# Does 1 qubit correspond to 2 bits?

In a lot of presentation I always see people say that $$n$$ qbit are approximately $$2^n$$ classical bit. Those talks where oriented for a broad audience, so they left out a lot of things. Deep down I felt that this couldn't be possible, but I didn't know anythings about qc so maybe this was the problem.

Now I've started learning qc (as a computer engineer) and I'm studying the concept of super dense coding, that it the base of the association $$n$$ qbit equal $$2^n$$ classical bits.(If it's not this let me know)

I've understood what the topic is about, but I still think that the association $$n$$ qbit equal $$2^n$$ classical bits is wrong or at least misleading.

Sending 2 bit to Bob, will require Alice to have 1 qbit and 1 entangled qbit where the other part of the eqbit is owned by Bob. Just by reading this is clear that 2 bit actually correspond to 1 qbit + 1eqbit, but reading online one common approach to say that 1qbit=2bit is by introducing a third part that is responsible to send and eqbit to Alice and Bob. Isn't this a flawed way to think? When someone say $$n$$ qbit equal $$2^n$$ bit, they are implicitly stating that there is a way to encode the information of $$2^n$$ bits in $$n$$ qbit, but if you actually study the theory is not like this.

Also saying that 1qbit +1eqbit =2 bit, is not too much different than saying 1qbit+1qbit=2qbit=2bit, because at the end of the day 1eqbit is just a qbit in a particular state. I know that differentiating between them is important since they are two different things, but physically we can see them as two object (two photons for example) that are in a different position, still they 'occupy the space of two object'. I also know that approximating 1 eqbit with 1 qbit is a strong affirmation, but stating that 1qbit = 2bit is stronger imo.

Is my way of thinking flawed? Why and where?

Also there is another thing that I couldn't understand on my own. In the textbook that I'm using (Quantum Computation and Quantum Information) one thing that they say is:

Suppose Alice and Bob initially share a pair of qubits in the entangled state..

Since the sharing of the eqbit and the sending of the qbit seems to happen in two different temporal window, they are able to store qbit? In essence I don't understand the temporal window of the algorithm. I understand how it works, but not when. Can you clarify this?

Note that this second question is related to the first, because I kind of understand the point of super dense coding, if you can send qbit at two different time and exploit quantum mechanics to send less qbit when needed, but if everything happen at the same time (the sending of the eqbit and the sending of the qbit) then I don't know the point of super dense coding.

• please, qubit, not qbit =) – glS Nov 30 '20 at 11:16

Lets start with notion that $$n$$ qubits are equivalent to $$2^n$$ classical bits. This is wrong. However, it is true that to describe a quantum state composed of $$n$$ qubits we need $$2^n$$ complex numbers since $$n$$ qubits state is superposition containing all combination of $$n$$ classical qubits ($$2^n$$). Writen by formula, $$n$$ qubits state is $$|q_0q_1...q_{n-1}\rangle = \sum_{i=0}^{2^n}a_{i}|i\rangle,$$ where $$i$$ represent basis states (e.g. $$|0...00\rangle$$, $$|0...01\rangle$$, $$|0...10\rangle$$, $$|0...11\rangle$$ etc. until $$|1...11\rangle$$) and $$a_{i} \in \mathbb{C}$$.

Concerning information contained in qubits. Any qubit can be described as $$|q\rangle = \alpha|0\rangle + \beta|1\rangle,$$ where $$\alpha, \beta \in \mathbb{C}$$. Since paraeters $$\alpha$$ and $$\beta$$ are complex numbers, theoretically a qubit can contain infinite amount of information because you need infinite amout of bits to represent complex number exactly. This is of course impossible in reality since computers have finite memory. Moreover and what is the most important, when you measure a qubit, it collapeses to either 0 or 1. So, in the end you have either 0 or 1, i.e. one classical bit. Therefore, one qubit contains one bit of information.

Regarding superdense coding. It is often said that in superdense coding you send only one qubit but in the end you have two classical bits. It is true that you send only one qubit but to have superdense coding, you need to have two entangled qubits before. So, effectively two qubits are needed for transmiting two classical bits.

• Thank you for the answer, you confirmed what I've deduced, but what can you tell me about the temporal window needed to send the eqbit in superdense coding?(second part of the question) – malloc Nov 29 '20 at 10:03
• @malloc: Could you please specify where in the book (I suppose it is Nielsen and Chuang) is this statement placed? – Martin Vesely Nov 30 '20 at 7:48
• yes on page 97. (2.3 Application: superdense coding) – malloc Nov 30 '20 at 14:13
• I know that at some point a qubit decay at |0>, so technically I cannot share a qubit and then, after some day for example, use it. On this assumption I don't see how we can share qubit fast enough so that the state doesn't collapse. – malloc Nov 30 '20 at 14:19
• @malloc: Superdense conding would be useful once you have a "Quantum Internet" and you will have as many entangled qubits pairs among all nodes of the "Internet". But until this time...yes, it is not very useful. – Martin Vesely Dec 1 '20 at 13:44

There are two different things at play here: (i) superdense coding and (ii) Holevo's bound.

1. Holevo's bound tells us that $$n$$ qubits can only store $$n$$ bits of information. See for example, this answer: How can the Holevo bound be used to show that $n$ qubits cannot transmit more than $n$ classical bits?
2. Superdense coding allows us to send 2 bits of information using a single qubit in the presence of preshared entanglement. This is not the same as 1 qubit being "equal" to 2 bits of information.
• Thank you for the answer, after studying a little more I've understood that I was confusing two arguments, but what I still don't understand is the part on superdense coding when they share the entangled qubit. Supposing that: 1) qubits cannot be stored 2) qubit decay to the |0> state after some time. I don't see how Alice and Bob can share and equbit and then use it to send 2 classical bits (after one day for example) – malloc Nov 30 '20 at 14:50
• @keisuke.akira Doesn't Holevo's bound say that you can never extract more than $1$ bit per qubit? – JSdJ Nov 30 '20 at 14:57
• @malloc The superdense coding protocol assumes that the ebit remains (maximally) entangled for as long as the protocol takes -- and of course as you guessed, if there is decoherence, then may no longer be true and the success probability of transmitting $2$ bits decreases. – keisuke.akira Nov 30 '20 at 16:06
• @malloc Btw, I'm not sure what you mean by "qubits cannot be stored"? – keisuke.akira Nov 30 '20 at 16:07
• @JSdJ Yes. The key idea is that if we want to store and retrieve classical information from qubits then the best we can do is $n$ qubits to $n$ bits. The answer I linked explains this in more detail. – keisuke.akira Nov 30 '20 at 16:09