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What is the conversion factor for qubits (qudits) to bits/bytes in classical information theory/computation theory?

I mean, how can we know how many "bits/bytes" process, e.g., a 60 qubit quantum computer (quamputer), are equivalent to classical bits (dits)?What about memories and velocities in "attaniable" quantum computers versus classical current memories and computer velocities(GHz)?

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    $\begingroup$ What if there is no such equivalence relation between them? $\endgroup$ – Ali Jul 17 '13 at 16:46
  • $\begingroup$ possible duplicate of Quantum Computing Power Advantages $\endgroup$ – Ali Jul 18 '13 at 5:29
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    $\begingroup$ It's one to one. To the extent that bits and qubits can be compared to one another, one bit equals one qubit. $\endgroup$ – Nathaniel Jul 18 '13 at 15:07
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    $\begingroup$ ...though having said that, qubits do sometimes have the feeling of being equal to two classical bits each. E.g. in quantum teleportation you have to transmit two classical bits in order to transfer the state of one qubit. It's not very straightforward. But if you have 10 qubits, the maximum amount of classical data you can store is 10 bits, so I think one-to-one is the best way to look at it. $\endgroup$ – Nathaniel Jul 18 '13 at 15:10
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    $\begingroup$ What's the conversion factor between automobiles and boats? If I can drive from Miami to Tampa in 4.5 hours, how long will it take me by boat? Similarly, if I can travel from Miami to Cuba in 12 hours by boat, how long will it take by automobile? $\endgroup$ – Peter Shor Apr 13 '17 at 19:27
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I think one concept to grasp here is quantum parallelism. GHz refers to how fast a computer does 1 computation (A billionth of a second) and classic computers do 1 computation at a time (or a few in multi-core computers). However in a quantum computer multiple computations can be done at once in parallel.

Ultimately when you input Qubits into a system they are identical to regular bits... and when you measure the Qubits they are the same as bits... the interesting property about Qubits is the algorithms you can perform on them because of the quantum properties (assuming you can bring them back to a form where the measured results become useful bits).

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  • $\begingroup$ Then, if (G)Hz are not a good way to compare the speed of quantum computing vs. its classical analogue, how could we compare their "speeds"? $\endgroup$ – riemannium Jul 18 '13 at 10:00
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The value of a qubit is, in effect, a complex number. To specify a complex number requires an infinite (and in fact uncountable) number of bits. So in terms of the information that's stored, a qubit is the equivalent of a large infinite number of bits.

You cannot, of course, extract all of that information. As Peter Shor points out in comments, Holevo's Theroem says that you can extract at most one bit.

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    $\begingroup$ This misses a very important fact about quantum information—Holevo's theorem says that one qubit can transmit at most one classical bit of information. It doesn't matter a damn how good an engineer you are ... you can only get one bit of information out of it. $\endgroup$ – Peter Shor Apr 13 '17 at 19:22
  • $\begingroup$ @PeterShor: Point well taken. I will edit accordingly. $\endgroup$ – WillO Apr 13 '17 at 21:12
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Qubit (Qudit) equivalence with classical bits:

$1 \; qubit=2\; classical\; bits$ and $1 \; qudit=D\; classical\; dits $

$N \; qubits=2^N\; classical\; bits$ and $N \; qudit=D^N\; classical\; dits $

As 1 byte= 8 bits=$2^8\;bits$, or $1\;bit= 2^{-3}\;bytes=0.125\;bytes$, we have also

$N \; qubits=2^{N-3}\; classical\; bytes$ and $N \; qudytes=D^{N-3}\; classical\; dits $

Of course, one could define also the qubyte (8 qubits) and the qudyte avoid the -3 extra exponent and keep parallelism with classical information theory. It is trivial to get the equivalence with kilobytes.

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  • $\begingroup$ This doesn't make much sense - even more so if you don't explain the way this mapping supposedly works. $\endgroup$ – Norbert Schuch Sep 25 '16 at 19:52

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