9

First, about teleportation, you say that you think quantum communication takes place in the protocol, but it doesn't. They only share an EPR pair they created together, hence the coordination and after, what Alice sends to Bob when communication takes place are classical bits, she sends the measured bits of the 2 qubits she has, so the only communication we ...


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 ...


6

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 ...


5

Suppose Alice wants to send Bob a 1000 bit message. To receive the message, Bob flips 1000 coins and writes down the results as 0s and 1s. About 50% of the random bits in the message that Bob generated are the same as in the intended message. Clearly this is an even better faster-than-light communication method than teleportation, because it succeeds 50% of ...


5

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


5

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

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 ...


4

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)/\...


3

Currently IBM Q does not support IF statement on real quantum processor. The IF can be used on simulator only. However, there is a theorem stating that quantum gates controlled by classical register can be replaced by those controlled by qubits before measurement. So, you can simply replace the statement measure q[1] -> c[1]; if (c == 1) x q[2]; by cx ...


3

I think faster than light communication is more stubborn than that :-) The principle of deferred measurement, to quote Nielsen and Chuang, page 186, is Measurements can always be moved from an intermediate stage of a quantum circuit to the end of the circuit; if the measurement results are used at any stage of the circuit then the classically controlled ...


3

As @Rammus points out this is just shorthand for taking the tensor product, rather than algebraically multiplying out as you may first assume. If we take the first term of $\phi_2$ and expand it out more explicitly with subscripts $A$ for Alice and $B$ for Bob: $|0_A0_A\rangle(\alpha|0_B\rangle + \beta|1_B\rangle) = |0_A0_A\rangle \otimes (\alpha|0_B\...


3

Information can't be transmitted faster-than-light using quantum teleportation protocol, because as you correctly noticed quantum teleportation involves classical communication between sender and receiver. The paper's abstract is saying about implementing quantum teleportation using a particular technology (silicon integrated optics) for the first time. ...


3

There is a mistake in design of the first circuit. Both $X$ and $Z$ gate work when value 1 is in classical register. Gate $X$ should work in case qubit $q_2$ is in state $|1\rangle$. Similarly $Z$ acts when $q_1$ in state $|1\rangle$. Also you have to deal with state when both $X$ and $Z$ have to act. In your case you conditioned both $X$ and $Z$ on c==1. ...


3

1) In this step, you connect a teleported qubit with entangled qubits between Alice and Bob. This means, Bob now has an "access" to the teleported qubit. 2) Here you get some information about the teleported qubit and "partially colapse" Bob's qubit according to a state of the teleported qubit. 3) In this last step you bring information about the ...


3

Perhaps the following diagram helps to show what acts where. I think you've got a little muddled.


3

In the teleportation protocol, the two parties share the entangled Bell State and it is implemented via a CNOT gate between the state to be sent (suppose Alice is sending the state) and the part of entangled Bell state Alice has. The CNOT gate creates another entangled state whose measurement Alice will send to the second party, say it's Bob, to perform the ...


2

There's a mistake. It's incorrect even for $a=0,b=0,U=I$. In this case the correct formula is $$ |\phi\rangle \otimes |B_{00}\rangle = \frac{1}{2} \bigg(|B_{00}\rangle \otimes |\phi\rangle + |B_{01}\rangle \otimes Z|\phi\rangle + |B_{10}\rangle \otimes X|\phi\rangle + |B_{11}\rangle \otimes XZ|\phi\rangle\bigg) $$ but their formula swaps $X$ and $Z$ in the ...


2

I think the code example closest to what you're looking for is task 1.4 of the Teleportation kata which gives the learner three qubits (the message qubit and the pair of qubits that Alice and Bob will entangle and share before teleportation) and asks them to transform the Bob's qubit into the message state. You'll notice that the test for this task doesn't ...


2

The top circuit uses classically controlled X and Z gates - gates that are conditioned on the measurement result. The bottom circuit uses quantum version of controlled X and Z gates - gates that are conditioned on the state of another qubit. Since each of the control qubits have been measured before applying CX and CZ gates, their state matches exactly the ...


2

The calculation of the resulting state of the described circuit: After applying the same circuit identities described in this answer (and here) to the connected question we will obtain a "simplified" circuit: Here are the links to the initial, first intermediate, second intermediate and final circuits presented in the Quirk. Whenever it was ...


2

Without that fixup, the state of Bob's qubit can be in any one of the states $\alpha |0\rangle + \beta |1\rangle$ (the state we actually wanted to teleport) or $\alpha |0\rangle - \beta |1\rangle$, $\beta |0\rangle + \alpha |1\rangle$ or $\beta |0\rangle - \alpha |1\rangle$ (the states we don't want to end up with). X and Z gates are applied to fix the ...


2

It sounds like the source of your misunderstanding is the following Bob is able to harness the information of a qubit which contains more information than regular classical bits This statement is not true. There is a theorem known as Holevo's theorem which states that the maximum information that can be obtained from a single qubit is one bit.


1

About checking. I do not quite understand why you can't check how the GHZ-state was teleported in the most ordinary ways: in addition to measuring the state immediately after teleportation, as well as by YYX, YXY, XYY, XXX measurements, you can inverse your GHZ-state and make sure that all 0 are obtained, e.g. like as for this with the most usual GHZ-state ...


1

I think it is worth trying to understand the circuit that the authors really want to implement: Here they produce the Bell state they want to teleport onto the final two qubits, and cluster state (personally, I wouldn't call it a cluster state after they've added the two extra hadamards). Then they do two single-qubit teleportation protocols to make the ...


1

The question reads like quantum encryption is a special example of quantum teleportation, which is not the case. Teleportation solves the following problem: Alice wants to send Bob a qubit, but has no way to physically transport it without destroying it. However, Alice and Bob already share an entangled state, and they can communicate classically. They ...


1

Thanks to Martin Vesely! Now I modify method 1 by using the correct c values. I think I got the right answer. My goal is to see the 1st bit entangled with the 4th bit. As can be seen below, 1st and 4th bits always have the same values. So they are entangled.


1

I contacted @fran-cabrera from the IBMQ team and he could reproduce the bug! The problem is with the visualization of the transpiled circuit, not the execution (the result should be correct). The team is working on solving it and they expect to deploy a fix at the end of the week. I ran you example in Qiskit and it seems to work IBMQ.load_account() ...


1

If you are interested only in the measurements of the system that comprises only 2nd and 3rd qubits, then you could compute the reduced density matrix $\rho_{23}$ on those qubits from the total density matrix $\rho_{1234}=|\psi\rangle \langle\psi|$, i.e. $$ \rho_{23} = \text{Tr}_{14}(\rho_{1234}) $$ Density matrix carry all information that is need to ...


1

Judging from your computations, the Bell states are $$|B_{00}\rangle = |\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ $$|B_{10}\rangle = |\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$$ $$|B_{01}\rangle = |\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$$ $$|B_{11}\rangle = |\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\...


1

"Assume that there is a setup with Bob which creates some sort of interference of the qubits he receives." I think you need to be more specific on this point. From Bob's perspective his qubit is simply a mixed state in the computational basis $$\rho_B=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}|1\rangle\langle1|$$ so he will not be able to perceive any ...


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