# Why is the Bloch sphere three-dimensional?

I can understand the intuition behind a two dimensional bloch circle, as it represents the probability distribution of a certain state vector. However, I fail to grasp what the third dimension adds to the visualization when it comes to probabilities. How does it help? What is it supposed to add to the model that the Bloch circle didn't?

Since length is always 1, then what's the point of adding a z axis?

A qubit is a two-level quantum system and hence it can be written as:

$$|\psi \rangle = \alpha |0\rangle + \beta|1\rangle$$

where $$|0 \rangle$$ and $$|1\rangle$$ are the computational basis and they defined as $$|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \ \ \ \ |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

and $$\alpha, \beta \in \mathbb{C}$$. So we have that $$|\psi \rangle \in \mathbb{C}^2$$ and you can think of $$\mathbb{C}^2$$ as $$\mathbb{R}^4$$. However, the overall phase in the quantum state doesn't matter, that is, $$|\psi\rangle$$ is equivalent to $$e^{i\phi} |\psi \rangle$$. Thus, you can factor this overall phase out of $$|\psi\rangle$$ and have only 3 parameters left. Essentially now you are considering the space $$\mathbb{R}^3$$. Along with the normalization requirement, that a quantum state must have unit norm, this forces it to be a unit-sphere, which we call the Bloch-Sphere.

• Why doesn't the overall phase of the quantum state matter? I'm not familiar with Euler's formula yet, I'll look into it. I'm trying to see how those two state vectors are equivalent. Oct 14 '20 at 19:17
• Suppose $|\psi_1 \rangle = \alpha |0\rangle + \beta |1 \rangle$ and $|\psi_2 \rangle = e^{i\phi} \alpha |0\rangle + e^{i\phi} \beta |1 \rangle$. Then the probability of observing the state $|0\rangle$ in $|\psi_1 \rangle$ is $|\alpha|^2$, which is the same as the probability of observing $|0\rangle$ in $|\psi_2$ since $|e^{i\phi} \alpha |^2 = |e^{i\phi}| |\alpha |^2 = 1\cdot |\alpha |^2$. Similarly, the probability of observing $|1\rangle$ in $|\psi_1 \rangle$ is the same as the probability of observing $|1\rangle$ in $| \psi_2 \rangle$ Oct 14 '20 at 22:35
• Adding to the above comment by @KAJ226, since the overall ("global") phase of the quantum state can not be determined via a physical measurement, it has no physical significance and might as well be factored out. Oct 15 '20 at 18:30
• @KAJ226 I get how the global phase factor doesn't need to be represented, so what are the other three parameters/dimensions supposed to mean in that case? Oct 18 '20 at 12:55
• @Sinestro38 There aren't three free parameters, there's two. We start with four, because two complex numbers is characterized by 4 parameters. But one is eliminated by global phase invariance, and another is eliminated by the constraint that the square amplitudes of the two complex numbers must sum to one. We're left with two, which happens to have the geometry of the surface of a 3-dimensional sphere ($S2$). While the $S2$ geometry technically can be represented on a two-dimensional picture, it's just easier to visualize it as the surface of a 3-dimensional sphere. Jun 10 at 16:13

You're not trying to visualise probabilities. You're trying to visualise the much richer structure afforded by probability amplitudes. For a single qubit, the probability amplitudes require 3 real parameters, hence three dimensional space.

The Bloch sphere represents state with the following probability amplitudes: $$|\psi\rangle = \cos(\theta/2)\ |0\rangle +e^{i\phi} \sin(\theta/2)\ |1 \rangle,$$ where $$\theta$$ and $$\phi$$ are the angles on the Bloch sphere. Each point on the sphere represents a different state. As mentioned by other answers these $$\cos(\theta/2)$$ and $$e^{i\phi} \sin(\theta/2)$$ are not probabilities of the state being in the state $$0$$ or $$1$$, rather they represents the probability amplitudes. To get a probability form a probability amplitude, we square the probability amplitudes. That is, $$\cos^2(\theta/2)$$ and $$\sin^2(\theta/2)$$ are the probabilities, and you see they add up to $$1$$.

You also notice that the angle $$\phi$$ does not show up in the expressions for probabilities. Does this mean that we can just ignore $$\phi$$ all together? No. If we could then indeed we wouldn't need a 2-dimensional sphere to represents the states.

The reason we cannot ignore the angle $$\phi$$ from our considerations is because of the superposition principle of quantum mechanics. Suppose we have, $$|\psi_1\rangle = \frac{1}{\sqrt{2}}( |0\rangle + |1 \rangle), \ \ \ \ \ \ \ \ \ |\psi_2\rangle = \frac{1}{\sqrt{2}}( |0\rangle +e^{i\phi} \ |1 \rangle.$$ Then the superposition principle tells us that we can make a third state out of their superposition, $$|\psi_3\rangle = N(|\psi_1\rangle + |\psi_2\rangle) = \sqrt{\frac{2}{3+\cos(\phi)}}\left(|0\rangle + \frac{1+e^{i\phi}}{2} |1\rangle \right) ,$$ where $$N$$ is some normalization factor written down explicitly in the second equality. Anyway, the point here is now $$\phi$$ is very important. For example the probability of the state $$|\psi_3 \rangle$$ being in the $$|1\rangle$$ state is, $$P(1) = \frac{1+\cos(\phi)}{3+\cos(\phi)}$$ which very much depends on $$\phi$$. In fact this probability ranges from $$0.5$$ when $$\phi = 0$$ to $$0$$ when $$\phi = \pi$$.