Is the Bloch sphere a bad way to visualize a qubit? When a qubits is in a quantum state it can be measured as $[0\rangle$ or $[1\rangle$. Then why does the Bloch sphere have these two states on antipodal sides of the spheres? If I want to plot the position of an object and their velocity, then I don't put position on one side of the axis and velocity on the other side. I would put them on different axis.
I read an explanation here, but you get some practical issues when you use the Bloch sphere. These different quantum states are projected on the same place on the Bloch sphere:

*

*$[0\rangle$

*$i[0\rangle$

*$-[0\rangle$

*$-i[0\rangle$
Why can't the real probabilities be projected like this?

Then in three dimensions you can add an axis $i[0\rangle$ and in four dimensions you add an axis $i[1\rangle$.
 A: This is a very good question that many people ask when they learn about the Bloch sphere. There are two important reasons
Physical reason
The most famous two state system in quantum mechanics is the spin 1/2 particle. When you draw the spin 1/2 states on the Bloch sphere something very nice happens, the points on the Bloch sphere correspond to the physical directions of the states. For example the $|\uparrow\,\rangle$ state corresponds to a point that points in the $+z$ direction. Similarly the state $|x+\rangle=\tfrac{ 1}{\sqrt{2}}(|\uparrow\,\rangle+|\downarrow\,\rangle)$ points in the $+x$ direction on the Bloch sphere. To be more precise for some state $|\psi\rangle$ the vector of expectation values given by $\langle\psi|\vec S|\psi\rangle=\langle \vec S\rangle =(\langle S_x\rangle,\langle S_y\rangle,\langle S_z\rangle)^T$ points in the same direction as the Bloch vector.
See also this picture:

Mathematical reason
Another reason that the Bloch representation is good is that states which are physically the same correspond to the same point on the Bloch sphere. This is also mentioned in the comments. In your diagram $|0\rangle$ and $-|0\rangle$ are antipodal points but they represent the same state up to an arbitrary phase. Remember that $|\psi\rangle$ and $e^{i\alpha}|\psi\rangle$ represent the same state. The Bloch sphere intuitively leaves out any global phase. To represent the full state on a Bloch sphere you would have to specify an additional phase for any point. In Steane's "Introduction to Spinors" this is represented by drawing a state as a little flag. The direction of the flag gives the point on the Bloch sphere and the angle that flag makes gives the global phase. So in this representation the states $|0\rangle$ and $-|0\rangle$ both point in the z-direction but the first one has its flag rotated to $\alpha=0$ and the second one has it rotated to $\alpha=\pi$ radians. Interestingly when you rotate a spin 1/2 particle by $360^\circ$ its state will pick up a minus sign and you would have to rotate it by another $360^\circ$ before it becomes exactly the same state.

