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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 secure. There are two distinct phases to a superdense coding protocol. In phase 1, Alice and Bob prepare a Bell state $(|00\rangle+|11\rangle)/\sqrt{2}$. This is ...

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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 so that the net effect is a quantum state is sent from the first party to the second without sending any quantum data. In superdense coding, the parties start ...

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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 of qubits in terms of bits. Bits, qubits and ebits work in qualitatively different ways, which in some contexts allows to draw an advantage. The closest thing ...

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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 photons $-$ prior to actual communication. Even if it's you who generate the entangled pair and share the entangled photon $-$ you could do it way before the time ...

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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: $$|\psi\rangle = \sum_{i=1}^d \lambda_i |i\rangle_A|i\rangle_B.$$ You are now asking how many (orthogonal) states A can implement by acting with a unitary on their side of the state. This number is upper bounded by ...

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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 answer: How can the Holevo bound be used to show that $n$ qubits cannot transmit more than $n$ classical bits? Superdense coding allows us to send 2 bits of ...

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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 physical implementation of the qubit. It could be transmission of a photon through free space or fiber or physical transportation of a device with the qubit ...

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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 input signals and transmitting the outputs to other nodes. To put it another way, network coding is an approach to distributing information using a communications ...

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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 application). In order to build such a TRNG, from a quantum circuit perspective, all you need is a single qubit, a Hadamard gate and a measurement. Although there might ...

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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 efficient, iteratively nesting them inside of each other would allow you to send more encoded qubits than the number of physical qubits you sent. Suppose you could ...

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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 matrices. However, if you want to identify which bits Alice and Bob each have, we often use a notation like $$(|0\rangle_A|0\rangle_B+|1\rangle_A|1\rangle_B)/\... 4 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 capability of producing single-qubit (non-entangled) states. These are sent to a server who can (locally) entangle them for the purposes of performing a measurement-... 4 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 yet, but they have the same reduced density matrices). I presume the reason for specifying the positivity of E in the question is to help you make the ... 4 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 double the available capacity. Superdense coding can turn a two-way quantum channel with bandwidth B (in both directions) into a one-way classical channel with ... 4 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 on my two qubits to steer the state into 5 orthogonal states. You can simply set up two standard superdense coding runs. So, this starts with each qubit of ... 4 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. This is because Qiskit numbers the bits in a string from right to left. If this convention seems odd to you, don’t worry. It seems odd to lots of other people ... 4 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}\...

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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: sending 2 classical bits of information uses up one pair, and this pair cannot be reused to send another 2 bits without extra steps (like you suggested, Bob ...

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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, quantum teleportation uses two classical bits of information to send a single qubit that is in an unknown quantum state. I suggest you to check IBM Q ...

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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 (mathematically this would be represented as tensor product of two single-qubit states). The best description of the individual qubits received by Alice and Bob is that ...

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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 $|\Psi\rangle\equiv\sum_{ij}\psi_{ij}|ij\rangle$, this means that $\sum_i \psi_{ij}\bar\psi_{kj}=\delta_{ik}/D$, with $D$ the dimension of each space ($D=2$ in your ...

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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 $\beta_{00}$: $$\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ Now you can employ superdense conding as usual. Note: matrix description of above mentioned ...

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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 general, if Alice has access to $n$ qubits and Bob has access to $m$ qubits and all $n+m$ qubits are entangled, then Alice can't transmit more than \$n+\text{...

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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 speed".

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Your understanding is almost correct. As said in the previous answer it is irrelevant that they are entangled. However, it is crucial for bob to have both to be aware of the actual state. This is a somewhat complicated argument to make but instant communication is impossible because Alice is not aware of what she is communicating and Bob isn’t aware of what ...

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Superdense coding is a protocol used to transmit two classical bits of information from one party to the other by using only a single qubit and no classical communication between them. It is a corollary of Holevo's theorem that 1 qubit can not hold more than 1 bit of information. A communication protocol which demonstrates the transmission of 1 classical bit ...

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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 tester prior to discussing multi-qubit entanglement in lecture 5. In the Elitzur-Vaidman tester, a single qubit in a superposition can be used to probe and ...

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