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I read the basic introductory information about qubits on Wikipedia:

There are two possible outcomes for the measurement of a qubit—usually 0 and 1, like a bit. The difference is that whereas the state of a bit is either 0 or 1, the state of a qubit can also be a superposition of both. [1]

and

The state of a three-qubit quantum computer is similarly described by an eight-dimensional vector $(a_{0},a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7})$ (or a one dimensional vector with each vector node holding the amplitude and the state as the bit string of qubits). [2]

Hence does it mean that qubit using superdense coding can achieve a double capacity with the possible number of combinations of $2^{2^n}$?

Sanchayan Dutta
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Kaspar Siricenko
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  • What do you mean by condense coding? – Sanchayan Dutta May 01 '18 at 22:02
  • I am sorry for the confusion. I meant superdense coding. – Kaspar Siricenko May 01 '18 at 22:37
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    Another thing is not clear to me. What do you mean by "double capacity"? Double with respect to what? – Sanchayan Dutta May 01 '18 at 23:09
  • For a system of n components, a complete description of its state in classical physics requires only n bits. Therefore information that classical n-bits can hold is $2^n$. – Kaspar Siricenko May 02 '18 at 02:42
  • So if superdense coding can achieve $2$ bits per qubit it means the ability to store information has to be either double amount of classical bits or $2^{2n}$ times of information. And the problem is that it states that 3 qubits have 8 vectors or in other terms, $2^3$ producing $2^{2^3}$ holding information, therefore, the total number of distinguishable messages for n-qubits can be $2^{2^n}$. Which doesn't connect to the first assumption of $2^{2n}$ – Kaspar Siricenko May 02 '18 at 03:01
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    how did you get $2^{2^n}$? Twice $2^n$ is $2^{n+1}$, and doubling the number of qubits you get $2^{2n}$. Both are very different from $2^{2^n}$. – glS May 02 '18 at 17:53
  • cross-posted on physics.SE – glS May 03 '18 at 09:07
  • @glS I wonder what our policy on cross-posting should be. I feel it is fine as long as the OP links the cross-posts to each other. – Sanchayan Dutta May 03 '18 at 12:25

1 Answers1

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The short answer is no: we don't double the capacity. It turns out it's not that quite simple. There is no general mathematical expression that gives you the storage (or processing power) of a number of qubits in terms of bits. Bits, qubits and ebits work in qualitatively different ways, which in some contexts allows to draw an advantage.

The closest thing to an answer to your question are the so-called Bennett's laws, four inequalities comparing the practical information contents of classical bits, quantum bits (or qubits) and entanglement bits (or ebits), reproduced here from wikipedia. The ⩾ signs are to be taken as "can do the job of":

  • 1 qubit ⩾ 1 bit (classical),
  • 1 qubit ⩾ 1 ebit (entanglement bit),
  • 1 ebit + 1 qubit ⩾ 2 bits (via superdense coding),
  • 1 ebit + 2 bits ⩾ 1 qubit (via quantum teleportation),

On the particular aspect of superdense coding, I refer you to the question "What are the real advantages of superdense coding?" and its answers.

agaitaarino
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