$ to be real by eliminating the factor of $e^{i\psi}$.

Thus we end up with:

$$\alpha = \cos(\theta/2)$$ and $$\beta=e^{i\phi}\sin(\theta/2)$$
Where $\theta$ can run from $0$ to $\pi$, and $\phi$ can run from $0$ to $2\pi$.

This practical simplification allows you to represent a qubit's state using just $2$ degrees of freedom on $3$-dimensional spherical surface having unit radius, which again can again efficiently be "drawn" on a $2$-dimensional surface, as shown in the following image.

**Mathematically, it is not possible to reduce the degrees of freedom any further, and so, I'd say there is no other "more efficient" geometrical representation of a single qubit than the Bloch sphere.**

_{Source: Wikipedia:Bloch_Sphere}

Thank you for your answer. Please, can you add a very brief description of how to represent a qubit (not qutrit) on a Majorana sphere? Then I'll mark this answer as accepted because it answers perfectly my question. – incud – 2018-04-09T06:51:54.697

@incud - Added another paper at the top that's a bit easier going and directly qubit oriented. – Rob – 2018-04-09T14:14:47.793