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

I'm unsure of this exact protocol you're referring to, but if you're interested in the "cost" or reliability of teleportation , then I have some information pertaining to the original setup. Teleportation is a protocol that uses local operations and classical communications to teleport a state $\rho_{\text{in}}$ from Alice (A) to Bob (B) who share ...


2

I guess the way that I'd start (aside from just getting a computer to do it!) is to remember that the Bell states form an orthonormal basis. So, you can ask, for example, about what the $|\Phi^+\rangle^{AD}$ component is: $$ \langle\Phi^+|^{AD}|\Phi\rangle^{ABCD}=-\frac12|\Phi^+\rangle^{BC}. $$ You do this for each of the four states, and you can use that to ...


2

No, it is not possible without communication. To see why, consider B and C, and just ignore A -- since they cannot communicate, for anything B and C can do A's presence is irrelevant. Then, B and C's measurement outcomes are completely uncorrelated (since they don't share any entanglement, or correlations, just each of them holds a maximally mixed state). ...


1

I think what you want is in this paper : Fusion-based quantum computation Look at the "Bell fusion" in Fig 2. Basically you measure $X_BX_C$ and $Z_B Z_C$ and that leaves $q_A q_D$ entangled.


1

You need to calculate $U=e^{-iHt}$. The trick to doing this is working out the eigenvectors of $H$: there's $|00\rangle$ and $|11\rangle$ with eigenvalues J, and $$ |\Psi^{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2} $$ with eigenvalues $(-1\pm 2)J$. In particular, notice that this means 3 of the eigenvalues are $J$. Hence, there are two eigenspaces of $H$,...


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

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.


Only top voted, non community-wiki answers of a minimum length are eligible