Are there natural geometrical representations for a qubit other than the Bloch sphere? [closed]

Question

The Bloch Sphere is a geometrical representation of the state space of a qubit system. I'm wondering if there are other natural geometrical representations one could use as alternatives to the Bloch sphere. For example, could you use a cube to geometrically represent the state space of a qubit system? If so, why is the Bloch sphere a better representation?

Answer 1

Here's a loose intuition for why the shape of the Bloch sphere is a sphere. (For something more rigorous, check out this answer.)

The bloch sphere represents a visualization (on a sphere) of the set of all possibilities that can be assigned to a qubit. To get an idea why the shape is a sphere, I will show that there are two independent angles that each can form circles - these circles represent the "equator" and "prime meridian".

First (assuming we have pure states), a qubit can be represented as

$cos(\theta) |0\rangle + sin(\theta) e^{i \phi} |1\rangle$

These two independent angles each can be plotted to form their own independent circles. Let's first cover how $\theta$ forms a circle.

(For a fixed $\phi = 0$), if we plot the probability amplitudes of the state as a vector ($|0\rangle$, in the x-direction, and $|1\rangle$ in the y-direction), you see that as we plot each posible $\theta$ value, we obtain a "circle" of possiblities. Also note that this circle follows the constraint requiring the probabilities to add up to 1: $|c_0|^2 + |c_1|^2$ = 1

Now, this constraint embedded in $\theta$ tells us how the real parts of our amplitudes are constrained, but we don't have any information about the imaginary parts of our amplitudes.

In polar form we can represent any complex number as $c = |c|e^{i\phi}$, and we can graph this on the "complex plane." If we vary $\phi$, but we hold |c| constant, then we see this form a circle. The key here is that to represent complex numbers, we add an additional dimension to represent values that take on i.

So now we see that we have formed two independent circles. These form an "equator" and "prime meridian" for the larger set of possiblities. The dimensions for the 3D plot forming a circle is {Real Part of $|0\rangle$, Real Part of $|1\rangle$, Imag Part of $|1\rangle$}

To get the entire sphere we can just find what each of these three dimensions are for all values of $\theta$ and $phi$. Let's work these out so we can plot an entire sphere.

Real Part of $|0\rangle$: $cos(\theta)$

Real Part of $|1\rangle$: $sin(\theta)*sin(\phi)$

Imag Part of $|1\rangle$: $cos(\theta)*sin(\phi)$

If you're familiar with polar coordinates this describes the equation for a circle with a radius of 1. The points on the surface of the sphere represent the set of all possible values that can be assigned to the probability amplitudes assigned a single qubit.

EDIT: I "fleshed out this answer" as sort-of requested in the comments. Also please note, the following explanation only covers "pure states" and a different discription would be required for mixed states.