Visualize representation of a qubit

Question

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.

Answer 1

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.

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Just for the sake of adding another example, how about $$ \frac{3}{5}|0\rangle+\frac{2(1-\sqrt{3}i)}{5}|1\rangle. $$