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10 votes

What are the real advantages of superdense coding?

TL;DR: While two qubits must be transmitted in total, in the instant where two bits are to be communicated, only one qubit has to be sent. The information being sent is masked, but it is not truly ...
DaftWullie's user avatar
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9 votes
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Why is super-dense coding called the inverse of quantum teleportation?

In quantum teleportation, one starts with an entangled state shared between two parties, and (after some messing at the sender's side), two classical bits are transmitted from one party to the other ...
DaftWullie's user avatar
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8 votes
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Does superdense coding allow to double the information capacity of a set of qubits?

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 ...
agaitaarino's user avatar
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7 votes

What is the practical interest of superdense coding?

No, you need to send only one photon (from the pair). The other party could generate entangled pair and send the entangled photon to you. Or it could be the third party that send both of you your ...
Danylo Y's user avatar
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Can superdense coding be made more efficient?

No, you can't send more than 2 bits per transmitted qubit. Ultradense Coding would allow FTL Signalling. The basic problem is that, if either teleportation or superdense coding was slightly more ...
Craig Gidney's user avatar
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6 votes

Controlling high-dimensional Hilbert spaces with a single qubit

Let me call you A and your friend B. You initially share a state $|\psi\rangle$. W.l.o.g., we can write this in its Schmidt decomposition: $$ |\psi\rangle = \sum_{i=1}^d \lambda_i |i\rangle_A|i\...
Norbert Schuch's user avatar
6 votes

Does 1 qubit correspond to 2 bits?

There are two different things at play here: (i) superdense coding and (ii) Holevo's bound. Holevo's bound tells us that $n$ qubits can only store $n$ bits of information. See for example, this ...
keisuke.akira's user avatar
5 votes
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How do you transmit a single qubit of an entangled qubit over?

It means moving the qubit from one place to another. The fact that the qubit happens to be entangled with some other qubit elsewhere is irrelevant. The technical details of transmission depend on the ...
Adam Zalcman's user avatar
5 votes

In what ways can qubits be used for applications that do not require entanglement?

One (obvious) application is the generation as True Random Number Generators, e.g. IDQ, or you can download some here Free True Random Numbers (please do not use these for security relevant ...
Fleeep's user avatar
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5 votes

What is the applicability of quantum network coding?

Network coding — both classical network coding, and quantum network coding — is an approach to distributing information by performing simple operations at nodes in a network, acting on ...
Niel de Beaudrap's user avatar
4 votes

What are the real advantages of superdense coding?

Superdense coding can be used to smooth out network utilization by "storing bandwidth". During low utilization, top up the traffic with EPR halves. During high utilization, burn the EPR halves to ...
Craig Gidney's user avatar
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4 votes
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Controlling high-dimensional Hilbert spaces with a single qubit

There are a total of 5 qubits. I have 2, and my friend has 3. Clearly the upper bound of classical bits I can send to my friend is 5. But is it possible to find 5 unitary operations I can perform ...
DaftWullie's user avatar
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4 votes

In what ways can qubits be used for applications that do not require entanglement?

Certainly not exhaustive, but to get the ball rolling... One possible application is blind quantum computation. In this, there is a user who wants to complete a computation, but only has the ...
DaftWullie's user avatar
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Eavesdropping in superdense coding: why can't the third party infer anything about the message?

Mathematically, this has nothing to do with the positivity of $E$. It doesn't really have anything to do with $E$ at all - it's a property of the Bell states themselves (you've probably not got there ...
DaftWullie's user avatar
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Superdense coding circuit returns wrong result

Bit ordering convention in Qiskit is reversed to what is common in many physics textbooks. From Qiskit textbook: The bit we flipped, which comes from qubit 7, lives on the far left of the string. ...
Yehuda Naveh's user avatar
4 votes
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Does 1 qubit correspond to 2 bits?

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 ...
Martin Vesely's user avatar
4 votes
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What does it look like to split an EPR pair?

