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A qubit is a quantum system in which the Boolean states 0 and 1 are rep- resented by a prescribed pair of normalised and mutually orthogonal quantum states labeled as ${|0⟩, |1⟩}$

According to [1]. Then a quantum register $\mid x_1x_2...x_n\rangle, x_i\in\{0,1\}$ is defined to be collection of n qubits.

Now I often see expressions like $\mid x_1, ... x_n \rangle$ where the $x_i$ belong to some $S \subset \mathbf{Z}$.

  1. Can the individual constituents $\mid x_i \rangle$ be called qubits even though they are non-binary?
  2. Would it be appropriate to call $\mid x_1, ... x_n \rangle$ a qubit register in this case?
  3. What is the physical interpretation of such a register?
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  • $\begingroup$ I think qubits are binary by definition, but there may be another to represent them I don't know of. $\endgroup$ – BrockenDuck Aug 27 '20 at 5:47
  • $\begingroup$ "Now I often see..." -- Where? $\endgroup$ – Norbert Schuch Aug 30 '20 at 9:39
  • $\begingroup$ please remember that each post should contain a single, focused question $\endgroup$ – glS Aug 31 '20 at 7:16
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The $|x_i\rangle$ you mention here are qudits, they are the generalization of qubits to base $d$ with $|S| = d$. It is categorized by a superposition of $d$ states, same way a qubit is described by the superposition of 2 states.
In base 3 it has a specific name as well, this is called qutrit.

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  • $\begingroup$ I see that it is a generalization in syntax, but what does it mean? What's the physical meaning of $\mid -12\rangle$? $\endgroup$ – gen Aug 27 '20 at 13:31
  • $\begingroup$ Also how can you consider the basis vector $\mid 0 \rangle$ to be a superposition of two states? Which two? $\endgroup$ – gen Aug 27 '20 at 13:50
  • $\begingroup$ @gen: First question: -12 is not natural number, hence it cannot appear in your definition. Second question: in $|0\rangle$ and $|1\rangle$ basis any qubit can be writen as $|q\rangle = \alpha|0\rangle + \beta|1\rangle$, where $\alpha, \beta \in \mathbb{C}$. So, $|0\rangle$ is superposition where $\alpha = 1$ and $\beta = 0$. This seems strange but the qubit is considered to be in superposition until it is measured. In this case, zero is always returned. $\endgroup$ – Martin Vesely Aug 28 '20 at 6:53
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    $\begingroup$ @MartinVesely thanks for your comment. First question: Yes, the definition was mistyped. I meant to write $S \subset \mathbf{Z}$. So I'm still interested what $\mid -12 \rangle$ means. Second: That's fair enough I guess. $\endgroup$ – gen Aug 29 '20 at 2:33

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