Bloch sphere is a physical representation of all possible qubit states.

Each qubit is in its essence a vector on Bloch’s sphere. Each vector on the sphere can be represented in two basis: θ and ϕ. The first is θ which is the angle between the vector and the z-axis. The second is ϕ which is the angle between the vector and the positive x-axis measuring counter-clockwise. Here is a visual representation of what was just described.  From the picture, it is clear that we can achieve all possible vectors in the Bloch sphere using these two angles, even with limits on them. θ is between 0 and π, inclusive. ϕ is between 0 and 2π, inclusive.

Now that Bloch sphere is understood, we can move on to the “equation” for a single qubit. As a side note, qubits are in the notation |> in order to distinguish them from normal bits.

$$|ψ\rangle = cos(\frac{θ}{2})|0\rangle + sin(\frac{θ}{2})e^{iϕ}|1\rangle$$

Now, one question which I (Eric Li) had was what did the x- and y-axes represent? As we all know, the z-axis represents \(|0\rangle\) and \(|1\rangle\) — these also are analogous to the North and South pole of Earth where the North is \(|0\rangle\) and South is \(|1\rangle\) — and thus, any vector that isn’t directly \(|0\rangle\) or \(|1\rangle\) would account for some superposition of both. In terms of the equation, \(|0\rangle\) would be \(\theta = 0\) and \(\phi\) doesn’t matter. Similarly, \(|1\rangle\) would be \(\theta = \pi\) and \(\phi\) would also be irrelevant. While on the z-axis, the z-axis is 0 and 1 and these represent actual values as a classical bit does, the other axes do not. However, that does not constitute that these other axes are insignificant. In fact, it is these other axes that play a role in some of the most useful quantum logic gates and algorithms. Now, we can move on to actually calculating what the two axes represent.

Starting with the positive x-axis, it is obvious that θ would be π/2. Similarly, ϕ is 0. By plugging these numbers into the aforementioned formula, the equation of the qubit is

$$|ψ\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$

Now, moving on to the negative x-axis, θ is still π/2, but ϕ is π. By plugging these numbers into the same formula, the equation of the qubit is

$$|ψ\rangle = \frac{1}{\sqrt{2}}(|0\rangle – |1\rangle)$$

Since one has an addition sign and one has a subtraction sign, the positive x-axis has been dubbed “positive” and the negative x-axis has been dubbed “negative.”

Now, onto the y-axis. The positive y-axis has θ = π/2 as well as ϕ = π/2. Thus, plugging in the values into the equation gives $$|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$$

Thus, the negative y-axis would be $$|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle – i|1\rangle)$$