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 $2^{2^n}$?
Please, can anyone confirm or deny my assumption with some human-readable clarification
Best Answer
If I'm reading these lecture notes correctly (linked from the Wikipedia article), then superdense coding allows you to transmit 2 bits of information using 1 qubit and a previously set up entangled state that you share with the recipient. More generally, with $n$ qubits and $n$ pre-shared entangled states, you can send $2n$ bits of information. The total number of distinguishable messages is thus $2^{2n}$, not $2^{2^n}$.
The 8 vectors in the second quote are not bits. They are eigenstates of a 3-bit quantum computer: $|000\rangle$, $|001\rangle$, $|010\rangle$, etc. You need an 8-dimensional vector of complex numbers (not ones and zeros) to describe the state since the quantum computer can exist in a superposition of memory states: $$\Psi = a_0|000\rangle + a_1|001\rangle + a_2|010\rangle + a_3|011\rangle + a_4|100\rangle + a_5|101\rangle + a_6|110\rangle + a_7|111\rangle$$ where $$\sum_i |a_i|^2 = 1.$$
While the superposition state of the quantum computer has uncountably infinite many states, if you read the data in the qubits, you will only get one of 8 possible eigenstates, so the number of bits is $\log_2 8 = 3.$