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

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?

• I don't think this question is unclear. The question is "Why is the bloch sphere a sphere, and not some other shape?" I provided an answer that was both upvoted and selected as an accepted answer, so I think it's clear that I was not incorrect in my interpretation of the question. Also considering that there is some discussion about if my accepted answer is even correct, I would think that this should be reopened for other answers. Apr 24 '19 at 19:05
• @StevenSagona This is not the question. The question is: "I'm wondering if there are other natural geometrical representations one could use as alternatives to the Bloch sphere." It is not asking why the sphere is a sphere. And I do think it is unclear. I could start writing an answer proposing a Sierpinski triangle and then elaborating why this is not particularly smart, as compared to a Bloch sphere. It is not clear what a good vs bad answer would be, because it does not have a clear scope. Apr 24 '19 at 20:31
• I think it's very reasonable to ask why the shape is a sphere (as opposed to a cube). The circular shape comes from the fact that the sum of the squares is one. The spherical shape is because of the need for an additional degree of freedom to describe (as discussed in my answer below). This information was likely not clear to the asker, and that's probably why (one year ago) the asker accepted my answer. I would think that your disagreement with the answer being correct is even further evidence that there is information pertinent to the asker's question that warrant discussion. Apr 24 '19 at 20:45

## 1 Answer

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.

• If the shape came from normalization, points inside the sphere should have norm <1. But they are normalized mixed states. Apr 5 '19 at 0:39
• I see your point. I think I'm still correct if you limit my answer to just pure states. Honestly I'm not sure what is what for mixed states. Maybe you can add a more complete answer involving mixed states? Apr 19 '19 at 22:40
• You are correct in that you make two correct statements. However, connecting them is completely arbitrary. Apr 19 '19 at 22:50
• I don't mean to be rude, but are you actually interested in me taking the time to clarify this? I wrote this to try to help people who are trying to learn, not to fight with people who already understand things. Apr 19 '19 at 23:13
• I don't see how you help people by providing incorrect explanations. Following your logic, there should be such a sphere picture for any state in $\mathbb C^d$, and points inside/outside the Bloch sphere would have norm >1 or <1, which is not the case. Apr 20 '19 at 10:43