# Tag Info

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Quantum gate teleportation is the act of being able to apply a quantum gate on the unknown state while it is being teleported. This is one of the ways in which measurement-based computation can be described using graph states. Usually, teleportation works by having an unknown quantum state $|\psi\rangle$ held by Alice, and two qubits in the Bell state $|\... 10 Gate teleportation is in principle a method that allows the creation of different gates from an available set of gates, by teleporting qubits through entangled states. An example of the use of this method, is the creation of the T gate from a Clifford set of gates in order to make the set universal. The construction in this particular case is done with the ... 8 In quantum teleportation, one starts with an entangled state shared between two parties, and (after some messing at the sender's side), two classical bits are transmitted from one party to the other so that the net effect is a quantum state is sent from the first party to the second without sending any quantum data. In superdense coding, the parties start ... 7 I don't know for sure how you would achieve fewer than two bits of classical communication for a teleportation, but here's one way that you could have a non-integer number: if you teleport a qudit with dimension$d$that is not a power of two. For each teleportation protocol, you'd have to send two dits of information, which you could represent in bits using ... 7 Whenever you have a quantum gate (like a CNOT) acting on some qubits but not others, it is assumed that the other qubits are acted on with the identity operator. This is done using the "Left Kronecker product" or the "tensor product". So the 8x8 matrix is made by applying CNOT to qubits 1 & 2 and the identity matrix to qubit 3: $$\begin{bmatrix} 1 ... 6 Let's define the shift and clock matrices (the generalisation of the Pauli X and Z matrices) as$$ X=\sum_{i=0}^{d-1}|i+1\text{ mod }d\rangle\langle i|\qquad Z=\sum_{i=0}^{d-1}\omega^i|i\rangle\langle i| $$where \omega=e^{2\pi \sqrt{-1}/d}. Now we can define a maximally entangled orthonormal basis (the equivalent of the Bell basis):$$ |\Psi_{ij}\rangle=(... 6 Your initial calculations are correct. When Alice performs her first measurement and gets a 0 outcome then, as you say, Alice and Bob are left sharing a two-qubit state $$|\Psi\rangle=\alpha|00\rangle+\beta|11\rangle$$ (you can safely ignore the measured qubit). The problem is the statement She can tell Bob: "Your half of the EPR is now the qubit I ... 6 A necessary and sufficient condition on the unitary$B$is that its columns all correspond to maximally entangled states. There also does not need to be any relationship between the two unitaries labeled$B$and$B^{-1}$in your figure: as long as you start with a maximally entangled state of systems 2 and 3, and then measure systems 1 and 2 with respect ... 5 The first two qubits stay on the sender's side, and since they are not entangled with the receiver's qubit, it's not necessary to mention them any further — we switched to discussing the receiver's side of the protocol, and all information from the first two qubits is now in measurement results. The superscripts denote the power of the operator applied: if ... 5 Alice receives a quantum state$|\psi\rangle$, which is an element of some basis$\mathcal{B}$, though she does not know what$\mathcal B$is. She then teleports this to Bob, who is told by someone else what$\mathcal B$is. Furthermore, it seems that specifically either$\mathcal{B} = \{ \lvert 0 \rangle, \lvert 1 \rangle \}$or$\mathcal{B} = \{ \lvert + \...

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I recently found a paper by Subhash Kak that introduces teleportation protocols that require lesser classical communication cost (with more quantum resource). I thought it'd be better to write a separate answer. Kak discusses three protocols; two of them use 1 cbit and the last one requires 1.5 cbits. But the first two protocols are in a different setting, ...

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I believe the current record is held by the Jian-Wei Pan group in China, who are able to generate entanglement via a satellite. The journal article is here, while there's plenty of media coverage that is a bit more accessible, e.g. New Scientist. This claims a distance of 1203km.

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The issue is that you are applying operations after measurement gates and this is currently not available on the real hardware. I think the hardware also does not support reset operations mid-way through a circuit at the moment. The best way forward is to keep running this on the simulator or try to find a different way of expressing the circuit such that ...

