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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 $2^{2^n}$?

Please, can anyone confirm or deny my assumption with some human-readable clarification

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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 $\log_2 8 = 3.$

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  • $\begingroup$ I am not sure about it. Yes, that was my first assumption, but then in theory, if there would be 3 qubits the answer had to be $2*3$ therefore 6 vectors that could be represented as 6 bits that can have $2^6$ combinations. But as you can see in my second quote. They clearly stated that 3 qubits can form 8 distinct vectors that can form 0 or 1 or in other words a bit. Hence it can mean that a total number of bits per n-qubits is $2^n$, therefore the total distinguishable messages are $2^{2^n}$. $\endgroup$ – Kaspar Siricenko May 2 '18 at 2:19
  • $\begingroup$ @KasparSiricenko Let me know if the addition to my answer helps. $\endgroup$ – Mark H May 2 '18 at 4:28
  • $\begingroup$ I feel very nice to see your interest. If I understood you right. That there are 8 eigenstates for 3 qubits in order to describe them. But it is exactly the same amount of states 3 classical bits can hold as well because the amount of distinct combinations of n-bits is $2^n$ $=>$ $2^3=8$. Does it mean that n-qubits can hold the same amount of data as the classical bits? And then why superdense coding states that 2 bits of data can be encoded to 1 qubit? $\endgroup$ – Kaspar Siricenko May 2 '18 at 18:28
  • $\begingroup$ @KasparSiricenko Two bits of data can be transmitted using a single qubit and a preshared entangled pair of qubits. In fact, the most amount of data that can be reliably packed into a single qubit is one bit. To transmit more data requires extra infrastructure (the entangled state). $\endgroup$ – Mark H May 2 '18 at 19:21

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