# Visualize representation of a qubit

I am new to quantum computing and I am learning about qubits.

I recently learnt a qubit can be represented by $$\alpha |0\rangle + \beta|1\rangle$$ where $$\alpha, \beta \in \mathbb{C}$$ and $$|\alpha|^2 + |\beta|^2=1$$. After a while, I realized most of the examples I have seen are simplified, in the sense that the imaginary component of both $$\alpha$$ and $$\beta$$ are zero. For something like $$\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$$ we can draw a unit circle to visualize it.

My question has two related parts:

1. Could you provide an instance of a qubit that has all non-zero coefficients in its probability amplitudes $$\alpha$$ and $$\beta$$, for both real and imaginary components?

2. How do we visualize that?

I am aware of Bloch sphere representation, but I would like to know if we can visualize such qubit without the use of Bloch sphere.

One certainly can give examples of the form you're asking for. Here's one: $$\frac{1}{\sqrt{2}}(e^{i\pi/3}|0\rangle+e^{-i\pi/3}|1\rangle).$$ However, note that you will essentially never see an example where both amplitudes are complex. This is because you can pull out one of the phases as a global factor, something like $$=\frac{e^{i\pi/3}}{\sqrt{2}}(|0\rangle+e^{-i2\pi/3}|1\rangle).$$ Moreover, this global phase factor has no observable consequences, so we just drop it $$\equiv\frac{1}{\sqrt{2}}(|0\rangle-\frac{1+i\sqrt{3}}{2}|1\rangle).$$
In terms of visualisation, the Bloch sphere that you mention is the only one that has the full capacity to uniquely plot every different single-qubit state. However, if you restrict to certain classes of state, that corresponds to a cross-section through the Bloch sphere, and hence just requires a unit circle. You've already seen one example of this, where the amplitudes are real, so you can write $$\cos\theta|0\rangle+\sin\theta|1\rangle$$ (this is the X-Z plane of the Bloch sphere), varying over $$\theta$$. Any other fixed phase $$\gamma$$ $$\cos\theta|0\rangle+\sin\theta e^{i\gamma}|1\rangle$$ gives you a different plane. But there are plenty of other options as well, such as $$\frac{1}{\sqrt{2}}(|0\rangle+e^{i\theta}|1\rangle)$$ which extracts the X-Y plane of the Bloch sphere.
Just for the sake of adding another example, how about $$\frac{3}{5}|0\rangle+\frac{2(1-\sqrt{3}i)}{5}|1\rangle.$$