A bit is a binary digit, typically 0 or 1.
Until a value is assigned (or a measurement is made) a bit is in a superposition of the entangled binary pair, is it not?
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2$\begingroup$ Entanglement is a property of multiqubit systems; if you have a single qubit there is nothing to be entangled. $\endgroup$– kludgCommented Jul 31, 2018 at 6:12
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4 Answers
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A bit, either 0 or 1, can certainly be thought of as a special case of being a qubit. However, that is not to say that anything capable of computing with classical bits is capable of computing with quantum bits.
If you have a bit, and don’t know if it’s a 0 or a 1, then how do you describe its state? You have to use Bayesian priors. If you have no idea which it is, you assign the two options equal probabilities. And it really is probabilities here. So it’s 50:50 being in 0 or 1. You can’t describe this as a pure quantum state, but you can use a density matrix: $$\rho=(|0\rangle\langle 0|+|1\rangle\langle 1|)/2.$$ This formalism also means that if you later learn something about what the bit value might be (perhaps as a result of reading other bits in a computation), you can update those probabilities using conditional probabilities and Bayes’ rule. Note that this also means it’s a subjective description of the state: it reflects your personal knowledge of the state of the bit, while somebody else, who perhaps prepared the bit, already knows what value it has. This is perhaps one way to see why it should be different from the pure state that you proposed, which should be an objective description that everyone would agree on (in that there is no uncertainty in the state; the only uncertainty is induced by the action of the measurement).
answered Jul 31, 2018 at 5:26
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No, superposition is not the same as uncertainty about an outcome.
If you throw a coin, the state of the coin is not $|0\rangle+|1\rangle$ before landing. If you really want to use the formalism of quantum mechanics to describe the uncertainty about the outcome, you have to describe it as a mixture of the two possible outcomes, that is, something like $|0\rangle\!\langle0|+|1\rangle\!\langle1|$. As also discussed in this other answer, coherent superpositions and mixtures are two entirely different beasts.
In particular, saying that the state of a coin (classical bit) is $|0\rangle+|1\rangle$ before the measurement implies the possibility of performing coherent operations on it, which is wrong: it is not possible, even in principle, to do something like measuring the state of the coin in the $|+\rangle$ basis.
In other words, the crucial difference is that even if you do not know in what state a bit is, it is still the case that the bit is in some definite state (which means no interferences), you just don't which one. On the other hand, the "uncertainty" associated with a qubit does not mean that there is something that we do not know about the state: the qubit can be fully characterised, and yet lead to uncertainty in (some) measurement outcomes.
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A bit is effectively
$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$$
where either $\alpha = 1, \beta = 0$ (the bit = 0) or $\alpha = 0, \beta = 1$ (the bit = 1).
So, yes, in the sense I think you mean, a classical bit can be represented with the quantum notation.
Remember for quantum mechanics to be a legitimate theory, it must describe all the phenomenons and behaviors that classical mechanics explains, which it does. Similar is true in this case.
Note that if you ever refer to a classical bit as quantum, you will probably confuse the heck out of everybody, which is why the distinction is made. Also different gates can apply, etc.
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1$\begingroup$ I just learned about Shannons (binary digits) & Hartleys (decimal digits). They represent 50% & 10% probability amplitudes respectively. Proceeding from a bit w/ 50% amplitude, a qubit would have a 25% probability amplitude (&/or distribution?) would it not? Is a bit just a 1-level quantum system? Are digits 0-level? $\endgroup$ Commented Jul 31, 2018 at 2:07
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1$\begingroup$ @meowzz let's start by clarifying terminology: bits are binary, working on base 2. Digits are decimal, working in base 10. You can't talk about binary digits. So, bits have 2 options (0 or 1) and digits have 10 values, 0 to 9. To represent these in a physical system (and they have to be represented in a physical system) they respectively need 2 or 10 levels. The distinction between the classical and quantum versions is that the classical versions can be in any of the distinct levels (or probabilistic mixtures), while the quantum ones can be in superpositions of any combination of levels. $\endgroup$ Commented Jul 31, 2018 at 6:34
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$\begingroup$ I think that this is misleading. When we talk of "bits" we are implicitly assuming information encoded classically. One can extract a bit of information from a qubit, but a bit is not "effectively" a coherent superposition of its two possible states. Saying that implies the possibility of interfering the two possible states of the bit, but if you can do that, then you are talking of qubits, not bits. $\endgroup$– glS ♦Commented Jul 31, 2018 at 10:18
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$\begingroup$ @DaftWullie are you possibly confusing "digit" with "decimal" here? $\endgroup$– glS ♦Commented Jul 31, 2018 at 10:21
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$\begingroup$ @glS let's say that I've over-egged it. I know that what I've said is going too far (but it is true that the origin of the word digit is to do with base 10). It is at least better to talk about the bits of a binary string rather than its digits, and that helps emphasise the close connection to the qubit, just as trits (base 3) would help emphasise connection to the qutrit. $\endgroup$ Commented Jul 31, 2018 at 13:51
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You are correct that if you set $\alpha=0$ or $\alpha=1$, then the information gain upon measurement will be none. There is no way to be surprised. You do need a some nontrivial $p$ for be the prior distribution, to get information gain. If you say the state is $\sqrt{p} \mid 0 \rangle + \sqrt{1-p} \mid 1 \rangle$ instead, then when you measure it in the computational basis you can have some surprise/information. Presumably let $p=\frac{1}{2}$ because you want a full bit, not less. One might play fast and loose between "less than or equal to 1 bit" vs "1 bit" because of implicit assumptions about the probability distribution.
