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]


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 $2^{2^n}$?

  • $\begingroup$ What do you mean by condense coding? $\endgroup$ May 1, 2018 at 22:02
  • $\begingroup$ I am sorry for the confusion. I meant superdense coding. $\endgroup$ May 1, 2018 at 22:37
  • 1
    $\begingroup$ Another thing is not clear to me. What do you mean by "double capacity"? Double with respect to what? $\endgroup$ May 1, 2018 at 23:09
  • $\begingroup$ 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 $2^n$. $\endgroup$ May 2, 2018 at 2:42
  • 2
    $\begingroup$ 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 $2^{2^n}$. $\endgroup$
    – glS
    May 2, 2018 at 17:53

1 Answer 1


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.

  • $\begingroup$ That's exactly what I wanted to understand! $\endgroup$ May 2, 2018 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.