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Based on my current understanding, a qubit is represented as a vector $(a, b)$ which satisfy $a^2 + b^2 = 1$. Classical bit one can be represented as $(0, 1)$ and bit zero can be represented as $(1, 0)$.

And then we can have things that cannot be described isomorphically in classical computing such as $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.

Then, I came to a realization where this is quite similar to trigonometry identity where $\sin^2(t) + \cos^2(t) = 1$.

Thus, I'm wondering if a qubit is a point $p$ that lies anywhere on the circumference of a unit circle, can a qubit $(a, b)$ be described as $(\sin(t), \cos(t))$ where $t$ is the angle measured from positive $x$-axis to the line formed between $p$ and center of the circle?

Is this analogy faulty?

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    $\begingroup$ you can include math using the $ delimiters. E.g. $a^2+b^2=1$ renders as $a^2+b^2=1$ $\endgroup$
    – glS
    May 8 '20 at 15:54
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    $\begingroup$ You're not quite right, but this concept is very similar to the Bloch sphere: en.wikipedia.org/wiki/Bloch_sphere $\endgroup$ May 9 '20 at 0:52
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That's not too far from the truth.

You can always describe the probabilities of a two-outcome event like that: you have $p_1+p_2=1$ and thus defining $c_i\equiv\sqrt{p_i}$ you observe that $(c_1,c_2)$ are distributed on a circle (more precisely, in the upper-right sector of one). More generally, if an event has $n$ possible outcomes, you can describe the corresponding probabilities as the squares of the components of points on (one sector of) a hypersphere: $p_i\equiv c_i^2$ with $(c_i)_{i=1}^n\equiv\boldsymbol c\in S^{n-1}$.

Note that this has nothing to do with quantum mechanics, it's just something that follows from the mathematics on any probabilistic description of an event.

The question is, why would you ever want to do this? The answer is that, in QM, it turns out that describing states in the terms of the coefficients $c_i$ (let's call these amplitudes) is simpler. However, it also turns out that using values $c_i\in[0,1]$, as is the case when you define these as square roots of probabilities, is not sufficient to fully describe quantum states. This can be fixed by "promoting" these coefficients to be complex numbers, $c_i\in\mathbb C$. In the case of a qubit, this amounts to the state being describable as a point on a sphere, rather than on a circle (the complex phases add an additional "phase angle" on top of the "angle" corresponding to the measurement probabilities).

In summary, yes you can naturally associate to every point on a circle (a hypersphere) a qubit (qudit). The inverse relation will not hold however: there are more quantum states than those describable in such a way. It might be worth noting that if you are only interested in what happens in a single measurement basis, then you have a one-to-one relationship between your circle and the possible outcome probabilities. This is to be expected, as if you do not work with different measurement bases, everything can be described with classical probability theory.

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What you have described is a probabilistic bit, and is interesting in its own right, but not 'equal' to a proper quantum bit. A quick search gives for instance this talk (which I have not seen myself).

Your analogy fails because a general qubit's state cannot be written using one, but rather two free (real) parameters/angles:

\begin{equation} |\psi\rangle = \cos(\frac{\theta}{2}) |0\rangle + e^{i\phi}\sin (\frac{\theta}{2})|1\rangle, \end{equation} where $\theta$ is divided by $2$ for reasons I briefly mention below.

  • $\theta$ is like the angle that you have described, which describes the relative 'population' of the $|0\rangle$ and $|1\rangle$ states.
  • $\phi$ is commonly referred to as the phase (or relative phase) of the qubit, and there's not really a classical analogy. It is of vital importants to the power of the qubit/quantum computer, and some would say that quantum computing is actually only about manipulating this phase is smart ways.

Combining these two parameters (and continuing on your intuitive visualtization of viewing the state on a unit circle) we can visualize the state of a qubit as a point on a sphere, specifically the Bloch sphere.

The $|0\rangle$ and $|1\rangle$ state are the north- and south pole (on the $Z$-axis) of the sphere, respectively, and altering $\theta$ from $0$ to $2\pi$ rotates the state from the north- to the south pole. (Since this is actually just a rotation of $\pi$, we divide $\theta$ by $2$.)

The phase $\phi$ is then the angle the state vector makes with the $X$-axis (in the $X$-$Y$ plane).

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