# Does superdense coding allow to double the information capacity of a set of qubits?

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 $2^{2^n}$?

• What do you mean by condense coding? – Sanchayan Dutta May 1 '18 at 22:02
• I am sorry for the confusion. I meant superdense coding. – Kaspar Siricenko May 1 '18 at 22:37
• Another thing is not clear to me. What do you mean by "double capacity"? Double with respect to what? – Sanchayan Dutta May 1 '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 2 '18 at 2:42
• 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 2 '18 at 17:53