The whole point of an EPR pair is that you cannot write (without losing some information) "This is what Alice has" and "This is what Bob has". Partial descriptions can be given using reduced density ...
DaftWullie's user avatar
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3 votes

Strange inequivalence between superdense coding and teleportation

Imagine we have some two-qubit state $\omega$ which Bob holds after the transmission of a qubit from Alice. For superdense coding, he's going to measure it in the Bell basis. Let's assume the answer ...
DaftWullie's user avatar
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Does sending a qubit using a quantum network through some conventional physical medium mean it is not with sender anymore after sending?

Yes, your understanding is correct: when Alice sends her qubit to Bob, she doesn't have that qubit any longer. Entangled pairs of qubits distributed between the parties are a resource that is used up: ...
Mariia Mykhailova's user avatar
3 votes

Eavesdropping in superdense coding: why can't the third party infer anything about the message?

This happens for any maximally entangled state $|\Psi\rangle$ and operator $E$. Indeed, a maximally entangled state is, by definition, one whose partial trace is the maximally mixed one. Writing $|\...
glS's user avatar
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3 votes

If Alice and Bob share a Bell state, can Alice send her individual qubit to a third party?

Q1) The qubits in the $|\Phi^+\rangle$ state are entangled - this means that (by definition) you can not represent the state of one of them individually without talking about the second one (...
Mariia Mykhailova's user avatar
3 votes
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What is the difference between quantum teleportation and quantum entanglement with dense coding?

Both teleportation and (super)dense coding use entanglement as a resource. I like to think that one is dual to the other, as the quantum circuits are, basically, inverses of each other. For example, ...
Mark Spinelli's user avatar
2 votes

Why is super-dense coding called the inverse of quantum teleportation?

Basically quantum teleportation is in facto the determinate side of super-dense coding. In superdense coding we fit two classical bits of information using fairly a single qubit. On the other hand, ...
Amit Kumar Jaiswal's user avatar
2 votes

In what ways can qubits be used for applications that do not require entanglement?

In lecture 4 of O'Donnell's series on quantum computing, he introduces the Elitzur-Vaidman bomb tester, which is an interesting application of the quantum Zeno effect. O'Donnell introduces the bomb ...
Mark Spinelli's user avatar
2 votes

Are superdense coding and teleportation just a prototype or the 'only' type?

As for superdense coding, you can design different schemes for $n, m$ qubits, but they are not giving any advantage over a simple scheme $1 \text{qubit} + 1 \text{ebit} \rightarrow 2 \text{cbits}$. In ...
Danylo Y's user avatar
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How is the qubit sent from Alice to Bob in superdense coding?

So to clarify, in your steps above, Alice and Bob each of one qubit of the entangled EPR pair in (1), Alice performs some gate on her qubit/half of the pair in (2) based on the two bits she wishes to ...
Chris E's user avatar
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2 votes

What is the implementation of the state in superdense coding?

If you apply $\mathrm{X}$ on second qubit, you will get state $$ \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle) $$ After that you can apply $\mathrm{Z}$ on second qubit as well to get Bell state $\beta_{...
Martin Vesely's user avatar
2 votes
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$2$ ebits $+$ $1$ bit $ = 2$ bits?

The protocol you describe is correct, but the resource estimation is wrong. Furthermore, something like superdense coding with purely classical bits is prohibited by the No-Signaling Principle. This ...
forky40's user avatar
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2 votes

What is meant by "double speed" in dense coding protocol?

Dense coding lets you send two classical bits for every qubit you transmit, assuming the two parties already share enough entanglement. This two for one trade-off is what's being called "double ...
DaftWullie's user avatar
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2 votes

How to modify the quantum circuit to do superdense coding with the state $|00\rangle-|11\rangle$?

It could be, yes, depending on whether you want the sender or the receiver to adjust their part of the protocol. Either of them applying a $Z$ gate to their half of the pair before proceeding with the ...
Mariia Mykhailova's user avatar

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