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

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 + \...


5

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 ...


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

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

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

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

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 ...


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

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

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 $|q\rangle|0\rangle \to |0\rangle|q\rangle$.


2

A trick you can use, in order to figure out the necessary fixups, is to see what happens when you condition the output display on the measurement results. Like this: See how the conditioned states are all pointing in different directions? Your goal is to fix that by, for each of the measurement cases, introducing controlled operations that make them all ...


2

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{...


2

I.e., if we have the entire initial state is written as follows $$| q \rangle \otimes | \beta_{00} \rangle =(\alpha | 0\rangle+\beta | 1 \rangle ) \otimes \frac{1}{\sqrt{2}}( |00\rangle+| 11 \rangle)$$ $$= \frac{1}{\sqrt{2}}(\alpha | 000\rangle+\alpha | 011 \rangle+\beta | 100 \rangle+\beta | 111 \rangle ),$$ then, in step three, we obtain the state (the ...


2

In conclusion, NO and NO, respectively. Quantum teleportation does not mean instantaneous disappearance and reappearance of information at another spatial point. Quantum teleportation in plain English means moving a quantum state from one place to another using quantum operations and quantum measurement, which are connected by classical as well as quantum ...


2

"Payload preperation gate" is just a fancy name for preparing the message with a more clear setting with the probability of 85.4% ∣0⟩ and 14.6% ∣1⟩ in this case. The chance of being a zero is less than that of one. A standard example of teleportation gives a 50% chance, in that case it is not a clear 0 or 1. The second H gate changes the relative phase of ...


2

Be sure you have executed the command "simulaqron set backend projectq" in your shell, in order to set the projectq backend. Moreover, in your Bob's code there is a reference to c1 which is undefined, in: qB=Bob.recvQubit(c1, "Alice")


2

I'm not sure what was the expected solution, but this also works. First of all, note that $$ I = |\phi^+\rangle\langle\phi^+| + |\phi^-\rangle\langle\phi^-| + |\psi^+\rangle\langle\psi^+| + |\psi^-\rangle\langle\psi^-|,$$ where $|\phi^+\rangle, |\phi^-\rangle, |\psi^+\rangle, |\psi^-\rangle$ are Bell states and $I$ is 4-dimensional identity operator. ...


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 ...


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 ...


1

Bob does not "see" the interference in the sense you assume. To "see" the interference (actually the states of his qubits) Bob needs to measure the qubits, and the measurement gives Bob no information whether Alice measured her qubits or not.


1

Thanks for help, it works. Here is a changed circuit and results on IBM Q (Vigo processor). Just note that q-bit $0.5|0\rangle + \sqrt {0.75}|1\rangle$ is teleported from q-bit q0 to q2. Angle in Ry gate is $2\pi/3$. ,


1

It seems like you are saying that when Alice sends/teleports her qubit to Bob, the first (left) qubit "becomes" the second (right) qubit. It's like you want to relabel the left (Bob) qubit with the right (Alice) qubit. However, the states that Alice and Bob have are already a product state and the qubits are not entangled. For example, Alice's first (left)...


1

You can obtain the effect of this sequence of gates as the matrix product: $$H T H |0\rangle = \frac{1}{\sqrt2} \begin{bmatrix} 1 & 1 \\ 1 & -1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & e^{i\pi/4}\end{bmatrix} \frac{1}{\sqrt2} \begin{bmatrix} 1 & 1 \\ 1 & -1\end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix}$$ Alternatively, if you ...


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