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Your circuit does not measure $q_2$ qubit after teleportation; I guess that is why teleportation of $|1\rangle$ qubit is shown incorrectly.

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Since your circuit is teleportation, $|C\rangle =|\varphi\rangle$ and since you measured $|11\rangle$ on $|AB\rangle$ the answer is $|ABC\rangle = |11\rangle|\varphi\rangle$. Now, let look why this is true. Firstly Hadamard and CNOT gate on second and third qubit prepares entangled Bell state $|\beta_{00}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)... 4 Photons travel fast, and there's often the option to transfer their entanglement to solid state. Of course, the advantage of transferring entanglement to a solid state qubit is that one is able to operate with it (one- and two-qubit gates, for example) with ease and efficiency, whereas it is very hard to effect two-qubit quantum gates on photons themselves, ... 4 I tested this out in Quirk by embedding each qutrit into two qubits, and I get a simular result to you, where in addition to the cyclic shift fixup and the phasing operation you need to transpose states 1 and 2. Presumably there's some simple change to the circuit that fixes this, such as picking a different F, but I did't check too hard to see if it was ... 4 measurements in every circuit can be postponed or never performed in a circuit while achieving the same functionality of the circuit That's correct. But if the circuit involves two parties, this process will introduce quantum operations between the two parties. It will require a quantum communication channel, so that the qubits can be shuttled back and ... 4 Think about the following sequence: $$XT^\dagger X=\left(\begin{array}{cc} e^{-i\pi/4} & 0 \\ 0 & 1 \end{array}\right)=e^{-i\pi/4}T.$$ So, that lets us write $$TXT^\dagger=e^{-i\pi/4}TTX=e^{-i\pi/4}SX$$ Up to some phase, you have the decomposition that you want. In this context, that phase should be an irrelevant global phase that you can ignore. 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 Quantum teleportation has been suggested to be possible to occur in photosynthetic reaction centers. https://doi.org/10.1007/BF03166259 And it could in principle be occuring all the time through photochemistry in biology. Though to no real purpose, however, a related purposeful effect is the avian compass which(is hypothesized) to operate through quantum ... 3 Cloning means the generation of$|q\rangle|q\rangle$from$|q\rangle|0\rangle$. This is not what happens in teleportation. Teleportation is kind of a swap operation, i.e. something like$|q\rangle|0\rangle \to |0\rangle|q\rangle$. 3 To answer this, contrast quantum teleportation with the swap gate. Ignoring everything in the middle, the effect of quantum teleportation is to get from state$|\phi00\rangle$to$|00\phi\rangle$. This can obviously be accomplished with a simple application of the swap gate. So, why do we care about quantum teleportation? You'd never run the quantum ... 3 Firstly, carefully read through the formal presentation of the protocol as described on Wikipedia. Secondly, there's nothing to prove as such here. It is evident from the teleportation protocol itself. Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e.$|q\rangle$. ... 3 I asked someone from IBM and got this answer: Teleportation can not be run on the IBM Q devices at the moment as no operations can be performed after a measurement. 3 Since the quantum teleportation circuit has three qbits, the matrix at each step is 8x8 and thus has 64 elements; this is pretty clunky to type out in its entirety, so I'll just walk you through step by step and you can derive the full matrix for a specific step if you want. Given a qbit we want to teleport:$|\psi\rangle = \begin{bmatrix} \alpha \\ \beta \...

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One way to do it is to build a sort of quantum IF statement. You have in quantum computing projector operators telling you whether a qubit is 0 or 1: $$P_0 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$ $$P_1 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$ Then we have the Z gate :  Z= \begin{pmatrix} 1 & 0 \\ 0 &...

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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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The simplest way to generalize teleportation is to just repeat it. If you have one EPR pair divided between Alice and Bob, teleportation allows you to move one qubit from Alice to Bob (or vice versa) by consuming the EPR pair and using a classical communication channel. If you have two EPR pairs, you can move two qubits by performing two independent ...

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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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